\input{"preamble.tex"}

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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Algebraic Number Theory
  }
    \\ {\normalsize Lectures by Paul Pollack. University of Georgia,
Spring 2021} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in Algebraic Number Theory taught by Paul Pollack
at the University of Georgia in Spring 2021. As such, any errors or
inaccuracies are almost certainly my own. } } \\
    \end{flushleft}
    \end{minipage}
    \hfill
    \begin{minipage}{.65\linewidth}
    \end{minipage}
  }







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2021-08-22 }
\end{flushleft}


\newpage

% Note: addsec only in KomaScript
\addsec{Table of Contents}
\tableofcontents
\newpage

\hypertarget{preface}{%
\section{Preface}\label{preface}}

See
\href{https://www.youtube.com/watch?v=JGLuFkqXIpI\&list=PLA0xtXqOUji8fjQysx4k8a6h-hOZ7x5ue}{Youtube
Video Playlist}.

\hypertarget{notation}{%
\section{Notation}\label{notation}}

\todo[inline]{Todo: definitions.}

\begin{itemize}
\tightlist
\item
  \(U({\mathbb{Z}}/k), ({\mathbb{Z}}/k)^{\times}\) denotes the group of
\item
  \({\mathbb{Z}}/k\): the integers modulo \(k\).
\item
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\):
  the algebraic closures of \({\mathbb{Z}}, {\mathbb{Q}}\) respectively.
\item
  \(K, L\) will generally denote fields, and most likely number fields
  (so extensions of \({\mathbb{Q}}\)) \(K/{\mathbb{Q}}\) will denote
  that \(K\) is a field extension of \({\mathbb{Q}}\), and
  \(K/L/{\mathbb{Q}}\) denotes a tower of extensions \(K/L\) and
  \(L/{\mathbb{Q}}\).
\item
  \([K: {\mathbb{Q}}]\) is the degree of the extension
  \(K/{\mathbb{Q}}\), i.e.~the dimension of \(K\) as a
  \({\mathbb{Q}}{\hbox{-}}\)vector space.
\item
  \(K(\alpha)\) will denote adjoining an element
  \(\alpha \in \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu\)
  to \(K\) to get a field extension \(K(\alpha)/K\).
\item
  \({\mathbb{Z}}_K \coloneqq\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu \cap K\),
  the algebraic integers in \(K\).
\item
  \(\operatorname{ff}(K)\) is the fraction field of \(K\).
\item
  \({\mathsf{Ab}}, \mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}\): the
  category of abelian groups.
\item
  \(\zeta_n\) will denote a primitive \(n\)th root of unity, so e.g.~for
  \(n\) prime one can take \(\zeta_n \coloneqq e^{2\pi i \over n}\).
\item
  \(H\leq G\): \(H\) is a subgroup of \(G\).
\end{itemize}

\hypertarget{diophantine-equations-lec.-1-thursday-january-14}{%
\section{Diophantine Equations (Lec. 1, Thursday, January
14)}\label{diophantine-equations-lec.-1-thursday-january-14}}

\hypertarget{intrologistics}{%
\subsection{Intro/Logistics}\label{intrologistics}}

See website for notes on books, intro to class.

\begin{itemize}
\item
  Youtube Playlist:
  \url{https://www.youtube.com/playlist?list=PLA0xtXqOUji8fjQysx4k8a6h-hOZ7x5ue}
\item
  Free copies of textbook:
  \url{https://www.dropbox.com/sh/rv5j222kn74bjhm/AABZ1qcR1rOnpaBsa5CL3P_Ea?dl=0\&lst=}
\end{itemize}

Paul's description of the course:

\begin{quote}
This course is an introduction to arithmetic beyond \({\mathbb{Z}}\),
specifically arithmetic in the ring of integers in a finite extension of
\({\mathbb{Q}}\). Among many other things, we'll prove three important
theorems about these rings:

\begin{itemize}
\tightlist
\item
  Unique factorization into ideals.
\item
  Finiteness of the group of ideal classes.
\item
  Dirichlet's theorem on the structure of the unit group.
\end{itemize}
\end{quote}

\hypertarget{motivation}{%
\subsection{Motivation}\label{motivation}}

\begin{remark}

The main motivation: solving \textbf{Diophantine equations},
i.e.~polynomial equations over \({\mathbb{Z}}\).

\end{remark}

\begin{example}[of a Diophantine equation]

Consider \(y^2 = x^3 + x\).

\begin{claim}

\((x, y) = (0, 0)\) is the only solution.

\end{claim}

To see this, write \(y^2 = x(x^2+1)\), which are relatively prime,
i.e.~no \(D\in {\mathbb{Z}}\) divides both of them. Why? If
\(d \divides x\) and \(d \divides x+1\), then
\(d\divides (x^2+1) + (-x) = 1\). It's also the case that both \(x^2+1\)
and \(x^2\) are squares (up to a unit), so \(x^2, x^2 + 1\) are
consecutive squares in \({\mathbb{Z}}\). But the gaps between squares
are increasing: \(1, 2, 4, 9, \cdots\). The only possibilities would be
\(x=0, y=1\), but in this case you can conclude \(y=0\).

\end{example}

\begin{example}[Fermat]

Consider \(y^2 = x^3-2\).

\begin{claim}

\((3, \pm 5)\) are the only solutions.

\end{claim}

Rewrite
\begin{align*}
x^3 = y^2+2 &= (y+ \sqrt{-2})(y - \sqrt{-2}) \\ 
&\in
{\mathbb{Z}}[\sqrt{-2}] \coloneqq\left\{{a+b\sqrt{-2} {~\mathrel{\Big|}~}a,b,\in {\mathbb{Z}}}\right\} \leq {\mathbb{C}}
.\end{align*}
This is a subring of \({\mathbb{C}}\), and thus at least an integral
domain. We want to try the same argument: showing the two factors are
relatively prime. A little theory will help here:

\begin{definition}[Norm Map]

\begin{align*}
N \alpha \coloneqq\alpha\mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu&& \text{for } \alpha\in {\mathbb{Z}}[ \sqrt{-2} ]
.\end{align*}

\end{definition}

\begin{lemma}[?]

Let \(\alpha, \beta \in {\mathbb{Z}}[\sqrt{-2}]\). Then

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(N(\alpha \beta) = N(\alpha) N(\beta)\)
\item
  \(N( \alpha) \in {\mathbb{Z}}_{\geq 0}\) and \(N(\alpha) = 0\) if and
  only if \(\alpha= 0\).
\item
  \(N(\alpha) = 1 \iff \alpha\in R^{\times}\)
\end{enumerate}

\end{lemma}

\begin{proof}[?]

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Missing, see video (10:13 AM).
\item
  \(N(\alpha) = a^2 + 2b^2 \geq 0\), so this equals zero if and only if
  \(\alpha= \beta= 0\)
\item
  Write
  \(1 = \alpha\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu\)
  if \(N(\alpha) = 1 \in R^{\times}\). Conversely if
  \(\alpha\in R^{\times}\) write \(\alpha \beta = 1\), then
  \begin{align*} 
  1 = N(1) = N(\alpha \beta) = N(\alpha ) N(\beta ) \in {\mathbb{Z}}_{\geq 0} 
  ,\end{align*}
  which forces both to be 1.
\end{enumerate}

\end{proof}

\begin{claim}

The two factors \(y \pm \sqrt 2\) are \emph{coprime} in
\({\mathbb{Z}}[\sqrt{-2}]\), i.e.~every common divisor is a unit.

\end{claim}

\begin{proof}[?]

Suppose \(\delta\divides y\pm \sqrt{-2}\), then
\(y + \sqrt{-2} = \delta \beta\) for some
\(\beta\in {\mathbb{Z}}[\sqrt{-2}]\). Take norms to obtain
\(y^2 + 2 = N \delta N \beta\), and in particular

\begin{itemize}
\tightlist
\item
  \(N \delta y^2 +2\)
\item
  \(\delta \divides (y+ \sqrt{-2} ) - (y - \sqrt{-2} ) = 2 \sqrt{-2}\)
  and thus \(N \delta \divides N(2 \sqrt{-2} ) = 8\).
\end{itemize}

In the original equation \(y^2 = x^3-2\), if \(y\) is even then \(x\) is
even, and \(x^3 - 2 \equiv 0-2 \operatorname{mod}4 \equiv 2\), and so
\(y^2 \equiv 2 \operatorname{mod}4\). But this can't happen, so \(y\) is
odd, and we're done: we have \(N \delta\divides 8\) which is even or 1,
but \(N \delta\divides y^2 +2\) which is odd, so \(N \delta = 1\).

\end{proof}

We can identify the units in this ring:
\begin{align*}
{\mathbb{Z}}[\sqrt{-2} ]^{\times}= \left\{{ a + b \sqrt{-2} {~\mathrel{\Big|}~}a^2 + 2b^2 = 1}\right\}
\end{align*}
which forces \(a^2 \leq 1, b^2 \leq 1\) and thus this set is
\(\left\{{\pm 1}\right\}\). So we have \(x^3 = ab\) which are relatively
primes, so \(a,b\) should also be cubes. We don't have to worry about
units here, since \(\pm 1\) are both cubes. So e.g.~we can write
\begin{align*}
y + \sqrt{-2} = (a + b \sqrt{-2} )^3 = (a^3-6ab^2) + (3a^2b -2b^3) \sqrt{-2}
.\end{align*}
Comparing coefficients of \(\sqrt{-2}\) yields
\begin{align*} 1 = b(3a^2b - 2b^2) \in {\mathbb{Z}}\implies b \divides 1
,\end{align*}
and thus \(b\in {\mathbb{Z}}^{\times}\),
i.e.~\(b\in \left\{{\pm 1}\right\}\). By cases:

\begin{itemize}
\item
  If \(b=1\), then \(1 = 3a^2 -2 \implies a^2 = 1 \implies a = \pm 1\).
  So
  \begin{align*}
  y = \sqrt{-2} = (\pm 1 + \sqrt{-2} )^3 = \pm 5 + \sqrt{-2}
  ,\end{align*}
  which forces \(y=\pm 5\), the solution we already knew.
\item
  If \(b = -1\), then \(1 = -(3a^2 - 1)\) which forces
  \(1=3a^2 \in {\mathbb{Z}}\), so there are no solutions.
\end{itemize}

\end{example}

\hypertarget{failure-of-unique-factorization}{%
\subsection{Failure of Unique
Factorization}\label{failure-of-unique-factorization}}

\begin{example}[where unique factorization fails]

Consider \(y^2 = x^3 - 26\). Rewrite this as
\begin{align*}
x^3 = y^2 + 26 = (y + \sqrt{-26} )(y - \sqrt{-26} )
,\end{align*}
then the same lemma goes through with \(2\) replaced by \(26\)
everywhere where the RHS factors are still coprime. Setting
\(y + \sqrt{-26} = (a + b \sqrt{-26} )^3\) and comparing coefficients,
you'll find \(b=1, a = \pm 3\). This yields \(x=35, y=\pm 207\). But
there are more solutions: \((x, y) = (3, \pm 1)\)! The issue is that we
used unique factorization when showing that \(ab\) is a square implies
\(a\) or \(b\) is a square (say by checking prime factorizations and
seeing even exponents). In this ring, we can have \(ab\) a cube with
\emph{neither} \(a,b\) a cube, even up to a unit.

\end{example}

\begin{question}

When does a ring admit unique factorization? Do you even \emph{need} it?

\end{question}

\begin{remark}

This will lead to a discussion of things like the \textbf{class number},
which measure the failure of unique factorization. In general, the above
type of proof will work when the class number is 3!

\end{remark}

\hypertarget{number-fields-lec.-2tuesday-january-19}{%
\section{Number Fields (Lec. 2,Tuesday, January
19)}\label{number-fields-lec.-2tuesday-january-19}}

\hypertarget{embeddings}{%
\subsection{Embeddings}\label{embeddings}}

\begin{remark}

Today: Ch.2 of the book, ``Cast of Characters''. Note that all rings
will be commutative and unital in this course.

Last time: looked at factorization in
\({\mathbb{Z}}[\sqrt 2], {\mathbb{Z}}[\sqrt{26}]\). Where do rings like
this come from?

\end{remark}

\begin{definition}[Number Field]

A \textbf{number field} is a subfield \(K \subseteq {\mathbb{C}}\)
\footnote{Some authors don't require \(K \subseteq {\mathbb{C}}\), but
  any finite extension of \({\mathbb{Q}}\) will embed into
  \({\mathbb{C}}\) so there's no harm in this extra requirement.} such
that \([K: {\mathbb{Q}}] < \infty\).

\end{definition}

\begin{example}[of number fields]

Examples of number fields include

\begin{itemize}
\tightlist
\item
  \({\mathbb{Q}}[\sqrt[3]{2}]\),
\item
  \({\mathbb{Q}}[\sqrt 2, \sqrt[5]{7}]\), or
\item
  \({\mathbb{Q}}(\theta)\) where \(\theta\) is a root of
  \(x^5 - x - 1\), which one can check is irreducible.
\end{itemize}

Note that the round vs.~square brackets here won't make a difference,
since we're adjoining \emph{algebraic} numbers.

\end{example}

\begin{proposition}[Degree equals number of embeddings for finite extensions]

Let \(K_{/{\mathbb{Q}}}\) be a finite extension, say of degree
\(n\coloneqq[K: {\mathbb{Q}}]\). Then there are \(n\) distinct
embeddings \footnote{An \textbf{embedding} is an injective ring
  morphism.} of \(K\) into \({\mathbb{C}}\)

\end{proposition}

\begin{proof}[of proposition]

We have \(K_{/{\mathbb{Q}}}\), which is necessarily separable since
\(\operatorname{ch}({\mathbb{Q}}) = 0\). By the primitive element
theorem, we can write \(K = {\mathbb{Q}}(\theta)\) where \(\theta\) is a
root of some degree \(n\) irreducible polynomial
\(f(x) \in {\mathbb{Q}}[x]\). Since \({\mathbb{C}}\) is algebraically
closed, \(f\) splits completely over \({\mathbb{C}}\) as
\begin{align*}
f = \prod_{i=1}^n (x- \theta_i) 
\end{align*}
with each \(\theta_i \in {\mathbb{C}}\) distinct since \(f\) was
irreducible and we're in characteristic zero. Then for each \(i\) there
is an embedding \(K = {\mathbb{Q}}[\theta]\) given by
\begin{align*}
\iota_i: {\mathbb{Q}}[\theta] &\hookrightarrow{\mathbb{C}}\\
g(\theta) &\mapsto g(\theta_i)
.\end{align*}
There are some easy things to check:

\begin{itemize}
\tightlist
\item
  This is well-defined: elements in \(K\) are polynomials in \(\theta\)
  but they all differ by a multiple of the minimal polynomial of
  \(\theta\),
\item
  This is an injective homomorphism and thus an embedding, and
\item
  For distinct \(i\) you get distinct embeddings: just look at the image
  \(\iota_i(\theta)\), these are distinct numbers in \({\mathbb{C}}\).
\end{itemize}

\end{proof}

\begin{definition}[Real and Nonreal embeddings]

Let \(K_{/{\mathbb{Q}}}\) be a finite extension of degree
\(n = [K : {\mathbb{Q}}]\). We'll say an embedding
\(\sigma:K \to {\mathbb{C}}\) is \textbf{real} if
\(\sigma(K) \subseteq {\mathbb{R}}\) , otherwise we'll say the embedding
is \textbf{nonreal}.

\end{definition}

\begin{remark}

If \(\sigma\) is a nonreal, then
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu\) is
a nonreal embedding, so these embeddings come in pairs. As a
consequence, the total number of embeddings is given by
\(n = r_1 + 2r_2\), where \(r_1\) is the number of real embeddings and
\(r_2\) is the number of nonreal embeddings.

\end{remark}

\begin{example}[of computing the number of real and nonreal embeddings]

Let \(K = {\mathbb{Q}}(\sqrt[3]{2})\). Here \(n=3\) since this is the
root of a degree 3 irreducible polynomial. Using the proof we can find
the embeddings: factor
\begin{align*}
x^3 - 2 = (x - \sqrt[3]{2})(x - \omega \sqrt[3]{2}) (x - \omega^2 \sqrt[3]{2})
.\end{align*}
where \(\omega = e^{2\pi i / 3}\) is a complex cube root of unity. We
can form an embedding by sending
\(\sqrt[3]{2} \to \omega^j \sqrt[3]{2}\) for \(j=0,1,2\). The case
\(j=0\) sends \(K\) to a subset of \({\mathbb{R}}\) and yields a real
embedding, but the other two will be nonreal. So \(r_1 = 1, r_2 = 1\),
and we have \(3 = 1 + 2(1)\), which is consistent.

\end{example}

\hypertarget{algebraic-closures}{%
\subsection{Algebraic Closures}\label{algebraic-closures}}

\begin{remark}

We've only been talking about fields, where unique factorization is
trivial since there are no primes. There are thus ``too many'' units in
fields when compared to the rings we were considering before, so we'll
restrict to subrings of fields. The question is: where is the
arithmetic? Given a number field \(K\), we want a ring
\({\mathbb{Z}}_K\) that fits this analogy:

\begin{center}
\begin{tikzcd}
{\mathbb{Q}}\ar[dd] & \leadsto & K\ar[dd] \\
\\
{\mathbb{Z}}& \leadsto & {\mathbb{Z}}_K = ?
\end{tikzcd}
\end{center}

\end{remark}

\begin{definition}[Algebraic Numbers]

Given \(\alpha\in {\mathbb{C}}\) we say \(\alpha\) is an
\textbf{algebraic number} if and only if \(\alpha\) is algebraic over
\({\mathbb{Q}}\), i.e.~the root of some polynomial in
\({\mathbb{Q}}[x]\).

\end{definition}

\begin{remark}

We know that if we define
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\coloneqq\left\{{\alpha\in {\mathbb{C}}{~\mathrel{\Big|}~}\alpha \text{ is algebraic over } {\mathbb{Q}}}\right\}\),
we can alternatively describe this as
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu= \left\{{ \alpha\in {\mathbb{C}}{~\mathrel{\Big|}~}[{\mathbb{Q}}(\alpha) : {\mathbb{Q}}] < \infty }\right\}\).
This is convenient because it's easy to see that algebraic numbers are
closed under sums and products, just using the ways degrees behave in
towers.

\end{remark}

\begin{corollary}[Every number field is a subfield of $\bar\QQ$]

\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\hookrightarrow{\mathbb{C}}\)
is a subfield and every number field is a subfield of
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\).

\end{corollary}

\begin{remark}

These are still fields, so lets define some interesting subrings.

\end{remark}

\begin{definition}[$\bar \ZZ$ ]

Define
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\coloneqq\left\{{ \alpha\in {\mathbb{C}}{~\mathrel{\Big|}~}\alpha\text{ is the root of a monic polynomial }f\in {\mathbb{Z}}[x]}\right\}\).

\end{definition}

\begin{theorem}[$\bar \ZZ$ is a ring]

\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
is a ring, and in fact a domain since it's a subring of
\({\mathbb{C}}\).

\end{theorem}

\begin{remark}

We'll use an intermediate criterion to prove this:

\end{remark}

\begin{proposition}[Integrality Criterion]

Let \(\alpha\in {\mathbb{C}}\) and suppose there is a finitely generated
\({\mathbb{Z}}{\hbox{-}}\)submodule of \({\mathbb{C}}\) with
\(\alpha M \subseteq M \neq 0\). Then
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\),
i.e.~\(\alpha\) is the root of a monic polynomial with integer
coefficients.

\end{proposition}

\begin{proof}[of integrality criterion]

Chasing definitions, take \(M\) and choose a finite list of generators
\(\beta_1, \beta_2, \cdots, \beta_m\) for \(M\). Then
\(\alpha M \subseteq M \implies \alpha \beta_i \in M\) for all \(M\),
and each \(\alpha \beta_i\) is a \({\mathbb{Z}}{\hbox{-}}\)linear
combination of the \(\beta_i\) . I.e. we have
\begin{align*}
\alpha 
\begin{bmatrix}
\beta_1 
\\
\vdots  
\\
\beta_n  
\end{bmatrix}
= 
\begin{bmatrix}
a_{11} & a_{12} & \cdots
\\
a_{21} &  a_{22} & 
\\
 \vdots &  &\ddots  
\end{bmatrix}
\begin{bmatrix}
\beta_1 
\\
\vdots  
\\
\beta_n  
\end{bmatrix}
\coloneqq A \vec{\beta}
,\end{align*}
where \(A \in \operatorname{Mat}(n\times m, {\mathbb{Z}})\). We can
rearrange this to say that
\begin{align*}
\qty{ \alpha I - A} 
\begin{bmatrix}
\beta_1 
\\
\vdots  
\\
\beta_n  
\end{bmatrix}
=
\mathbf{0}
.\end{align*}
Not all of the \(\beta_i\) can be zero since \(M\neq 0\), and thus
\(\alpha I - A\) is singular and has determinant zero, so
\(\operatorname{det}(x I - A)\Big|_{x=\alpha} = 0\). We have
\begin{align*}
x\operatorname{id}- A = 
\begin{bmatrix}
x - a_{1,1} &  & &
\\
&  x - a_{2, 2} & & 
\\
&  & \ddots &
\\
& &  & x - a_{m, m}
\end{bmatrix}
,\end{align*}
where the off-diagonal components are constants in \({\mathbb{Z}}\)
coming from \(A\). Taking the determinant yields a monic polynomial: the
term of leading degree comes from multiplying the diagonal components,
and expanding over the remaining minors only yields terms of smaller
degree. So \(\operatorname{det}(x I - A) \in {\mathbb{Z}}[x]\) is monic.

\end{proof}

\begin{proof}[of theorem]

We want to show that
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
is a ring, and it's enough to show that

\begin{itemize}
\tightlist
\item
  \(1\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\),
  which is true since \(x-1\) is monic.
\item
  It's closed under addition \((+)\) and multiplication \((\cdot)\).
\end{itemize}

Note that the first property generalizes to
\({\mathbb{Z}}\subseteq \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\),
since \(x-n\) is monic for any \(n\in {\mathbb{Z}}\). For the second,
let
\(\alpha, \beta \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
Define \(M \coloneqq{\mathbb{Z}}[\alpha, \beta]\), then it's clear that
\((\alpha + \beta)M \subseteq M\) and \((\alpha \beta)M \subseteq M\)
since \({\mathbb{Z}}[\alpha, \beta]\) are polynomials in
\(\alpha, \beta\) and multiplying by these expression still yields such
polynomials. It only remains to check the following:

\begin{claim}

\(M\) is finitely-generated.

\end{claim}

\begin{proof}[?]

Let \(\alpha\) be a root of \(f \in {\mathbb{Z}}[x]\) and \(\beta\) a
root of \(g\), both monic with \(\deg f = n, \deg g = m\). We want to
produce a finite generating set for
\(M\coloneqq{\mathbb{Z}}[\alpha, \beta]\), and the claim is that the
following works:
\(\left\{{ \alpha^i \beta^j}\right\} _{\substack{0\leq i < n \\ 0 \leq j < m} }\),
i.e.~every element of \(M\) is some \({\mathbb{Z}}{\hbox{-}}\)linear
combination of these.

\hfill\break

Note that this is clearly true if we were to include \(n, m\) in the
indices by collecting terms of any polynomial in \(\alpha, \beta\), so
the restrictions are nontrivial. It's enough to show that for any
\(0 \leq I, J \in {\mathbb{Z}}\), the term \(\alpha^I \beta^J\) is a
\({\mathbb{Z}}{\hbox{-}}\)linear combination of the restricted elements
above. Divide by \(f\) and \(g\) to obtain
\begin{align*}
x^I &= f(x) q(x) + r(x) \\
x^J &= g(x) \tilde q(x) \tilde r(x)
\end{align*}
where \(r(x) = 0\) or \(\deg r < n\) and similarly for \(\tilde r\),
where (importantly) all of these polynomials are in \({\mathbb{Z}}[x]\).

\hfill\break

We're not over a field: \({\mathbb{Z}}[x]\) doesn't necessarily have a
division algorithm, so why is this okay? The division algorithm only
requires inverting the leading coefficient, so in general \(R[x]\)
admits the usual division algorithm whenever the leading coefficient is
in \(R^{\times}\). Now plug \(\alpha\) into the first equation to obtain
\(\alpha^I = r(\alpha)\) where \(\deg r < n\), which rewrite
\(\alpha^I\) as a sum of lower-degree terms. Similarly writing
\(\beta^J = r(\beta)\), we can express
\begin{align*}
\alpha^I \beta^J = r(\alpha) r(\beta)
,\end{align*}
which is what we wanted.

\end{proof}

\end{proof}

\begin{remark}

We've just filled in another part of the previous picture:

\begin{center}
\begin{tikzcd}
{\mathbb{Q}}\ar[dd] & K\ar[dd] & \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\ar[dd]\\
\\
{\mathbb{Z}}& {\mathbb{Z}}_K & \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu
\end{tikzcd}
\end{center}

\end{remark}

\hypertarget{rings-of-integers-and-fraction-fields}{%
\subsection{Rings of Integers and Fraction
Fields}\label{rings-of-integers-and-fraction-fields}}

\begin{definition}[Ring of Integers]

Define
\({\mathbb{Z}}_K = \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap K\),
the \textbf{ring of integers} of \(K\). Note that this makes sense since
the intersection of rings is again a ring.

\end{definition}

\begin{remark}

Why not just work in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)?
It doesn't have the factorization properties we want, e.g.~there are no
irreducible elements. Consider \(\sqrt 2\), we can factor is into two
non-units as \(\sqrt{2} = \sqrt{\sqrt 2} \cdot \sqrt{\sqrt 2}\), noting
that \(\sqrt 2\) is not a unit, and it's easy to check that if \(a\) is
not a unit then \(\sqrt a\) is not a unit. So this would yield
arbitrarily long factorizations, and a non-Noetherian ring. The
following is a reality check, and certainly a property we would want:

\end{remark}

\begin{proposition}[The ring of integers of $\QQ$ is $\ZZ$ ]

\({\mathbb{Z}}_{\mathbb{Q}}= {\mathbb{Z}}\).

\end{proposition}

\begin{proof}[of proposition]

\(\subseteq\): Easy, since
\({\mathbb{Z}}\subseteq \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
and \({\mathbb{Z}}\subseteq {\mathbb{Q}}\), and is thus in their
intersection \({\mathbb{Z}}_{\mathbb{Q}}\) .

\hfill\break

\(\supseteq\) : Let
\(\alpha\in {\mathbb{Z}}_{\mathbb{Q}}= {\mathbb{Q}}\cap\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
, so \(\alpha\) is a root of
\(x^n - a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \in {\mathbb{Z}}[x]\). We
know \(\alpha= a/b\) with \(a,b \in {\mathbb{Z}}\), and we can use the
rational root test which tells us that \(a\divides a_0\) and
\(b\divides 1\), so \(b = \pm 1\) and
\(\alpha = a/\pm 1 = \pm a \in {\mathbb{Z}}\) and thus
\(\alpha \in {\mathbb{Z}}\).

\end{proof}

\begin{remark}

We'll want to study \({\mathbb{Z}}_K\) for various number fields \(K\),
but we'll need more groundwork.

\end{remark}

\begin{proposition}[Easy criterion to check if an integer is algebraic]

Let
\(\alpha \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\),
then
\begin{align*} \alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\iff \min_ \alpha \in {\mathbb{Z}}[x], \end{align*}
where \(\min_ \alpha(x)\) is the unique monic irreducible polynomial in
\({\mathbb{Q}}[x]\) which vanishes at \(\alpha\).

\end{proposition}

\begin{proof}[?]

\(\impliedby\): Trivial, if the minimal polynomial already has integer
coefficients, just note that it's already monic and thus
\(\alpha \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
by definition.\\

\(\implies\): Why should the minimal polynomial have \emph{integer}
coefficients? Choose a monic \(f(x) \in {\mathbb{Z}}[x]\) with
\(f(\alpha) = 0\), using the fact that
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
, and factor \(f(x) = \prod_{i=1}^n (x- \alpha_i) \in {\mathbb{C}}[x]\).
Note that each
\(\alpha_i \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
since they are all roots of \(f\) (a monic polynomial in
\({\mathbb{Z}}[x]\)). Use the fact that \(\min_ \alpha(x)\) divides
every polynomial which vanishes on \(\alpha\) over \({\mathbb{Q}}\), and
thus divides \(f\) (noting that this still divides over
\({\mathbb{C}}\)). Moreover, every root of \(\min_ \alpha(x)\) is a root
of \(f\), and so every such root is some \(\alpha_i\).

\hfill\break

Now factor \(\min_ \alpha(x)\) over \({\mathbb{C}}\) to obtain
\(\min_ \alpha(x) = \prod_{i=1}^m (x - \beta_i)\) with all of the
\(\beta_i \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
What coefficients appear after multiplying things out? Just sums and
products of the \(\beta_i\), so all of the coefficients are in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
. Thus
\(\min_ \alpha(x) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu[x]\).
But the coefficients are also in \({\mathbb{Q}}\) by definition, so the
coefficients are in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap{\mathbb{Q}}= {\mathbb{Z}}\)
and thus \(\min_ \alpha(x) \in {\mathbb{Z}}[x]\).

\end{proof}

\begin{example}[Showing an integer is not algebraic using minimal polynomials]

\(\sqrt{5}/ 3 \not\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
since \(\min_ \alpha(x) = x^2 - 5/9 \not\in {\mathbb{Z}}[x]\), so this
is not an algebraic integer.

\end{example}

\begin{proposition}[$\ff(\ZZ_K) = K$ ]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
  has
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\)
  as its fraction field, and
\item
  For any number field \(K\), the fraction field of \({\mathbb{Z}}_K\)
  is \(K\).
\item
  If
  \(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\)
  then
  \(d \alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
  for some \(d\in {\mathbb{Z}}^{\geq 0}\)
\end{enumerate}

Moreover, both (a) and (b) follow from (c).

\end{proposition}

\begin{remark}

Thus the subring is ``big'' in the sense that if you allow taking
quotients, you recover the entire field. That \(c\implies a,b\): suppose
you want to write
\(\alpha \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\)
as \(\alpha=p/q\) with
\(p,q \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
Use \(c\) to produce
\(d\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\),
then just take \(d\alpha /d\). The same argument works for \(b\).

\end{remark}

\begin{exercise}[?]

Prove the proposition!

\end{exercise}

\begin{proposition}[The ring $\bar{\ZZ}$ contains all roots of monic polynomials with integer coefficients]

Suppose
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{C}}\mkern-1.5mu}\mkern 1.5mu\)
and \(\alpha\) is a root of a monic polynomial in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu [x]\).
Then
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).

\end{proposition}

\begin{remark}

This says that if a number \(\alpha\) is the root of a monic polynomial
whose coefficients are \emph{algebraic} integers, then \(\alpha\) itself
is an algebraic integer coefficients. This corresponds to the fact that
integral over integral implies integral in commutative algebra.

\end{remark}

\begin{exercise}[Prove the proposition.]

Prove this! One can use the integrality criterion (slightly
challenging), or alternatively Galois theory.

\end{exercise}

\hypertarget{quadratic-fields-lec.-3-thursday-january-21}{%
\section{Quadratic Fields (Lec. 3, Thursday, January
21)}\label{quadratic-fields-lec.-3-thursday-january-21}}

\begin{remark}

Today: roughly corresponds to chapter 3 in the book. Goal: do all of the
big theorems in the setting of quadratic number fields, then redo
everything for general number fields.

\end{remark}

\hypertarget{quadratic-number-fields}{%
\subsection{Quadratic Number Fields}\label{quadratic-number-fields}}

\begin{remark}

Simplest case: \({\mathbb{Q}}\), a degree 1 number field, so the next
simplest case is degree 2.

\end{remark}

\begin{definition}[Quadratic Number Fields]

A field \(K\) is a \textbf{quadratic number field} if and only if \(K\)
is a number field and \([K: {\mathbb{Q}}] = 2\).

\end{definition}

\begin{remark}

Some notation: if \(d\in {\mathbb{R}}^{\times}\), then \(\sqrt d\) means
the \emph{positive} square root of \(d\) if \(d \geq 0\), and if \(d<0\)
this denotes \(i\sqrt{{\left\lvert {d} \right\rvert}}\).

\end{remark}

\begin{proposition}[Quadratic fields are parameterized by squarefree integers]

If \(K\) is a quadratic number field, then
\(K = {\mathbb{Q}}(\sqrt{d})\) for some squarefree \footnote{\emph{Squarefree}
  means not divisible by \(n^2\) for any \(n > 1\in {\mathbb{Z}}\), or
  equivalently not divisible by the square of any primes.} integer
\(d\in {\mathbb{Z}}\). Moreover, this \(d\) is uniquely determined by
\(K\), so all quadratic number fields are parameterized by the set of
squarefree integers.

\end{proposition}

\begin{proof}[of proposition, existence]

\textbf{Existence}: Since \([K: {\mathbb{Q}}] = 2\), we have
\(K\supsetneq {\mathbb{Q}}\) so pick
\(\alpha\in K\setminus{\mathbb{Q}}\) then \(K = {\mathbb{Q}}(\alpha)\).
Note that we could also furnish this \(\alpha\) from the primitive
element theorem, although this is overkill here. So \(\alpha\) is a root
of some degree 2 \(p\in {\mathbb{Q}}[x]\), and by scaling coefficients
we can replace this by \(p\in {\mathbb{Z}}[x]\). So write
\(p(x) = Ax^2 + Bx + C\), in which case we can always write
\begin{align*}
\alpha = {-B \pm \sqrt{B^2 - 4AC} \over 2A} 
\end{align*}
where \(A\neq 0\), since this would imply that
\(\alpha\in{\mathbb{Q}}\). Writing \(\Delta\coloneqq B^2 - 4AC\), we
have \(K = {\mathbb{Q}}(\alpha) = {\mathbb{Q}}(\sqrt{\Delta})\). This is
close to what we want -- it's \({\mathbb{Q}}\) adjoin some integer --
but we'd like that integer to be squarefree.

\hfill\break

Now let \(f\in {\mathbb{Z}}^{\geq 0}\) be chosen such that
\(f^2 \divides \Delta\) and \(f\) is as large as possible, i.e.~the
largest square factor of \(\Delta\). Writing \(\Delta = f^2 - d\) where
\(d\) is whatever remains. Then \(d\) must be squarefree, otherwise if
\(d\) had a square factor bigger than 1, say \(d = r^2 d'\), in which
case \(f^2 r^2 > f^2\) would be a larger factor of \(\Delta\). So \(d\)
is squarefree, and \(\Delta = f \sqrt d\) and thus
\({\mathbb{Q}}(\Delta) = {\mathbb{Q}}(\sqrt{d})\).

\textbf{Uniqueness}: Well use some extra machinery.

\end{proof}

\hypertarget{norm-and-trace}{%
\subsection{Norm and Trace}\label{norm-and-trace}}

\begin{definition}[Norm and Trace]

Let \(K\) be a number field with \(K_{/{\mathbb{Q}}}\) Galois. For each
\(\alpha\in K\) define
\begin{align*}
N(\alpha) &\coloneqq\prod_{\sigma\in { \mathsf{Gal}} (K_{/{\mathbb{Q}}})} \sigma(\alpha) && \text{the norm} \\
\operatorname{Tr}(\alpha) &\coloneqq\sum_{\sigma\in { \mathsf{Gal}} (K_{/{\mathbb{Q}}})} \sigma(\alpha) && \text{the trace}
.\end{align*}

\end{definition}

\begin{remark}

Why use these kind of sum at all? Applying any element in the Galois
group just permutes the elements. Note that
\(N( \alpha), \operatorname{Tr}( \alpha)\) are
\(G(K_{/{\mathbb{Q}}}){\hbox{-}}\)invariant, and thus rational numbers
in \({\mathbb{Q}}\). The norm is multiplicative, and the trace is
additive and in fact \({\mathbb{Q}}{\hbox{-}}\)linear:
\(\operatorname{Tr}(a \alpha + b \beta) = a \operatorname{Tr}( \alpha) + b \operatorname{Tr}( \beta)\)
for all \(\alpha, \beta\in K\) and all \(a,b \in {\mathbb{Q}}\).

\end{remark}

\begin{remark}

What do the norm and trace look like for a quadratic field? We can write
\(K = \left\{{a + b \sqrt d {~\mathrel{\Big|}~}a,b \in {\mathbb{Q}}}\right\}\)
and there is a unique (non-identity) element
\(g\in { \mathsf{Gal}} (K_{/{\mathbb{Q}}})\) with
\(\sigma(a + b \sqrt d) = a - b \sqrt{d}\). We'll refer to this
automorphism as \textbf{conjugation}. We can compute
\begin{align*}
N(a + b \sqrt{d} ) &= a^2 - db^2 \\
\operatorname{Tr}(a + b \sqrt{d} ) &= 2a
.\end{align*}

\end{remark}

\begin{proof}[of proposition, uniqueness continued]

Returning to the proof, suppose otherwise that
\(K = {\mathbb{Q}}(\sqrt{d_1} ) = {\mathbb{Q}}( \sqrt{d_2} )\) with
\(d_1\neq d_2\) squarefree integers. Note that they must have the same
sign, otherwise one of these extensions would not be a subfield of
\({\mathbb{R}}\). We know \(\sqrt{d_1} \in {\mathbb{Q}}( \sqrt{d_2} )\)
and thus \(\sqrt{d_1} = a + b \sqrt{d_2}\) for some
\(a, b\in {\mathbb{Q}}\).

\hfill\break

Taking the trace of both sides, the LHS is zero and the RHS is \(2a\)
and we get \(a=0\) and \(\sqrt{d_1} = b \sqrt{d_2}\). Write \(b = u/v\)
with \(u,v\in {\mathbb{Q}}\). Squaring both sides yields
\(v^2 d_1 = u^2 d_2\). Let \(p\) be a prime dividing \(d_1\); then since
\(d_1\) is squarefree there is only one copy of \(p\) occurring in its
factorization. Moreover there are an even number of copies of \(p\)
coming from \(v^2\), thus forcing \(d_2\) to have an odd power of \(p\).
This forces \(p\divides d_2\), and since this holds for every prime
factor \(p\) of \(d_1\), we get \(d_1 \divides d_2\) since \(d_1\) is
squarefree. The same argument shows that \(d_2 \divides d_1\), so
they're the same up to sign: but the signs must match and we get
\(d_1 = d_2\).

\end{proof}

\begin{remark}

Note that this results holds for every squarefree number not equal to 1.
If \(K = {\mathbb{Q}}( \sqrt{d} )\), what is the ring of integers
\({\mathbb{Z}}_K\)? Some more machinery will help here.

\end{remark}

\hypertarget{the-field-polynomial}{%
\subsection{The Field Polynomial}\label{the-field-polynomial}}

\begin{definition}[The Field Polynomial of an Element]

Assume \(K_{/{\mathbb{Q}}}\) is a Galois number field and for
\(\alpha\in K\) define the \textbf{field polynomial of \(\alpha\)} as
\begin{align*}
\varphi_{\alpha}(x) \coloneqq\prod_{ \sigma\in { \mathsf{Gal}} (K_{/{\mathbb{Q}}})} \qty{ x - \sigma(\alpha)}
.\end{align*}

\end{definition}

\begin{remark}

For the same reasons mentioned for the norm/trace, we get
\(\varphi_{\alpha} \in {\mathbb{Q}}[x]\), and moreover
\(\varphi_{ \alpha } (\alpha) = 0\). When is
\(\alpha\in {\mathbb{Z}}_K\)? We have the following criterion:

\end{remark}

\begin{proposition}[The field polynomial detects integrality]

\begin{align*}
\alpha\in {\mathbb{Z}}_K \iff \varphi_{ \alpha } (x) \in {\mathbb{Z}}[x]
.\end{align*}

\end{proposition}

\begin{proof}[of proposition]

\(\impliedby\): This is easy, since if \(\varphi_\alpha\) is a monic
polynomial with integer coefficients, meaning that \(\alpha\) is an
algebraic integer and thus in \({\mathbb{Z}}_K\).

\hfill\break

\(\implies\): If \(\alpha \in {\mathbb{Z}}_K\) then it's the root of
some monic polynomial in \({\mathbb{Z}}[x]\), and the same is true for
\(\sigma(\alpha)\) and thus each
\(\sigma(\alpha) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
So
\(\varphi_{ \alpha}(x) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu[x]\).
We said \(\varphi_{ \alpha}\) has coefficients in \({\mathbb{Q}}\) too,
and thus in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap{\mathbb{Q}}= {\mathbb{Z}}\).
So the problem is reduced to finding out when \(\varphi_{\alpha}(x)\)
has integer coefficients.

If \(\deg(K_{/{\mathbb{Q}}}) = n\), then
\begin{align*}
\varphi_{ \alpha}( \alpha) = \prod x- \sigma(\alpha) = x^n - \operatorname{Tr}(\alpha)x^{n-1} + \cdots + (-1)^n N( \alpha)
.\end{align*}
If \(n=2\), these are the only terms, and so if \(K\) is a quadratic
number field then \(\alpha\in K\) is in \({\mathbb{Z}}_K\) if and only
if \(\operatorname{Tr}( \alpha), N(\alpha) \in {\mathbb{Z}}\).

\end{proof}

\begin{example}[of nonintuitive rings of integers]

Let \(K = {\mathbb{Q}}( \sqrt{5} )\), then is it true that
\({\mathbb{Z}}_K = {\mathbb{Z}}[\sqrt{5} ]\)? Since
\(1, \sqrt{5} \in {\mathbb{Z}}_K\), we have \(\supseteq\) since
\(1, \sqrt{5}\) are algebraic. The answer is \textbf{no}: take
\(\alpha\coloneqq{1 + \sqrt{5} \over 2}\), then \(N( \alpha) -4/4 = -1\)
and \(\operatorname{Tr}( \alpha) = 1\). These are integers, so
\(\alpha\in {\mathbb{Z}}_K\), and in fact \(\alpha\) is a root of
\(x^2 - x - 1 \in {\mathbb{Z}}[x]\).

\end{example}

\hypertarget{classification-of-mathbbz_k}{%
\subsection{\texorpdfstring{Classification of
\({\mathbb{Z}}_K\)}{Classification of \{\textbackslash mathbb\{Z\}\}\_K}}\label{classification-of-mathbbz_k}}

\begin{theorem}[Classification of $\ZZ_K$ for quadratic fields]

Let \(K = {\mathbb{Q}}( \sqrt{d} )\) be a quadratic number field. Then

\begin{itemize}
\item
  If \(d = 2,3 \operatorname{mod}4\), then
  \({\mathbb{Z}}_K = \left\{{ a + b \sqrt{d} {~\mathrel{\Big|}~}a, b\in {\mathbb{Z}}}\right\}\).
\item
  If \(d=1 \operatorname{mod}4\), then
  \({\mathbb{Z}}_K = \left\{{ {1 + b \sqrt{d} \over 2} {~\mathrel{\Big|}~}a,b\in {\mathbb{Z}},\, a\equiv b \operatorname{mod}2}\right\}\).
\end{itemize}

\end{theorem}

\begin{remark}

For \(d=1\), if \(a, b\) are even then we just recover the \(d=2,3\)
case, so we're picking up extra elements from when \(a,b\) both odd.

\end{remark}

\begin{proof}[?]

Let \(\alpha\in K\) and write \(\alpha = A + B \sqrt{d}\) with
\(A, B\in {\mathbb{Q}}\).

\begin{exercise}[?]

Check that \(N( \alpha), \operatorname{Tr}( \alpha) \in {\mathbb{Z}}\)
for both cases.

\end{exercise}

Assuming now that
\(N( \alpha), \operatorname{Tr}( \alpha) \in {\mathbb{Z}}\), then
\(A^2 - dB^2 \in {\mathbb{Z}}\). Multiply this by 2 to get
\((2A)^2 - d(2B)^2 \in 4{\mathbb{Z}}\). Recalling that
\(\operatorname{Tr}( \alpha) = 2 A\), we have
\((2A)^2 \in {\mathbb{Z}}\) and thus \(d(2B)^2 \in {\mathbb{Z}}\) as
well. The claim now is that \(2B \in {\mathbb{Z}}\): we know
\(2B\in {\mathbb{Q}}\). If \(2B\not\in {\mathbb{Z}}\), then the
denominator has some prime factor. This prime factor appears twice in
\((2B)^2\), and \(d(2B)^2 \in {\mathbb{Z}}\) then means that two copies
of \(p\) appear in \(d\) in order to cancel -- however, we assumed \(d\)
was squarefree. We now know that \(A, B \in {1\over 2}{\mathbb{Z}}\), so
write \(A = (1/2)a'\) and \(B = (1/2)b'\). Thus
\begin{align*} 
\alpha= (1/2)a' + (1/2)b' \sqrt{d} 
\implies
N( \alpha) = ((a')^2 - d(b')^2) / 4 \in {\mathbb{Z}}
.\end{align*}
So the numerator is a multiple of 4, which yields
\((a')^2 \equiv d(b')^2 \operatorname{mod}4\). We proceed by cases.

\hfill\break

\textbf{Case 1:} \(d = 2,3 \operatorname{mod}4\). If \(b'\) is odd then
\((b')^2 = 1\operatorname{mod}4\), which holds for any odd number. But
then \((a')^2 = d(b')^2 = d \operatorname{mod}4\), which is a problem --
squares modulo 4 can only be \(0\) or \(1\). This is a contradiction, so
\(b'\) must be even. Then \((b')^2 \operatorname{mod}4 = 0\), which
forces \(a' \equiv 0 \operatorname{mod}4\) and \(a'\) must be even. But
if \(a', b'\) are both even, \((1/2)a', (1/2)b'\in {\mathbb{Z}}\) and we
obtain \(\alpha\in {\mathbb{Z}}+ \sqrt{d} {\mathbb{Z}}\) .

\hfill\break

\textbf{Case 2:} If \(d\equiv 1 \operatorname{mod}4\), then
\((a')^2 \equiv (b')^2 \operatorname{mod}4\). We can conclude that
\(a', b'\) are either both odd or both even, otherwise we'd get
\(0\equiv 1 \operatorname{mod}4\), and thus we can write
\(a' \equiv b' \operatorname{mod}2\). But this was exactly the condition
appearing in the theorem.

\end{proof}

\begin{remark}

Let \(K\) be a quadratic number field. Then we can reformulate the
previous results as:

\begin{align*}
{\mathbb{Z}}_K = 
\begin{cases}
{\mathbb{Z}}[ \sqrt{d} ] &  d \equiv 2,3 \operatorname{mod}4
\\
{\mathbb{Z}}\left[{1 + \sqrt{d} \over 2}\right] & d \equiv 1 \operatorname{mod}4.
\end{cases}
\end{align*}

We've also shown that \({\mathbb{Z}}_K\) is a free
\({\mathbb{Z}}{\hbox{-}}\)module of rank 2, with basis either
\(\left\{{ 1, \sqrt{d} }\right\}\) or
\(\left\{{ 1, {1 + \sqrt{d} \over 2 } }\right\}\).

\end{remark}

\begin{remark}

What is true for general number fields? Important theorem:
\({\mathbb{Z}}_K\) is always a free \({\mathbb{Z}}{\hbox{-}}\)module,
i.e.~there always exists an \emph{integral basis}. Surprisingly, the
it's not always true that \({\mathbb{Z}}_K = {\mathbb{Z}}[\ell]\) for
\(\ell\) a single element.

\end{remark}

\hypertarget{failure-of-unique-factorization-lec.-4-wednesday-january-27}{%
\section{Failure of Unique Factorization (Lec. 4, Wednesday, January
27)}\label{failure-of-unique-factorization-lec.-4-wednesday-january-27}}

\hypertarget{revisiting-a-counterexample-to-unique-factorization}{%
\subsection{Revisiting a Counterexample to Unique
Factorization}\label{revisiting-a-counterexample-to-unique-factorization}}

\begin{remark}

Today roughly corresponds to chapter 4: ``Paradise Lost''! Setup: \(K\)
is a quadratic field, a degree 2 extension of \({\mathbb{Q}}\), which
can be written as \(K = {\mathbb{Q}}(\sqrt{d})\) with \(d\) squarefree.
Last time, we completely described \({\mathbb{Z}}_K\) (the algebraic
integers in \(K\)):
\begin{align*}
{\mathbb{Z}}_K = 
\begin{cases}
{\mathbb{Z}}[ \sqrt{d} ] &  d \equiv 2,3 \operatorname{mod}4
\\
{\mathbb{Z}}\left[{1 + \sqrt{d} \over 2}\right] & d \equiv 1 \operatorname{mod}4.
\end{cases}
\end{align*}
We saw that the second admitted a different description as
\(\left\{{ {a + b \sqrt{d} \over 2}}\right\}\) where \(a,b\) are either
both even or both odd. Note that we can do interesting arithmetic in
\({\mathbb{Z}}_K\), but it's not necessarily well-behaved:
\({\mathbb{Z}}_K\) is not always a UFD.

\end{remark}

\begin{example}[A counterexample to unique factorization]

Letting \(d=-5\), we have \({\mathbb{Z}}_K = {\mathbb{Z}}[ \sqrt{-5} ]\)
where \(6\) factors in two ways:
\begin{align*}
6 = (1 + \sqrt{5} )(1 - \sqrt{-5} ) = (2)(3) = 6
.\end{align*}

Note that this isn't quite enough to show failure of unique
factorization, e.g.~we can factor \(16 = (4)(4) = (2)(8)\). Here you
should check that all 4 factors are irreducible, and that the factors on
the right aren't unit multiples of the ones on the left. For example,
\(21 = (-7)(-3) = (7)(3)\), but the factors only differ by the unit
\(-1\in {\mathbb{Z}}^{\times}\). The key to checking all of those: the
\textbf{norm map}:
\begin{align*}
N(a + b \sqrt{-5} ) = (a + b \sqrt{-5} ) (a - \sqrt{-5} ) = a^2 + 5b^2
.\end{align*}
where the second factor was the \emph{conjugate}, i.e.~the image of the
element under the nontrivial element of the Galois group of
\(K_{/{\mathbb{Q}}}\). If \(a + b \sqrt{-5} \in {\mathbb{Z}}_K\), then
\(N(a + b \sqrt{-5} \in {\mathbb{Z}}_{\geq 0})\) and is equal to zero if
and only if \(a + b \sqrt{-5} = 0\). Moreover, this is a unit if and
only if its norm is 1, \footnote{\(\impliedby\): If the norm is 1, the
  conjugate is the inverse. For the reverse direction, the argument was
  more complicated, and reduced to showing norms of units are \(\pm 1\),
  and positivity forces it to be \(1\).} i.e.~\(a^2 + 5b^2 = 1\), which
forces \(b=0\) and \(a=\pm 1\). So
\(U({\mathbb{Z}}[ \sqrt{-5} ] ) = \left\{{\pm 1}\right\}\).

We'll show one of the factors is irreducible, \(1 + \sqrt{-5}\). Recall
that \(x\in R\) a domain is \textbf{irreducible} if and only if whenever
\(x = ab\), one of \(a,b\) is a unit. It itself is not a unit, since
\(N(1 + \sqrt{-5 }) = 6 \neq 1\). So suppose
\(1 + \sqrt{-5} = \alpha \beta\). Then
\begin{align*}
6 = N(\alpha \beta ) = N( \alpha) N( \beta)
,\end{align*}
and so up to reordering, we have \(N \alpha = 2, N \beta= 3\). Writing
\(\alpha= a + b \sqrt{-5}\) and taking norms yields \(2 = a^2 + 5b^2\),
which has no solutions: considering the equation \(\operatorname{mod}5\)
yields \(2\equiv a^2\), but \(2\) is not a square in
\({\mathbb{Z}}/5{\mathbb{Z}}\). \(\contradiction\)

Note that the only other way of factoring \(6\) is \(6=(1)(6)\), and
taking norms shows that one factor is a unit. So if we assume
\(\alpha, \beta\) aren't units, both \(N \alpha, N \beta > 1\), which
leads to the previous situation. By similar arguments, all 4 factors are
irreducible.

To see that the LHS factors aren't unit multiples of the RHS factors, we
can use the fact that the units are \(\pm 1\), and multiplying the LHS
by \(\pm 1\) can't yield \(2\) or \(3\). So this is a genuine
counterexample to unique factorization.

\end{example}

\hypertarget{factorization-theory}{%
\subsection{Factorization Theory}\label{factorization-theory}}

\begin{remark}

What went wrong in the previous example? We'll use a big of terminology
from an area of algebra called \emph{factorization theory}. Many
concepts related to divisibility can be discussed in this language!

\end{remark}

\begin{definition}[Monoid]

A \textbf{monoid} is a nonempty set with a commutative associative
binary operation \(\cdot\) with an identity \(1\). We say a monoid is
\textbf{cancellative} if and only if whenever
\(\alpha \beta= \beta \alpha\) or \(\beta \alpha = \gamma \alpha\) then
\(\beta = \gamma\).

\end{definition}

\begin{definition}[Terminology for Cancellative Monoids]

Let \(M\) be a cancellative monoid. Then

\begin{itemize}
\item
  \(\alpha\divides \beta\) if and only if \(\beta= \alpha \gamma\) for
  some \(\gamma\).
\item
  \(\epsilon\) is a \textbf{unit} if \(\epsilon\divides 1\).
\item
  \(\alpha , \beta\) are \textbf{associates} if
  \(\alpha = \epsilon \beta\) for some unit \(\epsilon\)
\item
  \(\pi\in M\) is \textbf{irreducible} if and only if \(\pi\) is a unit
  and whenever \(\pi= \alpha \beta\) then either \(\alpha\) or \(\beta\)
  is a unit.
\item
  \(\pi \in M\) is \textbf{prime} whenever \(\pi\divides \alpha \beta\)
  then \(\pi\divides \alpha\) or \(\pi\divides \beta\).
\item
  \(\delta \in M\) is a greatest common divisor of \(\alpha, \beta\) if
  and only if \(\delta\) is a common divisor that is divisible by every
  other common divisor.
\item
  \(M\) is a \textbf{unique factorization monoid} if and only if every
  nonunit element in \(M\) factors uniquely (up to order and associates)
  as a product of irreducibles.
\end{itemize}

\end{definition}

\begin{remark}

Given \(R\) an integral domain, then \(R\setminus\left\{{0}\right\}\)
with multiplication is a cancellative monoid. Moreover,
\(R\setminus\left\{{0}\right\}\) is a unique factorization monoid if and
only if \(R\) is a UFD.

\end{remark}

\begin{question}

How do you show something is a UFD?

\end{question}

\begin{answer}

Recall how this proof went for \({\mathbb{Z}}\):

\begin{itemize}
\tightlist
\item
  Use existence of a division algorithm.
\item
  Prove Euclid's lemma: every irreducible is prime.
\item
  Use factorization into irreducibles and proceed by induction, writing
  out two factorizations and cancelling things out in a combinatorial
  way.
\end{itemize}

So we'd like

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  To know that irreducibles are prime, and
\item
  Everything to factor into irreducibles.
\end{enumerate}

\end{answer}

\begin{definition}[Atomic]

For \(M\) a cancellative monoid, \(M\) is \textbf{atomic} if every
nonunit element of \(M\) is a product of irreducibles.

\end{definition}

\begin{proposition}[Monoids have unique factorization iff atomic and irreducibles are prime]

Let \(M\) be a cancellative monoid, then \(M\) is a UFM if and only if
\(M\) is atomic and every irreducible is prime in \(M\).

\end{proposition}

\begin{proof}[of proposition]

Omitted -- no new ideas when compared to proof of unique factorization
in \({\mathbb{Z}}\).

\end{proof}

\begin{remark}

Note that in \({\mathbb{Z}}\), working in \({\mathbb{Z}}_{\geq 0}\) is
useful because the only positive unit is \(1\), and so any elements
differing by a unit are in fact equal. Can we emulate this for
cancellative monoids? The answer is yes, by modding out by the
equivalence relation of being equivalent up to a unit.

\end{remark}

\begin{definition}[Reduced Monoid]

Define \(M_{{ \text{red} }}\coloneqq M/\sim\) where
\(a\sim b \iff a-b\in M^{\times}\). The operation on \(M\) descends to
well-defined operation on \(M_{{ \text{red} }}\), and irreducibles and
primes are the same in \(M\) and \(M_{{ \text{red} }}\).

\end{definition}

\begin{example}[of a more familiar reduced monoid]

This is supposed to look like \({\mathbb{Z}}_{\geq 0}\), where
\(-7\in M \mapsto 7 \in M_{{ \text{red} }}\).

\end{example}

\begin{proposition}[A monoid has unique factorizations iff its reduced monoid does]

\(M\) is a UFM if and only if \(M_{{ \text{red} }}\) is a UFM if and
only if every element of \(M_{{ \text{red} }}\) factors uniquely as a
product of irreducibles, up to order.

\end{proposition}

\begin{remark}

What did this buy us? We didn't have to worry about associates in the
above statement, and the only unit is 1.

\end{remark}

\begin{remark}

Why isn't \({\mathbb{Z}}[ \sqrt{-5} ]\) is UFD? It doesn't have enough
elements to make unique factorization work!

\end{remark}

\begin{example}[of common refinements]

In \({\mathbb{Z}}^+\), write \(210 = 21\cdot 10 = 14 \cdot 15\). These
two factorizations differ but admit a common refinement to
\((7\cdot 3)(2\cdot 5) = (7\cdot 2)(3\cdot 5)\), where it becomes clear
that these factorizations are equal up to ordering. This is
\textbf{Euler's Four Number Theorem}, which turns out to be equivalent
to unique factorization.

\end{example}

\begin{theorem}[Characterization of unique factorization monoids]

Let \(M\) be a cancellative atomic reduced monoid. Then \(M\) is a UFM
if and only if whenever \(\alpha, \beta, \gamma, \delta \in M\) such
that \(\alpha \beta = \gamma \delta\), there are
\(\rho, \sigma, \tau, \nu\) with
\begin{align*}
\alpha &= \rho \sigma \\
\beta &= \tau \nu \\
\gamma &= \rho \tau \\
\delta &= \sigma \nu
.\end{align*}

Note that plugging these in on the LHS and RHS respectively yield the
same factors, just reordered.

\end{theorem}

\begin{proof}[of theorem]

Omitted, exercise in chasing definitions. The interesting part is that
you can go backward!

\end{proof}

\begin{remark}

Let
\(M_{{ \text{red} }} \coloneqq\qty{{\mathbb{Z}}[ \sqrt{5} ] \setminus\left\{{0}\right\}}_{{ \text{red} }}\),
motivated by the fact that \({\mathbb{Z}}[ \sqrt{-5} ]\) is not a UFD if
\({\mathbb{Z}}[ \sqrt{-5} ] \setminus\left\{{0}\right\}\) is not a UFM,
or equivalently its reduction is not a UFM. Then \(M\) is a not a UFM.
Noting that \(M\) is reduced under an equivalence relation, write
\(\left\langle{ \alpha}\right\rangle\) for the class of \(\alpha\) in
\(M\) for any \(\alpha\in {\mathbb{Z}}[ \sqrt{-5} ]\).

Our original counterexample for unique factorization now reads
\begin{align*}
\left\langle{ 1 + \sqrt{-5} }\right\rangle \left\langle{ 1 - \sqrt{5} }\right\rangle = \left\langle{2}\right\rangle \left\langle{3}\right\rangle
.\end{align*}
This is still a counterexample since these pairs admit no common
refinement.

Why are there ``not enough elements'' in \({\mathbb{Z}}[ \sqrt{-5} ]\)?
Recall that for integral domains (as rings), two elements differ by a
unit precisely when they generate the same ideal. So we can think of
elements of \(M_{{ \text{red} }}\) as nonzero principal ideals of \(M\),
which we'll write as
\(\mathop{\mathrm{Prin}}( {\mathbb{Z}}[ \sqrt{-5} ])\). To make this set
of ideals into a monoid, one define
\(\left\langle{ \alpha }\right\rangle \left\langle{ \beta }\right\rangle= \left\langle{ \alpha \beta }\right\rangle\),
where it's easy to check that this is well-defined. So the failure of
unique factorization is a failure of factorization in this set of
ideals. We can embed this in a larger collection of ideals by just
deleting the word ``principal'', which will restore unique
factorization.

\end{remark}

\begin{definition}[Multiplication of Ideals]

Let \(R\) be a commutative ring (always with 1). If
\(I, J {~\trianglelefteq~}R\) are ideals, we define
\begin{align*}
IJ \coloneqq\left\langle{ \left\{{ \alpha_i \beta_i {~\mathrel{\Big|}~}\alpha_i \in I, \beta_i \in J }\right\}}\right\rangle = \left\{{ \sum \alpha_i \beta_i {~\mathrel{\Big|}~}\alpha_i \in I, \beta_i \in J}\right\}
.\end{align*}

If \(R\) is a domain, define the monoid \(\operatorname{Id}(R)\) the
collection of nonzero ideals of \(R\) with the above multiplication.

\end{definition}

\begin{remark}

Note that the naive definition
\(IJ \coloneqq\left\{{ij{~\mathrel{\Big|}~}i\in I, j\in J}\right\}\) is
not necessarily an ideal, since it may not be closed under addition.
Taking the smallest ideal containing all products fixes this.

\end{remark}

\begin{proposition}[If $R$ is a domain, then $\Id(R)$ is a monoid]

Let \(R\) be a commutative ring. Then

\begin{itemize}
\tightlist
\item
  Multiplication \(\cdot\) for ideals is commutative,
\item
  Multiplication \(\cdot\) for ideals is associative,
\item
  The identity is \(\left\langle{ 1 }\right\rangle= R\),
\item
  Multiplication distributes over addition of ideals,
  i.e.~\(I(J+K) = IJ + IK\),
\item
  \(IJ \subseteq I \cap J\),
\item
  If \(I = \left\langle{ \alpha_1, \cdots, \alpha_j }\right\rangle\) and
  \(J = \left\langle{ \beta_1, \cdots, \beta_k }\right\rangle\) then
  \(IJ = \left\langle{ \alpha_1 \beta_1, \cdots, \alpha_j \beta_k }\right\rangle\)
  is generated by all of the \(jk\) pairwise products,
\item
  If \(R\) is a domain and \(I, J\) are nonzero then \(IJ\) is nonzero,
\end{itemize}

As a consequence, \(\operatorname{Id}(R)\) is a monoid when \(R\) is a
domain.

\end{proposition}

\begin{remark}

So instead of working in
\(\mathop{\mathrm{Prin}}( {\mathbb{Z}}[\sqrt{-5} ])\), we'll work in
\(\operatorname{Id}({\mathbb{Z}}[\sqrt{-5} ])\). The claim is that we
can refine our bad factorizations. Define

\begin{itemize}
\tightlist
\item
  \(I\coloneqq\left\langle{ 1 + \sqrt{-5} , 2 }\right\rangle\)
\item
  \(I'\coloneqq\left\langle{ 1 - \sqrt{-5} , 2 }\right\rangle\)
\item
  \(J\coloneqq\left\langle{ 1 + \sqrt{-5} , 3 }\right\rangle\)
\item
  \(J'\coloneqq\left\langle{ 1 - \sqrt{-5} , 3 }\right\rangle\)
\end{itemize}

Then

\begin{itemize}
\tightlist
\item
  \(IJ = \left\langle{ 1 + \sqrt{-5} }\right\rangle\)\\
\item
  \(I'J' = \left\langle{ 1 - \sqrt{-5} }\right\rangle\)\\
\item
  \(JJ' = \left\langle{ 3 }\right\rangle\)
\item
  \(II' = \left\langle{ 2 }\right\rangle\)
\end{itemize}

We can then write
\begin{align*}
\left\langle{ 1 + \sqrt{-5} }\right\rangle \left\langle{ 1 - \sqrt{-5} }\right\rangle = \left\langle{ 2 }\right\rangle \left\langle{ 3 }\right\rangle \implies (IJ)(I'J') = (II')(JJ')    
,\end{align*}
where the same terms are occurring in a different order. For an example
of how to work these out, let's compute \(IJ\). We get
\begin{align*}
IJ 
&= \left\langle{ (1 + \sqrt{-5} )^2, 3(1 + \sqrt{-5} ), 2(1 + \sqrt{-5} ), 6 }\right\rangle  \\
&= \left\langle{ 1 + \sqrt{-5} }\right\rangle \left\langle{ 1 + \sqrt{-5} , 3, 2, 1 - \sqrt{-5} }\right\rangle \\  
&= \left\langle{ 1 + \sqrt{-5} }\right\rangle \left\langle{ 1 }\right\rangle  \\
&= \left\langle{ 1 + \sqrt{-5} }\right\rangle
,\end{align*}
using the fact that \(3-2=1\) is in the ideal on the second line.

\hfill\break

We'll see later that this process allows you to recover unique
factorization in \({\mathbb{Z}}_K\) for any number field \(K\).

\end{remark}

\hypertarget{euclidean-quadratic-fields-lec.-5-thursday-january-28}{%
\section{Euclidean Quadratic Fields (Lec. 5, Thursday, January
28)}\label{euclidean-quadratic-fields-lec.-5-thursday-january-28}}

\hypertarget{setup}{%
\subsection{Setup}\label{setup}}

\begin{remark}

Today: roughly corresponds to Chapter 5. In a first algebra course, one
shows that if \(R\) is a Euclidean domain, then the arithmetic of \(R\)
is very interesting:

\begin{itemize}
\tightlist
\item
  \(R\) is a PID, and as a consequence
\item
  \(R\) is a UFD
\end{itemize}

\end{remark}

\begin{definition}[Euclidean Domain]

A domain \(R\) is \textbf{Euclidean} if and only if there is a function
\(\varphi R\setminus\left\{{0}\right\}\to {\mathbb{Z}}^{\geq 0}\) such
that for all \(a,b\in R\) with \(b\neq 0\) there are \(q, r\in R\) with
\(a = bq + r\) with \(r=0\) or \(\varphi(r) < \varphi(b)\). \(\varphi\)
is referred to as a \textbf{Euclidean function}.

\end{definition}

\begin{example}[Examples of Euclidean functions]

\envlist

\begin{itemize}
\tightlist
\item
  For \(R={\mathbb{Z}}\), one can take
  \(\varphi({-}) \coloneqq{\left\lvert {{-}} \right\rvert}\).
\item
  \(R = {\mathbb{F}}[t]\) for \({\mathbb{F}}\) a field with
  \(\varphi({-}) = \deg({-})\).
\end{itemize}

\end{example}

\begin{remark}

Given a number field \(K\), does \({\mathbb{Z}}_K\) have nice
factorization, i.e.~is it a UFD? Not always, as we saw last time. If it
were Euclidean, then yes!

\end{remark}

\begin{question}

Which quadratic fields \(K\) have a Euclidean ring of integers
\({\mathbb{Z}}_K\)?

\end{question}

\begin{definition}[Euclidean and Norm-Euclidean Number Fields]

If \(K\) is a quadratic field, then

\begin{itemize}
\item
  \(K\) is \textbf{Euclidean} if and only if \({\mathbb{Z}}_K\) is a
  Euclidean domain,
\item
  \(K\) is \textbf{norm-Euclidean} if and only if \({\mathbb{Z}}_K\) is
  Euclidean with respect to
  \(\varphi({-}) \coloneqq{\left\lvert {N({-})} \right\rvert}\).
\end{itemize}

\end{definition}

\begin{proposition}[Characterization of norm-Euclidean quadratic fields]

Let \(K\) be a quadratic field. Then \(K\) is norm-Euclidean if and only
if for all \(\beta\in K\) there is a \(\gamma\in {\mathbb{Z}}_K\) such
that \({\left\lvert { N(\beta- \gamma) } \right\rvert} < 1\). In other
words, \(K\) is norm-Euclidean if and only if every element can be
approximated by an element in \({\mathbb{Z}}_K\).

\end{proposition}

\begin{proof}[of proposition]

\(\impliedby\): Let \(a,b \in {\mathbb{Z}}_k\) with \(b\neq 0\). Define
\(\beta\coloneqq a/b \in K\), then by assumption choose \(\gamma\) such
that \({\left\lvert {N \qty{ {a\over b} - \gamma} } \right\rvert}< 1\).
Multiplying both sides by \(N(b)\) and using the fact that
\(N({-}), {\left\lvert {{-}} \right\rvert}\) are multiplicative, we have
\begin{align*}
{\left\lvert {N(a - b \gamma)} \right\rvert} < {\left\lvert {N(b)} \right\rvert} 
.\end{align*}
Then \(a = bq + r \coloneqq b \gamma + (a - b \gamma)\).

\end{proof}

\hypertarget{norm-euclidean-imaginary-quadratic-fields}{%
\subsection{Norm-Euclidean Imaginary Quadratic
Fields}\label{norm-euclidean-imaginary-quadratic-fields}}

\begin{remark}

Suppose \(K = {\mathbb{Q}}( \sqrt{d} )\) with \(d<0\) squarefree, so we
can write
\begin{align*}
K = \left\{{ a + b \sqrt{d} {~\mathrel{\Big|}~}a,b, \in {\mathbb{Q}}}\right\}
= \left\{{ a + b i\sqrt{{\left\lvert {d} \right\rvert}} {~\mathrel{\Big|}~}a,b, \in {\mathbb{Q}}}\right\} \subseteq {\mathbb{C}}
.\end{align*}
Geometrically, this is a dense subset of \({\mathbb{C}}\), so it's not
easy to draw. But we can draw \({\mathbb{Z}}_K\) -- what does it look
like? We know that \(d=2,3 \operatorname{mod}4\) then
\({\mathbb{Z}}_K = \left\{{a + b \sqrt{d} {~\mathrel{\Big|}~}a,b,\in {\mathbb{Z}}}\right\}\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
    \draw [thick, gray,-latex] (-5, 0) -- (7, 0);% Draw x axis
    \draw [thick, gray,-latex] (0, -5) -- (0, 5);% Draw y axis

    \clip (-5,-5) rectangle (10cm,10cm); % Clips the picture...
    %\pgftransformcm{1}{0.6}{0.7}{1}{\pgfpoint{0cm}{0cm}}
          % This is actually the transformation matrix entries that
          % gives the slanted unit vectors. 
    \draw[style=help lines,dashed] (-14,-14) grid[step=2cm] (6,5);
          % Draws a grid in the new coordinates.
          %\filldraw[fill=gray, fill opacity=0.3, draw=black] (0,0) rectangle (2,2);
              % Puts the shaded rectangle
    \foreach \x in {--3,-2,-1,0,1,2}{% Two indices running over each
      \foreach \y in {-2,-1, ..., 2}{% node on the grid we have drawn 
        \node[draw,circle,inner sep=2pt,fill] at (2*\x,2*\y) {};
            % Places a dot at those points
      }
    }
        \node[draw,red, circle,inner sep=2pt,fill] at (0, 0) {};

    \draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] (6.5,2.0) -- (6.5,0.1) node [black,midway,xshift=25pt] {\footnotesize $\sqrt{{\left\lvert {d} \right\rvert}}$};
    \draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] (6.5,-0.1) -- (6.5,-2) node [black,midway,xshift=25pt] {\footnotesize $\sqrt{{\left\lvert {d} \right\rvert}}$};

\end{tikzpicture}
}
\end{figure}

When \(d \equiv 1 \operatorname{mod}4\), we have
\({\mathbb{Z}}_k = \left\{{ {1\over 2}(a + b \sqrt{d} ) {~\mathrel{\Big|}~}a,b,\in {\mathbb{Z}},\, a\equiv b \operatorname{mod}2}\right\}\).
On the real axis, if \(b=0\) then \(a\) is an even integer and
\(\left\{{(1/2)a}\right\}\) is all integers. To get the remaining
elements, we don't just shift up and down: setting \(b=1\) yields
elements that look like \((1/2)a + \sqrt{d}\) where \(a\) is odd, so we
get the following:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
    \draw [thick, gray,-latex] (-5, 0) -- (7, 0);% Draw x axis
    \draw [thick, gray,-latex] (0, -5) -- (0, 5);% Draw y axis

    \clip (-5,-5) rectangle (10cm,10cm); % Clips the picture...
    %\pgftransformcm{1}{0.6}{0.7}{1}{\pgfpoint{0cm}{0cm}}
          % This is actually the transformation matrix entries that
          % gives the slanted unit vectors. 
    \draw[style=help lines,dashed] (-14,-14) grid[step=2cm] (6,5);
          % Draws a grid in the new coordinates.
          %\filldraw[fill=gray, fill opacity=0.3, draw=black] (0,0) rectangle (2,2);
              % Puts the shaded rectangle
    \foreach \x in {--3,-2,-1,0,1,2}{% Two indices running over each
      \foreach \y in {-2,-1, ..., 2}{% node on the grid we have drawn 
        \node[draw,circle,inner sep=2pt,fill] at (2*\x,2*\y) {};
            % Places a dot at those points
      }
    }
\foreach \x in {--2,-3,-1,0,1,2,-2}{% Two indices running over each
      \foreach \y in {-2,-1, ..., 2}{% node on the grid we have drawn 
        \node[draw, circle,inner sep=2pt,fill] at (2*\x+1,2*\y+1) {};
            % Places a dot at those points
      }
    }
        \node[draw,red, circle,inner sep=2pt,fill] at (0, 0) {};

    \draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] (6.5,1.0) -- (6.5,0.1) node [black,midway,xshift=25pt] {\footnotesize ${1\over 2} \sqrt{{\left\lvert {d} \right\rvert}}$};
    \draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] (6.5,-0.1) -- (6.5,-1) node [black,midway,xshift=25pt] {\footnotesize ${1\over 2} \sqrt{{\left\lvert {d} \right\rvert}}$};
\draw [decorate,decoration={brace,amplitude=10pt},xshift=-4pt,yshift=0pt] (6.5,1.95) -- (6.5,1.05) node [black,midway,xshift=25pt] {\footnotesize ${1\over 2} \sqrt{{\left\lvert {d} \right\rvert}}$};

\end{tikzpicture}
}
\end{figure}

Now we can think of the criterion for an imaginary quadratic field to be
norm-Euclidean: what does it mean to be within norm 1 of an element of
\({\mathbb{Z}}_K\)? If \(z\in K\), we can write
\(N(z) = z \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu = {\left\lvert {z} \right\rvert}^2\),
and thus reformulate our criterion: \(K\) is norm-Euclidean if and only
if for all \(\beta\in K\) there exists a \(\gamma\in {\mathbb{Z}}_K\)
such that \({\left\lvert {\beta- \gamma} \right\rvert}<1\). Note this
this is the familiar geometric distance in \({\mathbb{C}}\).

\end{remark}

\begin{example}[?]

\({\mathbb{Q}}(i)\) is norm-Euclidean: the ring of integers is
\({\mathbb{Z}}(i)\), which is the integer lattice in \({\mathbb{C}}\).
Note one can cover \({\mathbb{C}}\) by open circles of radius 1:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
    \draw [thick, gray,-latex] (-5, 0) -- (7, 0);% Draw x axis
    \draw [thick, gray,-latex] (0, -5) -- (0, 5);% Draw y axis

    \clip (-5,-5) rectangle (10cm,10cm); % Clips the picture...
    %\pgftransformcm{1}{0.6}{0.7}{1}{\pgfpoint{0cm}{0cm}}
          % This is actually the transformation matrix entries that
          % gives the slanted unit vectors. 
    \draw[style=help lines,dashed] (-14,-14) grid[step=2cm] (6,5);
          % Draws a grid in the new coordinates.
          %\filldraw[fill=gray, fill opacity=0.3, draw=black] (0,0) rectangle (2,2);
              % Puts the shaded rectangle
    \foreach \x in {--3,-2,-1,0,1,2}{% Two indices running over each
      \foreach \y in {-2,-1, ..., 2}{% node on the grid we have drawn 
        \node[draw,circle,inner sep=2pt,fill] at (2*\x,2*\y) {};
            % Places a dot at those points
      }
    }
        \node[draw,red, circle,inner sep=2pt,fill] at (0, 0) {};

\filldraw[dotted, fill=blue!50, draw=blue, very thick, fill opacity=0.2] (2,0) circle (2cm);
\filldraw[dotted, fill=blue!50, draw=blue, very thick, fill opacity=0.2] (0,0) circle (2cm);
\filldraw[dotted, fill=blue!50, draw=blue, very thick, fill opacity=0.2] (-2,0) circle (2cm);
\node at (5, 1) {$\cdots$};
\end{tikzpicture}
}
\end{figure}

Continuing this way, every point with rational coordinates can be
covered by some open disc of radius 1:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
    \draw [thick, gray,-latex] (-5, 0) -- (7, 0);% Draw x axis
    \draw [thick, gray,-latex] (0, -5) -- (0, 5);% Draw y axis

    \clip (-5,-5) rectangle (10cm,10cm); % Clips the picture...
    %\pgftransformcm{1}{0.6}{0.7}{1}{\pgfpoint{0cm}{0cm}}
          % This is actually the transformation matrix entries that
          % gives the slanted unit vectors. 
    \draw[style=help lines,dashed] (-14,-14) grid[step=2cm] (6,5);
          % Draws a grid in the new coordinates.
          %\filldraw[fill=gray, fill opacity=0.3, draw=black] (0,0) rectangle (2,2);
              % Puts the shaded rectangle
    \foreach \x in {--3,-2,-1,0,1,2}{% Two indices running over each
      \foreach \y in {-2,-1, ..., 2}{% node on the grid we have drawn 
        \node[draw,circle,inner sep=2pt,fill] at (2*\x,2*\y) {};
\filldraw[dotted, fill=blue!50, draw=blue, very thick, fill opacity=0.2] (2*\x,2*\y) circle (2cm);

            % Places a dot at those points
      }
    }
        \node[draw,red, circle,inner sep=2pt,fill] at (0, 0) {};


\node at (5, 1) {$\cdots$};
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{remark}

Note that this doesn't work for arbitrary \(d\), since the distance
between the horizontal lines grows with \(d\). It's not hard to work out
the exact list where everything \emph{is} covered:

\end{remark}

\begin{theorem}[When quadratic fields are norm-Euclidean]

\(K\) is norm-Euclidean if and only if
\(d\in \left\{{-1,-2,-3,-7,-11}\right\}\).

\end{theorem}

\begin{corollary}[When rings of integers are PIDs/UFDs]

For these \(d\), \({\mathbb{Z}}_K\) is a PID and thus a UFD.

\end{corollary}

\begin{remark}

So we've classified all norm-Euclidean imaginary quadratic fields. What
about removing the word ``norm''? We restricted to
\({\left\lvert {N({-})} \right\rvert}\) because there was a particularly
nice geometric interpretation, whereas being Euclidean involves a
mysterious \(\varphi\). Remarkably, it can be done, and it's the same
list!

\end{remark}

\begin{theorem}[Motzkin]

For \(K\) an imaginary quadratic field, \(K\) is Euclidean if and only
if \(d\in \left\{{-1,-2,-3,-7,-11}\right\}\).

\end{theorem}

\begin{remark}

If \({\mathbb{Z}}_K\) were never a PID in these cases, we could
immediately conclude it wasn't Euclidean either. But there are values of
\(d\) not on this list for which \({\mathbb{Z}}_K\) is a PID,
e.g.~\(d=-19\). Since \(-19 \equiv 1 \operatorname{mod}4\), one can
write
\({\mathbb{Z}}_K = {\mathbb{Z}}\left[ {1 + \sqrt{-19} \over 2}\right]\),
and by Motzkin's theorem this is a PID which is not a Euclidean domain.

\end{remark}

\begin{remark}

We'll prove this theorem! First we need a few lemmas.

\end{remark}

\begin{lemma}[Most imaginary quadratic fields have only two units]

Let \(K\) be an imaginary quadratic field, then
\(U({\mathbb{Z}}_K) = \left\{{\pm 1 }\right\}\) except if \(d=-1, -3\).

\end{lemma}

\begin{proof}[of lemma (Important!)]

We know that the units \(u\) satisfy
\({\left\lvert {N(u)} \right\rvert} = 1\), and for imaginary quadratic
fields norms are non-negative, so we know \(N(u) = 1\). What are the
solutions this equation? Suppose \(d=2,3 \operatorname{mod}4\), then we
can write \(\alpha = a + b \sqrt{d}\) with \(a,b\in {\mathbb{Z}}\) and
\(1 = N \alpha = a^2 - db^2 = a^2 + {\left\lvert {d} \right\rvert}b^2\).
If \({\left\lvert {d} \right\rvert}= 1\) then this will have four
solutions: \((a, b) = (\pm 1, 0), (0, \pm 1)\).

\hfill\break

Otherwise if \({\left\lvert {d} \right\rvert}> 1\) then \(b=0\) and
\(a^2=1 \implies a = \pm 1\) and thus \(\alpha = \pm 1\). So in this
case, the only units are \(\pm 1\), unless
\({\left\lvert {d} \right\rvert} = 1\). But the only negative squarefree
integer of absolute value 1 is \(-1\).

Suppose \(d \equiv 1 \operatorname{mod}4\). In this case, we need
\begin{align*}
1 = {a^2 + {\left\lvert {d} \right\rvert} b^2 \over 4} \implies a^2 + {\left\lvert {d} \right\rvert}b^2 = 4  
.\end{align*}
Note that \(d<0\) is \(1\operatorname{mod}4\), so it's possible that
\(d=-3\) -- but this was one of the exceptions in the theorem, so assume
otherwise. Thus \({\left\lvert {d} \right\rvert}\geq 7\), which forces
\(b=0 \implies a^2 = 4 \implies a = \pm 2\). Then \(\alpha= \pm 1\).

\end{proof}

\begin{remark}

For the excluded cases, the units can be explicitly computed. When
\(d=-1\), \(U({\mathbb{Z}}[i]) =\left\{{\pm 1, \pm i}\right\}\),
yielding 4 units. When \(d=-3\),
\begin{align*}
U\qty{{\mathbb{Z}}\left[ {1 + \sqrt{-3} \over 2 } \right]} = \left\{{\pm 1, {\pm 1 \pm \sqrt{-3} \over 2}}\right\}
,\end{align*}
yielding 6 units. Note that in the first case, these are exactly the 4th
roots of unity, and in the second case these are the sixth roots. This
is a general phenomenon that will appear again!

\end{remark}

\begin{lemma}[Norm of generator of principal ideal equals size of quotient]

Let \(K\) be any quadratic field and \(\alpha\in {\mathbb{Z}}_K\). Then
\(\# {\mathbb{Z}}_k / \left\langle{ \alpha }\right\rangle = {\left\lvert {N( \alpha )} \right\rvert}\).

\end{lemma}

\hypertarget{proof-of-motzkins-theorem}{%
\subsection{Proof of Motzkin's
Theorem}\label{proof-of-motzkins-theorem}}

\begin{proof}[of Motzkin's Theorem]

We want to show that being Euclidean implies \(d=-1,-2,-3,-7,-11\).
Suppose \({\mathbb{Z}}_K\) is Euclidean with respect to \(\varphi\).
Choose \(\beta\in {\mathbb{Z}}_K\) nonzero and not a unit with
\(\varphi(\beta)\) minimal among all such \(\beta\).

\begin{claim}

\begin{align*}
\# {\mathbb{Z}}_K / \left\langle{ \beta }\right\rangle \leq 3 
.\end{align*}

\end{claim}

\begin{proof}[of claim]

For any \(\alpha\in {\mathbb{Z}}_K\) and consider it in the quotient.
Since \({\mathbb{Z}}_K\) is Euclidean, we can write
\(\alpha = \beta + \gamma+ \rho\) where either \(\rho=0\) or
\(\varphi(\rho ) < \varphi (\beta )\). How can the second possibility
occur? \(\beta\) was chosen to have a minimal \(\varphi\) value, so the
only smaller elements are units. So \(\rho = 0\) or \(\rho\) is a unit.
Reducing \(\operatorname{mod}\beta\), we obtain
\(\alpha= \rho \operatorname{mod}\beta\), and hence
\(\# {\mathbb{Z}}_K / \left\langle{ \beta }\right\rangle \leq 1 + \# U({\mathbb{Z}}_K)\)
where the 1 comes from the zero element and everything else in the
quotient has a representative that is a unit. This is bounded above by
\(3\) when \(d\neq -1, -3\), which is one of the exclusions in the
theorem.

\end{proof}

Now we have \(N( \beta) \leq 3\) and this can be solved -- if \(d\) is
large, these solutions are widely distributed. If
\(d = 2, 3 \operatorname{mod}4\) then \(\beta= a + b \sqrt{d}\) with
\(a, b \in ZZ\) and \(a^2 + {\left\lvert {d} \right\rvert}b^2 \leq 3\).
We can assume \({\left\lvert {d} \right\rvert}> 3\), since \(d=-1, -2\)
are excluded. Then \(b=0\) is forced, and \(a = 0, \pm 1\). But why
can't \(\beta=0, \pm 1\)? It was chosen to be minimal among
\emph{nonzero nonunits}. \(\contradiction\)

If \(d \equiv 1 \operatorname{mod}4\), then
\(\beta= {a + b \sqrt{d} \over 2}\) where
\(a,b\in {\mathbb{Z}},\, a\equiv b \operatorname{mod}2\). Then
\begin{align*}
{a^2 + {\left\lvert {d} \right\rvert}b^2 \over 4} \leq 3 \implies a^2 + {\left\lvert {d} \right\rvert}b^2 \leq 12  
.\end{align*}
Now considering that
\(-d\equiv 1 \operatorname{mod}4 \implies -d \in \left\{{ -3, -7, -11, \cdots}\right\}\),
the first three of which are on our list of exclusions. So we can assume
\({\left\lvert {d} \right\rvert}\geq 15\), which forces \(b=0\), \(a\)
must be even, and \(a^2 \leq 12\). So
\(a=0, \pm 2 \implies \beta= 0, \pm 1\). \(\contradiction\)

\end{proof}

\begin{remark}

What's the story for real quadratic fields? We understand norm-Euclidean
ones, although the proofs aren't nearly as simple. Things worked out
nicely here because we had circles in the plane; in the real case these
end up being complicated hyperbolas. One can prove that if \(d > 73\)
then \(K \coloneqq{\mathbb{Q}}(\sqrt d)\) is not norm-Euclidean. What
are the Euclidean real quadratic fields? The situation is much
different, and there are two open conjectures.

\end{remark}

\begin{conjecture}

For real quadratic fields \(K\), \({\mathbb{Z}}_K\) is a PID for
infinitely many \(d> 0\). We don't even know about to prove there are
just infinitely many \emph{number} fields satisfying this condition! We
believe this is true since it happens a positive proportion of the time
experimentally.

\end{conjecture}

\begin{conjecture}

If \({\mathbb{Z}}_K\) is a PID, then \({\mathbb{Z}}_K\) is Euclidean
with respect to some norm function. This is a consequence of a certain
generalization of the RH. This is not true for imaginary quadratic
fields. Why is it different here? The unit group plays a large role, and
is infinite here. The real conjecture is that for \(K\) any number
field, if \({\mathbb{Z}}_K\) is a PID with infinitely many units then
\({\mathbb{Z}}_K\) is Euclidean.

\end{conjecture}

\begin{remark}

There has been some progress, a result along the lines of there being at
most two exceptions, but we don't know if those exceptions exist.

\end{remark}

\hypertarget{ideal-theory-and-quadratic-fields-lec.-6-tuesday-february-02}{%
\section{Ideal Theory and Quadratic Fields (Lec. 6, Tuesday, February
02)}\label{ideal-theory-and-quadratic-fields-lec.-6-tuesday-february-02}}

\hypertarget{prime-factorization-in-operatornameidmathbbz_k}{%
\subsection{\texorpdfstring{Prime Factorization in
\(\operatorname{Id}({\mathbb{Z}}_K)\)}{Prime Factorization in \textbackslash operatorname\{Id\}(\{\textbackslash mathbb\{Z\}\}\_K)}}\label{prime-factorization-in-operatornameidmathbbz_k}}

\begin{remark}

Today: roughly chapter 6. Recall that if \(R\) is a domain, we defined
\(\operatorname{Id}(R)\) as the set of nonzero ideals of \(R\), which is
a monoid. We want to get to the following theorem:

\end{remark}

\begin{theorem}[Fundamental Theorem of Ideal Theory (for Quadratic Fields)]

Let \(K\) be a quadratic field, then
\(\operatorname{Id}({\mathbb{Z}}_K)\) is a UFM.

\end{theorem}

\begin{remark}

This means that everything factors into irreducibles, and when you have
unique factorization, irreducible is the same as prime. Note that
``prime'' here means in this monoidal sense -- does this match up with
the existing notion of a prime ideal? I.e. that \(\mathfrak{p}\) is
prime if and only if \(R/ \mathfrak{p}\) is a domain, or
\(ab\in \mathfrak{p}\implies a,b \in \mathfrak{p}\)?

\end{remark}

\begin{proposition}[Prime in monoids equals prime in rings for $\Id(\ZZ_K)$]

``Prime'' in the usual ring-theoretic sense is equivalent to ``prime''
in the monoidal sense for \(\operatorname{Id}({\mathbb{Z}}_K)\).

\end{proposition}

\begin{remark}

Recall though that \(\operatorname{Id}({\mathbb{Z}}_K)\) is a reduced
monoid, so the only unit is the identity, so uniqueness is just up to
ordering and doesn't involve additional units. We can restart the unique
factorization theorem:

\end{remark}

\begin{proposition}[$\Id(\ZZ_K)$ has prime factorization]

Every nonzero ideal of \({\mathbb{Z}}_K\) factors uniquely as a product
of prime ideals in \({\mathbb{Z}}_K\).

\end{proposition}

\begin{remark}

Can we explicitly understand what ideals of \({\mathbb{Z}}_K\) look like
for quadratic fields?

\end{remark}

\begin{definition}[Standard Bases of Ideals]

Let \(K = {\mathbb{Q}}(\sqrt{d})\), so
\({\mathbb{Z}}_K = {\mathbb{Z}}[\sqrt d]\) if
\(d=2,3 \operatorname{mod}4\) or
\({\mathbb{Z}}\left[ {1 + \sqrt{d} \over 2} \right]\) if
\(d=1 \operatorname{mod}4\). Define \(\tau = \sqrt{d}\) or
\((1 + \sqrt{d}) /2\) accordingly and write
\({\mathbb{Z}}_k = {\mathbb{Z}}[\tau]\).

\end{definition}

\begin{remark}

Note that \(\left\{{ 1, \tau }\right\}\) is a
\({\mathbb{Z}}{\hbox{-}}\)basis of \({\mathbb{Z}}_K\). An ideal of
\({\mathbb{Z}}_K\) is a submodule as a \({\mathbb{Z}}{\hbox{-}}\)module,
which is free, so any ideal is free of rank at most 2. Can we write down
such a basis?

\end{remark}

\begin{lemma}[Existence of $\tau$]

Let \(I\) be a nonzero ideal of \({\mathbb{Z}}_K\), then \(I\) contains
a nonzero integer and an element of the form \(a + b \tau\) where
\(b\neq 0\).

\end{lemma}

\begin{proof}[of lemma]

How to produce a nonzero rational integer: let \(\alpha\in I\) be
nonzero and take the norm. Then \(N \alpha\) is a nonzero integer, and
since
\(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu\in {\mathbb{Z}}_K\)
we have
\(N \alpha = \alpha \mkern 1.5mu\overline{\mkern-1.5mu \alpha\mkern-1.5mu}\mkern 1.5mu \in I\).
Now since \(\tau\in {\mathbb{Z}}_K\) and \(I\) absorbs multiplication we
can set \(b \coloneqq N \alpha(\tau) \in I\).

\end{proof}

\begin{remark}

We wanted a nice description of bases for ideals -- here it is!

\end{remark}

\begin{proposition}[Existence of a standard basis for an ideal]

Let \(I {~\trianglelefteq~}{\mathbb{Z}}_K\) be a nonzero ideal. Choose
\(n\in {\mathbb{Z}}^+\) such that
\(I \cap{\mathbb{Z}}= n{\mathbb{Z}}\).\footnote{Why does this \(n\)
  exist? Every ideal in \({\mathbb{Z}}\) is of the form
  \(n{\mathbb{Z}}\), and it's easy to check \(I \cap{\mathbb{Z}}\) is an
  ideal in \({\mathbb{Z}}\) since its an ideal of \({\mathbb{Z}}_K\)
  intersected with \({\mathbb{Z}}\). How do we know it's not the zero
  ideal? This is exactly given by the last lemma.} Choosing
\(B \in {\mathbb{Z}}^+\) such that
\(\left\{{ b\in {\mathbb{Z}}{~\mathrel{\Big|}~}a + b \tau\in I \text{ for some } a \in {\mathbb{Z}}}\right\} = B{\mathbb{Z}}\).\footnote{The
  LHS is the set of coefficients of \(\tau\), which is an ideal of
  \({\mathbb{Z}}\), and we can take it to be positive since the LHS is
  not the zero ideal by the lemma.} Since \(B\) is in the LHS, pick
\(A\in {\mathbb{Z}}\) with \(A + B\tau \in I\). Then
\(\left\{{n, A+B\tau}\right\}\) is a \({\mathbb{Z}}{\hbox{-}}\)basis for
\(I\).

\hfill\break

Any such basis is referred to as a \textbf{standard basis} for \(I\)

\end{proposition}

\begin{remark}

Note that this is only determined up to \(A \operatorname{mod}n\).

\end{remark}

\begin{proof}[of proposition]

Take any element in \(I\), which can be represented as \(a + b \tau\),
we want to show that this can be expressed in terms of the proposed
basis. Note that \(B\divides b\) by its definition, since \(B\)
generated the ideal of \(\tau\) coefficients. So write \(b = Bs\), then
\begin{align*}
( a + b \tau) - (A + b \tau)s \in {\mathbb{Z}}\cap I = \left\langle{ n }\right\rangle 
.\end{align*}
So write this difference as \(nr\) for some \(r\in {\mathbb{Z}}\), then
rearranging yields
\begin{align*}
a + b \tau = nr + (A + B \tau)s
,\end{align*}
which is a \({\mathbb{Z}}{\hbox{-}}\)linear combination of the standard
basis elements. Uniqueness is easy and follows from the fact that every
element in \({\mathbb{Z}}_K\) has a unique representation in terms of
\(1, \tau\).

\end{proof}

\hypertarget{ideal-norms}{%
\subsection{Ideal Norms}\label{ideal-norms}}

\begin{remark}

In the previous section, we used the fact that for
\(a\in {\mathbb{Z}}_k\), the number of elements in
\({\mathbb{Z}}_K / \left\langle{ n }\right\rangle\) is
\({\left\lvert { N a } \right\rvert}\). That will be a consequence of
the theory we develop here.

\end{remark}

\begin{definition}[Norm of an ideal]

If \(I{~\trianglelefteq~}{\mathbb{Z}}_K\) is a nonzero ideal, define the
\textbf{norm of \(I\)} as
\(N(I) = {\left\lvert {{\mathbb{Z}}_K / I} \right\rvert}\).

\end{definition}

\begin{remark}

It's not completely obvious, but this quotient is always finite. We can
use the fact that \(I\leq {\mathbb{Z}}_K\) is a
\({\mathbb{Z}}{\hbox{-}}\)submodule of rank exactly 2. It's then a
general fact from algebra that \(A/B\) is finite when
\(\operatorname{rank}(A) = \operatorname{rank}(B)\), and there are ways
of figuring out the number of elements (see normal forms).

\end{remark}

\begin{proposition}[Norms can be computed in terms of a basis with respect to $\tau$]

Suppose that \(I{~\trianglelefteq~}{\mathbb{Z}}_K\) is a nonzero ideal
and let \(n, A+B \tau\) be a standard basis for \(I\). Then
\(N(I) = nB \in {\mathbb{Z}}^+\).

\end{proposition}

\begin{proof}[of proposition]

Check that
\(\left\{{ a + b \tau {~\mathrel{\Big|}~}0\leq a \leq n,\, 0 \leq b \leq B}\right\}\)
is a complete and irredundant set of representatives for
\({\mathbb{Z}}_K/I\).

\end{proof}

\begin{remark}

So given a standard basis, it's easy to compute norms! What does this
have to do with the previous notion of norms for elements?

\end{remark}

\begin{theorem}[The ideal that the norm generates]

Let \(I {~\trianglelefteq~}{\mathbb{Z}}_K\) be nonzero and define
\(\mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu = \left\{{ \mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu{~\mathrel{\Big|}~}\alpha \in I }\right\} {~\trianglelefteq~}{\mathbb{Z}}_K\).
Then
\(I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu = \left\langle{ N(I) }\right\rangle\).

\end{theorem}

\begin{lemma}[The $\tau$ coefficient divides the remaining coefficient]

Let \(n\) be as above and let \(A + B \tau\) be a standard basis for
\(I\). Then \(B\divides n\) and \(B\divides A\).

\end{lemma}

\begin{proof}[of lemma]

Recall that \(B\) was a generator for \(\tau\) components of elements of
\(I\), so we just need to find an element of \(I\) with \(\tau\)
component \(n\), and \(n \tau \in I\) works. Now compute
\((A + B \tau) \tau\in I\). This is equal to
\begin{align*}
A \tau + B \tau^2
.\end{align*}
Note that this could in principle be done in cases: if
\(\tau = \sqrt{d}\), the quantity \(Bd\) would be an integer and \(A\)
would be the \(\tau\) coordinate. Then since \(B\) divides every
\(\tau\) coefficient, we'd be done. But let's try this in a more unified
way: we know \(\tau\) is a root of a monic degree 2 polynomial, namely
\((x - \tau) (x - \mkern 1.5mu\overline{\mkern-1.5mu\tau\mkern-1.5mu}\mkern 1.5mu) = x^2 - \operatorname{Tr}( \tau)x + N( \tau)\),
and thus we can write
\begin{align*}
\tau^2 = \operatorname{Tr}( \tau) \tau - N( \tau)
.\end{align*}
Substituting yields
\begin{align*}
(A + B \tau) \tau
&= A \tau + B \tau^2 \\
&= A \tau + B ( \operatorname{Tr}( \tau) \tau - N( \tau) ) \\
&= - B N( \tau) + (A + B \operatorname{Tr}( \tau ) ) \tau
.\end{align*}
The coefficient of \(\tau\) must be a multiple of \(B\), which forces
\(B\divides A\).

\end{proof}

\begin{proof}[of theorem]

Let \(n, A + B \tau\) be a standard basis for \(I\). Then
\(I = \left\langle{ n, A+ B \tau }\right\rangle\), which is a generating
set as a \({\mathbb{Z}}_K{\hbox{-}}\)module since they generate \(I\)
over \({\mathbb{Z}}\) and subset containment both ways can be readily
checked. We can then write
\(\mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu = \left\langle{ n, A+B \mkern 1.5mu\overline{\mkern-1.5mu\tau \mkern-1.5mu}\mkern 1.5mu}\right\rangle\),
since conjugating ordinary integers doesn't change them. Using the
lemma, we can write
\begin{align*}
I &= \left\langle{ Bn', BA' + B \tau }\right\rangle\\
\mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu &= \left\langle{ Bn', BA' + B \mkern 1.5mu\overline{\mkern-1.5mu\tau\mkern-1.5mu}\mkern 1.5mu }\right\rangle
.\end{align*}

We can factor out a \(B\) to get
\begin{align*}
I &= \left\langle{ B }\right\rangle \left\langle{ n', A' + \tau}\right\rangle \\  
\mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu &= \left\langle{ B }\right\rangle \left\langle{ n', A' + \mkern 1.5mu\overline{\mkern-1.5mu\tau\mkern-1.5mu}\mkern 1.5mu }\right\rangle 
.\end{align*}

Now multiplying the two yields
\begin{align*}
I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu = \left\langle{ B^2 }\right\rangle \left\langle{ (n')^2, n'(A' + \mkern 1.5mu\overline{\mkern-1.5mu \tau\mkern-1.5mu}\mkern 1.5mu ), n'(A' + \tau), N(A' + \tau) }\right\rangle  
.\end{align*}
It's tempting to factor out \(n'\), but it isn't obviously in the last
factor. But it is! Note that
\(N(A' + \tau) \in \left\langle{ A' + \tau, n' }\right\rangle\) and thus
\(B N(A' + gt) \in \left\langle{ B }\right\rangle \left\langle{ A' + \tau, n' }\right\rangle = I\).
But the first expression is an ordinary integer, i.e.~in
\(I \cap{\mathbb{Z}}= \left\langle{ n }\right\rangle\) and thus a
multiple of \(n\). So \(Bn' = n \divides BN(A' + \tau)\), and thus
\(n' \divides N(A' + \tau)\). So we can rewrite
\begin{align*}
I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu 
&= \left\langle{ B^2 }\right\rangle \left\langle{ n' }\right\rangle \left\langle{ n', A' + \mkern 1.5mu\overline{\mkern-1.5mu \tau\mkern-1.5mu}\mkern 1.5mu , A' + \tau, { N(A' + \tau) \over n'} }\right\rangle   \\
&= \left\langle{ B^2 n' }\right\rangle 
\left\langle{ n', A' + \mkern 1.5mu\overline{\mkern-1.5mu \tau\mkern-1.5mu}\mkern 1.5mu , A' + \tau, { N(A' + \tau) \over  n'} }\right\rangle   
.\end{align*}

We can now note that \(B^2 n' = B^2(n/B) = nB = N(I)\). We've thus shown
that
\begin{align*}
I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu 
= \left\langle{ N(I) }\right\rangle 
\left\langle{ n', A' + \mkern 1.5mu\overline{\mkern-1.5mu \tau\mkern-1.5mu}\mkern 1.5mu , A' + \tau, { N(A' + \tau) \over  n'} }\right\rangle   
.\end{align*}
We'd really like the second term to just be
\(\left\langle{ 1 }\right\rangle\). Note that this factor contains some
integers: \(n', N(A' + \tau)/n'\), and
\((A' + \mkern 1.5mu\overline{\mkern-1.5mu\tau\mkern-1.5mu}\mkern 1.5mu) + (A' + \tau) = \operatorname{Tr}(A' + \tau)\).
So let
\begin{align*}
J \coloneqq\left\langle{ n, N(A' + \tau)/n', \operatorname{Tr}(A' + \tau) }\right\rangle {~\trianglelefteq~}{\mathbb{Z}}
,\end{align*}
then it's enough to show
\(J = \left\langle{ 1 }\right\rangle {~\trianglelefteq~}{\mathbb{Z}}\).
Why? If so, \(1\) is a \({\mathbb{Z}}{\hbox{-}}\)linear combination of
these elements, but every \({\mathbb{Z}}{\hbox{-}}\)linear combination
is also a \({\mathbb{Z}}_K{\hbox{-}}\)linear combination. Every such
combination will be in the original ideal appearing in
\(I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu\),
which we want to show is the unit ideal. We can write
\(J = d{\mathbb{Z}}\) where \(d\in {\mathbb{Z}}^+\) and suppose toward a
contradiction that \(d>1\).

\hfill\break

Consider \(\alpha \coloneqq(A' + \tau) / d \in K\). Taking the trace is
\({\mathbb{Q}}{\hbox{-}}\)linear, so
\(\operatorname{Tr}( \alpha) = (1/d) \operatorname{Tr}(A' + \tau) \in {\mathbb{Z}}\).
This follows because the trace \(\operatorname{Tr}(A' + \tau)\) is in
\(J\), thus a multiple of \(d\). We can also compute
\(N \alpha = N(A' + \tau) / d^2\) using that
\(d\mkern 1.5mu\overline{\mkern-1.5mud\mkern-1.5mu}\mkern 1.5mu = d^2\)
since \(d\) is rational.

\hfill\break

The claim is that \(N \alpha\) is also an integer: since
\(N(A' + \tau)/n', \operatorname{Tr}(A' + \tau)\) are in \(J\), \(d\)
divides both. So we know that
\(d^2 \divides (n') (N(A' + \tau) / n') = N(A' + \tau)\), which forces
\(N \alpha\in {\mathbb{Z}}\). So we know
\(N \alpha, \operatorname{Tr}\alpha \in {\mathbb{Z}}\), which forces
\(\alpha\in {\mathbb{Z}}_K\) since \(\alpha\) is a root of
\(x^2 - \operatorname{Tr}(\alpha) + N \alpha\). But \(\alpha\) can't be
in \({\mathbb{Z}}_K\), since these consist only of
\({\mathbb{Z}}{\hbox{-}}\)linear combinations of \(1, \tau\) -- however
here the coefficient of \(\tau\) is \(1/d \not \in {\mathbb{Z}}\), and
thus \(\alpha = A'/d + (1/d) \tau \not\in {\mathbb{Z}}_K\).

\end{proof}

\begin{remark}

This is a long proof! It's nice in that it's direct, but less nice in
that it required some clever steps. When we do the case for general
number fields, we'll be able to use a more conceptual approach that
avoids some of these computations. Many other facts fall out of these
theorem -- in fact, there are nice results as long as
\(I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu\) is a
principal ideal.

\end{remark}

\hypertarget{fundamental-theorem-of-ideal-theory-lec.-7-thursday-february-04}{%
\section{Fundamental Theorem of Ideal Theory (Lec. 7, Thursday, February
04)}\label{fundamental-theorem-of-ideal-theory-lec.-7-thursday-february-04}}

\hypertarget{norms-multiplicativity-and-computations}{%
\subsection{Norms: Multiplicativity and
Computations}\label{norms-multiplicativity-and-computations}}

\begin{remark}

Today: roughly chapter 6. Goal: establish unique factorization of ideals
for quadratic fields. Let \(K = {\mathbb{Q}}(\sqrt d)\) be a quadratic
field and we let
\begin{align*}
\tau = 
\begin{cases}
\sqrt d & d \equiv 2,3 \operatorname{mod}4  
\\
{1 + \sqrt{d} \over 2} & d \equiv 1 \operatorname{mod}4.
\end{cases}
\end{align*}
In this case we saw that
\({\mathbb{Z}}_K = {\mathbb{Z}}+ {\mathbb{Z}}\tau = {\mathbb{Z}}[\tau]\).
We defined the norm of a nonzero ideal
\(I{~\trianglelefteq~}{\mathbb{Z}}_K\) by
\(N(I) = {\left\lvert {{\mathbb{Z}}_K/I} \right\rvert}\). The main
theorem was that
\(I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu = \left\langle{ N(I)}\right\rangle\),
so the two definitions of norms are closely related. Some corollaries of
this theorem:

\end{remark}

\begin{corollary}[The norm is multiplicative]

Let \(I, J {~\trianglelefteq~}{\mathbb{Z}}_K\) be nonzero, then
\(N(IJ) = N(I) N(J)\).

\end{corollary}

\begin{proof}[of corollary]

Note that
\(IJ \mkern 1.5mu\overline{\mkern-1.5muIJ\mkern-1.5mu}\mkern 1.5mu = \left\langle{ N(IJ) }\right\rangle\)
on one hand, and on the other hand we can write this as
\begin{align*}
IJ \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu = I\mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu J \mkern 1.5mu\overline{\mkern-1.5muJ\mkern-1.5mu}\mkern 1.5mu = \left\langle{ N(I) }\right\rangle \left\langle{ N(J) }\right\rangle = \left\langle{ N(I) N(J) }\right\rangle
.\end{align*}
So we know that
\(\left\langle{ N(IJ) }\right\rangle = \left\langle{ N(I) N(J) }\right\rangle\)
in \({\mathbb{Z}}_K\), how can we conclude that the generators are the
same? In a domain, they are the same up to a unit, so
\begin{align*}
{ N(IJ) \over N(I) N(J) } \in U({\mathbb{Z}}_K)
,\end{align*}
and thus this quotient is in
\({\mathbb{Z}}_K \cap{\mathbb{Q}}= {\mathbb{Z}}\). The same argument
shows that its reciprocal is also in \({\mathbb{Z}}\), so the ratio must
be a unit in \({\mathbb{Z}}\) and we have \(N(IJ) = \pm N(I) N(J)\). But
the norm counts something, so both sides must be positive.

\end{proof}

\begin{remark}

Note that if we knew \(I, J\) were comaximal, we could appeal to the
Chinese Remainder Theorem, but we don't need any assumptions on the
ideals for this proof.

\end{remark}

\begin{corollary}[Computing norms of principal ideals]

Let \(\alpha\in {\mathbb{Z}}_K \setminus\left\{{0}\right\}\), then
\begin{align*}
N( \left\langle{ \alpha }\right\rangle ) = {\left\lvert {N(\alpha)} \right\rvert}  = {\left\lvert {\alpha\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu} \right\rvert} 
.\end{align*}

\end{corollary}

\begin{proof}[of corollary]

We have
\begin{align*}
\left\langle{ N( \left\langle{ \alpha }\right\rangle  )}\right\rangle 
\left\langle{ \alpha }\right\rangle \left\langle{ \mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu}\right\rangle 
&=
\left\langle{ \alpha }\right\rangle \left\langle{ \mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu}\right\rangle \\
&=
\left\langle{ \alpha\mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu}\right\rangle\\
&= \left\langle{ N \alpha }\right\rangle 
.\end{align*}
By the same argument as the previous corollary, this can only be true if
the generators are the same up to sign, so
\(N( \left\langle{ \alpha }\right\rangle = \pm N \alpha\).

\end{proof}

\hypertarget{unique-factorization-for-ideals}{%
\subsection{Unique Factorization for
Ideals}\label{unique-factorization-for-ideals}}

\begin{lemma}[Principal Multiple Lemma]

Let \(I {~\trianglelefteq~}{\mathbb{Z}}_K\setminus\left\{{0}\right\}\),
then there is another ideal
\(\tilde I {~\trianglelefteq~}{\mathbb{Z}}_K\setminus\left\{{0}\right\}\)
such that \(I \tilde I\) is principal.

\end{lemma}

\begin{proof}[of Principal Multiple Lemma]

Take
\(\tilde I \coloneqq\mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu\),
since we proved that this is principal and generated by the norm.

\end{proof}

\begin{remark}

Why write this down when it's weaker than the previous theorem? In the
next proofs, we'll only really use that \(I\) times something is
principal.

\end{remark}

\begin{lemma}[Cancellation in the Monoid of Ideals]

Suppose that \(IJ = IJ'\) is an equation of nonzero ideals in
\({\mathbb{Z}}_K\) with \(I\) principal. Then \(J = J'\), so principal
ideals can be cancelled from both sides.

\end{lemma}

\begin{definition}[Dilation of Ideals]

If \(I {~\trianglelefteq~}{\mathbb{Z}}_K\), for any \(\alpha\in K\)
define
\begin{align*}
\alpha I \coloneqq\left\{{ \alpha \beta {~\mathrel{\Big|}~}\beta \in I }\right\} 
,\end{align*}
i.e.~scale or dilate the ideal \(I\) by the factor \(\alpha\). We'll
refer to this as the \textbf{dilation of \(I\) by \(\alpha\)}.

\end{definition}

\begin{remark}

Is this still an ideal in \({\mathbb{Z}}_K\)? It still contains zero, is
still closed under addition, and still absorbs multiplication by
elements in \(O_K\) -- however, it may not be a subset of
\({\mathbb{Z}}_K\), since we can dilate by any element in \(K\). For
example, for \(I \coloneqq 2{\mathbb{Z}}\) take \(\alpha\coloneqq 1/5\).
These are referred to as \textbf{fractional ideal}, i.e.~a
\({\mathbb{Z}}_K{\hbox{-}}\)submodule of \(K\). It is an ideal in
\({\mathbb{Z}}_K\) when it is contained in \({\mathbb{Z}}_K\).

\end{remark}

\begin{proof}[of cancellation in the monoid of ideals]

Write \(I = \left\langle{ \alpha }\right\rangle\). Then
\(\left\langle{ \alpha }\right\rangle J = \left\langle{ \alpha }\right\rangle J'\),
however the RHS is equal to the dilated ideal \(\alpha J'\) and the LHS
is \(\alpha J\). So dilate both sides by \(1/ \alpha\) to get
\(J = J'\).

\end{proof}

\begin{remark}

This was the easy case, when \(I\) was principal. What if \(I\) is not
principal?

\end{remark}

\begin{proposition}[The monoid $\Id(\ZZ_K)$ is Cancellative]

If \(IJ = IJ'\) then \(J = J'\), with no assumptions on \(I\).

\end{proposition}

\begin{proof}[?]

Choose \(\tilde I\) using the previous lemma and multiply it to both
sides to obtain
\begin{align*}
(I \tilde I) J = (I \tilde I) J'
.\end{align*}
Then since \(I\tilde I\) is principal, it can be cancelled using the
previous lemma.

\end{proof}

\hypertarget{proving-unique-factorization}{%
\subsubsection{Proving Unique
Factorization}\label{proving-unique-factorization}}

\begin{theorem}[To divide is to contain]

Let \(I, J\) be nonzero ideals of \({\mathbb{Z}}_K\), then
\begin{align*}
I \divides J \iff I \supseteq J
.\end{align*}

\end{theorem}

\begin{proof}[of theorem]

\(\implies\): This is true in any ring! If \(I\divides J\), then
\(J = IM\) where \(M {~\trianglelefteq~}{\mathbb{Z}}_K\), and by
definition \(IM \subseteq I\) and so \(J \subseteq I\).

\hfill\break

\(\impliedby\): Suppose \(I \supseteq J\), we then want to find
\(B {~\trianglelefteq~}{\mathbb{Z}}_K\) with \(J = IB\). We'll proceed
by pretending we had such a \(B\) and seeing what it must be! If \(B\)
satisfies this equation, pick \(\tilde I\) where
\(I\tilde I = \left\langle{ \alpha}\right\rangle\), then
\begin{align*}
\tilde I J = \tilde I I B = \left\langle{ \alpha }\right\rangle B = \alpha B 
.\end{align*}
From here we can solve for \(B\) by dilating by \(1/ \alpha\), so
\(B = \alpha ^{-1} (\tilde I J)\). If we make this definition, does it
work?

\hfill\break

First, do we have \(B \subseteq {\mathbb{Z}}_K\)? This amounts to check
that \(\tilde I H \subseteq \left\langle{ \alpha }\right\rangle\). This
is true, using the assumption \(J \subseteq I\), since
\(\tilde I J \subseteq \tilde I I = \left\langle{ \alpha }\right\rangle\).
So \(B\) is not a fractional ideal, and is an honest ideal of
\({\mathbb{Z}}_K\). We can also check that
\begin{align*}
IB 
&= I( \alpha ^{-1} \tilde I J) \\
& = \alpha^{-1}(I \tilde I J) \\
&= \alpha^{-1}( \left\langle{ \alpha }\right\rangle J ) \\
&= \alpha ^{-1} ( \alpha J) \\
&= J
,\end{align*}
using that dilation commutes with ideal multiplication.

\end{proof}

\begin{remark}

We now want to prove that \(\operatorname{Id}({\mathbb{Z}}_K)\) is a
UFM. If it's cancellative, we just need to check factorization into
irreducibles and that irreducibles are prime, i.e.~the analog of
Euclid's lemma.

\end{remark}

\begin{remark}

We'll use the fact that \(\operatorname{Id}({\mathbb{Z}}_K)\) is a
\emph{reduced} monoid, i.e.~the only unit is the identity
\(\left\langle{ 1 }\right\rangle\), the entire ring. This follows from
the fact that the product of ideals is contained in both factors, so
each factor would contain 1 and thus be the entire ring. We'll proceed
in two steps:

\end{remark}

\begin{proposition}[Unique Factorization]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Every element of \(\operatorname{Id}({\mathbb{Z}}_K)\) factors into
  irreducibles in \(\operatorname{Id}({\mathbb{Z}}_K)\), and
\item
  (Euclid's Lemma) Irreducibles in \(\operatorname{Id}({\mathbb{Z}}_K)\)
  are prime.
\end{enumerate}

\end{proposition}

\begin{remark}

We'll use the fact that it's reduced to avoid having to say ``non-unit
element'' in (1), since we have only one unit and we'll think of it as
the empty product.\\

How do you prove (1)? The same way you prove it for the integers:
suppose you have a smallest counterexample. That can't be prime, since a
product of 1 prime is an allowable factorization, so this factors into a
product of two smaller things which necessarily can \emph{not} be
counterexamples by minimality. So the smaller factors break up into
primes -- but then so does their product, the original counterexample,
contradiction. The tricky part here is choosing what ``smaller'' should
mean.

\end{remark}

\begin{proof}[of 1]

If not, choose \(I\) of smallest norm where \(I\) has no such
factorization. Then \(I \neq \left\langle{ 1 }\right\rangle\) since by
convention this does factor as the product of zero irreducibles, and
\(I\) is not irreducible since irreducibles count as their own
factorization. So we can factor \(I = AB\) with
\(A, B \neq \left\langle{ 1 }\right\rangle\). Taking norms yields
\begin{align*}
N(I) = N(AB) = N(A) N(B)
.\end{align*}
We'd like the norms of \(A, B\) to be smaller, since then we could apply
the inductive hypothesis. The only obstruction to this would be if
\(N(A) = 1\) and \(N(B) = N(I)\). But having norm 1 means that
\(A = \left\langle{ 1 }\right\rangle\), since this means the quotient
has one element, forcing it to be the zero ring. So everything in the
ring is zero mod the ideal, i.e.~in the ideal. So
\(1 < N(A), N(B) < N(I)\). Since \(I\) was the smallest counterexample,
both \(A\) and \(B\) can be factored into irreducibles, but then
concatenating the two factorizations yields a factorization for \(AB\).
\(\contradiction\)

\end{proof}

\begin{proof}[of 2: Euclid's Lemma]

Suppose \(P\) is irreducible in \(\operatorname{Id}({\mathbb{Z}}_K)\)
and suppose \(P \divides IJ\), we want to show \(P\) divides one of
these two. Suppose \(P\notdivides I\), then \(P\) does not contain \(I\)
and \(P+I \supsetneq P\). This means that \(P+I\) is a \emph{proper
divisor} of \(P\), i.e.~it divides \(P\) but is not equal to \(P\). But
\(P\) was irreducible, so \(P+I\) is a unit, which forces
\(P + I = \left\langle{ 1 }\right\rangle\). Multiplying by \(J\) yields
\(PJ + IJ = J\). We said that \(P \divides IJ\) by assumption, so
\(IJ = PA\) for some nonzero ideal \(A\). So
\begin{align*}
J 
&= PJ + IJ \\
&= PJ + PA \\ 
&= P(J + A)
,\end{align*}
which shows that \(P\divides J\).

\end{proof}

\begin{remark}

Now running the exact same proof as for \({\mathbb{Z}}\) yields unique
factorization.

\end{remark}

\begin{exercise}[?]

Let \(P\) be a nonzero ideal of \(\operatorname{Id}({\mathbb{Z}}_K)\),
then \(P\) is monoidally prime in \(\operatorname{Id}({\mathbb{Z}}_K)\)
if and only if \(P\) is prime in the usual sense of prime ideals.

\begin{quote}
\emph{Hint: use ``to divide is to contain''.}
\end{quote}

\end{exercise}

\hypertarget{preview-ramification}{%
\subsection{Preview: Ramification}\label{preview-ramification}}

\begin{remark}

This chapter is about understanding prime ideals in quadratic number
rings, i.e.~\({\mathbb{Z}}_K\) for quadratic fields. What are the
building blocks of the nonzero prime ideals?

\end{remark}

\begin{definition}[Prime ideal above a prime number]

Let \(P\) be a nonzero prime ideal, then \(P\) \textbf{lies above} the
rational prime \(p\) if and only if
\(P \supseteq \left\langle{ p }\right\rangle\). Equivalently,
\(p\in P\), or \(P\divides \left\langle{ p }\right\rangle\).

\end{definition}

\begin{theorem}[Lying above unique primes]

Every nonzero prime ideal of \({\mathbb{Z}}_K\) lies above a unique
rational prime \(p\).

\end{theorem}

\begin{proof}[of theorem]

Consider \(P \cap{\mathbb{Z}}{~\trianglelefteq~}{\mathbb{Z}}\). Tracing
through the definitions, if \(P\) is a prime ideal in
\({\mathbb{Z}}_K\), then this intersection is also prime in
\({\mathbb{Z}}\). Moreover \(P \cap Z \neq \left\{{ 0 }\right\}\), since
we can take any nonzero element \(\alpha \in P\), then
\(0\neq \alpha\mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu\in {\mathbb{Z}}\)
and since \(P\) absorbs multiplication, this is still in \(P\). The
nonzero prime ideals of \({\mathbb{Z}}\) are of the form
\(n{\mathbb{Z}}\) with \(n\) prime, so
\(P \cap{\mathbb{Z}}= p{\mathbb{Z}}\) for some prime \(p\). But then
\(p\in P\) and \(P\) lies above \(p\). Why is this unique? If \(P\) lies
above \(q\), we would have \(q\in P \cap{\mathbb{Z}}= p {\mathbb{Z}}\)
and thus \(p\divides q\). But since these are both primes, \(p=q\).

\end{proof}

\begin{remark}

If we want to figure out all of the prime ideals \(P\) of
\({\mathbb{Z}}_K\), we should see how \(\left\langle{ p }\right\rangle\)
factors, since each \(P\) shows up as a factor of some
\(\left\langle{ p }\right\rangle\). Thus the major question will be:
given \(p\), how does \(\left\langle{ p }\right\rangle\) factor into
prime ideals in \({\mathbb{Z}}_K\)?

\end{remark}

\hypertarget{prime-ideals-of-mathbbz_k-lec.-8-tuesday-february-09}{%
\section{\texorpdfstring{Prime Ideals of \({\mathbb{Z}}_K\) (Lec. 8,
Tuesday, February
09)}{Prime Ideals of \{\textbackslash mathbb\{Z\}\}\_K (Lec. 8, Tuesday, February 09)}}\label{prime-ideals-of-mathbbz_k-lec.-8-tuesday-february-09}}

\hypertarget{dedekind-kummer-mirroring}{%
\subsection{Dedekind-Kummer Mirroring}\label{dedekind-kummer-mirroring}}

\begin{remark}

Today: chapter 7. Let \(K\) be a quadratic number field. Recall that if
\(P {~\trianglelefteq~}{\mathbb{Z}}_K\) is a prime then \(P\)
\textbf{lies above} \(p\in {\mathbb{Z}}\) if
\(P \supseteq \left\langle{ p }\right\rangle\). Equivalently,

\begin{itemize}
\tightlist
\item
  \(P\) contains \(p\), or
\item
  \(P \divides \left\langle{ p }\right\rangle\)
\end{itemize}

Last time we saw that every \(P\) lies above a unique \(p\). The
following diagram illustrates the situation:

\begin{center}
\begin{tikzcd}
    K && {{\mathbb{Z}}_K} && P \\
    \\
    {\mathbb{Q}}&& {\mathbb{Z}}&& p
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=1-3, to=3-3]
    \arrow[no head, from=1-5, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJLIl0sWzAsMiwiXFxRUSJdLFsyLDAsIlxcWlpfSyJdLFsyLDIsIlxcWloiXSxbNCwwLCJQIl0sWzQsMiwicCJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzQsNSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

If we want to determine all of the primes \(P\), we should consider
factoring all of the ideals \(\left\langle{ p }\right\rangle\) into
prime ideals of \({\mathbb{Z}}_K\). We have unique factorization for
prime ideals, so we can write
\(\left\langle{ p }\right\rangle = P_1 \cdots P_g\). Taking norms yields
\begin{align*}
N( \left\langle{ p }\right\rangle ) = \prod N(P_i) 
,\end{align*}
where we can identify the LHS as \(p^2\), since the norm for principal
ideals is the square of the generating element. Alternatively, we can
check the size of \({\mathbb{Z}}_K/ \left\langle{ p }\right\rangle\).
Note that \({\mathbb{Z}}_K\) is a free \({\mathbb{Z}}{\hbox{-}}\)module
on 2 generators, and we take both coordinates mod \(p\) to get
\(({\mathbb{Z}}/p{\mathbb{Z}})^2\). Since none of the terms on the RHS
are the unit ideal, none have norm 1, and we make the following
definition based on the possible cases:

\begin{definition}[Inert, Split, and Ramified Primes]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(g=1\) and \(P_1 = \left\langle{ g }\right\rangle\) and
  \(\left\langle{ p }\right\rangle\) is prime. In this case we say \(p\)
  \textbf{is inert}.
\item
  If \(g=2\) and \(P_1 \neq P_2\), then we say \(p\) \textbf{is split}.
\item
  If \(g=2\) and \(P_1 = P_2\)\textless{} then we say \(p\) \textbf{is
  ramified}.
\end{enumerate}

\end{definition}

Let \(K = {\mathbb{Q}}( \sqrt{d} )\) and \(\tau\) as usual. We can
compute its minimal polynomial:
\begin{align*}
\min_\tau(x) = 
\begin{cases}
x^2 - d & d \equiv 2,3 \operatorname{mod}4  
\\
x^2 - x + \qty{1-d \over 4} & d \equiv 1 \operatorname{mod}4.
\end{cases}
\end{align*}

\end{remark}

\begin{theorem}[Dedekind-Kummer, Prime Factorization Mirroring Theorem]

Let \(p\in {\mathbb{Z}}\) be prime. Then the factorization of
\(\left\langle{ p }\right\rangle\) into prime ideals in
\({\mathbb{Z}}_K\) mirrors the factorization of \(\min_\tau(x)\) into
irreducibles mod \(p\), i.e.~in \({\mathbb{F}}_p[x]\). If
\(\min_\tau(x)\) is irreducibles, then \(p\) is inert. Otherwise,
\begin{align*}
\min_\tau(x) \equiv (x-a)(x-b) \operatorname{mod}p
\end{align*}
for some \(a, b\in {\mathbb{Z}}\), since this is a monic quadratic. In
this case \(\left\langle{ p }\right\rangle= P_1 P_2\) where

\begin{itemize}
\tightlist
\item
  \(P_1 \coloneqq\left\langle{ p, \tau - a }\right\rangle\),
\item
  \(P_2 \coloneqq\left\langle{ p, \tau - b}\right\rangle\),
\end{itemize}

and both ideals have norm \(p\). Finally,
\(P_1 = P_2 \iff a\equiv b \operatorname{mod}p\).

\end{theorem}

\begin{example}[of inert, split, and ramified cases]

Let \(K = {\mathbb{Q}}( \sqrt{5} )\), then \(\tau= \sqrt{-5}\) and
\(\min_\tau(x) = x^2 + 5\). We can check how this factors modulo small
primes
\begin{align*}
x^2 + 5 = (x+1)^2 \in {\mathbb{F}}_2[x]
,\end{align*}
and we're in the ramified case. In this case,
\begin{align*}
\left\langle{ 2 }\right\rangle = \left\langle{ 2, \sqrt{ -5} -1 }\right\rangle^2  
.\end{align*}
We also have
\begin{align*}
x^2 + 5 \equiv (x-1)(x+1) \in {\mathbb{F}}_3[x]
,\end{align*}
which is the split case, so
\begin{align*}
\left\langle{ 3 }\right\rangle= \left\langle{ 3, \sqrt{-5} -1 }\right\rangle \left\langle{ 3, \sqrt{-5} + 1 }\right\rangle   
.\end{align*}
Taken mod 5, we have
\begin{align*}
x^2 + 5 \equiv x^2 \in {\mathbb{F}}_5[x]
,\end{align*}
so
\begin{align*}
\left\langle{ 5 }\right\rangle = \left\langle{ 5, \sqrt{-5} }\right\rangle^2 = \left\langle{ \sqrt{-5} }\right\rangle ^2 
.\end{align*}
Similarly,
\begin{align*}
x^2 + 5 \text{ is irreducible } \in {\mathbb{F}}_{11}[x]
,\end{align*}
so \(\left\langle{ 11 }\right\rangle\) is inert.

\end{example}

\begin{lemma}[Characterization of $\ZZ_K$ as a quotient of a polynomial ring]

There is a surjective morphism
\begin{align*}
{\mathbb{Z}}[x] &\to {\mathbb{Z}}_K = {\mathbb{Z}}[ \tau ] \\
f( \alpha) &\mapsto f( \tau)
,\end{align*}
so by the first isomorphism theorem,
\begin{align*}
{\mathbb{Z}}[x] / \left\langle{ \min_\tau(x) }\right\rangle \cong {\mathbb{Z}}_K 
.\end{align*}

\end{lemma}

\begin{proof}[of Dedekind-Kummer mirroring]

Note that
\({\mathbb{Z}}_K / \left\langle{ p }\right\rangle = {\mathbb{Z}}[ \tau] / \left\langle{ p }\right\rangle\),
and using the lemma, this is isomorphic to
\({\mathbb{Z}}[x] / \left\langle{ \min_{ \tau} (x), p }\right\rangle \cong {\mathbb{F}}_p[x] / \left\langle{ \min_\tau(x) \operatorname{mod}p }\right\rangle\).
In this case, if \(\min_\tau\) is irreducible mod \(p\), then the
quotient is a field. Why? The numerator is a polynomial ring over a
field and the denominator is generated by an irreducible, and a PID mod
an irreducible is always a field. Thus
\(\left\langle{ p }\right\rangle\) must be a maximal ideal by
considering the first expression above, and maximals are prime here, so
\(p\) is inert.

Now suppose it's not irreducible, so
\begin{align*}
\min_\tau(x) = (x-a)(x-b) \operatorname{mod}p
.\end{align*}
Define \(P_1, P_2\) as in the theorem. Why are these of norm \(p\)?
Consider
\begin{align*}
{\mathbb{Z}}_K/P^1 
&\cong { {\mathbb{Z}}[x] / \left\langle{ \min_\tau(x) }\right\rangle \over \left\langle{ p, x-a }\right\rangle } \\
&\cong {\mathbb{Z}}/p{\mathbb{Z}}[x] / \left\langle{ \min_\tau(x), x-a}\right\rangle \\
&\cong {\mathbb{Z}}/p{\mathbb{Z}}[x] / \left\langle{ \min_\tau(x), x-a}\right\rangle \\
&\cong {\mathbb{Z}}/p{\mathbb{Z}}[x] / \left\langle{ x-a}\right\rangle && \text{since } x-a \divides \min_\tau(x)\\
&\cong {\mathbb{Z}}/p{\mathbb{Z}}
.\end{align*}
So \(P_1\) is maximal and thus prime, and moreover \(N(P_1) = p\) since
there are \(p\) elements in \({\mathbb{Z}}/p{\mathbb{Z}}\). The same
argument works for \(P_2\). Now multiplying them yields
\begin{align*}
P_1 P_2 
&= \left\langle{ p, p(\tau - a), p (\tau - b), (\tau -a)(\tau -b) }\right\rangle
.\end{align*}
Note that
\begin{align*}
\min_\tau(x) &\equiv (x-a)(x-b) \operatorname{mod}p \\
\implies
\min_\tau(x) &= (x-a)(x-b) + pG(x)
\end{align*}
for some \(G\in {\mathbb{Z}}[x]\). Plugging in \(\tau\), the LHS is
zero, while on the RHS yields \(\cdots + pG(\tau)\). This last term is
\(pr\) for some \(r\in R\), which is zero mod
\(p \in {\mathbb{Z}}[\tau] = {\mathbb{Z}}_K\) So \(p\) now divides every
term in the generating set above, and since to contain is to divide, we
have \(P_1 P_2 \subseteq \left\langle{ p }\right\rangle\) and
\(\left\langle{ p }\right\rangle \divides P_1 P_2\). Write
\(P_1 P_2 = \left\langle{ p }\right\rangle I\) for some ideal \(I\),
taking norms yields
\begin{align*}
N(P_1) N(P_2) = N( \left\langle{ p }\right\rangle) N(I) 
.\end{align*}
The LHS is \(p^2\) as shown above, and the RHS is \(p^2 N(I)\) which
forces
\(N(I) = 1 \iff I = \left\langle{ 1 }\right\rangle = {\mathbb{Z}}_K\)
(the entire ring)

\hfill\break

We now want to show \(P_1 = P_2 \iff a\equiv b \operatorname{mod}p\).
The reverse direction is clear, since generators in \(P_1, P_2\) can be
adjusted by \(p\) without changing the ideal. Conversely, suppose
\(P_1 = P_2\). Then \(P_1\) contains \(\tau - a, \tau - b\), and thus
their difference \(a-b = (\tau -b ) - (\tau - a) \in P_1\). Moreover
\(p\in P_1\), and so \(P_1\) contains the \({\mathbb{Z}}\) ideals
generated by \(p\) and \(a-b\) and thus \(\gcd(p, a-b)\). If
\(a \not\equiv b\operatorname{mod}p\), this greatest common divisor must
be 1, forcing \(1\in P_1\). This is a contradiction since \(P_1\) is
prime and thus can't be the unit ideal, so
\(a \equiv b \operatorname{mod}p\).

\end{proof}

\begin{question}

Can we be more explicit about how \(\min_\tau\) factors?

\end{question}

\begin{proposition}[Characterization of inert/split/ramified primes]

Let \(p\) be an odd prime, then

\begin{itemize}
\tightlist
\item
  \(p\) is inert \(\iff d\) is not a square \(\operatorname{mod}p\),
\item
  \(p\) splits \(\iff d\) is a nonzero square \(\operatorname{mod}p\),
\item
  \(p\) ramifies \(\iff d \equiv 0 \operatorname{mod}p\).
\end{itemize}

\end{proposition}

\begin{proposition}[Inert/Split/Ramified primes for quadratic fields]

\envlist

\begin{itemize}
\tightlist
\item
  \(d \equiv 5 \operatorname{mod}8 \implies\) 2 is inert.
\item
  \(d \equiv 1 \operatorname{mod}8 \implies\) 2 is split.
\item
  \(d \equiv 2, 3 \operatorname{mod}4 \implies\) 2 is ramified.
\end{itemize}

\end{proposition}

\begin{remark}

The proof follows from looking at how \(\min_\tau(x)\) factors
\(\operatorname{mod}2\), and there aren't many possibilities.

\end{remark}

\hypertarget{units-in-mathbbz_k}{%
\subsection{\texorpdfstring{Units in
\({\mathbb{Z}}_K\)}{Units in \{\textbackslash mathbb\{Z\}\}\_K}}\label{units-in-mathbbz_k}}

\begin{remark}

Roughly chapter 8. For the imaginary quadratic case, we can write down
the unit group explicitly.

\end{remark}

\begin{proposition}[Imaginary quadratic fields have at most 6 units]

If \(d<0\) (i.e.~the imaginary quadratic case) then
\({\left\lvert { U({\mathbb{Z}}_J)} \right\rvert} \leq 6\).

\end{proposition}

\begin{remark}

``Usually'' \(U({\mathbb{Z}}_K) = \left\{{ \pm 1 }\right\}\). Here
``usually'' means there are only two exceptions:

\begin{itemize}
\item
  For \({\mathbb{Q}}( \sqrt{-1} )\) then the units are
  \(\left\{{ \pm 1, \pm i }\right\}\).
\item
  For \(d=-3\), there were 6 units.
\end{itemize}

In every other case, there are only two.

\end{remark}

\begin{proposition}[Existence of the fundamental unit]

Suppose \(d>0\), then there is a unit
\(\epsilon_0 > 1 \in {\mathbb{Z}}_K\) such that
\(U({\mathbb{Z}}_K) = \left\{{ \pm \epsilon_0 ^k {~\mathrel{\Big|}~}k\in {\mathbb{Z}}}\right\}\).
Moreover \(\epsilon_0\) is unique, and we'll refer to this as the
\textbf{fundamental unit}.

\end{proposition}

\begin{corollary}[The unit group is infinite for real quadratic fields]

When \(d>0\), \(U({\mathbb{Z}}_K)\) is infinite and in fact isomorphic
to \({\mathbb{Z}}/2{\mathbb{Z}}\oplus {\mathbb{Z}}\).

\end{corollary}

\begin{remark}

Here the \({\mathbb{Z}}/2{\mathbb{Z}}\) corresponds to the \(\pm\) and
the \({\mathbb{Z}}\) to the exponent.

\end{remark}

\begin{example}[of the fundamental unit]

\envlist

\begin{itemize}
\item
  For \(d=2\), we have \({\varepsilon}_0 = 1 + \sqrt{2}\). This is a
  unit because it has inverse \(\sqrt{2} -1\).
\item
  For \(d=43\), it turns out that \({\varepsilon}_0 = 531 \sqrt{43}\).
\end{itemize}

\end{example}

\begin{lemma}[Computation of norm of the fundamental unit]

Let \(\epsilon\in {\mathbb{Z}}_K\), then
\(\epsilon \in U({\mathbb{Z}}_K) \iff N( \epsilon) = \pm 1\).

\end{lemma}

\begin{remark}

Note that norms were positive in the imaginary quadratic case, but can
be negative for real quadratics.

\end{remark}

\begin{proof}[of computation of norm]

\(\impliedby\): This means
\(\epsilon \mkern 1.5mu\overline{\mkern-1.5mu \epsilon \mkern-1.5mu}\mkern 1.5mu = \pm 1\),
so one of
\(\pm \mkern 1.5mu\overline{\mkern-1.5mu\epsilon\mkern-1.5mu}\mkern 1.5mu\)
is the inverse.

\(\implies\): Write \(\epsilon \epsilon ^{-1} = 1\) and take norms of
both sides.

\end{proof}

\begin{remark}

Our strategy: show that the group of positive units
\(U({\mathbb{Z}}_K)^+\) is infinite cyclic. If we get a generator
\({\varepsilon}_0 > 1\), replace it with its reciprocal, and note that
we don't want \({\varepsilon}_) = 1\) since this wouldn't yield an
infinite group. If we can generate all of the positive units, all of the
negative units are negatives of positive units. How we'll do this: we'll
look at the map
\begin{align*}
\log: {\mathbb{G}}_m({\mathbb{R}}^+) \xrightarrow{} {\mathbb{G}}_a({\mathbb{R}})
.\end{align*}
and consider the image \(\log( U( {\mathbb{Z}}_K)^+)\), which will be an
infinite cyclic subgroup of \({\mathbb{G}}_a({\mathbb{R}})\).

\end{remark}

\begin{proposition}[The log subgroup is discrete]

The subgroup \(\log( U ({\mathbb{Z}}_K)^+ )\) is discrete, i.e.~it has
finite intersection with \([-X, X] \subseteq {\mathbb{R}}\) for every
\(X>0\).

\end{proposition}

\begin{proof}[of proposition]

It's enough to show finite intersection with \([0, X]\) for all \(X>0\).
Why? Any subgroup \(H\leq {\mathbb{G}}_a({\mathbb{R}})\) is symmetric
about 0, i.e.~\(a\in H \iff -a\in H\), and so having finite intersection
with the positive interval implies finite intersection with both. So let
\(\epsilon \in U({\mathbb{Z}}_K)^+\) with
\(\log( \epsilon) \in [0, X]\), we'll show there are only finitely many
choices for \(\epsilon\), since every \(\log({\varepsilon})\) correspond
to a point in the intersection.

\begin{claim}

Write \(\epsilon = u + v \sqrt{d}\) with \(u,v \in {\mathbb{Z}}\) or
\({1\over 2}{\mathbb{Z}}\), then \(u, v \geq 0\).

\end{claim}

If we have this, we're done since \(\log( u + v \sqrt{d}) \leq X\).
Exponentiating yields \(u + v\sqrt {d} \leq e^X\), and so we must have
\(u, v \leq e^X\). But there are only finitely many possibilities, since
these are integers or half-integers.

\begin{proof}[of claim]

We have \(\epsilon \geq 1\) since \(u, v \geq 0\). There are now two
cases:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(N( \epsilon) = 1\). In this case,
  \(\epsilon \mkern 1.5mu\overline{\mkern-1.5mu\epsilon \mkern-1.5mu}\mkern 1.5mu= 1\)
  and so \(\epsilon ^{-1} \epsilon\). We can write
  \(u = (1/2)( \epsilon + \mkern 1.5mu\overline{\mkern-1.5mu\epsilon\mkern-1.5mu}\mkern 1.5mu) = (1/2)(\epsilon + \epsilon ^{-1} ) > 0\).
  Similarly,
  \begin{align*} 
    v = (\epsilon - \mkern 1.5mu\overline{\mkern-1.5mu\epsilon\mkern-1.5mu}\mkern 1.5mu)/2 \sqrt{d} = (\epsilon - \epsilon ^{-1} ) / 2 \sqrt{d} \geq 0
    .\end{align*}
\item
  \(N(\epsilon) = -1\). This case proceed similarly.
\end{enumerate}

\end{proof}

This completes the proof.

\end{proof}

\begin{question}

What do the discrete subgroups of \({\mathbb{G}}_a({\mathbb{R}})\) look
like?

\end{question}

\begin{answer}

Some examples are
\(\left\{{0}\right\}, {\mathbb{Z}}, \lambda {\mathbb{Z}}\) for
\(\lambda \in {\mathbb{R}}\), etc. It turns out that these are the only
ones. Knowing that these must be the image of the log map, if we're in
the \(\alpha{\mathbb{Z}}\) case we're fine because this is infinite
cyclic, but the case \(\left\{{ 0 }\right\}\) is an issue: this would
mean that the only positive unit is \(e^0 = 1\), and the only units are
\(\pm 1\). So we just need to show that there are units other than
\(\pm 1\).

\end{answer}

\hypertarget{units-in-mathbbz_k-lec.-9-monday-february-15}{%
\section{\texorpdfstring{Units in \({\mathbb{Z}}_K\) (Lec. 9, Monday,
February
15)}{Units in \{\textbackslash mathbb\{Z\}\}\_K (Lec. 9, Monday, February 15)}}\label{units-in-mathbbz_k-lec.-9-monday-february-15}}

\hypertarget{review}{%
\subsection{Review}\label{review}}

\begin{remark}

Today: chapter 8. We'll continue with the statements from last time:

\end{remark}

\begin{proposition}[Subgroups of $\RR$ are either discrete or infinite cyclic]

A discrete subgroup \(\Lambda \leq {\mathbb{R}}\) is either \(0\) or
infinite cyclic, where \emph{discrete} means having finite intersection
with every interval \([-x, x]\).

\end{proposition}

\begin{proof}[of proposition]

Suppose \(\Lambda \neq 0\), then we can choose a smallest positive
element \(\alpha\in \Lambda\). Why does this exist? There are only
finitely many elements in \([0, \alpha]\), so there is a smallest, and
we could replace \(\alpha\) with it. The claim is that
\(\Lambda = {\mathbb{Z}}\alpha\). The reverse containment is clear
because the RHS is necessarily a subgroup. Toward a contradiction,
suppose there is some \(\beta\in \Lambda\setminus{\mathbb{Z}}\alpha\)
with \(n \alpha < \beta < (n+1) \alpha\) for some \(n\in {\mathbb{Z}}\).
This can't happen: subtracting \(n\) from both sides yields
\begin{align*}
0 < \beta - n \alpha < \alpha
,\end{align*}
where the middle term is necessarily in \(\Lambda\), contradicting
minimality of \(\alpha\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{6pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/NumberTheory/sections/figures}{2021-02-15_21-44.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{proof}

\begin{remark}

Recall that to show the theorem we wanted, it was enough to show
\(\log U({\mathbb{Z}}_K)^+\) is an infinite cyclic subgroup of
\({\mathbb{G}}_a({\mathbb{R}})\). We proved that this was a discrete
subgroup. If this were just the zero element, the only possible units
would be \(\pm 1\), so it suffices to find a unit
\(\epsilon \in U({\mathbb{Z}}_K)\) with \(\epsilon>0\) and
\(\epsilon\neq 1\).

\end{remark}

\hypertarget{an-aside-diophantine-approximation}{%
\subsection{An Aside: Diophantine
approximation}\label{an-aside-diophantine-approximation}}

\begin{remark}

Let \(\alpha\in {\mathbb{R}}\) and let \(Q \in {\mathbb{Z}}^+\). How
well can we approximate \(\alpha\) with a fraction with denominator
bounded by \(Q\)?

\end{remark}

\begin{theorem}[Dirichlet's Approximation Theorem]

There is a \(q \leq Q \in {\mathbb{Z}}^+\) with
\begin{align*} 
{\left\lVert {q \alpha } \right\rVert} \leq {1 \over Q+1} 
,\end{align*}
where \({\left\lVert {{-}} \right\rVert}\) denotes the distance to the
nearest integer.

\end{theorem}

\begin{remark}

The way to think about this inequality: if the LHS is close to an
integer \(p\), then \(\alpha\) is close to \(p/q\).

\end{remark}

\begin{proof}[of Dirichlet's theorem]

Chop the interval into \(Q+1\) pieces, and think of the inequality as a
condition on the fractional part of \(\alpha\), denoted
\(\left\{{ qa }\right\} \coloneqq qa - {\left\lfloor qa \right\rfloor}\in [0, 1)\).
Note that if \(\left\{{ qa }\right\} \in [0, 1/Q+1)\) or \([Q/Q+1, Q)\)
for some \(q\), then we are done. If not, it must land in one of the
\(q-1\) middle intervals
\begin{align*}
[1/Q+1, 2/Q+1), 
[2/Q+1, 3/Q+1), 
\cdots,
[Q-1/Q+1, Q/Q+1)
\end{align*}
for all all \(q\leq Q\). But we have \(Q\) choices for \(q\) and only
\(Q-1\) intervals, so there are two values of \(q\) with fractional part
in the same interval. So choose these, say \(q_1<q_2 \leq Q\), and
consider \(q \coloneqq q_2 - q_1\). Since
\(\left\{{q_1 \alpha}\right\}, \left\{{q_2 \alpha}\right\}\) are in the
same interval, we have \(\left\{{q \alpha}\right\} \in [0, 1/Q+1)\),
putting it close to an integer.

\end{proof}

\begin{corollary}[Infinitude of elements of bounded norm]

There are infinitely many pairs of positive integers \((p, q)\) such
that
\begin{align*}
{\left\lvert {p^2 - dq^2} \right\rvert}\leq 1 + 2 \sqrt{d}  
,\end{align*}
where \(d\) was the squarefree integer for which
\(K = {\mathbb{Q}}( \sqrt{d} )\).

\end{corollary}

\begin{remark}

Note that the RHS does not depend on \(p\) or \(q\), and only depends on
the field. Moreover, this proof is also true with the 1 removed.

\end{remark}

\begin{proof}[of corollary]

Using Dirichlet's approximation theorem, choose \(Q \in {\mathbb{Z}}^+\)
and \(1\leq q\leq Q\) such that
\begin{align*}
{\left\lVert { q \sqrt{d} } \right\rVert} \leq {1 \over Q+1}
,\end{align*}
then there is a \(p \in {\mathbb{Z}}\) such that
\begin{align*}
{\left\lvert { p - q \sqrt{d} } \right\rvert} \leq {1 \over Q+1}
.\end{align*}
We know \(q\) is positive by Dirichlet's theorem, and \(p\) is positive
since \(q\sqrt{d} \geq \sqrt{d} \geq 1\), and the distance from \(p\) to
\(q\) is at most \(1/2\). We can now check
\begin{align*}
{\left\lvert {p^2 - dq^2} \right\rvert} 
&= {\left\lvert {p - q \sqrt{d}} \right\rvert} {\left\lvert {p + q \sqrt{d}} \right\rvert}    \\
&= {\left\lvert {p - q \sqrt{d}} \right\rvert} {\left\lvert {(p- q \sqrt{d} ) + 2q \sqrt{d} } \right\rvert}    \\
&\leq {\left\lvert {p - q \sqrt{d}} \right\rvert} {\left\lvert {p- q \sqrt{d}} \right\rvert} + {\left\lvert {2q \sqrt{d} } \right\rvert}    \\
&\leq {1\over Q+1} \qty{ {1\over Q+1} + 2Q \sqrt{d} }\\
&= \qty{1\over Q+1}^2 + {2Q \over Q+1} \sqrt{d} \\
&< 1 + 2\sqrt{d}
,\end{align*}
where we've applied the triangle inequality and used the bound twice.
How do we know that this results in infinitely many distinct pairs?
Things could also go wrong if the same pairs resulted from all but
finitely many choices of \(Q\). However, the bound from Dirichlet's
theorem prevents this: any pair \((p, q)\) can arise for ay most
\emph{finitely} many starting values for \(Q\). Pick a \(Q\), then
produce \(q\) satisfying the bound. Then
\({\left\lVert { q \sqrt{d} } \right\rVert} \neq 0\) since \(\sqrt{d}\)
is irrational, and thus the LHS is some positive irrational number. For
a fixed \(q\), choosing \(Q'\) big enough can make the RHS smaller than
the LHS, meaning that \(q\) can not occur for that value of \(Q'\) or
anything larger. In other words, we're using
\begin{align*}
{\left\lVert { q \sqrt{d} } \right\rVert} \leq {1\over Q+1} \overset{Q \to \infty } \to 0
,\end{align*}
and there can't be any infinite sequences of \(Q_i\) yielding the same
fixed \(q\), since the RHS would go to zero while the LHS does not.

\end{proof}

\begin{remark}

Choosing a pair \((p, q)\) as above, we'll have
\(p + q \sqrt{d} \in {\mathbb{Z}}_K\) and
\begin{align*}
{\left\lvert { N( p + q \sqrt{d} ) } \right\rvert} 
= {\left\lvert { p^2 - d q^2 } \right\rvert} \\
< 1 + 2 \sqrt{d} 
.\end{align*}
So we have many elements in \({\mathbb{Z}}_K\) whose norm is bounded,
which will force the existence of a nontrivial unit.

\end{remark}

\begin{lemma}[Finitely many ideals of bounded norm]

For all real \(x> 0\) there are finitely many nonzero ideals
\(I{~\trianglelefteq~}{\mathbb{Z}}_K\) with
\(N(I) \coloneqq{\left\lvert { {\mathbb{Z}}_K/I } \right\rvert} \leq x\).

\end{lemma}

\begin{proof}[of lemma]

Suppose \(N(I) \coloneqq m \leq x\) with \(m \in {\mathbb{Z}}^+\); it's
enough to show that for each \(m\) there are at most finitely many
\(I\), since there are only finitely many values of \(m \leq x\). View
\({\mathbb{Z}}_K / I\) as a group under addition, so by Lagrange every
element has order dividing \(m\). We can check
\(m = 1 + 1 + \cdots + 1\), which must be the identity in
\({\mathbb{Z}}_K/I\). So \(m\in I\), and since to contain is to divide,
\(I \divides \left\langle{ m }\right\rangle\). But
\(\left\langle{ m }\right\rangle\) has only finitely many ideal
divisors. Why? This is because there is unique prime factorization, and
just like \(n = \prod p_i^{n_i}\) in the integers, \(n\) has
\(\sum n_i < \infty\) possible divisors.

\end{proof}

\begin{proof}[There exists a nontrivial unit]

We now want to show that there exists a unit \(\epsilon>0\) that is not
equal to 1. Consider all ideals
\(I_{p, q} \coloneqq\left\langle{ p + q \sqrt{d} }\right\rangle\) where
\((p, q)\) is a pair of positive integers such that
\({\left\lvert {p^2 - dq^2} \right\rvert} < 1 + 2 \sqrt{d}\). Taking
norms amounts to taking absolute values of generators, so
\begin{align*}
N(I_{p, q} ) < 1 + 2 \sqrt{d}
\end{align*}
for all \(p, q\). By the last lemma, this means there are only finitely
many different ideals. On the other hand, there are infinitely many such
pairs, so infinitely many pairs give rise to the same ideal. Pick two
pairs \((p, q)\) and \((p', q')\) such that
\(\left\langle{ p + q \sqrt{d} }\right\rangle = \left\langle{ p' + q' \sqrt{d} }\right\rangle\).
If two ideals are equal, the generators differ by a unit, and so
\begin{align*}
(p + q \sqrt{d} ) = {\varepsilon}(p' + q' \sqrt{d} ), && {\varepsilon}\in U({\mathbb{Z}}_K)
.\end{align*}
Everything in sight is positive, so solving for \({\varepsilon}\) yields
\({\varepsilon}> 0\). But \({\varepsilon}\neq 1\), since the pairs would
have to have been the same by comparing coefficients in the expression
above.

\end{proof}

\begin{remark}

This gives us the fundamental unit. How do we actually find it? See the
book -- use continued fractions! It's not surprising they'd come up,
since they provide a more constructive proof of Dirichlet's
approximation theorem.

\end{remark}

\begin{example}[of the fundamental unit]

Take \(d=2\), what is \({\varepsilon}_0\)? We have
\(U({\mathbb{Z}}_K) = \left\{{ \pm {\varepsilon}_0 ^k {~\mathrel{\Big|}~}k\in {\mathbb{Z}}}\right\}\),
and so if we just look at positive units, the smallest power such that
\({\varepsilon}_0^k > 1\) will just be equal to \({\varepsilon}_0\). So
we're really looking for the smallest unit greater than 1. We proved
that if \({\varepsilon}_0 = u + v \sqrt{d}\), then \(u, v \geq 0\), and
if \({\varepsilon}_0 > 1\) is strict then \(u, v > 0\) is strict as
well. We also know that \(u, v \geq 1\), using that
\({\mathbb{Z}}_K = {\mathbb{Z}}[\sqrt{2} ]\). Luckily enough,
\(1 + \sqrt{2}\) is a unit, and so \({\varepsilon}_0 = 1 + \sqrt{2}\).

\end{example}

\hypertarget{class-groups-and-the-class-number}{%
\subsection{Class Groups and the Class
Number}\label{class-groups-and-the-class-number}}

\begin{remark}

This is now chapter 9. Let \(K\) be a quadratic field.

\end{remark}

\begin{definition}[Dilation Equivalence]

If \(I, J\) are nonzero ideals of \({\mathbb{Z}}_K\), we say \(I, J\)
are \textbf{dilation equivalent} if there exists a
\(\lambda\in K^{\times}\) such that \(I = \lambda J\).

\end{definition}

\begin{remark}

It's easy to check that this is an equivalence relation, so we'll use
\(I \approx J\).

\end{remark}

\begin{definition}[Class Group]

The \textbf{class group} of \({\mathbb{Z}}_K\) is defined as
\begin{align*}
{ \operatorname{Cl}} ({\mathbb{Z}}_K) \coloneqq\operatorname{Id}({\mathbb{Z}}_K)/\approx
.\end{align*}

\end{definition}

\begin{remark}

A priori this is just a set, but we can descent the monoid structure to
define a group multiplication. We define \([I] [J] = [IJ]\), and it's
easy to check that this is well-defined on equivalence classes. The
identity is \([ \left\langle{ 1 }\right\rangle]\), and for inverses we
can use the fact that
\([I \mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu] = [\left\langle{ N(I) }\right\rangle] = N(I) [ \left\langle{ 1 }\right\rangle ]\).
In fact, any \(J\) for which \(IJ\) is principal serves as an inverse
for \(I\). So the inverses come from the \emph{Principal Multiple
Lemma}, and a similar story will go through for general number fields.

\end{remark}

\begin{remark}

This is an abelian group, wouldn't it be nice if it were finite? This is
one of the big theorems of number theory:
\({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is finite. We can thus define
the following:

\end{remark}

\begin{definition}[Class Number]

The \textbf{class number} of \(K\) is defined as:
\begin{align*}
h_k \coloneqq{\left\lvert { { \operatorname{Cl}} ({\mathbb{Z}}_K) } \right\rvert}
.\end{align*}

\end{definition}

\begin{lemma}[Comparison bound between element norm and ideal norm]

There is a constant \(C\) depending on \(K\) such that for every
\(I \in \operatorname{Id}({\mathbb{Z}}_K)\) there is a nonzero
\(\alpha\in I\) such that
\begin{align*}
{\left\lvert {N \alpha} \right\rvert}\leq C N(I) 
.\end{align*}
In fact, one can take
\begin{align*}
C \coloneqq 1 + \operatorname{Tr}(\tau) + {\left\lvert {N(\tau)} \right\rvert} 
.\end{align*}

\end{lemma}

\begin{remark}

The norm of \(I\) is a natural thing to compare \(N \alpha\) to, since
\(I \divides \left\langle{ \alpha }\right\rangle\) and thus
\(N(I) \divides N(\left\langle{ \alpha }\right\rangle)\), so there's no
hope of the LHS being smaller than \(N(I)\).

\end{remark}

\begin{proof}[?]

Look at all elements \(a + b \tau \in {\mathbb{Z}}_K\) such that
\(0 \leq a,b, \sqrt{ N(I) }\). How many elements does this yield?
Precisely \(\qty{ {\left\lfloor \sqrt{N(I)} \right\rfloor} +1 }^2\).
Note that this is strictly larger than \(N(I)\), using
\({\left\lfloor x \right\rfloor} > x-1\) for any \(x\). Then going to
the quotient by \(I\), there are exactly \(N(I)\) elements, two elements
reduce to the same element of \({\mathbb{Z}}_K/I\). So their difference
is in \(I\), so we get something of the form \(a' + b' \tau\) where
\(a;, b; \in {\mathbb{Z}}\) (where they could now be negative), but are
bounded by
\begin{align*}
- \sqrt{N(I)}
\leq a', b' \leq 
 \sqrt{N(I)} 
.\end{align*}

The claim is now that the given value of \(C\) in the theorem works:
\begin{align*}
{\left\lvert {N( \alpha)} \right\rvert}
&=
{\left\lvert {N( a' + b' \tau)} \right\rvert} \\
&=
{\left\lvert {(a' + b' \tau) (a' + b' \mkern 1.5mu\overline{\mkern-1.5mu\tau\mkern-1.5mu}\mkern 1.5mu)} \right\rvert} \\
&= {\left\lvert {(a')^2 + a'b' \operatorname{Tr}(\tau) + (b')^2 N \tau} \right\rvert} \\
&\leq 
{\left\lvert {a'} \right\rvert}^2 + {\left\lvert {a'} \right\rvert} {\left\lvert {b'} \right\rvert} \operatorname{Tr}(\tau) + {\left\lvert {b'} \right\rvert}^2 N(\tau) \\
&\leq C N(I)
,\end{align*}
where we've used \(a', b' \leq \sqrt{N(I)}\) and collected terms in the
last step.

\end{proof}

\begin{proposition}[Class representatives of small norm]

Every ideal class contains an ideal \(I\) of norm \(N(I) \leq C\).

\end{proposition}

\begin{corollary}[Class numbers are finite]

\(h_K < \infty\).

\end{corollary}

\begin{remark}

Why is this true? There are only finitely many ideals with this norm
bound, and this says every ideal class belongs to this finite set.

\end{remark}

\begin{proof}[of proposition]

Since we're working with a group, it suffices to work with inverses,
since these still run over all elements. It's enough to show that for
every \(I \in \operatorname{Id}({\mathbb{Z}}_K)\), we can write
\([I]^{-1}= [J]\) for some \(J\) satisfying \(N(J) \leq C\). Choose a
nonzero \(\alpha\in I\) with
\({\left\lvert {N( \alpha )} \right\rvert}\leq C N(I)\). Since
\(\alpha\in I\) we know that
\(I \divides \left\langle{ \alpha }\right\rangle\), so we can write
\(\left\langle{ \alpha }\right\rangle = IJ\) for some ideal \(J\). We
have
\([I] [J] = [IJ] = [ \left\langle{ \alpha }\right\rangle ] = [ \left\langle{ 1 }\right\rangle ]\),
since all principal ideals are dilation-equivalent to
\(\left\langle{ 1 }\right\rangle\). This means that \([J] = [I] ^{-1}\),
and our hope is that it has small norm. Taking norms in
\(\left\langle{ \alpha }\right\rangle = IJ\) yields
\begin{align*}
{\left\lvert {N \alpha} \right\rvert} 
&= N(I) N(J) \\
\implies N(J) 
&= { {\left\lvert {N \alpha} \right\rvert} \over N(I) } \\
&\leq { C N(I) \over N(I)} \\
&= C
.\end{align*}

\end{proof}

\begin{example}[?]

What we'll look at next:
\({ \operatorname{Cl}} ( {\mathbb{Z}}[ \sqrt{-5} ])\). We know this does
not have unique factorization, and the claim is that the class group is
nontrivial. If it were, every ideal would be dilation-equivalent to
\(\left\langle{ 1 }\right\rangle\), making every ideal principal, and
every PID is a UFD. Here we'll have \(C=6\).

\hfill\break

One could try to write down all ideals of norm bounded by 6, but instead
lets consider how they factor into primes. Every ideal of norm at most 6
factors into prime ideals, whose norm is also bounded by 6. So this
factors into prime ideals lying above \(2,3,5\), since any ideal lying
above a prime \(p\) has norm \(p\) or \(p^2\), and we need
\(p, p^2 < 6\) here. We've worked out all such primes before, coming
from the \emph{prime factor mirroring theorem}:

\begin{itemize}
\tightlist
\item
  \(\left\langle{ 2 }\right\rangle= P_1^2,\, P_1 \coloneqq\left\langle{ 2, 1 + \sqrt{-5} }\right\rangle\)
\item
  \(\left\langle{ 3 }\right\rangle = P_2 P_3, P_2 \coloneqq\left\langle{ 3, 1 - \sqrt{-5} }\right\rangle, P_3 \coloneqq\left\langle{ 1 + \sqrt{-5} }\right\rangle\)
\item
  \(\left\langle{ 5 }\right\rangle= P_4^2, P_4 \coloneqq\left\langle{ \sqrt{-5} }\right\rangle\)
\end{itemize}

This allows us to conclude that
\begin{align*}
{ \operatorname{Cl}} ({\mathbb{Z}}[ \sqrt{-5} ]) = \left\langle{ [P_1], [P_2], [P_3], [P_4] }\right\rangle
.\end{align*}
In fact, since \(P_4\) is principal we can leave it out.

\end{example}

\hypertarget{class-groups-lec.-10-thursday-february-18}{%
\section{Class Groups (Lec. 10, Thursday, February
18)}\label{class-groups-lec.-10-thursday-february-18}}

\hypertarget{computing-class-groups}{%
\subsection{Computing Class Groups}\label{computing-class-groups}}

\begin{remark}

Last time: we defined an equivalence relation on nonzero ideals of
\({\mathbb{Z}}_K\), namely \(I \approx J \iff I = \alpha J\) for some
\(\alpha \in K^{\times}\). We then defined the \textbf{class group}
\begin{align*}
{ \operatorname{Cl}} ({\mathbb{Z}}_K) \coloneqq\operatorname{Id}({\mathbb{Z}}_K) / \approx
.\end{align*}
We saw that ideal multiplication descends to a well-defined group
structure on ideal classes. Since ideal multiplication is commutative,
this is an abelian group, and moreover it is finite.

\end{remark}

\begin{example}[Computing the Class Group]

Let \(K = {\mathbb{Q}}(d)\) where \(d \coloneqq\sqrt{ -5 }\). We saw
that every ideal class is represented by an element with bounded norm.
Applying it to this specific value of \(d\), every element is
represented by \([I]\) where \(N(I) \leq 6\). If we have such an ideal,
it will factor into primes, and thus the class will factor into prime
classes. Thus the group is actually \emph{generated} by prime ideals of
norm at most 6. Any such ideal will lie above a prime \(p \leq 6\), so
\(p=2,3,5\). We saw
\begin{align*}
\left\langle{ 2 }\right\rangle = P_1^2 && P_1 \coloneqq\left\langle{ 2, 1 + \sqrt{-5} }\right\rangle \\
\left\langle{ 3 }\right\rangle = P_2 P_3 && P_2 \coloneqq\left\langle{ 3, 1 + \sqrt{-5} }\right\rangle,\, P_3 \coloneqq\left\langle{ 3, 1 - \sqrt{-5} }\right\rangle    \\
\left\langle{ 5 }\right\rangle = P_4^2 && P_4 \coloneqq\left\langle{ \sqrt{-5} }\right\rangle.  
.\end{align*}
We conclude that
\({ \operatorname{Cl}} ({\mathbb{Z}}_K) = \left\langle{ P_1, \cdots, P_4 }\right\rangle\).
What are the relations? Consider \(P_4\), and note that
\(\left\langle{ \sqrt{-5} }\right\rangle \approx \left\langle{ 1 }\right\rangle\)
since \(P_4 = \sqrt{-5} \left\langle{ 1 }\right\rangle\). A similar
argument works for any principal ideal, and we can throw out \(P_4\).
Consider \(P_2\) and \(P_3\). Since
\(\left\langle{ 3 }\right\rangle \approx \left\langle{ 1 }\right\rangle\),
we have \(P_2 = P_3 ^{-1}\), so we can also throw out \(P_2\), since we
don't need to include the inverse of a generator. Recall that there is a
factorization
\begin{align*}
\left\langle{ 1 - \sqrt{-5} }\right\rangle 
=
\left\langle{ 2, 1 - \sqrt{-5} }\right\rangle 
\left\langle{ 3, 1 - \sqrt{-5} }\right\rangle  
= P_1 P_3
\end{align*}
and so these are inverses and we can get rid of \(P_3\). So
\({ \operatorname{Cl}} ({\mathbb{Z}}_K) = \left\langle{ [P_1] }\right\rangle\),
which is a cyclic group. The generator has to have order 1 or 2, since
\(P_1^2 = \left\langle{ 1 }\right\rangle\). The claim is that the order
is 2: otherwise, it would be trivial, making the class group trivial,
which would imply that \({\mathbb{Z}}_K\) is a PID. Why? This implies
that every \(I \in \operatorname{Id}({\mathbb{Z}}_K)\) is dilation
equivalent to the unit ideal, so
\(I = \alpha \left\langle{ 1 }\right\rangle\) for some
\(\alpha\in K^{\times}\). But since \(I\) is an ideal in
\({\mathbb{Z}}_K\), this forces \(\alpha\in {\mathbb{Z}}_K\) and
\(I = \left\langle{ \alpha }\right\rangle\). This is a contradiction,
since every PID is a UFD, and \({\mathbb{Q}}( \sqrt{-5} )\) has
non-unique factorization. So we can write
\({ \operatorname{Cl}} ({\mathbb{Z}}_K) \cong {\mathbb{G}}_a({\mathbb{Z}}/2{\mathbb{Z}})\).

\end{example}

\begin{remark}

What is the class group useful for? We'll tie this into Diophantine
equations.

\end{remark}

\begin{example}[of using the class group to solve Diophantine problems]

Solve the following equation in \({\mathbb{Z}}\):
\begin{align*}
y^2 + 5 = x^3
.\end{align*}
Recall that we originally tried to do this by factoring the left-hand
side and appealing to unique factorization in a number field to deduce
that various factors were powers. However, we don't have unique
factorization. Although we can write
\(( y + \sqrt{-5} ) (y - \sqrt{-5} ) = x^3\), it's not clear that this
is helpful. The fix will be to go to
\(\operatorname{Id}({\mathbb{Z}}_K)\), which does have unique
factorization, where we'll also be able to use facts about the class
group. We can turn this into an equation in ideals:
\begin{align*}
\left\langle{ y + \sqrt{-5} }\right\rangle  
\left\langle{ y - \sqrt{-5} }\right\rangle  
= 
\left\langle{ x }\right\rangle^3 && \in \operatorname{Id}({\mathbb{Z}}[\sqrt{-5} ]) 
.\end{align*}
The original strategy was to show the left-hand factors were coprime in
order to deduce they were both cubes. We'll try to show these ideals are
coprime in the monoid sense, then since their product is a cube they'll
have to be cubes. This uses the fact that this is a reduced unique
factorization monoid, so being a cube up to a unit is not something we
have to worry about here.

\begin{claim}

There is no common prime ideal that divides both factors, using unique
factorization.

\end{claim}

\begin{proof}[?]

Suppose toward a contradiction that \(P\) is prime and divides both.
Using that ideal norms are multiplicative,
\(N(P) \divides N( \left\langle{ y + \sqrt{-5} }\right\rangle) = y^2 + 5\).
We also know \(P\) contains both factors, so it contains
\((y + \sqrt{-5} ) - (y - \sqrt{-5} ) = 2 \sqrt{-5}\), so
\(N(P) \divides N( \left\langle{ 2 \sqrt{-5} }\right\rangle ) = 20\).
Thus \(N(P) \divides \gcd(y^2 + 5, 20)\) in \({\mathbb{Z}}\). This is
impossible!

\begin{itemize}
\item
  \(y\) is necessarily even for the original equation to be true. If
  \(y\) is odd, take the equation \(\operatorname{mod}8\): an odd
  squared is \(1\operatorname{mod}8\), so
  \(y^2 +5 \equiv 6 \operatorname{mod}8\), which is not a cube in
  \({\mathbb{Z}}/8\) since any cube is \(0\operatorname{mod}8\).
\item
  \(5\) can not divide \(y\). If so, \(5\) would divide the left-hand
  side and thus the right-hand side, which forces \(5\divides x\) since
  \(5\) is prime. Then \(5^3 \divides x^3\), meaning
  \(5^3 \divides y^2 + 5\). In this case, \(5^2 \divides y^2 + 5\), and
  if \(5\divides y\) then \(5^2 \divides y^2\), so we'd need
  \(5^2 \divides 5\).
\end{itemize}

These together imply that \(\gcd(y^2 + 5, 25) = 1\). This
\(N(P) \divides 1\), forcing \(P = \left\langle{ 1 }\right\rangle\), a
contradiction.

\end{proof}

Thus we can write
\begin{align*}
\left\langle{ y + \sqrt{-5} }\right\rangle  &= I^3 \\
\left\langle{ y - \sqrt{-5} }\right\rangle  &= J^3
.\end{align*}

In the previous argument, we wrote out \((a + b \sqrt{-5} )^3\),
expanded, and compared coefficients. Here we have an equation in ideals,
and we can't do something similar unless \(I, J\) are principal. This is
in fact the case: we'll restrict our attention to the class group. The
left-hand side is the unit ideal, since it is principal. So we can write
\([I]^3 = [J]^3 = e\), but we also know
\({ \operatorname{Cl}} ({\mathbb{Z}}_K) \cong {\mathbb{G}}_a({\mathbb{Z}}/2{\mathbb{Z}})\),
so this can only happen if \([I] = [J] = e\) and \(I, J\) must be
principal. So we can write
\(I = \left\langle{ a + b \sqrt{-5}}\right\rangle\) for some
\(a, b \in {\mathbb{Z}}\). Thus
\begin{align*}
\left\langle{ y + \sqrt{-5} }\right\rangle = \left\langle{ (a + b \sqrt{-5} )^2 }\right\rangle \implies y + \sqrt{-5} = \pm 1 \qty { a + b \sqrt{-5} }^3  
,\end{align*}
using the fact that they differ by a unit but the only units in
\({\mathbb{Z}}[ \sqrt{-5} ]\) are \(\pm 1\). The original proof now goes
through, comparing coefficients of \(\sqrt{-5}\). This will force
\(b = \pm 1\), then plug things back in to find \(a\), then \(y\), then
\(x\). The conclusion is that there are no solutions.

\end{example}

\begin{remark}

The critical takeaway: unique factorization failed, but the structure of
the class group saved us! We crucially used that it had no elements of
order 3. See the book for a general theorem about equations
\(y^2 + d = x^3\). Ideal theory gives us a way to study Diophantine
equations.

\end{remark}

\hypertarget{the-class-group-as-a-measure-of-non-unique-factorization}{%
\subsection{The Class Group as a Measure of Non-unique
Factorization}\label{the-class-group-as-a-measure-of-non-unique-factorization}}

\begin{remark}

This is chapter 10. This statement shows up in talks: it's more of a
vague sentiment than an actual theorem, but we'll discuss a way to make
it precise.

\end{remark}

\begin{theorem}[Class number 1 iff UFD]

Recall that the class number is defined as
\(h_K \coloneqq\# { \operatorname{Cl}} ({\mathbb{Z}}_K)\). Then
\begin{align*}
h_K = 1 \iff {\mathbb{Z}}_K \text{ is a UFD}
.\end{align*}

\end{theorem}

\begin{proof}[of theorem]

\(\implies\): Every ideal is equivalent to the unit ideal, so every
ideal is principal and PID implies UFD.

\hfill\break

\(\impliedby\): Note that this is subtle: this is the claim that
\({\mathbb{Z}}_K\) is a UFD \(\implies {\mathbb{Z}}_K\) is a PID, which
isn't true for general rings (e.g.~\({\mathbb{Z}}[x]\)). Suppose
\({\mathbb{Z}}_K\) is a UFD, then it's enough to show that every prime
ideal is principal. Let \(P\) be prime, then \(P\) lies above some
ordinary prime \(p\), so \(P \divides \left\langle{ p }\right\rangle\).
We can factor
\(\left\langle{ p }\right\rangle= \left\langle{ \prod_{i=1}^k \pi_i }\right\rangle = \prod_{i=1}^k \left\langle{ \pi_i }\right\rangle\)
for some \(\pi_i\) irreducible. A prime ideal dividing a product, by
unique factorization, must divide a factor, so
\(P \divides \left\langle{ \pi_i }\right\rangle\) for some \(i\). In a
UFD, irreducibles are prime, so \(\left\langle{ \pi_i }\right\rangle\)
is a prime ideal, so we have a prime ideal dividing a prime ideal. By
unique factorization, this forces
\(P = \left\langle{ \pi_i }\right\rangle\), make \(P\) principal.

\end{proof}

\begin{question}

Can anything be said if \(h_K = 2\), even though we know
\({\mathbb{Z}}_K\) is not a UFD?

\end{question}

\begin{theorem}[Carlitz]

\(h_K = 2 \iff\) in \({\mathbb{Z}}_K\), any two factorizations of
nonzero nonunit \(\alpha\) into irreducibles have the same number of
terms.

\end{theorem}

\begin{remark}

For example, in \({\mathbb{Z}}[ \sqrt{-5} ]\) we have
\(6 = (2)(3) = (1 + \sqrt{-5} )(1 - \sqrt{-5} )\), which have the same
number of factors. To prove this theorem, we'll first need a lemma:

\end{remark}

\begin{lemma}[Class Number 2 implies 2 factors]

Suppose \(h_K = 2\), and suppose \(\pi \in {\mathbb{Z}}_K\) is an
irreducible that is not prime (which is possible in a non-UFD). Then
factoring \(\left\langle{ \pi }\right\rangle = P_1 P_2\) involves
exactly two prime ideals \(P_1, P_2\) in \({\mathbb{Z}}_K\).

\end{lemma}

\begin{proof}[of lemma]

Write \(\left\langle{ \pi }\right\rangle= \prod_{i=1}^g P_i\), we then
want to show \(g=2\). We have \(g\geq 2\), since otherwise this would be
a prime ideal, which would make \(\pi\) a prime element. The claim is
that none of the \(P_i\) can be principal. Suppose toward a
contradiction \(P_1 = \left\langle{ \rho }\right\rangle\). Note that
multiplying ideals yields smaller sets, so the right-hand side is a
subset of \(\left\langle{ \rho }\right\rangle\), as is the left-hand
side, and so \(\rho \divides \pi\). Since the \(P_i\) were principal
prime ideals, \(\rho\) is prime and thus irreducible (since prime
\(\implies\) irreducible for any domain), so \(\rho = u \pi\) for some
unit. Thus they generate the same ideal, and
\(P_1 = \left\langle{ \rho }\right\rangle = \left\langle{ \pi }\right\rangle\).
But then \(\pi\) generates a prime ideal, make \(\pi\) prime, a
contradiction.

So none of the \(P_i\) are principal. Look at this equation in the class
group. The left-hand side is the identity, and the right-hand side are
all non-identity elements a group of order 2. So \([P_1][P_2] = e\),
making \(P_1 P_2 = \left\langle{ \omega }\right\rangle\) principal. Then
\(\left\langle{ \pi }\right\rangle\subseteq \left\langle{ \omega }\right\rangle\)
and so \(\omega \divides \pi\). Moreover, \(\omega\) is not a unit since
the product of two prime ideals is not the unit ideal. Since \(\pi\) is
irreducible, this makes \(\omega= u \pi\) and thus
\(\left\langle{ \omega }\right\rangle = \left\langle{ \pi }\right\rangle\).
If this were the case, we could cancel in the original equation:
\begin{align*}
\left\langle{ \pi }\right\rangle &= (P_1 P_2)P_3 \cdots P_g = \left\langle{ \pi }\right\rangle P_3 \cdots P_g   \\
\implies \left\langle{ 1 }\right\rangle = P_3 \cdots P_g 
,\end{align*}
but this is a product of prime ideals resulting in the unit ideal. This
can only happen if there are no terms in this product, so \(g=2\).

\end{proof}

\begin{proof}[of theorem ($\implies$)]

Suppose \(h_K = 2\). We already know \({\mathbb{Z}}_K\) is not a UFD by
the previous theorem. The nontrivial part is showing factorization into
nonzero nonunits of the same number of terms. Instead of working with
factorization of elements, we'll work with factorization of their
principal ideals into principal ideals generated by irreducibles, which
will obviate the need to worry about units. We thus want to show that
any principal ideal
\(P \neq \left\langle{ 0 }\right\rangle, \left\langle{ 1 }\right\rangle\)
has all of its factorizations into principal ideals generated by
irreducibles the same length.

\begin{example}[?]

Much like the previous example, we have
\begin{align*}
\left\langle{ 6 }\right\rangle = \left\langle{ 1 + \sqrt{-5} }\right\rangle \left\langle{ 1 - \sqrt{-5} }\right\rangle = \left\langle{ 2 }\right\rangle \left\langle{ 3 }\right\rangle     
.\end{align*}

\end{example}

Suppose that \(P\) factors as
\begin{align*}
P = 
\prod_{i=1}^k \left\langle{ \pi_i }\right\rangle
=
\prod_{j=1}^\ell \left\langle{ \rho_j }\right\rangle
&&
\pi_i, \rho_j \text{ irreducible}
,\end{align*}
we'd then like to show that \(k=\ell\).

\begin{observation}

If \(\pi_1\) is prime, we can use that
\(\left\langle{ \pi_1 }\right\rangle\divides \prod \left\langle{ \rho_j }\right\rangle\),
and would thus have to divide (say)
\(\left\langle{ \rho_1 }\right\rangle\) up to relabeling. Since
everything is irreducible, if \(\pi_1 \divides \rho_1\) then
\(\left\langle{ \pi_1 }\right\rangle= \left\langle{ \rho_1 }\right\rangle\),
meaning we can cancel.

\end{observation}

So after cancellation, we can suppose that all the \(\pi_i, \rho_j\) and
none are prime. Consider the number of prime ideals that show up after
factoring all of the principal ideals on either side. By the lemma, any
irreducible that's not prime factors into two primes, so we get \(2k\)
primes on the left-hand side (not necessarily distinct) and \(2\ell\) on
the right-hand side. But the factorization into primes \emph{is} unique,
so \(2k=2\ell\) and \(k=\ell\).

\end{proof}

\begin{remark}

See the book for the other direction!

\end{remark}

\begin{theorem}[Landau]

Every ideal class contains infinitely many prime ideals.

\end{theorem}

\begin{remark}

This is an analytic theorem! The proof is similar to how Dirichlet
proved the infinitude of primes in arithmetic progressions, which
involves \(L{\hbox{-}}\)functions.

\end{remark}

\begin{remark}

What about \(h_K \geq 2\)? We'll introduce a way of measuring how bad
unique factorization fails in a ring, the notion of \emph{elasticity}.

\end{remark}

\hypertarget{elasticity}{%
\subsection{Elasticity}\label{elasticity}}

\begin{definition}[Elasticity of a Ring]

Let \(\alpha\in {\mathbb{Z}}_K\) where \(\alpha\neq 0\) and is not a
unit. Define
\begin{align*}
\rho(\alpha) \coloneqq{ L( \alpha) \over S( \alpha) }
,\end{align*}
where \(L( \alpha)\) is the number of terms in the longest\footnote{There
  is a way to factor that maximizes the number of irreducibles
  appearing, and there are not arbitrarily long factorizations.}
factorization of \(\alpha\) and \(S( \alpha )\) is the shortest number
of terms. This measures how far away from unique the factorization of
\(\alpha\) is. Now define the \textbf{elasticity} of \({\mathbb{Z}}_K\)
as
\begin{align*}
\rho(K) \coloneqq\sup_{\alpha} \rho( \alpha)
.\end{align*}

\end{definition}

\begin{remark}

Note that \(h_K = 1, 2\iff \rho(K) = 1\), and
\(h_K > 2 \implies \rho(K) > 1\).

\end{remark}

\begin{theorem}[Elasticity in terms of the Davenport constant]

For \(h_K \geq 2\),
\begin{align*}
\rho(K) = {1\over 2} D( { \operatorname{Cl}} ( {\mathbb{Z}}_K))
,\end{align*}
where \(D(G)\) is the \textbf{Davenport constant} of the finite abelian
group \(G\): the smallest number \(D\) such that every sequence of \(D\)
elements of \(G\) contains a nonempty subsequence whose product is the
identity. This is a function from combinatorial group theory.

\end{theorem}

\begin{exercise}[bounding the Davenport constant]

Show that \(D(G) \leq {\left\lvert {G} \right\rvert}\).

\end{exercise}

\begin{fact}

\(D(G) \to \infty\) as \({\left\lvert {G} \right\rvert} \to \infty\).

\end{fact}

\begin{corollary}[?]

If \(h_K \to \infty\) for a sequence of number fields, then
\(\rho(K) \to \infty\).

\end{corollary}

\begin{remark}

This just follows from the above facts, since \(h_K \to \infty\) means
the size of the group
\(G \coloneqq{ \operatorname{Cl}} ( {\mathbb{Z}}_K)\) goes to infinity,
which is a constant times \(\rho(K)\). So as the class group gets
larger, factorization gets worse.

\end{remark}

\hypertarget{prime-producing-polynomials-and-unique-factorization-lec.-11-tuesday-february-23}{%
\section{Prime Producing Polynomials and Unique Factorization (Lec. 11,
Tuesday, February
23)}\label{prime-producing-polynomials-and-unique-factorization-lec.-11-tuesday-february-23}}

\begin{remark}

Today: chapters 11 and 12.

\end{remark}

\hypertarget{ch.-11-prime-producing-polynomials-and-unique-factorization}{%
\subsection{Ch. 11: Prime Producing Polynomials and Unique
Factorization}\label{ch.-11-prime-producing-polynomials-and-unique-factorization}}

\begin{remark}

18th century observation by Euler about the following polynomial:
\begin{align*}
f(x) \coloneqq x^2 -x + 41 
.\end{align*}
Goldbach proved that it's impossible for any polynomial
\(g\in {\mathbb{Z}}[x]\) to have \emph{every} output prime. Euler noted
that this \(f\) produces quite a few: for \(x=1, \cdots, 40\), the
output \(f(x)\) is prime, but \(f(41) = 41^2 - 41 + 41 = 41^2\) is not.
Let's define a variant: for \(q\) a positive integer, set
\begin{align*}
f_q(x) \coloneqq x^2 - x + q
.\end{align*}
Note that \(f_q(q) = q^2\), so eventually the output is composite. We'll
say \(f_q\) is \textbf{optimal} if \(f_q(x)\) is prime for all integers
\(0 < x < q\). As an example, \(q=41\) was optimal.

\end{remark}

\begin{theorem}[Rabinowitz]

Let \(q \geq 2 \in {\mathbb{Z}}^{> 0}\) and let
\(d = \Delta(f_q) = 1-4q\) be the discriminant of \(f_q\). Assume that
\(d\) is squarefree, then \(f_q\) is optimal if and only if
\(\mathbb{Z}\left[ { {1 + \sqrt{d}\over 2 }} \right]\) is a UFD.
\footnote{Note that this is equal to \({\mathbb{Z}}_K\) when
  \(K\coloneqq{\mathbb{Q}}( \sqrt{d} )\).}

\end{theorem}

\begin{example}[of a ring of integers that is a UFD]

For \(q=41, d = -163\) and thus
\(\mathbb{Z}\left[ { {1 + \sqrt{-163} \over 2}} \right]\) is a UFD.

\end{example}

\begin{proof}[$\impliedby$]

\begin{quote}
Big idea: uses that \(\min_\tau(x) = f_q(x)\) and remembering that how
\(\min_\tau(x) \operatorname{mod}p\) factors is exactly how
\(\left\langle{ p }\right\rangle\) factors into prime ideals.
\end{quote}

Assume \(\mathbb{Z}\left[ { \tau } \right]\) is a UFD, where
\(\tau\coloneqq{1 + \sqrt{d} \over 2 }\). Toward a contradiction,
suppose \(f_q(x)\) is composite for some \(0<x<q\). We can write
\begin{align*}
f_q(x) = x^2 - x + q = (x - \tau)( x - {\overline{{ \tau}}} ) = \min_{ \tau}(x)_{/{\mathbb{Q}}}
.\end{align*}
By considering how this function increases, we can conclude
\(1<q<f_q(x) < f_q(q) = q^2\). Let \(p\) be the least prime factor of
\(f_q(x)\), which is necessarily bounded by \(\sqrt{ f_q(x) }\), so
\(p<q\). Since \(p\divides f_q(x)\), we have \(x\in {\mathbb{Z}}\) as a
root of \(f\operatorname{mod}p\). So \(\min_\tau\) has a root modulo
\(p\). Recall that studying how \(\left\langle{ p }\right\rangle\)
factors into ideals of \({\mathbb{Z}}_K\) involved studying how
\(\min_\tau\) factors mod \(p\). Since we've shown it has a root mod
\(p\), it breaks into two linear factors. So
\(\left\langle{ p }\right\rangle = P_1 P_2\) as prime ideals of norm
\(p\). By assumption, \(\mathbb{Z}\left[ {[} \right]\tau]\) is a UFD and
the ring of integers of a number field, and by an earlier theorem, is
thus also a PID (noting that this is not generally the case). So
\(P_1, P_2\) are principal, and we can write
\begin{align*}
P_1 = \left\langle{ a + b \tau }\right\rangle\implies p = N(p_1) = N( a + b \tau) = a^2 + ab + qb^2
.\end{align*}
Completing the square yields
\begin{align*}
\cdots = (a + b/2)^2 + (q-1/4)b^2
.\end{align*}
Note that \(b\neq 0\), since this would yield \(p = a^2\) in the first
equation and \(a, p \in {\mathbb{Z}}\) with \(p\) prime. So both terms
in the second equation are non-negative, and the second is positive
because \(b>1\), so \(p \geq q- 1/4\). Since \(p, q\in {\mathbb{Z}}\) we
can strengthen this to \(p \geq q\). But \(p\) was the \emph{least}
prime factor of \(f_q(x) < q^2\) which was composite, so this is a
contradiction. \(\contradiction\)

\end{proof}

\begin{remark}

The forward direction is harder here.

\end{remark}

\begin{proof}[$\implies$]

We'll prove something stronger. Assume \(f_q(x)\) is prime whenever
\begin{align*}
1 \leq x \leq {1\over 2} \sqrt{ {\left\lvert {d} \right\rvert} \over 3 } + {1 \over 2}
,\end{align*}
then we'll prove that \({\mathbb{Z}}_K\) is a PID and hence a UFD. Note
that this is stronger because the range is smaller than \(0<x<q\).

\begin{claim}

\(p\) is inert for all
\(p \leq \sqrt{ {\left\lvert {d} \right\rvert}\over 3 }\) (so the prime
ideal \(\left\langle{ p }\right\rangle\) remains prime).

\end{claim}

\begin{proof}[?]

If not, \(\min_\tau\) has a root \(\operatorname{mod}p\). Recalling that
\(\min_\tau(x) = f_q(x) = x^2 - x + q\), if this has one root then it
has two which sum to \(-b = -(-1) = 1\), where one of them satisfies
\begin{align*}
1 \leq x \leq {1\over 2} \sqrt{ {\left\lvert {d} \right\rvert} \over 3 } + {1\over 2}
.\end{align*}
Why? If the other root \(x = r\) with \(1<r<p\) doesn't satisfy this,
then the first root is \(p+1-r\) and will satisfy this. Then
\(p\divides f_q(x)\), but this is a problem! This forces
\begin{align*}
p = f_q(x) = x^2 - x + q \geq q > \sqrt{{\left\lvert {d} \right\rvert} \over 3 } \geq p
.\end{align*}
This contradicts \(f_q(x)\) being prime. \(\contradiction\)

\end{proof}

So assuming \(f_q(x)\) is prime for
\(1 \leq x \leq {1 \over 2 } \sqrt{{\left\lvert {d} \right\rvert}\over 3 } + {1 \over 2}\),
we showed that every ``small'' prime up to
\(\sqrt{{\left\lvert {d} \right\rvert}\over 3 }\) is inert. Suppose
\(P\) is a prime ideal above \(p\), then since \(p\) is inert,
\(P = \left\langle{ p }\right\rangle\) is generated by a prime. But
we'll just use a slightly weaker conclusion: \(P\) is principal.

\begin{theorem}[When the class group is generated by small primes]

Let \(d\) be a negative squarefree integer with
\(d \equiv 1 \operatorname{mod}4\) (such as the \(d\) we are looking
at). Then \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is generated as a
group by \([P]\) where \(P\) runs over all prime ideals above primes
\(p \leq \sqrt{{\left\lvert {d} \right\rvert}\over 3 }\).

\end{theorem}

Given this theorem, we are done: in our situation, all such \([P]\) are
trivial in the class group since they are principal, which makes
\({ \operatorname{Cl}} ({\mathbb{Z}}_K) = 1\) and every ideal is
principal.

\end{proof}

\hypertarget{proof-of-rabinowitzs-theorem}{%
\subsection{Proof of Rabinowitz's
Theorem}\label{proof-of-rabinowitzs-theorem}}

\begin{remark}

It just remains to prove the above theorem. We'll use the following:

\end{remark}

\begin{proposition}[Almost Euclidean Domains]

Take the same assumptions on \(d\) as above. Then for each
\(\theta \in K = {\mathbb{Q}}( \sqrt{d} )\), there is a positive integer
\(t \leq \sqrt{{\left\lvert {d} \right\rvert}\over 3 }\) and a
\(\xi \in {\mathbb{Z}}_K\) with norm \(N(t\theta - \xi) < 1\).

\end{proposition}

\begin{remark}

This is slightly technical. In words: for any element in your quadratic
field, you can approximate it by an \emph{integer} of your field,
possibly after a small \(t\) dilation. Note that we saw a similar
condition for the Euclidean algorithm, namely that \(t=1\) always
sufficed.

\end{remark}

\begin{proof}[of proposition]

Write \(\theta = a + b \tau\) where we don't necessarily know
\(\theta \in {\mathbb{Z}}_K\) (although this would make the statement
trivial), but \(a, b \in {\mathbb{Q}}\). We want to find an appropriate
\(t\) where \(\xi \coloneqq A+ B \tau\) for \(A, B \in {\mathbb{Z}}\).
Multiplying the inequality out using the definition of the norm results
in
\begin{align*}
\qty{ (ta - A) + \qty{tb - B \over 2} }^2
+ {\left\lvert {d} \right\rvert} \qty{ tb - B \over 2 }^2<1
.\end{align*}
We'll start by making the second term small by making \(tb\) close to an
integer (where \(b\) is fixed) and choosing \(B\) to be that closest
integer. We can choose
\(t \leq \sqrt{{\left\lvert {d} \right\rvert}\over 3 }\) with
\({\left\lVert {tb} \right\rVert} < 1 / \sqrt{{\left\lvert {d} \right\rvert}\over 3 }\),
where the norm is the distance to the nearest integer. Why can we do
this? This is Dirichlet's approximation theorem, where we could choose
\(t\leq N\) such that this norm was bounded by \(1/(N+1)\), and we can
take
\(N \coloneqq{\left\lfloor \sqrt{{\left\lvert {d} \right\rvert}\over 3 } \right\rfloor}\).
How do we choose \(A, B\)? Choose \(B\) such that
\({\left\lVert {tb} \right\rVert}\) satisfies the above inequality to
obtain
\begin{align*}
B\in {\mathbb{Z}},\quad {\left\lvert {tb - B} \right\rvert}< 1 / \sqrt{{\left\lvert {d} \right\rvert}/3 }  
.\end{align*}
Then considering the second term in the original equation, we get
\begin{align*}
{\left\lvert {d} \right\rvert} \qty{ tb -B \over 2 }^2 < {\left\lvert {d} \right\rvert}\qty{1\over 4}\qty{1 \over {\left\lvert {d} \right\rvert}/3} = {3\over 4} 
,\end{align*}
so it suffices now to choose \(A\) such that the first term is bounded
by \(1/4\). Why can we do this? We have control over \(A\), so we can
simply choose it freely to shift the inner quantity into the interval
\([-1/2, 1/2]\), i.e.~
\begin{align*}
{\left\lvert { ta - A + \qty{tb- B \over 2} } \right\rvert} \leq {1\over 2}
,\end{align*}
and then squaring yields the desired bound.

\end{proof}

\begin{proof}[of theorem]

We have \(d \equiv 1 \operatorname{mod}4\) and we want to show that the
class group is generated by small primes
\(p\leq D\coloneqq\sqrt{ {\left\lvert {d} \right\rvert} / 3 }\). Let
\(I\) be a nonzero ideal of \({\mathbb{Z}}_K\), we'll find a nonzero
ideal \(J\) such that \([I] = [J]\) and \(J\) is a product of primes
above \(p\) (which have the appropriate upper bound). In this case,
\([J]\) and thus \([I]\) will factor as the product of those primes, and
is thus in the subgroup generated by small primes. Fix \(\beta\in I\)
nonzero of minimal norm.

\begin{claim}

For any \(\alpha \in I\) there is a \(t\leq D\) with
\(\left\langle{ t \alpha }\right\rangle \in \left\langle{ \beta }\right\rangle\).

\end{claim}

\begin{remark}

How to think about this: when the field is Euclidean with respect to the
norm, it is a PID. How do you find generators? By taking a nonzero
element \(\beta\) of minimal norm in the ideal. Then any element of
\(I\) would be in \(\beta\). Here we have an almost-Euclidean property,
where elements of \(I\) can be hit with a small dilation to land in this
principal ideal.

\end{remark}

\begin{proof}[of claim]

So apply the last element to the element \(\alpha/ \beta\) to pick
\(t\leq D\) and \(\xi \in {\mathbb{Z}}_K\) with
\(N( t \qty{\alpha\over \beta} < 1\). Multiplying through by
\(N( \beta)\) yields
\begin{align*}
N ( t \alpha - \beta\xi ) < N( \beta )
.\end{align*}
Note that the inner term on the left-hand side is in \(I\), since
\(\alpha, \beta\in I\). This is an element of \(I\) of norm less than
the norm \(\beta\), but by minimality this can only happen if
\(t \alpha - \beta\xi =0\) and thus
\(t \alpha = \beta\xi \in \left\langle{ \beta }\right\rangle\).

\end{proof}

Now let \(T \coloneqq{\left\lfloor D \right\rfloor} !\), where we take
the factorial. Then for any \(\alpha\in I\) we have
\(T \alpha \in \left\langle{ \beta }\right\rangle\). Why? We already
know \emph{some} factor of \(T\) is a multiple of \(\beta\), and
multiplying by other factors doesn't take it out of the ideal. Since
this was true for every \(\alpha\in I\) we have
\(TI \subseteq \left\langle{ \beta }\right\rangle = \beta{\mathbb{Z}}_K\).
Define \(J \coloneqq\qty{T/ \beta} I\), where noting that
\(T\in {\mathbb{Z}}^{> 0}\), this is just a dilation of \(I\). Then
\(J \subseteq {\mathbb{Z}}_K\), since
\begin{align*}
TI \subseteq \beta{\mathbb{Z}}_K \overset{\cdot \beta^{-1}}{\implies } \beta^{-1}T I \subseteq \beta^{-1}\beta {\mathbb{Z}}_K = {\mathbb{Z}}_K
.\end{align*}
Moreover, \(J {~\trianglelefteq~}{\mathbb{Z}}_K\) is an ideal, since
it's a dilation of an ideal, \([J] = [I]\) since they're related by
dilation, and \(J\) contains \(\qty{T / \beta } \beta = T\). Since to
contain is to divide, we have
\(J \divides \left\langle{ T }\right\rangle\). Recall that \(T\) was a
product of integers, so however \(\left\langle{ T }\right\rangle\)
factors into prime ideals, every such prime ideal will lie above an
actual prime no bigger than \(D\). You can factor
\(\left\langle{ T }\right\rangle\) by first factoring \(T\) into prime
integers, then break them up into prime ideals, all of which would have
norm bounded by \(D\).

\end{proof}

\begin{remark}

\end{remark}

\begin{remark}

This proves Rabinowitz's theorem.\\

This says that being an optimal prime is entirely equivalent to a
certain ring being a UFD. Are there optimal examples other than
\(q=41\)? It turns out that there are \emph{no} optimal \(f_q\) for
\(q>41\), which is not easy to prove. This didn't happen until the 20th
century, by folks interested in the UFD side of this statement:

\end{remark}

\begin{theorem}[Baker-Heegner-Stark]

\({\mathbb{Z}}_K\) is not a UFD if
\(K \coloneqq{\mathbb{Q}}( \sqrt{d} )\) with \(d\) squarefree with
\(d<-163\).

\end{theorem}

\begin{remark}

So remarkably, there are \emph{not} infinitely many examples for which
the ring of integers is a UFD. Thus the class number only takes on the
value 1 for finitely many fields. What about for 2? This also only
happens finitely often. In fact, for \emph{any} fixed \(h\), there are
only finitely many imaginary quadratic fields with class number \(h\).
This follows from the fact that
\({ \operatorname{Cl}} ({\mathbb{Q}}( \sqrt{d} )) \approx d^{1/2}\),
which increases as \(d\) does. It's still hard to determine for a given
\(h\) which values of \(d\) appear, partially because the last statement
is \emph{ineffective}\footnote{Note that this may not be true as of
  2020! See Griffin, M., \& Ono, K. (2020). Elliptic curves and lower
  bounds for class numbers. Journal of Number Theory, 214, 1-12.} in the
sense that there aren't constants to put into the asymptotic statement.

\end{remark}

\begin{remark}

What about real quadratic fields? An expert on this will be joining us
here at UGA starting Fall 2021. The situation is expected to be very
different, and a conjecture is the following:

\end{remark}

\begin{conjecture}

The real quadratic field \({\mathbb{Q}}(\sqrt{d})\) is a UFD most of the
time.

\end{conjecture}

\hypertarget{lattice-points}{%
\subsection{Lattice Points}\label{lattice-points}}

\begin{remark}

This corresponds to chapter 12. Everything we've done up until now has
been for quadratic fields. After this chapter, we'll start anew and
rebuild everything for general number fields.

\end{remark}

\begin{definition}[Lattice Point]

A \textbf{lattice point} in \({\mathbb{R}}^n\) is a point in
\({\mathbb{Z}}^n\).

\end{definition}

\begin{question}

Given a region \(R \subseteq {\mathbb{R}}^n\), how many lattice points
does it contain? I.e., how large is the sum
\begin{align*}
\sum_{\mathbf{v} \in {\mathbb{Z}}^n} \chi_R(\mathbf{v})
.\end{align*}

\end{question}

\begin{remark}

A first guess might be that this is approximately
\(\operatorname{vol}(R)\). To see why, one can try just choosing to
count any squares for which the lower-left point is contained in \(R\)
and adding up the areas:

\pgfmathsetseed{12}

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\draw[step=1.0,black,thin] (0.5,0.5) grid (9,7);

 \draw [shade,
        top color=blue,
        bottom color=white,
        fill opacity=.1,
        decoration={random steps,segment length=2cm,amplitude=.75cm},
        decorate,
        rounded corners=.3cm]
     (2, 3) -- (4,6)   -- (6,6) -- (7.6,1.5) -- (4, 1) -- (2, 3);

\node[fill=black, circle, inner sep=0.1cm] at (4,2) {};
\node[fill=black, circle, inner sep=0.1cm] at (5,2) {};
\node[fill=black, circle, inner sep=0.1cm] at (6,2) {};
\node[fill=black, circle, inner sep=0.1cm] at (7,2) {};

\node[fill=black, circle, inner sep=0.1cm] at (3,3) {};
\node[fill=black, circle, inner sep=0.1cm] at (4,3) {};
\node[fill=black, circle, inner sep=0.1cm] at (5,3) {};
\node[fill=black, circle, inner sep=0.1cm] at (6,3) {};
\node[fill=black, circle, inner sep=0.1cm] at (7,3) {};


\node[fill=black, circle, inner sep=0.1cm] at (3,4) {};
\node[fill=black, circle, inner sep=0.1cm] at (4,4) {};
\node[fill=black, circle, inner sep=0.1cm] at (5,4) {};
\node[fill=black, circle, inner sep=0.1cm] at (6,4) {};
\node[fill=black, circle, inner sep=0.1cm] at (7,4) {};


\node[fill=black, circle, inner sep=0.1cm] at (4,5) {};
\node[fill=black, circle, inner sep=0.1cm] at (5,5) {};
\node[fill=black, circle, inner sep=0.1cm] at (6,5) {};

\draw[draw=black, fill=green, fill opacity=0.1] (4, 2) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (5, 2) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (6, 2) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (7, 2) rectangle ++(1,1);

\draw[draw=black, fill=green, fill opacity=0.1] (3, 3) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (4, 3) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (5, 3) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (6, 3) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (7, 3) rectangle ++(1,1);

\draw[draw=black, fill=green, fill opacity=0.1] (3, 4) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (4, 4) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (5, 4) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (6, 4) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (7, 4) rectangle ++(1,1);


\draw[draw=black, fill=green, fill opacity=0.1] (4, 5) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (5, 5) rectangle ++(1,1);
\draw[draw=black, fill=green, fill opacity=0.1] (6, 5) rectangle ++(1,1);

\node at (2.25, 2.25) {$R$};

\node at (8.15, 5.45) {$\approx\mathrm{vol}(R)$};

\node at (2.25, 6.25) {${\mathbb{Z}}^n \subseteq {\mathbb{R}}^n$};

\end{tikzpicture}
}
\end{figure}

This isn't exactly right, but would become closer as \(R\) grew larger,
and the correction term comes from edge effects. For
\(R \subseteq {\mathbb{R}}^n\) and \(t\in {\mathbb{R}}\), define the
dilation
\begin{align*}
tR \coloneqq\left\{{ t\mathbf{x} {~\mathrel{\Big|}~}\mathbf{x} \in R }\right\} 
.\end{align*}

\end{remark}

\begin{theorem}[The number of lattice points in a region is asymptotically the volume]

Let \(R\) be a region in \({\mathbb{R}}^n\) which is \emph{Riemann
measurable}.\footnote{This means that \(\chi_R\) should be Riemann
  integrable, i.e.~the bounded region is contained in a rectangle, and
  integrals over such rectangles converges to what we'll call the
  volume.} Then the number of lattice points satisfies
\begin{align*}
{1\over t^n} \sum_{\mathbf{v} \in {\mathbb{Z}}^n} \chi_{tR} (\mathbf{v})
\overset{t\to \infty }\to \operatorname{vol}(R)
.\end{align*}

\end{theorem}

\begin{proof}[of theorem]

Notice that the left-hand side can be written as
\begin{align*}
{1\over t^n} \sum_{\mathbf{v} \in {\mathbb{Z}}^n} \chi_{tR} (\mathbf{v})
=
{1\over t^n} = \sum_{\mathbf{w} \in t^{-1}{\mathbb{Z}}^n} \chi_R(\mathbf{w})
.\end{align*}
This has the effect of making the squares partitioning
\({\mathbb{R}}^n\) finer, the right-hand side is literally the Riemann
sum for
\begin{align*}
\int \chi_R(\mathbf{w}) \,d \mathbf{w} \coloneqq\operatorname{vol}(R)
.\end{align*}

\end{proof}

\begin{remark}

Note that there is a small technicality since \(t\) can take on
non-integer values, but the limiting behavior is the same. Next time:
we've seen that the number of lattice points is sometimes
well-approximated by volume, but it's possible to have regions of
unbounded volume with no lattice points, e.g.~by taking a large ball and
deleting all lattice points. It would be nice to have a theorem which
guarantee when a region will have lattice points, and Minkowski's
theorem will be one such theorem we'll look at next time.

\end{remark}

\hypertarget{lattice-points-lec.-12-monday-march-01}{%
\section{Lattice Points (Lec. 12, Monday, March
01)}\label{lattice-points-lec.-12-monday-march-01}}

\hypertarget{minkowski-version-1}{%
\subsection{Minkowski (Version 1)}\label{minkowski-version-1}}

\begin{remark}

Basic heuristic from last time: counting the lattice points in a region
\(R\) should be approximately \(\operatorname{vol}(R)\). We turned this
into a theorem for certain regions:

\end{remark}

\begin{theorem}[Lattice points with volume after scaling]

Let \(R \subseteq {\mathbb{R}}^n\) be a bounded region with a
well-defined with respect to the Riemann integral. Letting \(L_{R}\) be
the number of lattice points in \(R\), we have
\begin{align*}
{1\over t^n} L_{tR} \overset{t\to \infty}\to \operatorname{vol}(R)
.\end{align*}

\end{theorem}

\begin{remark}

Most of today: Minkowski's theorem, which will guarantee a lattice point
under some conditions.

\end{remark}

\begin{theorem}[Minkowski, Version 1]

Let \(R \subseteq {\mathbb{R}}^n\) be a bounded region that is

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Convex, so any line segment connecting two points in \(R\) is entirely
  contained within \(R\), and
\item
  Symmetric about \(\mathbf{0}\), so
  \(\mathbf{x}\in R\implies -\mathbf{x}\in R\).
\end{enumerate}

If \(\operatorname{vol}(R) > 2^n\), then \(R\) contains a nonzero
lattice point.

\end{theorem}

\begin{remark}

Any circle/ball or ellipse will be an example. Note that \(2^n\) is
sharp, i.e.~this theorem does not hold for a smaller constant: take the
square \((-1, 1) \times(-1, 1) \subseteq {\mathbb{R}}^2\), which has
volume 4 but only contains the origin as a lattice point.

\end{remark}

\begin{proof}[of Minkowski Version 1]

Note that any such region already contains \(\mathbf{0}\), since
containing \(\mathbf{x}\) and \(-\mathbf{x}\) plus convexity implies
containing the line between them, which passes through \(\mathbf{0}\).
By assumption \(\operatorname{vol}(R) > 2^n\), and hence
\((1/t^n)L_{tR} > 2^n\) for \(t\) large enough. So set \(t=m\) for some
\(m>> 1\in {\mathbb{Z}}^{\geq 0}\), this yields \(L_{mR} > (2m)^n\).
Consider \({\mathbb{Z}}^n\) and taking all coordinates
\(\operatorname{mod}2m\). This yields \((2m)^n\) equivalence classes of
points, so by the pigeonhole principle there exist
\(\mathbf{v}_1 \neq \mathbf{v}_2 \in mR\) such that
\((\mathbf{v}_1 - \mathbf{v}_2)/2m \in {\mathbb{Z}}^n\), and the claim
is that this is the lattice point we want.

\hfill\break

Note that this is nonzero, why is it in the region \(R\)? By definition,
\((1/m)\mathbf{v}_1 \in R\) and \((-1/m)\mathbf{v}_2 \in R\) using the
symmetric assumption. The midpoint between these is precisely the
previous point, and this is in \(R\) by convexity.

\end{proof}

\hypertarget{minkowski-version-2}{%
\subsection{Minkowski (Version 2)}\label{minkowski-version-2}}

\begin{definition}[Lattice]

A \textbf{lattice} in \({\mathbb{R}}^n\) is the
\({\mathbb{Z}}{\hbox{-}}\)span of a collection of
\({\mathbb{R}}{\hbox{-}}\)linearly independent vectors in
\({\mathbb{R}}^n\).

\end{definition}

\begin{example}[$n=2$]

\envlist

\begin{itemize}
\tightlist
\item
  \(\Lambda \coloneqq{\mathbb{Z}}{\left[ {1, 0} \right]}^t + {\mathbb{Z}}\left\{{0, 1}\right\}^t\).
  This tiles the plane by squares.
\item
  \(\Lambda \coloneqq{\mathbb{Z}}{\left[ {1, 1} \right]}^t + {\mathbb{Z}}\left\{{1, 3}\right\}^t\).
  Note that this now tiles the plane by parallelograms:
\end{itemize}

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
  \draw[thin,gray!40] (-1,-1) grid (4,4);
  \draw[<->] (-1,0)--(4,0) node[right]{$x$};
  \draw[<->] (0,-1)--(0,4) node[above]{$y$};
  \draw[line width=1pt,blue,-stealth](0,0)--(1,1) node[anchor=south west]{$\boldsymbol{u}$};
  \draw[line width=1pt,purple,-stealth](1,1)--(2,4) node[anchor=south west]{};
  \draw[line width=1pt,purple,-stealth](0,0)--(1,3) node[anchor=north east]{$\boldsymbol{v}$};
\draw[line width=1pt,blue,-stealth](1,3)--(2,4) node[anchor=south west]{};

\draw [draw=black, fill=purple, opacity=0.4]
       (0,0) -- (1,1) -- (2,4) -- (1,3) -- cycle;
\end{tikzpicture}
}
\end{figure}

\begin{itemize}
\tightlist
\item
  \(\Lambda \coloneqq{\mathbb{Z}}{\left[ {1, 0} \right]}^t\), which
  recovers \({\mathbb{Z}}\subseteq {\mathbb{R}}\). This is not a full
  lattice, since it lies in a proper subspace of \({\mathbb{R}}^2\).
\end{itemize}

Note that this will always result in a free abelian group on \(n\)
generators. Why not define a lattice this way? Here's a non-example:

\begin{itemize}
\tightlist
\item
  \(\Lambda \coloneqq{\mathbb{Z}}{\left[ {\sqrt{2} , 0} \right]}^t + {\mathbb{Z}}\left\{{1, 0}\right\}^t\),
  which are not linearly independent over \({\mathbb{R}}\). This yields
  a dense set of points on the real axis in \({\mathbb{R}}^2\), and is
  still a free abelian group of rank 2. We'll see later that no lattice
  can be dense, and in fact they must always be discrete.
\end{itemize}

\end{example}

\begin{remark}

It may not be obvious that a lattice has a uniquely determined number of
generating elements. This turns out to be true: if
\(\Lambda = \sum_{i=1}^d {\mathbb{Z}}\mathbf{v}_i\) with
\(\left\{{ \mathbf{v}_i }\right\}_{i=1}^d\) linearly independent over
\({\mathbb{R}}\), then
\begin{align*}
\Lambda\otimes_{\mathbb{Z}}{\mathbb{R}}\cong \sum_{i=1}^d {\mathbb{R}}\mathbf{v}_i \cong {\mathbb{R}}^d 
,\end{align*}
which is now an \({\mathbb{R}}{\hbox{-}}\)vector vector space of real
dimension \(d\). Noting that
\(\dim_{\mathbb{R}}(\Lambda\otimes_{\mathbb{Z}}{\mathbb{R}})\) doesn't
depend on the choice of basis, any different choice of generating set
for \(\Lambda\) must have the same number of generators.

\end{remark}

\begin{definition}[Full Lattices]

If \(d=n\), we'll call \(\Lambda\) a \textbf{full lattice}.

\end{definition}

\begin{definition}[Fundamental Parallelepiped]

Note that if \(\Lambda\) is full, then
\(\Lambda= \sum_{i=1}^n {\mathbb{Z}}\mathbf{v}_i\) with the
\(\mathbf{v}_i\) linearly independent over \({\mathbb{R}}\). We define
the \textbf{fundamental parallelepiped} as the set
\begin{align*}
\left\{{ \sum_{i=1}^n c_n \mathbf{v}_n {~\mathrel{\Big|}~}0 \leq c_i \leq 1}\right\} 
.\end{align*}
Note that this depends on the choice of generating set.

\end{definition}

\begin{example}[of a fundamental parallelepiped]

For \(\mathbf{v}_1 = {\left[ {1, 1} \right]}^t\) and
\(\mathbf{v}_2 = {\left[ {1, 3} \right]}\), we get the parallelogram
shown in the earlier figure. Note that \(\Lambda\) is also generated by
\(\mathbf{v}_1 = \left\{{1, 1}\right\}, \mathbf{w}_2 = {\left[ {0, 2} \right]}\),
but this generates a different parallelepiped:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
  \draw[thin,gray!40] (-1,-1) grid (4,4);
  \draw[<->] (-1,0)--(4,0) node[right]{$x$};
  \draw[<->] (0,-1)--(0,4) node[above]{$y$};
  \draw[line width=1pt,blue,-stealth](0,0)--(1,1) node[anchor=south west]{$\mathbf{v}_1$};
  \draw[line width=1pt,purple,-stealth](1,1)--(2,4) node[anchor=south west]{};
  \draw[line width=1pt,purple,-stealth](0,0)--(1,3) node[anchor=south east]{$\mathbf{v}_2$};
\draw[line width=1pt,blue,-stealth](1,3)--(2,4) node[anchor=south west]{};

\draw[line width=1pt,green,-stealth](0,0)--(0,2) node[anchor=south east]{$\mathbf{w}_2$};

\draw [draw=black, fill=purple, opacity=0.4]
       (0,0) -- (1,1) -- (2,4) -- (1,3) -- cycle;

\draw [draw=black, fill=green, opacity=0.2]
       (0,0) -- (1,1) -- (1,3) -- (0,2) -- cycle;
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{proposition}[The volume of the fundamental parallelotope is a lattice invariant]

In general, one gets a parallelotope whose volume is an invariant of the
lattice itself.

\end{proposition}

\begin{proof}[of proposition]

How do we compute this? Let
\(M_v = {\left[ { \mathbf{v}_1^t, \cdots, \mathbf{v}_n^t} \right]}\) be
the linear transformation obtained by placing all of the generating
vectors into the columns of a matrix, and consider scaling the unit cube
\(C = [0, 1]^n\). Then letting \(P\) be the fundamental parallelepiped,
we have \(P = M_v C\) and so
\begin{align*}
\operatorname{vol}(P) = \operatorname{vol}(M_v C) = {\left\lvert { \operatorname{det}(M_v) } \right\rvert} \operatorname{vol}(C) = \operatorname{det}(M_v)
,\end{align*}
using that the volume of the standard cube is 1. So it suffices to check
that if \(\mathbf{v}_i\) and \(\mathbf{w}_j\) generate the same lattice
\(\Lambda\), then \(\operatorname{det}M_v = \operatorname{det}M_w\). Why
is this true? If they generate the same lattice, every \(\mathbf{v}_i\)
is a \({\mathbb{Z}}{\hbox{-}}\)linear combination of the
\(\mathbf{w}_j\), and similarly every \(\mathbf{w}_j\) can be written as
a linear combination of the \(\mathbf{v}_i\). So we get
\begin{align*}
M_v = M_w A
\qquad 
M_w = M_v A'
\end{align*}
for some matrices \(A, A'\). We can thus write
\begin{align*}
M_v = M_vA' A
.\end{align*}
. Since the \(\mathbf{v}_i\) are linearly independent, \(M_v\) is
invertible, so right-multiplying yields \(AA' = I\) and taking
determinants yields
\begin{align*}
1 = \operatorname{det}(A')\operatorname{det}(A)
.\end{align*}
Noting that \(A, A' \in \operatorname{Mat}(n\times n, {\mathbb{Z}})\),
their determinants must be integers, which forces
\(\operatorname{det}(A) = \operatorname{det}(A') = \pm 1\). Taking
determinants in the original equation yields
\begin{align*}
\operatorname{det}(M_v) = \operatorname{det}(A) \operatorname{det}(M_w) = \pm \operatorname{det}(M_w)
,\end{align*}
and taking absolute values yields the result.

\end{proof}

\begin{definition}[Covolume of a Lattice]

We'll call the common value
\({\left\lvert {\operatorname{det}M_v} \right\rvert}\) for any choice of
generating set \(\left\{{ \mathbf{v}_i }\right\}\) the \textbf{covolume}
of \(\Lambda\):
\begin{align*}
\operatorname{covol}( \Lambda) \coloneqq{\left\lvert { \operatorname{det}M_v } \right\rvert}
.\end{align*}

\end{definition}

\begin{theorem}[Minkowski (Version 2)]

Let \(\Lambda\) be a full lattice in \({\mathbb{R}}^n\), and let
\(R \subseteq {\mathbb{R}}^n\) be a region that is convex and symmetric
about zero. Assume that
\begin{align*}
\operatorname{vol}(R) > 2^n \operatorname{covol}( \Lambda)
.\end{align*}
Then \(R\) contains a nonzero \(\mathbf{v} \in \Lambda\).

\end{theorem}

\begin{remark}

Taking \(\Lambda\coloneqq{\mathbb{Z}}^n\) recovers the first version.
Idea of proof: any full lattice is the image of the standard lattice
under some linear transformation.

\end{remark}

\begin{proof}[of Minkowski Version 2]

Let \(\mathbf{v}_1, \cdots, \mathbf{v}_n\) be \(n\) generators for
\(\Lambda\), then define a linear transformation
\begin{align*}
T: {\mathbb{R}}^n &\to {\mathbb{R}}^n \\
\mathbf{v}_i &\mapsto \mathbf{e}_i
,\end{align*}
which takes each generator to the corresponding standard basis vector.
The \(T \Lambda = {\mathbb{Z}}^n\) is the standard lattice, and
\begin{align*}
\operatorname{vol}(T(R)) = {\left\lvert { \operatorname{det}(T) } \right\rvert} \operatorname{vol}(R) = {\operatorname{vol}(R) \over \operatorname{covol}( \Lambda) } > 2^n
,\end{align*}
noting that
\(T^{-1}= {\left[ {\mathbf{v}_1^t, \cdots, \mathbf{v}_n^t} \right]}\).
Now applying the original Minkowski theorem we get a nonzero point
\begin{align*}
\mathbf{x}\in {\mathbb{Z}}^n \cap T(R)
\implies T^{-1}\mathbf{x} \in \Lambda \cap R
.\end{align*}

\end{proof}

\hypertarget{application-the-4-square-theorem}{%
\subsubsection{Application: The 4 Square
Theorem}\label{application-the-4-square-theorem}}

\begin{theorem}[4 Square Theorem (Lagrange)]

Every positive integer is a sum of 4 squares of integers.

\end{theorem}

\begin{lemma}[Finding sums of squares in $\ZZ/m$]

Let \(m \in {\mathbb{Z}}^{>0}\) be squarefree, then there are
\(A, B \in {\mathbb{Z}}\) such that
\begin{align*}
A^2 + B^2 + 1 \equiv 0 \operatorname{mod}m
,\end{align*}
i.e.~\(-1\) is always the sum of two squares in the ring
\({\mathbb{Z}}/m\).

\end{lemma}

\begin{proof}[of lemma]

We're trying to solve an equation \(\operatorname{mod}m\), and by the
CRT it suffices to solve it for every prime power dividing \(m\), and
since \(m\) is squarefree, all prime powers occur with exponent 1. So it
suffices to consider \(m=p\) a prime. We can further assume \(p\) is
odd, since if \(p=2\) we can take \(A=1, B=2\). Consider the following
two subsets of \({\mathbb{Z}}/p\):
\begin{align*}
S_1 &\coloneqq\left\{{ A^2 \operatorname{mod}p }\right\} \\
S_2 &\coloneqq\left\{{ -1-B^2 \operatorname{mod}p }\right\} 
.\end{align*}
Note that \(\# S_1 = {p+1 \over 2}\), since the number of nonzero
squares is half the number of elements, so \({p-1\over 2}\), and we add
in zero. Similarly \(\# S_2 = \# S_1\) since it can be obtained from
\(S_1\) by sending \(x\mapsto -1-x\). Note that
\({p+1\over 2} > {p\over 2}\), but
\({\left\lvert {{\mathbb{Z}}/p} \right\rvert} = p\), so these two sets
can't be disjoint. So there is some
\(A^1 = -1-B^2 \operatorname{mod}p\).

\end{proof}

\begin{proof}[of the 4 Square Theorem]

Suppose \(m\) is squarefree. Choose \(A, B\) as in the lemma, so
\(A^2 + B^2 \equiv -1 \operatorname{mod}m\), and define
\(\gamma \coloneqq A+ Bi \in {\mathbb{Z}}[i]\). Let
\begin{align*} \Lambda \coloneqq\left\{{ (\alpha, \beta) \in {\mathbb{Z}}[i] {~\mathrel{\Big|}~}\alpha\equiv \beta \gamma \operatorname{mod}m }\right\} .\end{align*}
Taking one such ordered pair in \(\Lambda\), we can apply complex
conjugation to obtain
\begin{align*}
\alpha\equiv \beta \gamma\operatorname{mod}m \implies {\overline{{ \alpha}}} {\overline{{ \beta}}} {\overline{{ \gamma}}} \operatorname{mod}{\overline{{m}}} = m
,\end{align*}
where we can immediately note that \({\overline{{m}}} = m\) since
\(m\in {\mathbb{Z}}\). Multiplying these two congruences yields
\begin{align*}
\alpha{\overline{{\alpha}}} 
= N( \alpha) 
\equiv N( \beta) N( \gamma) \operatorname{mod}m
\equiv - N( \beta) \operatorname{mod}m
,\end{align*}
and so we have \(N( \alpha) + N( \beta) \equiv 0 \operatorname{mod}m\).
But these are Gaussian integers, so writing
\(\alpha = a + bi, \beta = c + di\) we obtain
\begin{align*}
a^2 + b^2 + c^2 + d^2 = \equiv 0 \operatorname{mod}m
.\end{align*}
Being congruent to \(0\operatorname{mod}m\) in \({\mathbb{Z}}[i]\) means
that both the real and imaginary parts are divisible by \(m\), but since
the left-hand side above is an ordinary integer, it has no imaginary
part. So \(m \divides a^2 + b^2 + c^2 + d^2\) for every element of
\(\Lambda\). Noting that \(\Lambda \subseteq {\mathbb{Z}}[i]\) we can
identify \(\Lambda \subseteq {\mathbb{Z}}^4 \subseteq {\mathbb{R}}^4\)
by pairing \(( \alpha, \beta) \rightleftharpoons(a,b,c,d)\).

\begin{claim}

\(\Lambda\subseteq {\mathbb{R}}^4\) is a full lattice, and after writing
a set of generators and computing the determinant, one finds that
\(\operatorname{covol}( \Lambda) = m^2\).

\end{claim}

\begin{quote}
See the book for a proof of this claim!
\end{quote}

Now let
\begin{align*}
R \coloneqq\left\{{ {\left[ {x,y,z,w} \right]} \in {\mathbb{R}}^4 {~\mathrel{\Big|}~}x^2 + y^2 + z^2 + w ^2 < 2m  }\right\}
,\end{align*}
which is a convex and centrally symmetric region. A multivariable
Calculus exercise shows \(\operatorname{vol}(R) = 2\pi^2 m^2\), and
\(2\pi^2 > 2^4 \operatorname{covol}( \Lambda)\). Applying Minkowski
version 2, there exists a nonzero point
\(\mathbf{x} \in R \cap\Lambda\), and thus its coordinates satisfy
\begin{align*}
0 < x^2 + y^2 + z^2 + w^2 < 2m
.\end{align*}
The middle term is then an integer that is a multiple of \(m\), forcing
it to be equal to \(m\).

\end{proof}

\begin{remark}

We made the assumption that \(m\) was squarefree, but we can write any
\(m\in {\mathbb{Z}}^{>0}\) as \(m = k^2 m'\) where \(m'\) is squarefree.
Then writing \(m' = x^2 + y^2 + z^2 + w^2\), we have
\begin{align*} 
m = (kx)^2 + (ky)^2 + (kz)^2 + (kw)^2 
.\end{align*}
There are other applications of Minkowski's theorem that tell you when
certain types of numbers are represented by special quadratic forms
(such as the above sum of squares). See Pete Clark's papers!

\end{remark}

\hypertarget{starting-over-with-general-number-fields-lec.-13-thursday-march-04}{%
\section{Starting Over with General Number Fields (Lec. 13, Thursday,
March
04)}\label{starting-over-with-general-number-fields-lec.-13-thursday-march-04}}

\hypertarget{recasting-old-definitions}{%
\subsection{Recasting Old Definitions}\label{recasting-old-definitions}}

\begin{remark}

This corresponds to chapter 13, ``Starting Over''. Idea: we've phrased
everything so far for quadratic fields, now we want to do everything for
general number fields. The basic objects and tools: norm and trace. If
\(K/{\mathbb{Q}}\) is a number field that's Galois, we define
\(N \alpha \coloneqq\prod_{ \sigma \in G} \sigma( \alpha)\) and
\(\operatorname{Tr}\alpha \coloneqq\sum_{ \sigma\in G } \sigma( \alpha)\).
We'll have to modify this for a general number field since there's not
an immediate candidate for the Galois group.

Setup: let \(K\) be a number field with \([K: {\mathbb{Q}}] = n\), then
recall that there exist \(n\) different embeddings
\(\sigma: K \hookrightarrow{\mathbb{C}}\).

\end{remark}

\begin{definition}[Field Polynomial]

If \(\alpha\in K\) then define its \textbf{field polynomial}
\begin{align*}
\varphi_{ \alpha}(x) \coloneqq\prod_{\sigma: K\hookrightarrow{\mathbb{C}}} (x - \sigma( \alpha) )
.\end{align*}

\end{definition}

\begin{proposition}[The field polynomial is monic, has rational coefficients, and is a power of the minimal polynomial]

This is a monic polynomial with \({\mathbb{C}}{\hbox{-}}\)coefficients,
and in fact \(\varphi_{ \alpha} \in {\mathbb{Q}}[x]\) and
\(\varphi_{ \alpha}(x) = \min_{ \alpha} (x)^n\) (the minimal polynomial
over \({\mathbb{Q}}\)) for some power \(n\), and the correct choice
turns out to be \(n \coloneqq[K: F[ \alpha] ]\).

\end{proposition}

\begin{remark}

Note that the first claim follows from the second since
\(\min_{ \alpha}(x) \in {\mathbb{Q}}[x]\).

\end{remark}

\begin{lemma}[Number of embeddings of subfields]

Let \(K\) be a number field with \([K: {\mathbb{Q}}] = n\) and
\(F\leq K\) a subfield with \([F: {\mathbb{Q}}] = r\). Note that
\(r\divides n\). Then every embedding
\(\tau: F\hookrightarrow{\mathbb{C}}\) extends in \(n/r\) ways to an
embedding \(\sigma: K \hookrightarrow{\mathbb{C}}\).

\end{lemma}

\begin{proof}[of lemma, sketch]

Standard field theory exercise: by the primitive element theorem, write
\(K = F( \theta)\) where \(\deg( \theta) = [K: F] = n/r\) over \(F\).
Since we're extending an embedding, it suffices to define what it does
to \(\theta\). If \(m(x) = \min_{ \theta}(x)\) over \(F\), then
\(\sigma( \theta)\) must be a root of \((\tau m)(x)\), where the latter
polynomial is taking \(m\) and applying \(\tau\) to each of the
coefficients. Note that this preserves the degree, so
\(\deg \tau m = n/r\), and there are \(n/r\) choices for
\(\sigma( \theta)\). Now the proof follows from checking that every
single root is a possibility.

\end{proof}

\begin{proof}[of proposition]

By definition, \(\varphi_{ \alpha}(x)\) is a product over embeddings
\(\sigma:K \hookrightarrow{\mathbb{C}}\), and each such \(\sigma\)
restricts to an embedding \(F[ \alpha] \hookrightarrow{\mathbb{C}}\), so
applying the lemma to \(F[ \alpha] \leq K\) yields
\begin{align*}
\varphi_{ \alpha}(x) 
&= \prod_{ \sigma: K\hookrightarrow{\mathbb{C}}} (x - \sigma( \alpha) ) \\
&= \prod_{ \tau: F[ \alpha] \hookrightarrow{\mathbb{C}}} \,\,\, \prod_{ \sigma \text{ s.t. } { \left.{{ \sigma}} \right|_{{ F[ \alpha ] }} } = \tau } (x - \sigma( \alpha) ) \\
&= \prod_{ \tau: F[ \alpha ] \hookrightarrow{\mathbb{C}}} (x - \tau( \alpha ) )^{n(\tau)} \\
&= \qty{ \prod_{ \tau: F[ \alpha] \hookrightarrow{\mathbb{C}}} (x - \tau( \alpha ) ) }^{[K: F(\alpha)]} \\
&= \min_ \alpha(x) ^{[ K : F( \alpha ) ] }
.\end{align*}
where

\begin{itemize}
\tightlist
\item
  We've first just reorganized the product by grouping,
\item
  Then we've used that all of the terms in the inner product must have
  the same value for \(\sigma( \alpha)\) since \(\alpha\in F[ \alpha]\)
  and this makes \(\sigma( \alpha) = \tau( \alpha)\),
\item
  We note that the exponent should be the number of terms in the inner
  product, i.e.~the number of \(\sigma\) extending \(\tau\),
  i.e.~\(n(\tau) = [K: F[ \alpha] ]\) since
  \(r = [F[ \alpha] : {\mathbb{Q}}]\) and \(n = [K: {\mathbb{Q}}]\),
\item
  The last equality follows from remarks in chapter 1.
\end{itemize}

\end{proof}

\begin{remark}

The field polynomial gives us a way to determine whether an element of a
number field is in its ring of integers:

\end{remark}

\begin{proposition}[Field polynomial has integer coefficients iff the element is an integer]

\begin{align*}
\alpha\in {\mathbb{Z}}_K \iff \varphi_{ \alpha}(x) \in {\mathbb{Z}}[x]
.\end{align*}

\end{proposition}

\begin{proof}[of proposition]

\(\impliedby\): This direction is easy, since having integer
coefficients, being monic, and having \(\alpha\) as a root since
\(x - \sigma( \alpha) = x - \alpha\) for some \(\sigma\). But this puts
\(\alpha \in {\mathbb{Z}}_K\) by definition.

\(\implies\): We proved that if \(\alpha\in {\mathbb{Z}}_K\) then
\(\min_ \alpha(x) \in {\mathbb{Z}}[x]\), and \(\varphi_{ \alpha}\) is
just a power of \(\min_ \alpha(x)\).

\end{proof}

\begin{definition}[Norm and Trace]

Write
\begin{align*}
\varphi_{ \alpha} (x) = x^n + \sum_{i=1}^n a_i x^i \in {\mathbb{Q}}[x]
,\end{align*}
we then define the \textbf{norm} and \textbf{trace}\footnote{These come
  from the trace and determinant of the map \(y \mapsto y\cdot x\) on
  \(L/K\), viewed as a \(K{\hbox{-}}\)linear map on \(L\).} respectively
as
\begin{align*}
N( \alpha) &\coloneqq(-1)^n a_0 \in {\mathbb{Q}}\\
\operatorname{Tr}( \alpha) &\coloneqq-a_{n-1} \in {\mathbb{Q}}
.\end{align*}
Note that if \(\alpha \in {\mathbb{Z}}_K\) then these are both in fact
in \({\mathbb{Z}}\).

\end{definition}

\begin{remark}

Note that \(-(1)^n a_0 = \prod r_i\) is the product of the roots of
\(\varphi_{ \alpha} (x)\) and \(-a_{n-1} = \sum r_i\), so equivalently
we can think of these as
\begin{align*}
N( \alpha ) &= \prod_{ \sigma: K \hookrightarrow{\mathbb{C}}} \sigma( \alpha) \\
\operatorname{Tr}( \alpha ) &= \sum_{\sigma: K \hookrightarrow{\mathbb{C}}} \sigma( \alpha) 
.\end{align*}
It's also the case that \(N({-})\) is multiplicative and
\(\operatorname{Tr}({-})\) is \({\mathbb{Q}}{\hbox{-}}\)linear.

\end{remark}

\hypertarget{discriminants}{%
\subsection{Discriminants}\label{discriminants}}

\begin{remark}

Let \(K\) be a number field and \([K : {\mathbb{Q}}] = n\). Pick an
arbitrary ordering of embeddings
\(\sigma_1, \cdots, \sigma_n: K \hookrightarrow{\mathbb{C}}\).

\end{remark}

\begin{definition}[Tuple Discriminant]

For any \(n{\hbox{-}}\)tuple \((w_1, \cdots, w_n) \in K^n\) define the
\textbf{tuple discriminant} as
\begin{align*}
\Delta(w_1, \cdots, w_n) \coloneqq\operatorname{det}( D_{w_1, \cdots, w_n} )^2 
\end{align*}
where
\begin{align*}
D_{w_1, \cdots, w_n}
=
\begin{bmatrix}
\sigma_1(w_1) & \cdots  & \sigma_1(w_n) \\
\sigma_2(w_1) & \cdots  & \sigma_2(w_n) \\
\vdots & \cdots & \vdots \\
\sigma_n(w_1) & \cdots  & \sigma_n(w_n) 
\end{bmatrix}
.\end{align*}

\end{definition}

\begin{remark}

Why square this? Permuting two columns changes the sign of the
determinant, which is just swapping the order of the embeddings. So
squaring keeps this invariant under relabeling the \(\sigma_i\). It
turns out that this is a rational number, since we can write
\begin{align*}
\Delta(w_1, \cdots, w_n) 
= \operatorname{det}(D) \operatorname{det}(D) 
\operatorname{det}(D^t D)
,\end{align*}
where
\((D^t D)_{ij} = \operatorname{Tr}(w_i w_j) \implies D^t D \in \operatorname{Mat}(n\times n, {\mathbb{Q}})\).
So taking the determinant yields a rational number, so
\(\Delta(w_1, \cdots, w_n) \in {\mathbb{Q}}\). Moreover if you start
with the \(w_i \in {\mathbb{Z}}_K\), then
\(D^t D \in \operatorname{Mat}(n\times n, {\mathbb{Z}})\) and thus
\(\Delta(w_1, \cdots, w_n) \in {\mathbb{Z}}\).

Why is this called the discriminant?

\end{remark}

\begin{theorem}[The discriminant detects $\QQ\dash$bases]

Let \(w_1, \cdots, w_n\in K\), then
\begin{align*}
\left\{{ w_1, \cdots, w_n }\right\} \text{ form a ${\mathbb{Q}}{\hbox{-}}$basis for $K$}
\iff
\Delta(w_1, \cdots, w_n) \neq 0
.\end{align*}

\end{theorem}

\begin{remark}

So this \emph{discriminates} between bases and non-bases.

\end{remark}

\begin{proof}[of theorem]

\(\impliedby\): Suppose \(\Delta(w_1, \cdots, w_n) \neq 0\). Note that
the \(n\) elements \(w_1, \cdots, w_n\) are \(n\) elements in an
\(n{\hbox{-}}\)dimensional \({\mathbb{Q}}{\hbox{-}}\)vector space, so
the only way they could fail to be a basis would be if there were a
linear dependence. But then considering the matrix \(D\) above, a
\({\mathbb{Q}}{\hbox{-}}\)linear dependence between the \(w_i\), this
translates to a corresponding dependence between the columns of \(D\),
which would yield the contradiction \(\operatorname{det}(D)^2 = 0\).

\hfill\break

\(\implies\): This is the harder part. Toward a contradiction suppose
\(w_1, \cdots, w_n\) are a \({\mathbb{Q}}{\hbox{-}}\)basis for \(K\) but
\(\Delta(w_1, \cdots, w_n) = \operatorname{det}(D^t D) = 0\). Then the
columns of \(D^t D\) are linearly dependent, so there are
\(c_i \in {\mathbb{Q}}\) not all zero such that
\begin{align*}
\sum_{j=1}^n c_j \operatorname{Tr}( w_i w_j) = 0 && \forall i=1, \cdots, n
.\end{align*}
Introduce an element
\(\beta \coloneqq\sum_{j=1}^n c_j w_j \in K^{\times}\), which is not
zero since not all of the \(c_j\) are zero and the \(w_i\) are a basis.
Using linearity of the trace, we can write
\begin{align*}
\operatorname{Tr}(w_i \beta) = 0 && \forall i=1, \cdots, n
.\end{align*}
Again using linearity, we actually have
\(\operatorname{Tr}( \alpha \beta) = 0\) for all \(\alpha\in K\) since
every \(\alpha\) is in the \({\mathbb{Q}}{\hbox{-}}\)span of the
\(w_i\), which are a basis. It's then perfectly fine to take
\(\alpha \coloneqq\beta^{-1}\), which forces
\(\operatorname{Tr}(1) = 0\). But we can compute directly that
\(\operatorname{Tr}(1) = n > 0\) since every embedding \(\sigma\) must
send 1 to 1.

\end{proof}

\hypertarget{integral-bases}{%
\subsection{Integral Bases}\label{integral-bases}}

\begin{theorem}[Integral Basis Theorem]

For \(K\) any number field of degree \(n\),
\({\mathbb{Z}}_K \in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\) is
free of rank \(n\).

\end{theorem}

\begin{observation}

Suppose
\(\mathbf{(}w_1, \cdots, w_n), (\theta_1, \cdots, \theta_n) \in K\)
where
\({\left[ {w_1, \cdots, w_n} \right]} = {\left[ { \theta_1, \cdots, \theta_n } \right]} M\)
for some matrix \(M\in \operatorname{Mat}(n\times n, {\mathbb{Q}})\).
Then
\begin{align*}
\Delta({ {w}_1, {w}_2, \cdots, {w}_{n}}) = \Delta({ {\theta}_1, {\theta}_2, \cdots, {\theta}_{n}}) \operatorname{det}(M)^2
.\end{align*}

\end{observation}

\begin{proof}[of observation]

Applying the embeddings \(\sigma_i\) yields an equality
\(D_{ { {w}_1, {w}_2, \cdots, {w}_{n}} }= D_{{ {\theta }_1, {\theta }_2, \cdots, {\theta }_{n}} } M\).
Now taking determinants and squaring yields the result.

\end{proof}

\begin{proof}[of integral basis theorem]

Choose \({ {w}_1, {w}_2, \cdots, {w}_{n}} \in {\mathbb{Z}}_K\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \({\Delta}({ {w}_1, {w}_2, \cdots, {w}_{n}} ) \neq 0\)
\item
  \({\left\lvert { {\Delta}({ {w}_1, {w}_2, \cdots, {w}_{n}} ) } \right\rvert}\)
  is minimal among those satisfying (1).
\end{enumerate}

Does this make sense? The claim is that if (1) is possible, then (1) and
(2) is also possible. This is because
\({\Delta}({ {w}_1, {w}_2, \cdots, {w}_{n}} ) \in {\mathbb{Z}}\), taking
absolute values makes it positive, and then we can minimize among the
positive integers occurring using the well-ordering principle. But we
can choose tuples satisfying (1): we can always choose a
\({\mathbb{Q}}{\hbox{-}}\)basis, and to get them down to
\({\mathbb{Z}}_K\) instead of \(K\), they can just be scaled by a
rational integer without changing that they form a basis.

\begin{claim}

Any tuple \({ {w}_1, {w}_2, \cdots, {w}_{n}}\) satisfying (1) and (2)
will be a \({\mathbb{Z}}{\hbox{-}}\)basis for \({\mathbb{Z}}_K\).

\end{claim}

How could this fail? No elements could have multiple representations as
a \({\mathbb{Z}}{\hbox{-}}\)basis, since they don't admit any in the
\({\mathbb{Q}}{\hbox{-}}\)basis. So it suffices to show
\({\operatorname{span}}_{\mathbb{Z}}\left\{{ { {w}_1, {w}_2, \cdots, {w}_{n}} }\right\} = {\mathbb{Z}}_K\).
If not, choose \(\alpha\in {\mathbb{Z}}_K\) not in their
\({\mathbb{Z}}{\hbox{-}}\)span -- it must still be in the
\({\mathbb{Q}}{\hbox{-}}\)span, so we can write \(\alpha= \sum c_i w_i\)
where the \(c_i\in {\mathbb{Q}}\). We can assume that
\(c_1\not \in {\mathbb{Z}}\) by renumbering. Now write
\begin{align*}
\beta \coloneqq\alpha - \left[ {c_1}\right]  w_1 \in {\mathbb{Z}}_K
,\end{align*}
where \(\left[ {{-}}\right]\) denotes taking the integer part. We can
write \(\beta = \left\{{c_1}\right\}w_1 + c_2 w_2 + \cdots + c_n w_n\).
Observe that the tuple
\begin{align*}
{\left[ { \beta, w_2, \cdots, w_n} \right]} 
= {\left[ { w_1 , w_2, \cdots, w_n} \right]} 
M,
&& 
M \coloneqq
\begin{bmatrix}
\left\{{c_1}\right\} & 0  & \cdots & \vdots \\
c_2 &  1 & \ddots & \vdots \\
c_n &  0 & 0 & 1
\end{bmatrix}
.\end{align*}
noting that the first column describes how to write \(\beta\) as a
linear combination of the \(w_i\). Taking discriminants yields
\begin{align*}
{\Delta}(\beta, w_2, \cdots, w_n) 
= {\Delta}( { {w}_1, {w}_2, \cdots, {w}_{n}} )\operatorname{det}(M)^2
= {\Delta}( { {w}_1, {w}_2, \cdots, {w}_{n}} )\left\{{c_1}\right\}^2
,\end{align*}
where we've computed the determinant using the fact that it is lower
triangular. Since \(c_1\not\in{\mathbb{Z}}\), we have
\(\left\{{c_1}\right\}\) nonzero, real, between 0 and 1. Since the
discriminant was nonzero, the right-hand side is nonzero and thus
neither is the left-hand side. We would then have
\begin{align*}
0 < 
{\left\lvert { {\Delta}( \beta, w_2, \cdots, w_n) } \right\rvert}
<
{\left\lvert { {\Delta}( { {w}_1, {w}_2, \cdots, {w}_{n}} ) } \right\rvert}
,\end{align*}
which contradicts the minimality of the \(w_i\). \(\contradiction\)

\end{proof}

\begin{remark}

Note that this is non-constructive, finding a basis is another question!

\end{remark}

\hypertarget{discriminant-of-number-fields}{%
\subsection{Discriminant of Number
Fields}\label{discriminant-of-number-fields}}

\begin{definition}[Discriminant of a Number Field]

Let \(K\) be a number field, then the \textbf{discriminant} of \(K\) is
defined as
\begin{align*}
{\Delta}_K \coloneqq{\Delta}({ {w}_1, {w}_2, \cdots, {w}_{n}}) 
,\end{align*}
where \(\left\{{ w_i }\right\}\) is any \({\mathbb{Z}}{\hbox{-}}\)basis
for \({\mathbb{Z}}_K\).

\end{definition}

\begin{remark}

Is this actually an invariant of \(K\), since we made a choice of basis?
Given two \({\mathbb{Z}}{\hbox{-}}\)bases for \({\mathbb{Z}}_K\), say
\({ {w}_1, {w}_2, \cdots, {w}_{n}}, { {\theta }_1, {\theta }_2, \cdots, {\theta }_{n}}\),
then
\begin{align*}
{\left[ { { {w}_1, {w}_2, \cdots, {w}_{n}}} \right]} = {\left[ { { {\theta }_1, {\theta }_2, \cdots, {\theta }_{n}} } \right]} M && M \in \operatorname{GL}(n, {\mathbb{Z}})
.\end{align*}
Hence
\begin{align*}
{\Delta}({ {w}_1, {w}_2, \cdots, {w}_{n}}) 
= {\Delta}( { {\theta }_1, {\theta }_2, \cdots, {\theta }_{n}} ) \operatorname{det}(M)^2
= {\Delta}( { {\theta }_1, {\theta }_2, \cdots, {\theta }_{n}} )
.\end{align*}
using that invertible matrices have unit determinants, which in
\({\mathbb{Z}}\) are just \(\pm 1\).

\end{remark}

\begin{remark}

Why do we care? The discriminant measures the complexity of the number
field and carries arithmetic information:

\begin{theorem}[Hermite]

For every \(X>0\), there are only finitely many number fields such that
\({\left\lvert { {\Delta}_K } \right\rvert} \leq X\).

\end{theorem}

\begin{remark}

Interesting question: how many are there as a function of \(X\)? This is
studied today by fixing a degree \(n\), and we have good answers for
\(n=2,3,4,5\), but it's still open to get an asymptotic formula for
\(n>5\). Note that our new faculty hire this year is an expert on these
kinds of questions!

\end{remark}

\begin{theorem}[Dedekind]

Taking a prime \(p\in {\mathbb{Z}}\), we have
\begin{align*}
p \text{ ramifies in } {\mathbb{Z}}_K \iff p \divides {\Delta}_K
,\end{align*}
where \textbf{ramification} occurs if when
\(\left\langle{ p }\right\rangle{~\trianglelefteq~}{\mathbb{Z}}_K\)
factors into prime ideals with a repeated prime factor. In particular,
\({\Delta}({-}) < \infty\), and so only finitely many such primes can
occur.

\end{theorem}

\end{remark}

\hypertarget{discriminants-and-norms-lec.-14-saturday-march-13}{%
\section{Discriminants and Norms (Lec. 14, Saturday, March
13)}\label{discriminants-and-norms-lec.-14-saturday-march-13}}

\begin{example}[of a discriminant]

Suppose \(K = {\mathbb{Q}}( \sqrt{d} )\) where \(d\) is squarefree. What
is its discriminant? We need a \({\mathbb{Z}}{\hbox{-}}\)basis of
\({\mathbb{Z}}_K\), for \(d=2,3 \operatorname{mod}4\) we can take
\((1, \sqrt{d} )\). Then we construct a matrix whose columns are the
different embeddings of each entry. The embeddings here are the identity
and complex conjugation, so we get
\begin{align*}
\Delta_K = \Delta(1, \sqrt{ d} )
= 
\operatorname{det}
\qty{
\begin{bmatrix}
1 & \sqrt{d} 
\\
 1 & -\sqrt{d} 
\end{bmatrix}
}^2
= (-2 \sqrt{d} )^2 = 4d
.\end{align*}
If \(d = 1 \operatorname{mod}4\), then we can take a basis
\((1, {1 + \sqrt{d} \over 2})\), and
\begin{align*}
\Delta_{K}
=
\qty{
\begin{bmatrix}
1 & {1 + \sqrt{d} \over 2}
\\
1 & {1 - \sqrt{d} \over 2}
\end{bmatrix}
}^2
= (- \sqrt{d} )^2 = d
.\end{align*}
So we have
\begin{align*}
\Delta_K = 
\begin{cases}
d & d = 1 \operatorname{mod}4 
\\
4d & d = 2,3 \operatorname{mod}4 .
\end{cases}
\end{align*}

\end{example}

\begin{remark}

Note that \(\Delta_{\mathbb{Q}}= 1\) if you trace through the
computation.

\end{remark}

\hypertarget{norms-of-ideals}{%
\subsection{Norms of Ideals}\label{norms-of-ideals}}

\begin{definition}[Norm of an ideal]

Let \(I {~\trianglelefteq~}{\mathbb{Z}}_K\) be a nonzero ideal, then
define \(N(I) \coloneqq\# {\mathbb{Z}}_K/I\).

\end{definition}

\begin{remark}

Note that this was finite in the quadratic field case since nonzero
ideals had a ``standard basis''. For general number fields, the ideals
can be more complicated, so we'll need another way to show finiteness.

\end{remark}

\begin{lemma}[Elements divide their norms]

Let \(\alpha\in {\mathbb{Z}}_K\), then \(\alpha\divides N( \alpha)\) in
\({\mathbb{Z}}_K\).

\end{lemma}

\begin{proof}[of lemma]

Write down the obvious thing and see that it works!
\begin{align*}
N \alpha 
&\coloneqq\prod_{ \sigma: K \hookrightarrow{\mathbb{C}}} \sigma( \alpha) \\
&= \alpha \prod_{ \substack{ \sigma: K \hookrightarrow{\mathbb{C}}\\ \sigma\neq \one_K } } \sigma( \alpha) \\
&\coloneqq\alpha C 
,\end{align*}
where we've used that one embedding is the identity and factored it out.
So it only remains to show that the \emph{cofactor} \(C\) (the product
term) is actually in \({\mathbb{Z}}_K\). It is
\(\mkern 1.5mu\overline{\mkern-1.5muZZ\mkern-1.5mu}\mkern 1.5mu\), since
\(\alpha\) was an algebraic integer, i.e.~a root of some monic
polynomial with integer coefficients. But then under every embedding,
\(\sigma( \alpha)\) is a root of the same monic polynomial, so each
\(\sigma( \alpha) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\),
as is their product since it's a ring. On the other hand, we can write
\(C = N \alpha/ \alpha\). Since \(N \alpha\) is a nonzero rational
integer and \(\alpha\in K\), and since \(K\) is a field, this quotient
is in \(K\). But then
\(C \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu \cap K = {\mathbb{Z}}_K\).

\end{proof}

\begin{proposition}[Nonzero ideals have finite norms in rings of integers]

For \(I{~\trianglelefteq~}{\mathbb{Z}}_K\) nonzero,
\begin{align*}
N(I) < \infty 
.\end{align*}

\end{proposition}

\begin{proof}[of proposition]

We start with principal ideals. Let \(m\in {\mathbb{Z}}^+\), then
\({\mathbb{Z}}_K \left\langle{ m }\right\rangle \coloneqq{\mathbb{Z}}_K /m {\mathbb{Z}}_K \cong_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}} {\mathbb{Z}}^n / m{\mathbb{Z}}^n \cong ({\mathbb{Z}}/m{\mathbb{Z}})^n\)
where we've forgotten the ring structure and are just considering it as
a \({\mathbb{Z}}{\hbox{-}}\)module . But this has size \(m^n < \infty\).

Now let \(\alpha\in I\) be nonzero and let \(m \coloneqq\pm N \alpha\),
choosing whichever sign makes \(m>0\). Since
\(\alpha\divides N \alpha\), so \(N \alpha = \ell \alpha\) is a multiple
of \(\alpha\). But \(\alpha\in I\) and \(I\) is an ideal, so
\(N \alpha\in I \implies m \in I\). Then (check!) the following map is
surjective:
\begin{align*}
{\mathbb{Z}}_K/ \left\langle{ m }\right\rangle &\twoheadrightarrow{\mathbb{Z}}_K/I \\
[ \alpha]_m &\mapsto [ \alpha]_I
,\end{align*}
where we've used \(m\in I\) for this to be well-defined. So
\(\# {\mathbb{Z}}_K /I \leq {\mathbb{Z}}_K / \left\langle{ m }\right\rangle = m^n < \infty\).

\end{proof}

\begin{theorem}[The norm is multiplicative]

For every pair \(I, J{~\trianglelefteq~}{\mathbb{Z}}_K\) nonzero,
\begin{align*}
N( IJ) + N(I) N(J)
.\end{align*}

\end{theorem}

\begin{proof}[that the norm is multiplicative]

Deferred!

\end{proof}

\begin{theorem}[Formula for norm of principal ideals]

For all \(\alpha\in {\mathbb{Z}}_K\) nonzero,
\begin{align*}
N( \left\langle{ \alpha }\right\rangle) = {\left\lvert { N ( \alpha ) } \right\rvert}
,\end{align*}
i.e.~the norm of a principal ideal is the absolute value of the norm of
the element-wise ideal.

\end{theorem}

\begin{remark}

This will follow from the following proposition:

\end{remark}

\begin{proposition}[Index = Determinant]

Let \(M \in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\) be free of rank \(n\)
and let \(H \leq M\). Then \(H\) is free of rank at most \(n\), so
suppose \(\operatorname{rank}_{\mathbb{Z}}H = n\). Suppose that
\(\omega_1, \cdots, \omega_n\) is a \({\mathbb{Z}}{\hbox{-}}\)basis for
\(M\) and \(\theta_1, \cdots, \theta_n\) a
\({\mathbb{Z}}{\hbox{-}}\)basis for \(H\). We can thus write
\({\left[ { \theta_1, \cdots, \theta_n} \right]} = {\left[ { \omega_1, \cdots, \omega_n } \right]} A\)
for some \(A \in \operatorname{Mat}(n\times n, {\mathbb{Z}})\). Then
\([M: H] = \#M/H = {\left\lvert { \operatorname{det}A } \right\rvert}\).

\end{proposition}

\begin{proof}[Sketch]

Idea: convert this problem about an arbitrary
\(M \in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\) to a problem about
\({\mathbb{Z}}^n\). We know \(M \cong {\mathbb{Z}}^n\), and if we send
the \(\omega_i\) to the standard basis vectors, this identifies
\(H \cong A {\mathbb{Z}}^n\). So
\(M/H \cong {\mathbb{Z}}^n/A{\mathbb{Z}}^n\), and it's easy to see that
\(\operatorname{det}A \neq 0\): if not, there would be a linear
dependence among the \(\theta_j\). Using \emph{Smith normal form}, we
can choose \(S, T \in \operatorname{GL}_n({\mathbb{Z}})\) with
\begin{align*}
SAT = \operatorname{diag}(a_1, \cdots, a_n) && a_i \in {\mathbb{Z}}
.\end{align*}
Since \(\operatorname{det}A \neq 0\), we have
\(\operatorname{det}S, \operatorname{det}T \neq 0\), and so all of the
\(a_i\) are nonzero. We can write
\({\mathbb{Z}}^n/A{\mathbb{Z}}^n \cong {\mathbb{Z}}^n/SAT{\mathbb{Z}}^n \cong \bigoplus_{i=1}^n {\mathbb{Z}}/a_i {\mathbb{Z}}\),
which has size
\(\prod {\left\lvert {a_i} \right\rvert} = {\left\lvert { \prod a_i } \right\rvert} = {\left\lvert { \operatorname{det}(SAT) } \right\rvert} = {\left\lvert { \operatorname{det}(A) } \right\rvert}\)
since \(S, T\) are invertible and thus have determinant \(\pm 1\).

\end{proof}

\begin{proof}[of formula for norm of principal ideals]

Let \(\omega_1, \cdots, \omega_n\) be a \({\mathbb{Z}}{\hbox{-}}\)basis
for \({\mathbb{Z}}_K\), then \(\alpha \omega_1, \cdots \alpha \omega_n\)
is a \({\mathbb{Z}}{\hbox{-}}\)basis for
\(\alpha{\mathbb{Z}}_K = \left\langle{ \alpha }\right\rangle\). Now to
compute \(\# {\mathbb{Z}}_K/ \left\langle{ \alpha }\right\rangle\), we
use the ``index equals determinant'' result: write
\begin{align*}
{\left[ { \alpha \omega_1, \cdots, \alpha \omega_n} \right]} = {\left[ { \omega_1, \omega_n} \right]} A \implies \# {\mathbb{Z}}_K / \left\langle{ a }\right\rangle = {\left\lvert { \operatorname{det}(A) } \right\rvert} 
,\end{align*}
we now just need to show that this is equal to
\({\left\lvert { N \alpha } \right\rvert}\). We'll proceed by taking
discriminants of tuples, applied to the first equation above. This
yields
\begin{align*}
{\Delta}( \alpha \omega_1, \cdots, \alpha \omega_n) 
&= 
{\Delta}( \omega_1, \cdots, \omega_n) 
\operatorname{det}(A)^2 \\
\implies \operatorname{det}(A)^2 
&= 
{
{\Delta}( \alpha \omega_1, \cdots, \alpha \omega_n) 
\over
{\Delta}( \omega_1, \cdots, \omega_n) 
} \\
&=
{
\operatorname{det}(D_{\alpha \omega_1, \cdots, \alpha \omega_n })^2
\over
\operatorname{det}( D_{\omega_1, \cdots, \omega_n} )^2
}
=
\qty{
\operatorname{det}(D_{\alpha \omega_1, \cdots, \alpha \omega_n })
\over
\operatorname{det}( D_{\omega_1, \cdots, \omega_n} )
}^2
.\end{align*}
Recall that these matrices were formed by taking the \(j\)th tuple
element for the \(j\)th column and letting the column entries be the
images under all embeddings. Just looking at the first rows in each,
we'll have
\begin{align*}
{\left[ { \sigma_1( \alpha \omega_1), \cdots, \sigma_1( \alpha \omega_n) } \right]} && 
{\left[ { \sigma_1( \omega_1), \cdots, \sigma_1( \omega_n) } \right]} 
.\end{align*}
In general, the \(i\)th row of the first matrix will be
\(\sigma_i( \alpha)\) times the \(i\)th row of the second matrix. But
then this ratio of determinants will be
\(\qty{ \prod_{i=1}^n \sigma_i( \alpha )}^2 \coloneqq(N \alpha)^2\). So
\(\operatorname{det}(A)^2 = (N \alpha)^2\), and taking square roots
yields the result.

\end{proof}

\hypertarget{ch.-14-integral-bases}{%
\subsection{Ch. 14: Integral Bases}\label{ch.-14-integral-bases}}

\begin{question}

Given \(K\) a number field, can you find an explicit
\({\mathbb{Z}}{\hbox{-}}\)basis for \({\mathbb{Z}}_K\)?

\end{question}

\begin{remark}

This depends on how one is given \(K\), and in general this is hard!
This is a question in algorithmic number theory. We'll focus on a
specific sub-problem.

\end{remark}

\begin{question}

Let \(K\) be a number field with \([K : {\mathbb{Q}}] = n\) and suppose
\(\theta_1, \cdots, \theta_n\) in \({\mathbb{Z}}_K\) are a
\({\mathbb{Q}}{\hbox{-}}\)basis for \(K\). Is there a simple condition
for when they form a \({\mathbb{Z}}{\hbox{-}}\)basis for
\({\mathbb{Z}}_K\)?

\end{question}

\begin{remark}

We know there is \emph{some} \({\mathbb{Z}}{\hbox{-}}\)basis for
\({\mathbb{Z}}_K\), so let
\({ { \omega}_1, { \omega}_2, \cdots, { \omega}_{n}}\) be one. Then
express the \(\theta\) in terms of the \(\omega\):
\begin{align*}
{\left[ { { { \theta}_1, { \theta}_2, \cdots, { \theta}_{n}} } \right]} &= {\left[ { { { \omega}_1, { \omega}_2, \cdots, { \omega}_{n}} } \right]} A
\\
\implies 
{\Delta}({ { \theta}_1, { \theta}_2, \cdots, { \theta}_{n}} ) 
&= 
{\Delta}({ { \omega}_1, { \omega}_2, \cdots, { \omega}_{n}} ) \operatorname{det}(A)^2
.\end{align*}
We can view \({\left\lvert { \operatorname{det}(A) } \right\rvert}\) as
the index of the subgroup generated by the \(\theta_i\) in the group
generated by the \(\omega_i\), so
\begin{align*}
{\left\lvert { \operatorname{det}(A) } \right\rvert} = [ {\mathbb{Z}}_K : H], && H\coloneqq{\operatorname{span}}_{\mathbb{Z}}\left\{{ \theta_i }\right\}
.\end{align*}
Thus
\begin{align*}
{\Delta}({ { \theta}_1, { \theta}_2, \cdots, { \theta}_{n}} ) 
&= 
{\Delta}({ { \omega}_1, { \omega}_2, \cdots, { \omega}_{n}} )
[ {\mathbb{Z}}_K: H]^2
.\end{align*}
We can thus form a simple condition for when \(H = {\mathbb{Z}}_K\):

\end{remark}

\begin{corollary}[A sufficient condition]

If \({\Delta}( { { \theta}_1, { \theta}_2, \cdots, { \theta}_{n}}\) is
squarefree, then \({ {\theta}_1, {\theta}_2, \cdots, {\theta}_{n}}\) are
a \({\mathbb{Z}}{\hbox{-}}\)basis of \({\mathbb{Z}}_K\).

\end{corollary}

\begin{remark}

Why? If the left-hand side is squarefree, then use that
\([{\mathbb{Z}}_K: H]^2\) divides the left-hand side to conclude it must
be 1. Note that this is \emph{not} necessary! We saw that for
\(d = 2,3 \operatorname{mod}4\) that \({\Delta}_K = 4d\), which is not
squarefree.

\end{remark}

\begin{example}[of finding bases]

Let \(K = {\mathbb{Q}}( \theta)\) where \(\theta\) is a root of
\begin{align*}
f(x) = x^5 - 3x^2 + 1
,\end{align*}
which is irreducible over \({\mathbb{Q}}\). This yields a degree 5
number field. We can look for an \(n{\hbox{-}}\)tuple of elements in
\({\mathbb{Z}}_K\) which is a \({\mathbb{Q}}{\hbox{-}}\)basis for
\({\mathbb{Z}}_K\)with a squarefree discriminant. A candidate would be
\(\left\{{ \theta^j {~\mathrel{\Big|}~}0\leq j \leq 4 }\right\}\), which
are all in \({\mathbb{Z}}_K\) since \(\theta\in {\mathbb{Z}}_K\) which
is closed under multiplication.

\begin{claim}

\begin{align*}
{\Delta}( 1, \theta, \theta^2, \theta^3, \theta^4) \text{ is squarefree}
.\end{align*}

\end{claim}

We have
\begin{align*}
{\Delta}( 1, \theta, \theta^2, \theta^3, \theta^4)
&\coloneqq\operatorname{det}( {\left[ { \sigma_i ( \theta^{j-1} ) } \right]} )^2 \\
&= \operatorname{det}( {\left[ { \sigma_i ( \theta )^{j-1} } \right]} )^2 && \text{ since the $\sigma_i$ are embeddings } \\
&= \prod_{1\leq i < j \leq 5} ( \sigma_j( \theta) - \sigma_i( \theta ) )^2 &&\text{since this is a Vandermonde matrix}\\
&= {\Delta}(f)
,\end{align*}
where this is the \emph{polynomial} discriminant. This can be computed
in a computer algebra system, and in this case it equals
\(-23119 = (-61)(379)\) which is squarefree. So this yields a
\({\mathbb{Z}}{\hbox{-}}\)basis for \({\mathbb{Z}}_K\),
i.e.~\({\mathbb{Z}}_K = {\mathbb{Z}}[ \theta]\). Note that
\({\Delta}_K = -23119\) as well, since it's the discriminant of
\emph{any} integral basis.

\end{example}

\begin{example}[of finding bases]

Let \(K = {\mathbb{Q}}( \alpha)\) where \(\alpha\) is a root of
\begin{align*}
f(x) = x^3 + x^2 - 3x + 8
.\end{align*}
We can try \(1, \alpha, \alpha^2\), and check
\begin{align*}
{\Delta}(1, \alpha, \alpha^2) = {\Delta}(f) = (-4)(503)
,\end{align*}
so we can't conclude this is a \({\mathbb{Z}}{\hbox{-}}\)basis. Going
back to the proof, we \emph{can} conclude that
\([{\mathbb{Z}}_K: H] ^2 \divides {\Delta}(1, \alpha, \alpha^2)\) where
\(H \coloneqq{\mathbb{Z}}+ {\mathbb{Z}}\alpha + {\mathbb{Z}}\alpha^2 = {\mathbb{Z}}[ \alpha ]\).
This allows us to conclude that \([{\mathbb{Z}}_K: H] = 1, 2\), so this
could still be an index 2 subgroup. If this happens,
\(\# {\mathbb{Z}}_K/H = 2\) and every element is annihilated by 2, so
\(2{\mathbb{Z}}_K \subseteq H = {\mathbb{Z}}[ \alpha ]\). This would
mean
\begin{align*}
{\mathbb{Z}}_K \subseteq {1\over 2} {\mathbb{Z}}[ \alpha ] 
= 
\left\{{ {c_0 + c_1 \alpha + c_2 \alpha^2 \over 2} {~\mathrel{\Big|}~}c_i \in {\mathbb{Z}}}\right\} 
.\end{align*}
So are there elements of \({\mathbb{Z}}_K\) of this form that are
\emph{not} in \({\mathbb{Z}}[ \alpha ]\)? If there's nothing of this
form in \({\mathbb{Z}}_K \setminus{\mathbb{Z}}[ \alpha ]\) then we can
conclude \({\mathbb{Z}}_K = {\mathbb{Z}}[ \alpha ]\). If there \emph{is}
something of this form in
\({\mathbb{Z}}_K \setminus{\mathbb{Z}}[ \alpha ]\), then
\({\mathbb{Z}}_K \supseteq {\mathbb{Z}}[\alpha]\). One can check that
\({\alpha+ \alpha^2 \over 2} \in {\mathbb{Z}}_K \setminus H\). So the
original candidate basis was wrong, but we can take
\(1, \alpha, {\alpha + \alpha^2 \over 2}\) instead, which is an integral
basis.

\end{example}

\begin{remark}

Why is this last part true? These are 3 elements of \({\mathbb{Z}}_K\)
that are still \({\mathbb{Q}}{\hbox{-}}\)linearly independent and
contains the \({\mathbb{Z}}{\hbox{-}}\)span of the previous 3 elements
defining \(H\). But the index of \(H\) was 2, so this forces it to be
everything. So \({\mathbb{Z}}_K \neq {\mathbb{Z}}[ \alpha]\), and in
fact Dedekind showed that \({\mathbb{Z}}_K \neq {\mathbb{Z}}[ \beta]\)
for \emph{any} choice of \(\beta\in {\mathbb{Z}}_K\). So cubic number
fields exhibit new behavior when compared to quadratic number fields!

\end{remark}

\begin{remark}

Next time: integral bases for cyclotomic fields.

\end{remark}

\hypertarget{cyclotomic-fields-lec.-15-saturday-march-13}{%
\section{Cyclotomic Fields (Lec. 15, Saturday, March
13)}\label{cyclotomic-fields-lec.-15-saturday-march-13}}

\begin{remark}

This is chapter 14 continued.

\end{remark}

\begin{definition}[Cyclotomic Fields]

A \textbf{cyclotomic field} is a number field \({\mathbb{Q}}( \zeta_m)\)
where \(\zeta_m \coloneqq e^{2\pi i / m}\), a primitive \(m\)th root of
1.

\end{definition}

\begin{remark}

The Kronecker-Webber theorem: any \emph{abelian extension}
\(K/{\mathbb{Q}}\) (so
\({ \mathsf{Gal}} (K/{\mathbb{Q}}) \in {\mathsf{Ab}}\)) is contained in
a cyclotomic extension, and every cyclotomic field is an abelian
extension. Given such a number field \(K = {\mathbb{Q}}( \zeta_m)\),
what is \({\mathbb{Z}}_K\)?

\end{remark}

\begin{theorem}[The ring of integers of a cyclotomic field is given by adjoining a primitive root of unity]

For \(K = {\mathbb{Q}}( \zeta_m)\),
\begin{align*}
{\mathbb{Z}}_K = {\mathbb{Z}}[ \zeta_m ]
.\end{align*}

\end{theorem}

\begin{remark}

The degree of any such \(K/{\mathbb{Q}}\) is \(\phi(m)\), and here
\(\phi(p) = p-1\). Also recall Eisenstein's criterion: if \(p\) divides
all of the coefficients of a polynomial \(f(x) \coloneqq\sum a_i x^i\)
but \(p^2\notdivides a_0\), then \(f\) is irreducible over
\({\mathbb{Q}}\).

\end{remark}

\begin{lemma}[The minimal polynomials of roots of unity]

The minimal polynomial of \(\zeta_p\) over \({\mathbb{Q}}\) is
\begin{align*}
\Phi_p(x) \coloneqq x^{p-1} + x^{p-2} + \cdots + x + 1
,\end{align*}
and so \([{\mathbb{Q}}(\zeta_p) : {\mathbb{Q}}] = p-1\).

\end{lemma}

\begin{proof}[of lemma, a sketch]

Note that \(\zeta_p\) is a root of \(\Phi_p\), since
\begin{align*}
\Phi_p(x) = {x^p-1 \over x - 1}
,\end{align*}
and \(\zeta_p\) is a root of the numerator of the right-hand side and
not of the denominator. This is irreducible by Eisenstein's criterion at
\(p\), using \(x\mapsto x+1\).

\end{proof}

\begin{proposition}[Eisenstein primes don't divide the extension degree]

Let
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
be an algebraic integer such that
\begin{align*}
\min_ \alpha (x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 && \in {\mathbb{Z}}[x]
\end{align*}
is Eisenstein at the prime \(p\). Let
\(K \coloneqq{\mathbb{Q}}( \alpha)\), a number field of degree \(n\).
Then
\begin{align*}
p\notdivides [{\mathbb{Z}}_K : {\mathbb{Z}}[ \alpha] ]
.\end{align*}

\end{proposition}

\begin{proof}[of proposition]

We first observe that \(\alpha^n\) is a multiple of \(p\) in
\({\mathbb{Z}}_K\). To see this, plug \(\alpha\) into the minimal
polynomial to get \(0 = \alpha^n + \cdots\) and solve for \(\alpha^n\)
to obtain
\begin{align*}
\alpha^n = -(a_{n-1} \alpha^{n-1} + \cdots + a_1 \alpha + a_0) \equiv 0 \operatorname{mod}p \text{ in } {\mathbb{Z}}_K 
,\end{align*}
and this is a multiple of \(p\) by the assumption on Eisenstein's
criterion. We want to show \(p\) doesn't divide
\(\# {\mathbb{Z}}_K/ {\mathbb{Z}}[ \alpha]\) as
\({\mathbb{Z}}{\hbox{-}}\)modules, identify the index as the size of
this quotient. It suffices to show that
\({\mathbb{Z}}_K/{\mathbb{Z}}[ \alpha]\) has no elements of order \(p\),
by applying Cauchy's theorem. If \(\beta\in {\mathbb{Z}}_K\) represents
an element of order \(p\) in the quotient, then
\(p \beta\in {\mathbb{Z}}[ \alpha]\) and so
\(p \beta = b_0 + b \alpha + \cdots + b_{n-1} \alpha^{n-1}\) for some
\(b_i \in {\mathbb{Z}}\). The order of \(\beta\) to be exactly \(p\), so
not all of the \(b_i\) are multiples of \(p\): otherwise one could
divide through by \(p\) and conclude \(\beta\in {\mathbb{Z}}[ \alpha]\),
making it zero in the quotient (and in particular, not of order \(p\) as
assumed). Suppose toward a contradiction that \(i\) is the smallest
index such that \(p\) does not divide \(b_i\). Then take this last
equation mod \(p\):
\begin{align*}
p \beta \equiv 0 \equiv b_i \alpha^{i} + \cdots + b_{n-1} \alpha^{n-1} \operatorname{mod}p
.\end{align*}
Now multiply by \(\alpha^{n-1-i}\) to obtain
\begin{align*}
0 \equiv b_i \alpha^{n-1} + \cdots \equiv b_i \alpha^{n-1} \operatorname{mod}p
,\end{align*}
where \(p\) divides all of the other terms since they all contain a
factor of \(\alpha^n \equiv 0 \operatorname{mod}p\). So
\(b_i \alpha^{n-1} /p \in {\mathbb{Z}}_K\), and by a previous theorem,
this forces \(N( b_i \alpha^{n-1} / p ) \in |ZZ\). But we can write
\begin{align*}
N \qty{ b_i \alpha^{n-1} \over p }
&=
N \qty{ b_i \over p} N( \alpha^{n-1} ) \\
&= \qty{b_i \over p}^{n} N( \alpha)^{n-1} \\
&= \qty{b_i \over p}^{n} \pm a_0 \\
&= \pm {b_i^n a_0^{n-1} \over p^n } \not\in {\mathbb{Z}}
.\end{align*}
where we've used that all embeddings fix rational numbers. But this is
not an integer, since by Eisenstein \(p^2\) does not divide \(a_0\). So
\(a_0^{n-1}\) contributes exactly \(n-1\) copies of \(p\), leaving a
\(p\) in the denominator, and \(p\notdivides b_i\) since we choose \(i\)
precisely to arrange for this. \(\contradiction\)

\end{proof}

\begin{remark}

Recall some facts about the discriminant: let \(F\) be a field and
\(f(x) \in F[x]\) monic. Then factor
\(f(x) = \prod_{i=1}^n (x - \alpha_i)\) over some splitting field. We
then define
\begin{align*}
{\Delta}(f) \coloneqq\prod_{i < j} ( \alpha_j - \alpha_i )^2
.\end{align*}
We won't discuss the theory, but we'll use a few facts.

\end{remark}

\begin{fact}

For each fixed \(n\) and all polynomials \(f\) of degree \(n\),
\({\Delta}(f)\) is given by a universal polynomial in the coefficients
of \(f\) with integer coefficients. For example, for \(n=2\) and
\(f(x) = x^2 + bx + c\), we have
\({\Delta}(f) = b^2 - 4c \in {\mathbb{Z}}[b,c]\). If \(n=3\) and
\(f(x) = x^3 + bx^2 + cx + d\), we have
\begin{align*}
{\Delta}(f) = 18bcd - 4b^3d + b^2c^2 - 4c^3 - 27d^2\in {\mathbb{Z}}[b,c,d]
.\end{align*}
So the discriminant is some polynomial expression in the coefficients,
which (more importantly) have \emph{integer} coefficients.

Some consequences:

\begin{itemize}
\tightlist
\item
  \({\Delta}(f) \in F\), despite the fact that the roots are generally
  not in \(F\) and are instead in some splitting field.
\item
  If \(F ={\mathbb{Q}}\) and \(f\in {\mathbb{Q}}[x]\) \emph{and} in fact
  \(f\in {\mathbb{Z}}[x]\), then \({\Delta}(f) \in {\mathbb{Z}}\).
\item
  If \(F = {\mathbb{Q}}\) and \(f\in {\mathbb{Z}}[x]\) with \(q\) some
  prime,
  \begin{align*}
  {\Delta}( f) \operatorname{mod}q = {\Delta}(f \operatorname{mod}q)
  ,\end{align*}
  where we first take the discriminant to land in \({\mathbb{Z}}\) and
  then reduce to \({\mathbb{F}}_q\), or we reduce
  \(f\in {\mathbb{Z}}[x]\) to
  \(f\operatorname{mod}q \in {\mathbb{F}}_q[x]\) and take the
  discriminant using some algebraic close of \({\mathbb{F}}_q\).
\end{itemize}

\end{fact}

\begin{proof}[That $\ZZ_K = \ZZ\adjoin{\zeta_p}$ for $K = \QQ(\zeta_p)$.]

To save space, we'll write \(\zeta \coloneqq\zeta_p\). We want to show
\(1, \zeta, \cdots, \zeta^{p-2}\) forms an integral basis, from last
time we have
\begin{align*}
{\Delta}( 1, \zeta, \cdots, \zeta^{p-2}) = 
{\Delta}_K 
[{\mathbb{Z}}_K : {\mathbb{Z}}[ \zeta ] ]^2 
\implies
[{\mathbb{Z}}_K : {\mathbb{Z}}[ \zeta ] ]^2 
\divides
{\Delta}( 1, \zeta, \cdots, \zeta^{p-2}) 
.\end{align*}

\begin{claim}

The right-hand side is a power of \(p\) (up to a sign), and hence so is
the left-hand side.

\end{claim}

We'll proceed by showing that the only prime that could divide the
right-hand side is \(p\). Suppose \(q\) divides the right-hand side,
i.e.~\(q \divides {\Delta}(x^{p-1} + \cdots + x + 1)\). So this is zero
mod \(q\), and thus
\({\Delta}( x^{p-1} + \cdots + x + 1 \operatorname{mod}q) \equiv 0\).
The discriminant was a product of roots, so it can only be zero if two
roots coincide, so there is a multiple root of
\(x^{p-1} + \cdots + x + 1 \operatorname{mod}q\) and thus also of
\(x^p - 1\). So \(x^p-1\) and its derivative \(px^{p-1}\) have a root in
common, and (check!) this can only happen if \(q=p\).

\hfill\break

So \([{\mathbb{Z}}_K : {\mathbb{Z}}[\zeta] ] = p^\ell\) for some
\(\ell\). Using that fact that
\({\mathbb{Z}}[ \zeta] \cong {\mathbb{Z}}[ \zeta - 1]\), we have
\([ {\mathbb{Z}}_K : {\mathbb{Z}}[ \zeta] ] = [ {\mathbb{Z}}_K: {\mathbb{Z}}[\zeta - 1] ]\).
But by the previous lemma, we know that the minimal polynomial of
\(\zeta_p - 1\) is \(\Phi(x+1)\), which is \(p{\hbox{-}}\)Eisenstein. So
by that lemma,
\(p\notdivides [{\mathbb{Z}}_K: {\mathbb{Z}}[ \zeta - 1]]\), which
forces \(\ell = 0\) and \({\mathbb{Z}}_K = {\mathbb{Z}}[ \zeta_p ]\).

\end{proof}

\begin{proof}[Sketch of the same proof for $K = \QQ(\zeta_m)$]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Do roughly the same proof for prime powers \(m = p^\ell\)
\item
  Show that if \(a, b\in {\mathbb{Z}}^{\geq 0}\) are coprime then
  \({\Delta}_a \coloneqq{\Delta}_{{\mathbb{Q}}( \zeta_a)}, {\Delta}_b \coloneqq{\Delta}_{{\mathbb{Q}}(\zeta_b)}\)
  are coprime.
\end{enumerate}

These are defined in terms of integral bases, and we're trying to prove
that something \emph{is} an integral basis, so how do you show this if
you don't know your basis is integral to begin with? Without knowing the
exact values of the discriminants, you can show
\({\Delta}_a \divides a^?\) divides some power of \(a\), and the same
for \(b\), and so \(a, b\) coprime will make \(a^?, b^?\) coprime as
well. This can be shown by computing the discriminant of a
\emph{candidate} integral bases rather than an actual one.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\item
  Use a key lemma: if \(K_1, K_2\) are number fields with coprime
  discriminants, then considering the composite field, we have
  \({\mathbb{Z}}_{K_1 K_2} = {\mathbb{Z}}_{K_1} {\mathbb{Z}}_{K_2}\), a
  composite ring.
\item
  If \(a, b\) are coprime, check that
  \({\mathbb{Q}}(\zeta_a) {\mathbb{Q}}(\zeta_b) = {\mathbb{Q}}(\zeta_{ab})\)
  and
  \({\mathbb{Z}}[ \zeta_a] {\mathbb{Z}}[ \zeta_b] = {\mathbb{Z}}[ \zeta_{ab} ]\).
\item
  Factor \(m = \prod_{i} p_i^{\ell_i}\) and apply steps (3) and (4)
  inductively.
\end{enumerate}

\end{proof}

\begin{remark}

The hard part is the lemma in (3). Also, questions about discriminants
tend to come up during oral exams that include algebraic number theory.

\end{remark}

\hypertarget{ideal-theory-in-general-number-rings-ch.-15}{%
\subsection{Ideal Theory in General Number Rings (Ch.
15)}\label{ideal-theory-in-general-number-rings-ch.-15}}

\begin{remark}

Here ``number rings'' means \({\mathbb{Z}}_K\) for \(K\) a general
number field. Let \(K\) be a number field with
\([K: {\mathbb{Q}}] = n\). We'd want

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\operatorname{Id}({\mathbb{Z}}_K)\) to be a UFM as a monoid,
\item
  \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is a finite group,
\end{enumerate}

Recall that we proved (1) and used it to deduce (2) for quadratic
fields, whereas for the general case we'll prove (2) and deduce (1). The
approach we'll take here is somewhat idiosyncratic -- the standard
treatment involves the theory of Dedekind domains, which uses a lot of
commutative algebra. This approach is more classic (circa 19th century,
very concrete), and we'll skip over less important details (e.g.~those
that are unlikely to show up on oral exams).

\end{remark}

\begin{definition}[Class Group of a Number Ring]

\begin{align*}
{ \operatorname{Cl}} ({\mathbb{Z}}_K) \coloneqq\operatorname{Id}({\mathbb{Z}}_K)/ \sim
,\end{align*}
where \(\sim\) denotes dilation equivalence.

\end{definition}

\begin{remark}

Our strategy:

\begin{itemize}
\tightlist
\item
  Prove \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is finite.
\item
  Prove \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is actually a group,
  i.e.~there are inverses, so that for every ideal there is another
  ideal such that their product is principal (the ``Principal Multiple
  Lemma'').
\item
  The remaining proofs from the quadratic field case go through almost
  word-for-word.
\end{itemize}

\end{remark}

\begin{proposition}[Dilations of elements are always close to integers]

There is a constant \(T = T(K)\) that only depends on \(K\) such that
for every \(\theta\in K\), there is a positive integer \(t \leq T\) and
a \(\xi \in {\mathbb{Z}}_K\) such that
\begin{align*}
{\left\lvert {N(t \theta - \xi ) } \right\rvert} < 1
.\end{align*}

\end{proposition}

\begin{remark}

I.e. anything in the field can be multiplied by a bounded integer to
make it close to something in the ring of integers. This proposition
came up for imaginary quadratic fields in the Rabinowitz criterion,
crucial for proving that the class group was generated by prime ideals
which lie above small primes.

\end{remark}

\begin{proof}[of proposition]

Omitted! See book, this proof wouldn't show up on an oral exam. This
uses Dirichlet's approximation criterion again, although in a different
way.

\end{proof}

\begin{theorem}[The class group is finite]

\begin{align*}
\# { \operatorname{Cl}} ({\mathbb{Z}}_K) < \infty 
.\end{align*}

\end{theorem}

\begin{proof}[that the class group is finite]

Very similar to how it goes for quadratic fields. As before, let
\(I \in \operatorname{Id}({\mathbb{Z}}_K)\) be nonzero and
\(\beta \in I\) nonzero with \({\left\lvert {N \beta} \right\rvert}\)
minimal.

\hfill\break

\begin{claim}

Let \(T\) be as in the proposition, then
\(T! I \subseteq \left\langle{ \beta }\right\rangle\).

\end{claim}

This follows from exactly the same argument as before.

\hfill\break

Now define \(J \coloneqq{T! \over \beta} I \subseteq {\mathbb{Z}}_K\),
which is a dilation of \(I\) and thus
\(J {~\trianglelefteq~}{\mathbb{Z}}_K\) as well. By definition,
\(I\sim J\), i.e.~\([I] = [J] \in \operatorname{Id}({\mathbb{Z}}_K)\),
and it's now enough to show that there are only finitely many
possibilities for \(J\), since then every class is equal to the class of
one of finitely many such \(J\). Since \(\beta\in I\), we can deduce
that \(T! \in J\) and thus
\(\left\langle{ T! }\right\rangle \subseteq J\). We'd like to say ``to
contain is to divide'' (as in the case of unique factorization) and
conclude \(J\divides T!\), which only has finitely many divisors.
However, we haven't proved this yet! We can use an algebra fact instead:
\begin{align*}
\left\{{\substack{
  \text{Ideals of ${\mathbb{Z}}_K$ }
  \\
  \text{containing } \left\langle{ T! }\right\rangle 
}}\right\}
&\rightleftharpoons
\left\{{\substack{
  \text{Ideals of } {\mathbb{Z}}_K / \left\langle{ T! }\right\rangle 
}}\right\}
,\end{align*}
so it's enough to show that the right-hand side is finite. This is
``obvious'', since
\(\# {\mathbb{Z}}_K / \left\langle{ T! }\right\rangle = (T!)^n\). This
comes from the fact that
\({\mathbb{Z}}_K \cong_{{\mathsf{Ab}}} {\mathbb{Z}}^n\), so as a
\({\mathbb{Z}}{\hbox{-}}\)module this is isomorphic to
\({\mathbb{Z}}^n / T! {\mathbb{Z}}^n \cong ({\mathbb{Z}}/T! {\mathbb{Z}})^n\),
so this is a finite ring and can thus only have finitely many
ideals.\footnote{In fact, we've already proved that
  \({\mathbb{Z}}_K / I\) for any nonzero ideal \(I\) is finite.}

\end{proof}

\begin{remark}

We now want to establish the cancellation law in
\(\operatorname{Id}({\mathbb{Z}}_K)\), then the principal multiple
lemma, and then everything else will follow as in the quadratic case.

\end{remark}

\hypertarget{ideal-theory-in-number-fields-continued-lec.-16-tuesday-march-30}{%
\section{Ideal Theory in Number Fields Continued (Lec. 16, Tuesday,
March
30)}\label{ideal-theory-in-number-fields-continued-lec.-16-tuesday-march-30}}

\hypertarget{setting-up-the-theory}{%
\subsection{Setting up the Theory}\label{setting-up-the-theory}}

\begin{remark}

We want to develop theorems of ideal theory for \({\mathbb{Z}}_K\) for
\(K\) a general number field, i.e.~factorization into prime ideals and
the finiteness of the class group. The strategy:

\begin{itemize}
\tightlist
\item
  Prove \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is a finite monoid,
\item
  Prove \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) has inverses and is
  thus a group, i.e.~every nonzero ideal can be multiplied by another
  ideal to become principal (principal multiple lemma),
\item
  Run proofs/corollaries as before.
\end{itemize}

Last time, we proved the first one.

\end{remark}

\begin{lemma}[When ideals are left identities under multiplication]

Let \(I, J \in \operatorname{Id}({\mathbb{Z}}_K)\), then if \(IJ = J\)
then \(I = \left\langle{ 1 }\right\rangle\).

\end{lemma}

\begin{remark}

Note that this is a special case of cancellation. To prove this, we'll
use that \({\mathbb{Z}}_K\) is Noetherian, i.e.~every ideal is finitely
generated as a \({\mathbb{Z}}_K{\hbox{-}}\)module. In fact,
\({\mathbb{Z}}_K \cong {\mathbb{Z}}^n\), so any ideal is free of rank
\(\leq n\) as a \({\mathbb{Z}}{\hbox{-}}\)module, hence finitely
generated as a \({\mathbb{Z}}{\hbox{-}}\)module, hence
finitely-generated as a \({\mathbb{Z}}_K{\hbox{-}}\)module since one can
use the same generators.

\end{remark}

\begin{proof}[of lemma]

Let \(J = \left\langle{ \beta_1, \cdots, \beta_n }\right\rangle\), then
since \(IJ = J\), for every \(j\) we can write
\(\beta_j = \sum_{i=1}^m A_{ij} \beta_i\). This means that there is some
matrix \(A\in \operatorname{Mat}(m\times m, I)\) with entries
\(A_{ij} \in I\) such that
\begin{align*}
{\left[ {\beta_1, \cdots, \beta_m} \right]}
=
{\left[ {\beta_1, \cdots, \beta_m} \right]}A
.\end{align*}
Then \(A-\one \mathbf{\beta} = 0\), making \(A-\one\) singular since not
all of the \(\beta_i\) were zero since they were generators of a nonzero
ideal. Now take the determinant mod \(I\), which yields
\begin{align*}
0 \equiv \operatorname{det}(A-\one) \equiv \operatorname{det}(-\one) \equiv \pm 1\operatorname{mod}I
,\end{align*}
but this can only occur if \(1\in I\), making
\(\left\langle{ 1 }\right\rangle= I\).

\end{proof}

\begin{lemma}[Right-cancellation when principal ideals are involved]

Let \(I, J{~\trianglelefteq~}{\mathbb{Z}}_K\), then if \(IJ = \beta J\)
with \(\beta \in {\mathbb{Z}}_K \setminus\left\{{0}\right\}\), we have
\(I = \left\langle{ \beta }\right\rangle\).

\end{lemma}

\begin{remark}

Note that the previous lemma is a special case of this where
\(\beta = 1\). One can then bootstrap the previous lemma to get this,
see the book.

\end{remark}

\begin{lemma}[Principal Multiple Lemma]

For all \(I\in \operatorname{Id}({\mathbb{Z}}_K)\) there is a
\(J \in \operatorname{Id}({\mathbb{Z}}_K)\) such that \(IJ\) is
principal.

\end{lemma}

\begin{proof}[of principal multiple lemma]

Consider
\([I], [I]^2, \cdots \in { \operatorname{Cl}} ({\mathbb{Z}}_K)\). By the
pigeonhole principal, there is some \(k, \ell\) such that
\([I^k] = [I]^\ell\), so \(I^k = \lambda I^\ell\) for some
\(\lambda \in K^{\times}\). Note that any nonzero element of \(K\) can
be written as \(k/n\) for \(k\in K\) and \(n\in {\mathbb{Z}}_K\). So we
can scale \(\lambda\) to put it in \({\mathbb{Z}}_K\), yielding
\(\lambda = \alpha/m\) where \(\alpha\in {\mathbb{Z}}_K\) and
\(m\in {\mathbb{Z}}^{\times}\). We then have
\begin{align*}
mI^k = \alpha I^\ell = (\alpha I^{\ell-k})I^k
.\end{align*}
We have enough to cancel the \(I^k\), and so
\(\left\langle{ m }\right\rangle= \alpha I^{\ell - k}\). Dilating both
sides by \(\alpha^{-1}\) yields
\(\left\langle{ m/\alpha }\right\rangle = I^{\ell-k}\). But this is a
power of \(I\) that is principal, so we can take
\(J \coloneqq I^{\ell-k-1}\).

\end{proof}

\begin{remark}

Note that the logical order in which these theorems are proved is
slightly reversed.

\end{remark}

\begin{corollary}[Class groups are finite and $\Id$ is a cancellative monoid]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is a group, and thus a
  finite abelian group.
\item
  \(\operatorname{Id}({\mathbb{Z}}_K)\) is cancellative. Just show one
  can cancel principal ideals (by dilation), and then in general you
  cancel by multiplying both sides principal and cancelling that
  principal ideal.
\end{enumerate}

To show unique factorization, we before showed factorization into
irreducibles first, then uniqueness as a consequence of Euclid's lemma.

\end{corollary}

\begin{lemma}[The monoid $\Id$ is atomic]

\(\operatorname{Id}({\mathbb{Z}}_K)\) is atomic, i.e.~every element
factors into irreducibles.

\end{lemma}

\begin{proof}[of lemma]

We'll proceed by induction on \(N(I)\), using that \(N(AB) \leq N(A)\)
for any \(A\) and so \(I\divides J \implies N(I) < N(J)\). Before we
used that \(N(AB) = N(A) N(B)\), but we haven't proved that here yet. We
also don't know that ``to divide is to contain'' here, but since
\(I\divides J\) and \(I\neq J\), we do obtain \(J \subsetneq I\). Hence
there is a surjection
\begin{align*}
{\mathbb{Z}}_K/J \to {\mathbb{Z}}_K/I
.\end{align*}
This has nontrivial kernel since \(I\setminus J \neq \emptyset\), so
\({\left\lvert {{\mathbb{Z}}_K/J} \right\rvert}> {\left\lvert {{\mathbb{Z}}_K/I} \right\rvert}\).

\end{proof}

\begin{lemma}[Analog of Euclid's Lemma]

Irreducibles in \(\operatorname{Id}({\mathbb{Z}}_K)\) are prime.

\end{lemma}

\begin{proof}[of lemma]

Same as before! Literally use the exact same words, we've set it up this
way.

\end{proof}

\begin{theorem}[The monoid $\Id$ is a unique factorization monoid]

\(\operatorname{Id}({\mathbb{Z}}_K)\) is a UFM, or equivalently every
nonzero ideal factors uniquely as a product of prime ideals.

\end{theorem}

\hypertarget{modern-approach}{%
\subsection{Modern Approach}\label{modern-approach}}

\begin{question}

What is the widest class of domains for which the previous theorem
holds?

\end{question}

\begin{definition}[Dedekind Domains]

Let \(R\) be a domain that is not a field (since ideals in fields are
uninteresting). Then \(R\) is a \textbf{Dedekind domain} if and only if

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(R\) is Noetherian,
\item
  \(R\) is integrally closed, so if \(K = \operatorname{ff}(R)\), then
  if \(\alpha\in K\) is a root of a monic polynomial in \(R[x]\) we have
  \(\alpha\in R\).\footnote{Compare to the classical rational root
    theorem.}
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{2}
\tightlist
\item
  Every nonzero prime ideal is maximal.
\end{enumerate}

\end{definition}

\begin{theorem}[Noether]

TFAE:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(R\) is a Dedekind domain,
\item
  Every nonzero ideal of \(R\) factors into prime ideals (not
  necessarily uniquely).
\item
  \begin{enumerate}
  \def\labelenumii{(\arabic{enumii})}
  \setcounter{enumii}{1}
  \tightlist
  \item
    along with uniqueness.
  \end{enumerate}
\end{enumerate}

\end{theorem}

\begin{proof}[of Noether's theorem]

Omitted, this is an exercise in commutative algebra.

\end{proof}

\begin{proposition}[Rings of integers are Dedekind domains]

For any number field \(K\), \({\mathbb{Z}}_K\) is a Dedekind domain.

\end{proposition}

\begin{proof}[of proposition]

We can check the definitions directly:

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  This is a consequence of the integral basis theorem.
\item
  Suppose \(\alpha \in K\) and is a root of a monic polynomial in
  \({\mathbb{Z}}_K[x]\), we then want to show
  \(\alpha\in {\mathbb{Z}}_K\). Then \(\alpha\) is a root of a monic
  polynomial in
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu[x]\),
  and by a previous proof, any monic polynomial with
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
  coefficients is itself in
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
  Since \(\alpha \in K\) as well, we have
  \(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}_K\mkern-1.5mu}\mkern 1.5mu \cap K \coloneqq{\mathbb{Z}}_K\).
\item
  Let \(P {~\trianglelefteq~}{\mathbb{Z}}_K\) be nonzero. Then
  \({\mathbb{Z}}_K/P\) is a domain, but any finite integral domain is a
  field and we know this is finite since \(N(P) < \infty\).
\end{enumerate}

\end{proof}

\hypertarget{norms-revisited}{%
\subsection{Norms Revisited}\label{norms-revisited}}

\begin{remark}

We left one theorem hanging when we discussed norms. We proved that
norms of ideals are finite, and \(N(P)\) for \(P\) principal is equal to
\(N(a)\) for \(a\) any generator. We haven't yet proved the following:

\end{remark}

\begin{theorem}[The norm is multiplicative]

\begin{align*}
N(IJ) = N(I) N(J) && \forall I, J \in \operatorname{Id}({\mathbb{Z}}_K)
.\end{align*}

\end{theorem}

\begin{remark}

If \(I, J\in \operatorname{Id}({\mathbb{Z}}_K)\), we have
\(\gcd(I, J) = I+J\), since this is the smallest ideal such that any
\(P\divides I, P\divides J\) must satisfy \(P\divides \gcd(I, J)\).

\end{remark}

\begin{proof}[that the norm is multiplicative]

It's enough to show \(N(IP) = N(I) N(P)\) for \(P\) prime, since every
\(J\) factors into primes and we can apply this result recursively. Now
\begin{align*}
N(IP) 
&= [{\mathbb{Z}}_K, IP] \\
&= [{\mathbb{Z}}_K: I] [I : IP] \\
&= N(I) [I: IP]
,\end{align*}
so it suffices to show \(N(P) = [I: IP]\).

\begin{claim}

This is true because \({\mathbb{Z}}_K \cong I/IP\) are isomorphic as
\({\mathbb{Z}}{\hbox{-}}\)modules.

\end{claim}

Choose \(\beta \in I\setminus IP\), using that \(I\) \emph{properly}
divides \(IP\) when it is properly contained. Define a map
\begin{align*}
\psi: {\mathbb{Z}}_K/p &\to I/IP \\
\alpha \operatorname{mod}p  &\mapsto \alpha \beta \operatorname{mod}IP
.\end{align*}
This is well-defined since for any two elements which differ by a
multiple of \(P\), multiplying by \(\beta \in I\) lands in \(IP\).

\begin{exercise}[?]

Check that this is a well-defined group morphism.

\end{exercise}

\hfill\break

\textbf{Injectivity}: Suppose that
\(\alpha\operatorname{mod}P \in \ker \psi\), so \(\alpha \beta\in IP\)
and
\(IP \divides \left\langle{ \alpha }\right\rangle \left\langle{ \beta }\right\rangle\).
Note that without the \(\alpha\) this would be false, so we're
critically using that \(\beta\) is in \(I\) but not \(IP\):
\(IP\not\divides \left\langle{ \beta }\right\rangle\) since
\(\beta \not\in IP\). So \(IP\) divides this product but not \(\beta\)
while \(I\) \emph{does} divide \(\beta\), this forces
\(P\divides m \left\langle{ \alpha }\right\rangle\). Then
\(\alpha\in P\) and \(\alpha\operatorname{mod}P = 0\), so
\(\ker \psi = 0\).

\hfill\break

\textbf{Surjectivity:} We might want to write
\(\Im(\psi) = \left\langle{ \beta }\right\rangle / IP\), but this
doesn't quite make sense since \(IP\) may not be a subgroup. This can be
fixed,
\(\operatorname{im}\psi = ( \left\langle{ \beta }\right\rangle + IP) / IP\).
But this equals \(\gcd( \left\langle{ \beta }\right\rangle, IP ) / IP\),
and this numerator is \(I\) since \(\beta\in I\) and
\(\beta\not\in IP\). So we have
\begin{align*}
\operatorname{im}(\psi) 
&= {\left\langle{ \beta }\right\rangle + IP \over  IP } \\
&= { \gcd( \left\langle{ \beta }\right\rangle, IP ) \over IP } \\
&= I/IP
.\end{align*}

\end{proof}

\begin{remark}

For quadratic fields, we could compute ideal norms by multiplying an
ideal \(I\) by its conjugate
\(\mkern 1.5mu\overline{\mkern-1.5muI\mkern-1.5mu}\mkern 1.5mu\) to get
a principal ideal generated by \(N(I)\). Here we don't know what
conjugates mean yet for a general number field -- one could try applying
all of the embeddings into \({\mathbb{C}}\) and taking a product, but
this may not yield an ideal in the same ring again. In particular, if
\(K\) isn't Galois, the embedding can land outside of \(K\) in
\({\mathbb{C}}\).

\end{remark}

\begin{definition}[Extending Ideals]

For \(R \subseteq S\) and \(I{~\trianglelefteq~}R\), define \(IS\) to be
the smallest ideal of \(S\) containing \(I\) (i.e.~take all
intersections), or equivalently take all finite \(S{\hbox{-}}\)linear
combinations of elements from \(I\).

\end{definition}

\begin{exercise}[Arithmetic of ideals]

Check that

\begin{itemize}
\tightlist
\item
  \((IJ)S = (IS)(JS)\),
\item
  \((\alpha R)S = \alpha S\).
\end{itemize}

\end{exercise}

\begin{theorem}[Norm is generated by product of conjugates]

Let \(I\in \operatorname{Id}({\mathbb{Z}}_K)\) and let \(L\) be the
Galois closure of \(K/{\mathbb{Q}}\). For each
\(\sigma: K\hookrightarrow{\mathbb{C}}\), the image \(\sigma(I)\) is an
ideal of \({\mathbb{Z}}_{\sigma(K)} \subseteq {\mathbb{Z}}_L\). Then
\begin{align*}
\prod_{\sigma: K\hookrightarrow{\mathbb{C}}} \sigma(I) {\mathbb{Z}}_L = N(I) {\mathbb{Z}}_L
.\end{align*}

\end{theorem}

\begin{remark}

This shows why the norm is multiplicative, and why
\(N( \left\langle{ \alpha }\right\rangle) = |N( \alpha ) |\).

\end{remark}

\hypertarget{applications-of-finiteness-of-class-group}{%
\subsection{Applications of Finiteness of Class
Group}\label{applications-of-finiteness-of-class-group}}

\begin{question}

Where do ideals come from?

\end{question}

\begin{remark}

They're meant to generalize multiples of an integer in \({\mathbb{Z}}\),
but not all ideals in a general number field are principal. However,
there is a way in which this is true for \({\mathbb{Z}}_K\) even when
it's not a PID.

\end{remark}

\begin{theorem}[Dedekind's theorem on the actuality of ideals]

Let \(K\) be a number field and
\(I{~\trianglelefteq~}\operatorname{Id}({\mathbb{Z}}_K)\). Then there is
a
\(\beta\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
such that
\(I = \beta \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu \cap K\),
or equivalently
\(I = \beta \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu \cap{\mathbb{Z}}_K\).

\end{theorem}

\begin{remark}

Example of a non-principal ideal: in \({\mathbb{Z}}[\sqrt{-5}]\), the
ideal \(I \coloneqq\left\langle{ 2, 1 + \sqrt{-5} }\right\rangle\) is
not principal, i.e.~not all such elements are given by multiples of some
element in \({\mathbb{Z}}[ \sqrt{-5} ]\). It turns out that instead this
is all multiples (in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu)\)
of \(\sqrt{2}\). So anything that's a multiple of \(\sqrt{2}\)
\emph{and} an algebraic integer that's in \({\mathbb{Z}}[\sqrt{-5}]\)
will be in \(I\) and vice-versa. So ideals are multiples of a single
element, provided you allow that element to be outside of
\({\mathbb{Z}}_K\) and in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
instead.

\end{remark}

\begin{lemma}[Ideals become principal after extending]

Let \(K\) be a number field and
\(I\in \operatorname{Id}({\mathbb{Z}}_K)\). Then there is a finite
extension \(L/K\) in which \(I{\mathbb{Z}}_L\) is principal. So any
ideal can be made principal after passing to some finite extension.

\end{lemma}

\begin{proof}[of lemma]

Let \(m \coloneqq\# { \operatorname{Cl}} ({\mathbb{Z}}_K)\). Then
\(I^m = \alpha{\mathbb{Z}}_K\) is principal since \(m\) is the order of
this group. Let \(\beta\coloneqq\sqrt[m]{\alpha} \in {\mathbb{C}}\) and
let \(L \coloneqq K( \beta)\). Here \(\beta\) is an algebraic integer
since it's an \(m\)th root of an algebraic integer. The claim is that
\(I{\mathbb{Z}}_L\) is principal. We have
\begin{align*}
(I {\mathbb{Z}}_L)^m
&= I^m {\mathbb{Z}}_L \\
&= ( \alpha{\mathbb{Z}}_K) {\mathbb{Z}}_L \\
&= \alpha{\mathbb{Z}}_L \\
&= \beta^m {\mathbb{Z}}_L \\
&= (\beta {\mathbb{Z}}_L)^m
.\end{align*}
But how can two ideals have the same \(m\)th power? By unique
factorization, they must be the same, so
\(I{\mathbb{Z}}_L = \beta {\mathbb{Z}}_L\).

\begin{quote}
To be continued.
\end{quote}

\end{proof}

\hypertarget{ch.-16-continued-thursday-may-13}{%
\section{Ch. 16, Continued (Thursday, May
13)}\label{ch.-16-continued-thursday-may-13}}

\hypertarget{actuality-of-ideals}{%
\subsection{Actuality of Ideals}\label{actuality-of-ideals}}

\begin{theorem}[Dedekind, Actuality of ideals]

If \(I\) is a nonzero ideal of \({\mathbb{Z}}_K\) for \(K\) a number
field, then there is an
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
such that
\begin{align*}
I = (\alpha \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu) \cap{\mathbb{Z}}_K
.\end{align*}

\end{theorem}

\begin{lemma}[?]

Let \(I\) be a nonzero ideal of \({\mathbb{Z}}_K\), then there is an
extension of number fields \(L/K\) such that \(I {\mathbb{Z}}_L\) is
principal ,say \(I {\mathbb{Z}}_L = \alpha{\mathbb{Z}}_L\).

\end{lemma}

\begin{proof}[Idea]

We've already proved this, but the idea was that we can take \(m\) to be
the order of the class group to obtain \(I^m = \beta {\mathbb{Z}}_K\).
Then if \(L = K( \beta^{1/m} )\) we will have
\(I{\mathbb{Z}}_L = \beta^{1/m} {\mathbb{Z}}_L\).

\end{proof}

\begin{remark}

How we'll use this to prove the theorem: the lemma shows that after
passing to a suitable extension we can find \(\alpha\), and the claim is
that this \(\alpha\) works. We'll need one more result:

\end{remark}

\begin{lemma}[Paul's up-down lemma]

Let \(K\) be a number field and \(I {~\trianglelefteq~}{\mathbb{Z}}_K\)
be a nonzero ideal. Let
\(R \leq \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
be a subring containing \({\mathbb{Z}}_K\), then
\begin{align*}
IR \cap{\mathbb{Z}}_K = I {~\trianglelefteq~}{\mathbb{Z}}_K
.\end{align*}

The picture: we can lift ideals from \({\mathbb{Z}}_K\) to \(R\) and
intersect them with \({\mathbb{Z}}_K\) to come back down, and the claim
is that this lands where you started:

\begin{center}
\begin{tikzcd}
    {\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu} \\
    \\
    R && IR \\
    \\
    {{\mathbb{Z}}_K} && I && {IR \cap{\mathbb{Z}}_K}
    \arrow[hook, from=5-1, to=3-1]
    \arrow[hook, from=3-1, to=1-1]
    \arrow["{({-}) \cdot R}", from=5-3, to=3-3]
    \arrow["{{-}\cap{\mathbb{Z}}_K}", from=3-3, to=5-5]
    \arrow[color={rgb,255:red,92;green,214;blue,214}, Rightarrow, no head, from=5-3, to=5-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCw0LCJcXFpaX0siXSxbMCwyLCJSIl0sWzAsMCwiXFxiYXJ7XFxaWn0iXSxbMiw0LCJJIl0sWzIsMiwiSVIiXSxbNCw0LCJJUiBcXGludGVyc2VjdCBcXFpaX0siXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsNCwiKFxcd2FpdCkgXFxjZG90IFIiXSxbNCw1LCJcXHdhaXQgXFxpbnRlcnNlY3QgXFxaWl9LIl0sWzMsNSwiIiwyLHsibGV2ZWwiOjIsImNvbG91ciI6WzE4MCw2MCw2MF0sInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=}{Link
to Diagram}
\end{quote}

\end{lemma}

\begin{proof}[?]

Follow your nose! The containment \(I \subseteq IR \cap{\mathbb{Z}}_K\)
is clear, since \(I\) is contained in both terms. For the reverse
containment, we'll first do it for \(I\) principal, and then reduce to
that case. So suppose \(I = \alpha {\mathbb{Z}}_K\), then
\(IR = \alpha R\). Take \(\beta \in IR \cap{\mathbb{Z}}_K\), then
\(\beta\over \alpha \in R \cap K \subseteq \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap K = {\mathbb{Z}}_K\).
Then \(\beta \in \alpha {\mathbb{Z}}_K\), which is what we wanted to
show.\\

We'll reduce to the principal case by using a familiar trick. Pick
\(m\in {\mathbb{Z}}^{\geq 0}\) so that \(I^m\) is principal, then
\(I^m R \cap{\mathbb{Z}}_K = I^m\) by the previous case. Taking
\(\gamma\in IR \cap{\mathbb{Z}}_K\), we'll show it must be in \(I\). We
know \(\gamma^m \in (IR)^m = I^m R\), and thus we have
\(\gamma^m \in I^m R \cap{\mathbb{Z}}_K = I^m\). Now using unique
factorization and to contain is to divide, this implies that
\(I^m \divides \left\langle{ \gamma ^m }\right\rangle= \left\langle{ \gamma }\right\rangle^m\).
So \(I \divides \left\langle{ \gamma }\right\rangle\) and thus
\(\gamma \in I\).

\end{proof}

\begin{proof}[of theorem]

Choose an extension \(L/K\) with
\(I{\mathbb{Z}}_L = \alpha{\mathbb{Z}}_M\) principal. consider
\(I\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
We have
\(I\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu= (I{\mathbb{Z}}_L)\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
-- this just says that extending \(I\) to
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
can be done in two steps, first extending to \({\mathbb{Z}}_L\).
Concretely, these extensions are linear combinations of elements in
\(I\) with coefficients in \({\mathbb{Z}}_L\), so extending in stages
yields the same thing. We have
\begin{align*}
I\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu
&= (I{\mathbb{Z}}_L) \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\\
&= ( \alpha {\mathbb{Z}}_L) \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\\
&= \alpha\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu
.\end{align*}
Now apply the up-down lemma to
\(R = \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
to obtain
\begin{align*}
I = I\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap{\mathbb{Z}}_K
,\end{align*}
which is equal to
\(\alpha\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\cap{\mathbb{Z}}_K\).

\end{proof}

\begin{remark}

Is is true that in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
that every ideal is principal? The answer is no, since this would force
it to be a UFD, but
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
has irreducibles at all since there are no irreducibles. One \emph{can}
prove that every finitely generated ideal of
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
is principal. One can also prove a variant of Gauss' lemma for
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\):
one can take the greatest common divisor to be the generator of a
principal ideal.

\end{remark}

\hypertarget{ch.-17-prime-decomposition-and-general-number-rings}{%
\section{Ch. 17: Prime Decomposition and General Number
Rings}\label{ch.-17-prime-decomposition-and-general-number-rings}}

\begin{definition}[lies above]

If \(P{~\trianglelefteq~}{\mathbb{Z}}_K\) is nonzero, then \(P\)
\textbf{lies above} a rational prime \(p\in {\mathbb{Z}}\) if
\(p\in P\). We also say \(p\) \textbf{lies below} \(P\).

The picture:

\begin{center}
\begin{tikzcd}
    K && {{\mathbb{Z}}_K} && P \\
    \\
    {\mathbb{Q}}&& {\mathbb{Z}}&& p
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=1-3, to=3-3]
    \arrow[no head, from=1-5, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJLIl0sWzAsMiwiXFxRUSJdLFsyLDAsIlxcWlpfSyJdLFsyLDIsIlxcWloiXSxbNCwwLCJQIl0sWzQsMiwicCJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzQsNSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{remark}

As in the quadratic field setting, every \(P\) lies above some uniquely
determined \(p\) in the sense that
\(P \cap{\mathbb{Z}}= p{\mathbb{Z}}\). The proof here goes through in
the same way. Note that since
\(p\in P \implies \left\langle{ p }\right\rangle \in P\), we have
\(\left\langle{ p }\right\rangle\divides P\). How do rational primes
decomposes in \({\mathbb{Z}}_K\)? I.e., how does
\(\left\langle{ p }\right\rangle\) factor into prime ideals?

\end{remark}

\begin{remark}

There are two main theorems: one describes how these prime ideals
factor, and the second (Dedekind-Kummer) will produce the factorization.

\end{remark}

\begin{theorem}[efg theorem]

Let \(K\) be a number field of degree \([K: {\mathbb{Q}}] = n\) and let
\(p\) be a rational prime. Factor
\begin{align*}
\left\langle{ p }\right\rangle = \prod_{i=1}^g P_ji{e_i}
,\end{align*}
where the \(P_i\) are distinct primes ideals and \(e_i > 0\). Then for
each \(i\) we have \(N(P_i) = p^{f_i}\) for some \(f_i > 0\), and
\begin{align*}
\sum_{i=1}^g e_i f_i = n
.\end{align*}

Before the proof, it's helpful to introduce some notation.

\end{theorem}

\begin{definition}[Residual degree and ramification index]

In the setup above, call \(f_i = f(P_i / p)\) the \textbf{residual
degree} of \(P_i/p\) and \(e_i = e(P_i/p)\) the \textbf{ramification
index}.

\end{definition}

\begin{proof}[of theorem]

Consider the homomorphisms
\begin{align*}
\phi: {\mathbb{Z}}&\to {\mathbb{Z}}_K/P_i \\
a &\mapsto a \operatorname{mod}{P_i}
.\end{align*}
We have \(\ker \varphi = P_i \cap{\mathbb{Z}}\), and since \(P\) lies
over \(p\), this equals \(p{\mathbb{Z}}\). So there is an induced
injection
\({\mathbb{F}}_p \coloneqq{\mathbb{Z}}/p{\mathbb{Z}}\hookrightarrow{\mathbb{Z}}_K/ P_i\),
where \(a\operatorname{mod}p \mapsto a \operatorname{mod}P_i\). Thus
\({\mathbb{Z}}_K/P_i \in { \mathsf{Vect} }_{{\mathbb{F}}_p}\) is an
\({\mathbb{F}}_p{\hbox{-}}\)vector space. We want to show the size of
the right-hand side, which is the size of \(P_i\), is a power of \(p\).
But any finite-dimensional vector space over \({\mathbb{F}}_p\) has
dimension \(p^\ell\) for some \(\ell\). Let
\(f_i \coloneqq\dim_{{\mathbb{F}}_p} ({\mathbb{Z}}_K/P_i)\), then
\(N(P_i) = \# {\mathbb{Z}}_K/P_i = p^{f_i}\).\\

Now proving that \(\sum e_i f_i = n\) is an easy corollary: go back to
the factorization and take norms of both sides. Noting
\(N( \left\langle{ p }\right\rangle ) = p^n\) since
\({\mathbb{Z}}_K \in {\mathsf{Ab}}_n^{\mathsf{Free}}\), we have
\begin{align*}
p^n 
&= N( \left\langle{ p }\right\rangle  ) \\
&= \prod_{i=1}^g  N(P_i)^{\ell_i} \\
&= \prod_{i=1}^g p^{e_i f_i} \\
&= p^{\sum_i e_i f_i}
,\end{align*}
and the result follows by comparing exponents.

\end{proof}

\begin{observation}

Each \({\mathbb{Z}}_K/P_i\) is a field. Since
\(P_i \in \operatorname{Spec}{\mathbb{Z}}_K\), so \({\mathbb{Z}}_K/P_i\)
is an integral domain. It's finite by the above argument, using that
norms are finite, so it's in fact a field. It contains
\({\mathbb{Z}}/p{\mathbb{Z}}\), and is thus an extension over it, and
the extension degree
\([{\mathbb{Z}}_K/P_i: {\mathbb{Z}}/p{\mathbb{Z}}] = f_i\). So \(f_i\)
are called the residual degrees since they're the degrees of extensions
of residue fields.

\end{observation}

\begin{definition}[Inert, split, ramified]

More terminology:

\begin{itemize}
\tightlist
\item
  If \(\left\langle{ p }\right\rangle\) is prime, the \(p\) is
  \textbf{inert}.
\item
  If \(g=n\), which forces \(e_i = f_i = 1\) for all \(i\), then \(p\)
  \textbf{splits completely}.
\item
  If \(e_i > 1\) for any \(i\), then \(p\) \textbf{ramifies}.
\end{itemize}

\end{definition}

\begin{remark}

How do we actually determine the prime factors?

\end{remark}

\begin{theorem}[Dedekind-Kummer]

Let \(K\) be a number field of degree \(n\) and suppose
\(K = {\mathbb{Q}}(\alpha)\) where
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
This can be done by using the primitive element and scaling by an
integer. Let \(p\) be a rational prime, and suppose
\(p\notdivides [ {\mathbb{Z}}_K: {\mathbb{Z}}[\alpha] ]\). Then the
prime ideal factorization of \(\left\langle{ p }\right\rangle\) mirrors
the factorization of \(\min_{\alpha}(x) \operatorname{mod}p\).\\

More precisely, supposing
\begin{align*}
\min_{ \alpha}(x) = \prod_{i=1}^g p_i(x)^{e_i} \operatorname{mod}p && p_i \in {\mathbb{Z}}[x]
,\end{align*}
where the \(p_i\) are the irreducible factors of \(\min_{ \alpha}(x)\)
in \({\mathbb{F}}_p[x]\). Then
\begin{align*}
\left\langle{ p }\right\rangle = \prod_{i=1}^g P_i ^{e_i} && P_i \coloneqq\left\langle{ p, p_i( \alpha) }\right\rangle 
.\end{align*}
These \(P_i \in \operatorname{Spec}{\mathbb{Z}}_K\) are distinct, and
the residual degrees are given by \(f_i = \deg(p_i)\).

\end{theorem}

\begin{remark}

So as long as this condition holds, we factor \(P\) by factoring a
minimal polynomial mod \(p\). Note that if one can find an
\(\alpha\in {\mathbb{Z}}_K\), the condition holds vacuously since the
index is 1. This always holds for quadratic fields, since
\({\mathbb{Z}}_K = {\mathbb{Z}}[\tau]\), so this recovers the
factorization statement in that setting. In an annoying twist of fate,
not every number field can be written as \({\mathbb{Z}}[ \alpha]\) for
some single \(\alpha\), and so this theorem necessarily excludes some
primes. Is there some easy way to check the divisibility condition? The
answer is yes, coming from the discriminant.

\end{remark}

\begin{proposition}[?]

Suppose \(K = {\mathbb{Q}}( \alpha)\) where
\(\alpha\in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
If \(p^2 \notdivides {\Delta}(\min_{\alpha} (x) )\), then
\(p\notdivides [{\mathbb{Z}}_K: {\mathbb{Z}}[\alpha]]\).

\end{proposition}

\begin{remark}

A while ago we showed that
\begin{align*}
{\Delta}( \min_{ \alpha}(x)) = {\Delta}(1, \alpha, \alpha^2, \cdots , \alpha^{n-1}) = {\Delta}_K \cdot [{\mathbb{Z}}_K: {\mathbb{Z}}[ \alpha] ]^2
.\end{align*}
Note that we're done if this is true: if \(p\) divides the index,
\(p^2\) divides the last term, meaning \(p^2\) divides the first.

\end{remark}

\begin{example}[?]

Let \(K = {\mathbb{Q}}( \alpha)\) where \(\alpha\) is a root of the
irreducible polynomial \(f(x) = x^3 - x^2 - 2x + 8\). Recall that we
used this example when looking at integral bases. By a computation,
\({\Delta}(f(x)) = -4 \cdot 503\), and the only squared prime dividing
this is \(p=2\). We can thus apply Dedekind-Kummer for all \(p\neq 2\).

\begin{itemize}
\item
  Mod 3, \(f\) is irreducible. So \(\left\langle{ 3 }\right\rangle\) is
  prime and \(3\) is inert.
\item
  Mod 5, \(f(x) \equiv (x+1)(x^2 - 2)\). So
  \(\left\langle{ 5 }\right\rangle= P_1 P_2\) where
  \(P_1 = \left\langle{ 5, \alpha+ 1 }\right\rangle\) and
  \(P_2 = \left\langle{ 5, \alpha^2 - 2 }\right\rangle\). Thus
  \(f_1 = \deg(x+1) = 1, f_2 = \deg(x^2 - 2) = 2\).
\item
  Mod 59, \(f(x) \equiv (x+11)(x+20)(x+25)\). So
  \(\left\langle{ 59 }\right\rangle = P_1 P_2 P_3\) where
  e.g.~\(P_1 = \left\langle{ 5, \alpha+ 11 }\right\rangle\) and so on.
  Note that \(g=3=n\), so \(59\) splits completely.
\item
  Mod 503, \(f(x) \equiv (x+259)(x+354)^2\). So
  \(\left\langle{ 503 }\right\rangle= P_1 P_2^2\), and thus \(503\)
  ramifies since an exponent is larger than 1.
\end{itemize}

Note that \({\Delta}(f(x)) = 0 \operatorname{mod}503\), so
\(f\in {\mathbb{F}}_{503}[x]\) has a repeated root in an extension. We
can run this backward to show that \(503\) is the only odd prime that
ramifies: Dedekind-Kummer says this can only happen if \(f\) has a
repeated root mod \(p\). In this case,
\({\Delta}(f) \operatorname{mod}p = 0\), so \(p\divides 4\cdot 503\).
Note that if you try to apply this theorem mod 2, this results in the
wrong answer!

\end{example}

\begin{lemma}[?]

Let \(K = {\mathbb{Q}}( \alpha)\) be a number field where
\(\alpha \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
Suppose \(p\notdivides [{\mathbb{Z}}_K: {\mathbb{Z}}[ \alpha] ]\). Then
the inclusion
\(\iota: {\mathbb{Z}}[\alpha] \hookrightarrow{\mathbb{Z}}_K\) induces an
isomorphism
\begin{align*}
\mkern 1.5mu\overline{\mkern-1.5mu\iota\mkern-1.5mu}\mkern 1.5mu: {{\mathbb{Z}}[\alpha] \over p {\mathbb{Z}}[ \alpha] } \xrightarrow{\sim} {\mathbb{Z}}_K/p{\mathbb{Z}}_K
.\end{align*}

\end{lemma}

\begin{proof}[?]

Start by showing
\(p {\mathbb{Z}}_K \cap{\mathbb{Z}}[ \alpha] = p {\mathbb{Z}}[ \alpha ]\).
The reverse containment is clear, so let
\(\beta\in p {\mathbb{Z}}_K \cap{\mathbb{Z}}[ \alpha ]\). Then
\(\beta/p \in {\mathbb{Z}}_K\), so look at its image in
\({\mathbb{Z}}_K/ {\mathbb{Z}}[ \alpha]\). The order of the image
divides \(p\), , since
\(p (\beta/p) = \beta \in {\mathbb{Z}}[ \alpha]\). But we assumed that
\(p\) doesn't divide the size of this quotient, forcing the order to be
1 and thus
\(\beta/p\in {\mathbb{Z}}[ \alpha] \implies \beta \in p{\mathbb{Z}}[ \alpha]\).

\end{proof}

\begin{remark}

The map
\(\mkern 1.5mu\overline{\mkern-1.5mu\iota\mkern-1.5mu}\mkern 1.5mu\) is
well-defined because
\(\iota( p {\mathbb{Z}}[ \alpha]) \subseteq p{\mathbb{Z}}_K\), and is
obviously a homomorphism. We want it to be an isomorphism, so what's the
kernel? We've just shown that
\(\ker \mkern 1.5mu\overline{\mkern-1.5mu\iota \mkern-1.5mu}\mkern 1.5mu= 0\),
and comparing cardinality makes it a surjection. Both have cardinality
\(p^n\), since both are free abelian groups of rank \(n\).

\end{remark}

\begin{remark}

This is useful because the Dedekind-Kummer theorem describes the
left-hand side, but we want to study the right-hand side instead. The
theorem says they're isomorphic!

\end{remark}

\hypertarget{ch.-17-dedekind-kummer-thursday-may-13}{%
\section{Ch. 17: Dedekind-Kummer (Thursday, May
13)}\label{ch.-17-dedekind-kummer-thursday-may-13}}

\begin{remark}

Big theorem, Dedekind-Kummer: we have a factorization of \(p\) for all
but finitely many primes \(p\). e supposed that
\(p \notdivides [{\mathbb{Z}}_K: {\mathbb{Z}}[ \alpha] ]\), and there
are only finitely many \(p\) that violate this condition. The claim was
that if \(\min_\alpha(x)\) factors into distinct irreducibles \(p_i(x)\)
in \({\mathbb{F}}_p[x]\), then setting
\(P_i \coloneqq\left\langle{ p, p_i( \alpha) }\right\rangle\). Then
there is a factorization of ideals
\(\left\langle{ p }\right\rangle = \prod P_i^{e_i}\) mirroring the
factorization of the minimal polynomial.

We had a lemma that under these hypotheses, the inclusion
\(\iota: {\mathbb{Z}}[ \alpha] \hookrightarrow{\mathbb{Z}}_K\) induces
an isomorphism on quotients
\begin{align*}
{\mathbb{Z}}[ \alpha] / p {\mathbb{Z}}[ \alpha] \xrightarrow{\sim} {\mathbb{Z}}_K / p{\mathbb{Z}}_K
.\end{align*}
So mod \(p\), these two are the same.

\end{remark}

\begin{proof}[of Dedekind-Kummer]

For notation, set \(m(x) \coloneqq\min_ \alpha(x)\). By assumption, we
have a factorization
\(m(x) = \prod_{i=1}^g p_i(x)^{e_i} \operatorname{mod}p\), and we set
\(P_i \left\langle{ p, p_i( g a) }\right\rangle\). We'll first show that
the \(P_i\) are prime ideals by modding to get a domain, and we'll
compute residue degrees. Considering
\begin{align*}
{\mathbb{Z}}_K/ P_i 
&= {\mathbb{Z}}_K / \left\langle{ p, p_i( \alpha) }\right\rangle \\
&\cong ({\mathbb{Z}}_k / \left\langle{ p }\right\rangle )/ \left\langle{ p_i( \alpha) \operatorname{mod}p }\right\rangle\\
&\cong ({\mathbb{Z}}[ \alpha] / \left\langle{ p }\right\rangle )/ \left\langle{ p_i( \alpha) \operatorname{mod}p }\right\rangle\\
&\cong {\mathbb{Z}}[ \alpha] / \left\langle{ p, p_i( \alpha) }\right\rangle \\
&\cong ( {\mathbb{Z}}[x] / \left\langle{ m(x) }\right\rangle ) / \left\langle{ p \operatorname{mod}m(x), p_i(x) \operatorname{mod}m(x) }\right\rangle \\
&\cong ( {\mathbb{Z}}[ \alpha] / \left\langle{ p }\right\rangle ) / \left\langle{ m(x) \operatorname{mod}p, p_i(x) \operatorname{mod}p }\right\rangle 
&\cong {\mathbb{F}}_p[x] / \left\langle{ p_i(x) \operatorname{mod}p }\right\rangle 
.\end{align*}
where we've used that \(p_i(x) \divides m(x)\) mod \(p\) by assumption.
This is a field of size \(p^{\deg p_i(x)}\), which proves that \(P_i\)
is prime of degree is \(f(P_i / p) = \deg p_i(x)\).

We'll now show the \(P_i\) are distinct, and in fact comaximal in the
sense that \(P_i + P_j = \left\langle{ 1 }\right\rangle\) for
\(i\neq j\). We've assumed that
\(p_i(x) \operatorname{mod}p, p_j(x) \operatorname{mod}p \in {\mathbb{F}}_p[x]\)
are distinct monic irreducibles in the PID \({\mathbb{F}}_p[x]\), so we
can find a linear combination equal to \(1\). So write
\(p_i(x) X(x) + p_j(x) Y(x) = 1 + pQ(x)\) for some polynomials
\(X,Y\in {\mathbb{F}}_p[x]\) and \(Q\in {\mathbb{Z}}[ x]\). Plug in
\(\alpha\) and mod out by
\(I \coloneqq\left\langle{ p, p_i( \alpha), p_j( \alpha) }\right\rangle\)
to get \(o \equiv 1 \operatorname{mod}I\). But then \(1\in I\) forces
\(I = \left\langle{ 1 }\right\rangle\) Thus
\begin{align*}
\left\langle{ 1 }\right\rangle 
&= \left\langle{ p, p_i( \alpha), p_j( \alpha) }\right\rangle \\
&= \left\langle{ p, p_i ( \alpha) }\right\rangle+ \left\langle{ p, p_j( \alpha) }\right\rangle \\
&= P_i + P_j
.\end{align*}

It remains to show that
\(\left\langle{ p }\right\rangle= \prod P_i^{e_i}\). Consider taking
powers of \(P_i\):
\begin{align*}
P_i^2 = \left\langle{ p^2,  p_i( \alpha), p_i(\alpha)^2 }\right\rangle 
\subseteq \left\langle{ p, p_i( \alpha) ^2 }\right\rangle  \\
P_i^3 \subseteq P_i \left\langle{ p_i, p_i (\alpha)^2}\right\rangle \subseteq \left\langle{ p_i, p_I( \alpha)^3 }\right\rangle  
.\end{align*}
Repeating this will show that
\(P_i^{e_i} \subseteq \left\langle{ p, p_i( \alpha)^{e_i} }\right\rangle\).
A similar argument will show
\begin{align*}
\prod P_i^{e_i}
\subseteq 
\prod \left\langle{ p, p_i( \alpha)^{e_i}  }\right\rangle \\\
\subseteq \left\langle{ p, \prod p_i (\alpha) ^{e_i} }\right\rangle 
.\end{align*}
Recall that we can write \(m(x) = \prod p_i(x)^{e_i} + pR(x)\) where
\(R \in {\mathbb{Z}}[x]\). Plugging in \(\alpha\) yields
\(0 \equiv \prod p_i( \alpha)^{e_i} \operatorname{mod}p\), so the
right-hand side is a multiple of \(p\), making the ideal above
redundant. Thus
\(\prod P_i^{e_i} \subseteq \left\langle{ p }\right\rangle\), and since
to contain is to divide, we can write
\begin{align*}
\prod P_i^{e_i} = \left\langle{ p }\right\rangle J
,\end{align*}
and we want to show \(J = \left\langle{ 1 }\right\rangle\).

Strategy: take norms. We know \(N(p_i)^{e_i} = p^{e_i \deg p_i}\), and
so
\begin{align*}
N\qty{ \prod P_i^{e_i} }
&= \prod N(P_i) ^{e_i} \\
&= p^{\sum e_i \deg p_i} \\
&= p^{\deg m(x)} \\
&= p^{[K: {\mathbb{Q}}] }
.\end{align*}

We also have
\(N( \left\langle{ p }\right\rangle ) = \# {\mathbb{Z}}_k / \left\langle{ p }\right\rangle= p^{[K: {\mathbb{Q}}]}\),
which forces \(N(J) = 1\)

\end{proof}

\begin{remark}

An recurring example in this class, due to Dedekind: let
\(K = {\mathbb{Q}}( \alpha)\) where \(\alpha\) is a root of the
irreducible polynomial \(x^3 +x^2 -2x + 8 \in {\mathbb{Q}}[x]\). We saw
that \({\mathbb{Z}}_K \neq {\mathbb{Z}}[ \alpha]\) but their index
divides 2, forcing it to be exactly 2. So the hypothesis of the
Dedkind-Kummer theorem are not satisfied, but is the conclusion still
true? It turns out that the answer is no, and 2 splits completely as
\(\left\langle{ 2 }\right\rangle = P_1 P_2 P_3\). The proof can be done
bare-hands, we won't do it here. This is incompatible with the
conclusion of the theorem: we would need the polynomial to factor into
three monic linear polynomials mod 2. But there are only 2 different
linear polynomials mod 2!

\end{remark}

\begin{remark}

Every ideal in \({\mathbb{Z}}_K\) can be generated by at most 2
elements. Or really, ``\(3/2\)'' elements -- look it up!

\end{remark}

\hypertarget{ch.-18-units-of-mathbbz_k}{%
\section{\texorpdfstring{Ch. 18: Units of
\({\mathbb{Z}}_K\)}{Ch. 18: Units of \{\textbackslash mathbb\{Z\}\}\_K}}\label{ch.-18-units-of-mathbbz_k}}

\begin{remark}

Setting up some notation: let \(K\) be a number field of degree \(n\),
and let

\begin{itemize}
\tightlist
\item
  \(r_1\) be the number of real embeddings, i.e.~their images are
  contained in \({\mathbb{R}}\),
\item
  \(r_2\) be \emph{half} the number of non-real embeddings, since they
  come in pairs by composing with complex conjugation.
\end{itemize}

Note that \(n = r_1 + 2r_2\). Label the real embeddings
\(\sigma_1, \cdots, \sigma_{r_1}\), and
\(\sigma_{r_1+1}, \cdots, \sigma_{r_1 + r_2}\) a set of non-real
embeddings, where we take one such nonreal embedding from each pair.

\end{remark}

\begin{question}

What is the structure of \({\mathbb{Z}}_K^{\times}\)?

\end{question}

\begin{remark}

For imaginary quadratic fields, the units were norm 1 in the ring of
integers. Usually there were just 2, \(\pm 1\), and in
\({\mathbb{Z}}[i]\) there were 4, and \({\mathbb{Q}}( \sqrt{-3} )\)
there were 6. All of these formed cyclic groups.

For real quadratic fields, there was a fundamental unit \(u\) of
infinite order, and all units were \(\pm u\), so the group is abstractly
\({\mathbb{Z}}\times{\mathbb{Z}}/2\). So everything was a root of unity
times a power of the fundamental unit.

Define
\begin{align*}
\mu_K = \left\{{ \zeta\in K {~\mathrel{\Big|}~}\zeta^n = 1 \text{ for some } n\in {\mathbb{Z}}^{> 0} }\right\} 
.\end{align*}
Clearly \(\mu_K \subseteq K^{\times}\), and in fact
\(\mu_k \subseteq \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\).
In fact, it forms a subgroup \(\mu_K \leq {\mathbb{Z}}_k^{\times}\). The
big theorem on the structure of units is the following, which says the
same thing as the real quadratic field case happens in general, just
with more fundamental units.

\end{remark}

\begin{theorem}[Dirichlet's Units Theorem]

There are elements
\({\varepsilon}_1, \cdots, {\varepsilon}_{r_1 + r_2 - 1} \in {\mathbb{Z}}_K^{\times}\)
of infinite order such that
\({\mathbb{Z}}_K^{\times}= \mu_K \prod_{i=1}^{r_1 + r_2 - 1} \left\langle{ {\varepsilon}_i }\right\rangle\)
as an internal direct product. So every unit has a unique decomposition
of this form.

Moreover, \(\# \mu_K < \infty\).

\end{theorem}

\begin{example}[?]

When is \({\mathbb{Z}}_K^{\times}\) finite? This happens iff
\(r_1 + r_2 = 1\), and there are a few cases:

\begin{itemize}
\tightlist
\item
  \(r_1=1, r_2=0\), and since \(n=r_1 + 2r_2 = 1\), this forces
  \(K = {\mathbb{Q}}\).
\item
  \(r_1 = 0, r_2 = 1 \iff n=2\), so this is a quadratic field with no
  real embeddings, so \(K\) is an imaginary quadratic field.
\end{itemize}

\end{example}

\begin{remark}

Our goal will be to prove the units theorem with \emph{some} number
\(g\), and we'll show in the next chapter that \(g = r_1 + r_2 - 1\) is
the right \(g\) to choose.

An outline of the proof:

\begin{itemize}
\item
  Define a homomorphism
  \begin{align*}
  \operatorname{Log}: K^{\times}&\to {\mathbb{R}}^{r_1 + r_2} \\
  \alpha &\mapsto {\left[ { 
  \log{\left\lvert {\sigma_1( \alpha) } \right\rvert}, 
  \cdots,
  \log{\left\lvert {\sigma_{r_1}( \alpha) } \right\rvert}, 
  2\log{\left\lvert {\sigma_{r_1+1}( \alpha) } \right\rvert}, 
  \cdots
  2\log{\left\lvert {\sigma_{r_1+r_2}( \alpha) } \right\rvert}, 
  } \right]}
  .\end{align*}
  We'll mostly consider its restriction to \({\mathbb{Z}}_K^{\times}\).
\item
  Prove
  \(\ker { \left.{{\operatorname{Log}}} \right|_{{{\mathbb{Z}}_K^{\times}}} } = \mu_K\)
  and is finite.
\item
  Prove \(\operatorname{Log}({\mathbb{Z}}_K)\) is a discrete subgroup of
  \({\mathbb{R}}^{r_1 + r_2}\).
\item
  Use the fact that such discrete subgroups are lattices.
\item
  Prove the units theorem with \(g\) defined as the rank of this
  lattice.
\item
  Finally prove the rank \(g\) is equal to \(r_1 + r_2 - 1\).
\end{itemize}

\end{remark}

\begin{lemma}[?]

Let \(M\in {\mathbb{R}}^{>0}\) and let \(\alpha \in {\mathbb{Z}}_K\). If
for all embeddings \(\sigma: K\hookrightarrow{\mathbb{C}}\) we have
bounds \({\left\lvert {\sigma( \alpha)} \right\rvert} \leq M\), then
\(\alpha\) is a root of a polynomial in
\begin{align*}
P_{n,m}
\coloneqq
\left\{{ x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \in {\mathbb{Z}}[x] {~\mathrel{\Big|}~}{\left\lvert {a_i} \right\rvert} \leq {n\choose i} M^{n-i} }\right\} 
.\end{align*}
Note that this is a finite set.

\end{lemma}

\begin{proof}[?]

Note that \(\alpha\) is a root of its \textbf{field polynomial}
\begin{align*}
f(x) \coloneqq\prod_{\sigma: K \hookrightarrow{\mathbb{C}}} (x - \sigma( \alpha))
.\end{align*}
Since \(\alpha \in {\mathbb{Z}}_K\), this polynomial
\(f \in {\mathbb{Z}}[x]\). So it suffices to show that
\(f\in P_{n, m}\). It's degree \(n\), the coefficients are integers, but
why do they satisfying the bound? Expanding the multiplication, there
are \({n\choose i}\) prducts each of which involves \(n-i\) of the
\(\sigma_i( \alpha)\), which is bounded.

\end{proof}

\begin{remark}

We'll first show the kernel of \(\operatorname{Log}\) is finite, then
it'll be easy to see it's \(\mu_K\).

\end{remark}

\begin{proposition}[?]

\begin{align*}
\# \ker { \left.{{ \operatorname{Log}}} \right|_{{{\mathbb{Z}}_K^{\times}}} } < \infty 
.\end{align*}

\end{proposition}

\begin{proof}[?]

Suppose \(\alpha\in {\mathbb{Z}}_K^{\times}\) and
\(\operatorname{Log}(\alpha) = \mathbf{0}\). This says that
\({\left\lvert { \sigma_i( \alpha)} \right\rvert} = 1\) for all
embeddings \(\sigma_i: K\hookrightarrow{\mathbb{C}}\). Applying the
lemma above with \(M=1\), \(\alpha\) is a root of a polynomials in
\(P_{n, 1}\), each of which has at most \(n\) roots. So there are only
finitely many possibilities for \(\alpha\).

\end{proof}

\begin{proposition}[?]

\begin{align*}
\ker { \left.{{ \operatorname{Log}}} \right|_{{{\mathbb{Z}}_K^{\times}}} } = \mu_K
.\end{align*}

\end{proposition}

\begin{proof}[?]

Start with \(\zeta\in \mu_K\), then under any \(\sigma\),
\(\sigma(\zeta)\) is a complex root of unity. So
\({\left\lvert {\sigma(\zeta)} \right\rvert} = 1\), and by the previous
result, \(\operatorname{Log}(\zeta) = \mathbf{0}\). So
\(\mu_K \subseteq \ker \operatorname{Log}\).

Conversely suppose \(\zeta\in \ker \operatorname{Log}\), which in
particular is a group. Suppose \(\# \ker \operatorname{Log}= n\), since
we know it's finite. Then \(\xi^n = e = 1\) in this group, making it an
\(n\)th root of unity.

\end{proof}

\begin{remark}

Note that \(\mu_K\) is always cyclic This follows because for \(K\) a
field, it's an exercise that any finite subgroup
\(H\leq {\mathbb{G}}_m(K)\), it is always cyclic. So we've written
\({\mathbb{Z}}_K = C_q \times {\mathbb{Z}}^{r_1 + r_2 - 1}\) as an
abstract group.

\end{remark}

\hypertarget{ch.-18-dirichlets-units-theorem-part-i-friday-may-21}{%
\section{Ch. 18: Dirichlet's Units Theorem Part I (Friday, May
21)}\label{ch.-18-dirichlets-units-theorem-part-i-friday-may-21}}

\begin{remark}

Some notation introduced last time: for \(K\) a number field,
\([K: {\mathbb{Q}}] = n\) the degree, set
\(\sigma_{1}, \cdots, \sigma_{r_1}\) to be the real embeddings, and
\(\sigma_{r_1 + 1}, \cdots, \sigma_{r_1 + r_2}\) (along with their
complex conjugates) to be the nonreal embeddings into \({\mathbb{C}}\).
Let
\(\mu_K \coloneqq\left\{{ z\in K {~\mathrel{\Big|}~}z^n = 1, n\in {\mathbb{N}}}\right\}\)
be the \(n\)th roots of unity Recall Dirichlet's unit theorem: there are
\({ {{\varepsilon}}_1, {{\varepsilon}}_2, \cdots, {{\varepsilon}}_{r_1+r_2-1}} \in {\mathbb{Z}}_K^{\times}\)
of infinite order such that
\begin{align*}
{\mathbb{Z}}_K^{\times}= \mu_K \times \prod_{i\leq r_1 + r_2 -1} \left\langle{ \mu_i }\right\rangle 
.\end{align*}
Moreover, \(\# \mu_K < \infty\).

\end{remark}

\begin{remark}[An outline of the proof]

Set \(d = r_1 + r_2\). We define a homomorphism
\begin{align*}
\operatorname{Log}: K^{\times}&\to {\mathbb{R}}^{d} \\
\alpha &\mapsto {\left[ { 
\log{\left\lvert {\sigma_i(\alpha)} \right\rvert}, 
\cdots,
\log{\left\lvert {\sigma_{r_1}(\alpha)} \right\rvert}, 
2 \log{\left\lvert {\sigma_{r_1+1}(\alpha)} \right\rvert}, 
\cdots,
2 \log{\left\lvert {\sigma_{r_1+r_2}(\alpha)} \right\rvert}, 
} \right]}
.\end{align*}
Note that we only used half of the non-real embeddings, but made up for
it by including a 2! We'll mostly be concerned with its restriction to
\({\mathbb{Z}}_K^{\times}\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Show that the restriction has finite kernel \(\mu_k\) -- we've
  completed this.
\item
  Show \(\operatorname{Log}({\mathbb{Z}}_K^{\times})\) is discrete as a
  subgroup of \({\mathbb{R}}^{d}\)
\item
  Use the fact that every discrete subgroup of \({\mathbb{R}}^d\) is a
  lattice, the \({\mathbb{Z}}{\hbox{-}}\)span of linearly independent
  vectors.
\item
  Prove the units theorem with \(d-1\) replaced by
  \(g = \operatorname{rank}\operatorname{Log}({\mathbb{Z}}_K^{\times})\).
\item
  Deferred to next chapter: show \(g = r_1 + r_2 - 1\).
\end{enumerate}

\end{remark}

\hypertarget{step-2-discreteness}{%
\subsection{Step 2: Discreteness}\label{step-2-discreteness}}

\begin{definition}[Discrete lattices]

Let \(d\in {\mathbb{Z}}^{\geq 0}\) and let
\(\Lambda \leq {\mathbb{R}}^d\) be a subgroup. Then \(\Lambda\) is
\textbf{discrete} iff for every \(R>0\), \(\Lambda \cap B_R(0)\) is a
finite set, where \(B_R(0)\) is a ball of radius \(R\) about 0.

\end{definition}

\begin{proposition}[?]

\(\operatorname{Log}({\mathbb{Z}}_K^{\times})\) is discrete in
\({\mathbb{R}}^d\) for \(d=r_1 + r_2\).

\end{proposition}

\begin{proof}[?]

Let \(R>0\), we'll show that there are only finitely many
\(\alpha \in {\mathbb{Z}}_K^{\times}\) for which
\(\operatorname{Log}(\alpha) \in B_R(0)\). This forces
\(\log{\left\lvert {\sigma(\alpha)} \right\rvert} \leq R\) for all
\(\sigma: K\hookrightarrow{\mathbb{C}}\). Thus
\({\left\lvert {\sigma(\alpha)} \right\rvert} \leq e^R\) for all such
embeddings. By a previous lemma, this makes \(\alpha\) a root of a
polynomial in some finite set \(P_{n,e^R}\) -- what this set is isn't
important, just recall that once you have a bound on the embeddings,
this gives a bound on the coefficients of the field polynomial.

\end{proof}

\hypertarget{step-3}{%
\subsection{Step 3}\label{step-3}}

\begin{theorem}[?]

Every discrete subgroup of \({\mathbb{R}}^d\) is a lattice.

\end{theorem}

\begin{proof}[?]

See the book!

\end{proof}

\hypertarget{step-3-1}{%
\subsection{Step 3}\label{step-3-1}}

\begin{theorem}[Weak units theorem]

Let \(g\) be the rank of
\(\operatorname{Log}{\mathbb{Z}}_K^{\times}\subseteq {\mathbb{R}}^{r_1 + r_2}\).
Then there are fundamental units
\({ {{\varepsilon}}_1, {{\varepsilon}}_2, \cdots, {{\varepsilon}}_{g}} \in {\mathbb{Z}}_K^{\times}\)
of infinite order such that
\begin{align*}
{\mathbb{Z}}_K^{\times}= \mu_K \times \prod_{i\leq g} \left\langle{ {\varepsilon}_i }\right\rangle 
.\end{align*}
Note that this is the same theorem from before, just with \(d\) replaced
by \(g\).

\end{theorem}

\begin{proof}[?]

Using that \(g\) is by definition the rank of the lattice
\(\operatorname{Log}{\mathbb{Z}}\), choose
\({ {{\varepsilon}}_1, {{\varepsilon}}_2, \cdots, {{\varepsilon}}_{g}}\in {\mathbb{Z}}_K^{\times}\)
such that the \(\operatorname{Log}({\varepsilon}_i)\) are
\({\mathbb{R}}{\hbox{-}}\)linearly independent and their
\({\mathbb{Z}}{\hbox{-}}\)span is the lattice. Let
\({\varepsilon}\in {\mathbb{Z}}_K^{\times}\) be any unit, then
\(\operatorname{Log}({\varepsilon}) \in \operatorname{Log}{\mathbb{Z}}_K^{\times}\),
so there are integers
\({ {n}_1, {n}_2, \cdots, {n}_{g}}\in {\mathbb{Z}}\) such that
\begin{align*}
\operatorname{Log}({\varepsilon}) = \sum n_i \operatorname{Log}({\varepsilon}_i)
.\end{align*}
Hence
\begin{align*}
\operatorname{Log}\qty{ {\varepsilon}/ \prod_{i\leq g} {\varepsilon}_i^{n_i} } = \mathbf{0}
.\end{align*}
So it's in the kernel and equal to some \(\zeta\in \mu_K\), so we can
write \({\varepsilon}= \zeta \prod_{i\leq g} {\varepsilon}_i^{n_i}\).
This representation of \({\varepsilon}\) is unique, in the sense that
\(\zeta\) and the \(n_i\) are uniquely determined. Why? For any other
representation, applying \(\operatorname{Log}\) would kill the \(\zeta\)
part and write
\(\operatorname{Log}({\varepsilon}) = \sum n_{j} {\varepsilon}_j\) in
the basis \({\varepsilon}_j\). So the \(n_i\) are unique, and \(\zeta\)
is just obtained by dividing, so it can be recovered uniquely as well.

\end{proof}

\begin{remark}

Our goal is to now show \(g=r_1 + r_2-1\), which is hard. It's easier to
obtain a bound instead.

\end{remark}

\begin{proposition}[Lattice rank bound]

Let
\(g = \operatorname{rank}\operatorname{Log}({\mathbb{Z}}_K^{\times})\),
then
\begin{align*}
g\leq r_1 + r_2 - 1
.\end{align*}

\end{proposition}

\begin{lemma}[?]

Let \(\Lambda \subseteq {\mathbb{R}}^d\) be a lattice, then
\(\operatorname{rank}\Lambda = \dim_{\mathbb{R}}V\) where \(V\) is the
smallest subspace of \({\mathbb{R}}^d\) containing \(\Lambda\).

\end{lemma}

\begin{proof}[?]

Let \(g\coloneqq\operatorname{rank}\Lambda\), and write
\(\Lambda= \sum_{i\leq g} {\mathbb{Z}}\mathbf{v}_i\) where the
\(\mathbf{v}_i\) are \({\mathbb{R}}{\hbox{-}}\)linearly independent.
Then \(V = \sum_{i\leq g} {\mathbb{R}}\mathbf{v}_i\), and
\(\dim_{\mathbb{R}}V = g = \operatorname{rank}\Lambda\).

\end{proof}

\begin{proof}[of proposition, lattice rank bound]

Let
\(\mathbf{w} = {\left[ {1, 1, \cdots, 1} \right]}\in {\mathbb{R}}^{r_1 + r_2}\).
If \(\alpha\in {\mathbb{Z}}_K^{\times}\), then dotting sums the
components, so
\begin{align*}
\mathbf{w} \cdot \operatorname{Log}\alpha = \sum_{i\leq r_1} \log {\left\lvert {\sigma_i( \alpha)} \right\rvert}
+ 2\sum_{r_1 + 1 \leq i \leq r_1 + r_2} \log {\left\lvert {\sigma_i( \alpha) } \right\rvert}
.\end{align*}
Note that
\({\left\lvert {\sigma_i(\alpha)} \right\rvert} = {\left\lvert {\mkern 1.5mu\overline{\mkern-1.5mu\sigma( \alpha)\mkern-1.5mu}\mkern 1.5mu} \right\rvert}\),
so because we have a 2 here, we could have written this as
\begin{align*}
\mathbf{w} \cdot \operatorname{Log}\alpha 
&= \sum_{\sigma:K\hookrightarrow{\mathbb{C}}} \log {\left\lvert {\sigma(\alpha)} \right\rvert}\\
&= \log\prod_{\sigma: K\hookrightarrow{\mathbb{C}}} {\left\lvert {\sigma( \alpha)} \right\rvert} \\
&= \log {\left\lvert { N ( \alpha)} \right\rvert} \\
&= 0
,\end{align*}
since \(\alpha\in {\mathbb{Z}}_K^{\times}\implies N( \alpha) = \pm 1\).
This says that
\(\operatorname{Log}({\mathbb{Z}}_K^{\times}) \subseteq \mathbf{w}^\perp\),
which has codimension \(1\) and thus dimension \(r_1 + r_2 - 1\). Now
\(V \subseteq \mathbf{w} ^\perp\). so its rank is bounded by this
dimension. So \(g\leq r_1 + r_2 - 1\).

\end{proof}

\hypertarget{ch.-20-unit-theorem-part-ii}{%
\section{Ch. 20: Unit Theorem, Part
II}\label{ch.-20-unit-theorem-part-ii}}

\begin{remark}

We still need to prove that this is an equality. We'll start with some
geometric preliminaries.

\end{remark}

\begin{definition}[The Minkowski Embedding]

The \textbf{Minkowski embedding} of \(K\) is the map
\begin{align*}
\iota: K &\to {\mathbb{R}}^{r_1} \times {\mathbb{C}}^{r_2} \\
\alpha &\mapsto {\left[ { 
\sigma_1(\alpha), \cdots, \sigma_{r_1}( \alpha),
\sigma_{r_1 + 1}( \alpha), \cdots, \sigma_{r_1 + r_2}( \alpha)
} \right]}
.\end{align*}
Identify \({\mathbb{C}}\cong {\mathbb{R}}^2\) using
\(a+bi \mapsto {\left[ {a, b} \right]}\), which identifies the codomain
above with
\({\mathbb{R}}^{r_1} \times {\mathbb{R}}^{2r_2} = {\mathbb{R}}^n\). So
we view \(\iota:K\to {\mathbb{R}}^n\).

\end{definition}

\begin{remark}

Note \(\iota\) is injective and \({\mathbb{Q}}{\hbox{-}}\)linear.

\end{remark}

\begin{proposition}[Extremely important]

\(\iota({\mathbb{Z}}_K)\) is a lattice of full rank in
\({\mathbb{R}}^n\).

\end{proposition}

\begin{proof}[?]

Why is this a lattice? We'll show it's a discrete subgroup.

Let \(R>0\), then if \(\alpha\in {\mathbb{Z}}_K\) and
\(\iota(\alpha) \in B_R(0)\), then
\({\left\lvert {\sigma( \alpha)} \right\rvert} \leq R\) for all
\(\sigma: K\hookrightarrow{\mathbb{C}}\). This forces \(\alpha\) to be a
root of one of finitely many polynomials in \(P_{n, R}\).

Now let \(g\coloneqq\operatorname{rank}\iota({\mathbb{Z}}_K)\) as a
lattice. This is a free abelian group of rank \(g\), so isomorphic to
\({\mathbb{Z}}^g\) since it's the \({\mathbb{Z}}{\hbox{-}}\)span of
\(g\) linearly independent elements. On the other hand, by the integral
basis theorem, \({\mathbb{Z}}_K\) is a free abelian group of rank \(n\).
Since \(\iota\) is injective, \(\iota({\mathbb{Z}}_K)\) is free of rank
\(n\), and a fact from algebra implies \(n=g\).

\end{proof}

\begin{remark}

Since \(\iota({\mathbb{Z}}_K\) is a full rank lattice,
\(\operatorname{covol}\iota({\mathbb{Z}}_K)\) is well-defined, where the
covolume is the volume of the fundamental parallelepiped spanned by any
generating set, and is measured by the absolute value of the determinant
of the matrix of basis elements. It turns out that the covolume is the
discriminant from before, but we don't need the exact value at the
moment.

\end{remark}

\begin{remark}

We'll now make some further reductions. Recall that
\(g\coloneqq\operatorname{rank}\operatorname{Log}({\mathbb{Z}}_K^{\times}) = \dim_{\mathbb{R}}V\),
where \(V\) is the smallest subspace of \({\mathbb{R}}^{r_1 + r_2}\)
containing \(\operatorname{Log}({\mathbb{Z}}_K^{\times})\). Recall the
rank-nullity theorem:
\begin{align*}
\dim_{\mathbb{R}}V + \dim_{\mathbb{R}}V^\perp = r_1 + r_2
\implies g = \dim_{\mathbb{R}}V = r_1 + r_2 - \dim_{\mathbb{R}}V^\perp
.\end{align*}
the goal is to show \(\dim_{\mathbb{R}}V^\perp = 1\). We know
\(\mathbf{w} \coloneqq{\left[ {1, 1, \cdots, 1} \right]} \in V^\perp\),
so it suffices to show every element of \(V^\perp\) is a multiple of
\(\mathbf{w}\). Suppose not, then there's an element of \(V^\perp\) of
the form \({\left[ {c_1, c_2 \cdots, c_j, 0} \right]}\) by subtracting a
suitable multiple of \(\mathbf{w}\). Not all of the \(c_i\) are zero,
since this would mean the original was a multiple of \(\mathbf{w}\).
Define a map
\begin{align*}
F: K^{\times}&\to {\mathbb{R}}\\
\alpha&\mapsto {\left[ {c_1, c_2, \cdots c_{r_1 + r_2 - 1}, 0 } \right]} \cdot \operatorname{Log}( \alpha)
.\end{align*}
Now note that \(F({\mathbb{Z}}_K^{\times}) = 0\), since
\(V \supseteq \operatorname{Log}({\mathbb{Z}}_K)\).

\end{remark}

\begin{proposition}[?]

Let \(c_1, c_2, \cdots, c_{r_1 + r_2 - 1} \in {\mathbb{R}}\) not all 0.
Then if \(F\) is defined as above,
\(F({\mathbb{Z}}_K^{\times}) \neq 0\).

\end{proposition}

\begin{remark}

This requires some delicate inequalities, and an application of
Minkowski's convex body theorem.

\end{remark}

\begin{lemma}[Key lemma]

There is a sequence of nonzero \(\alpha\in K\) such that

\begin{itemize}
\tightlist
\item
  \({\left\lvert {F(\alpha)} \right\rvert} \to \infty\), but
\item
  \({\left\lvert {N( \alpha)} \right\rvert} < C_1 = C_1(K, c_1, \cdots, c_{r_1 + r_2 - 1} )\)
  not depending on \(\alpha\).
\end{itemize}

\end{lemma}

\begin{proof}[of proposition, assuming the key lemma]

Note that there are only finitely many ideals of \({\mathbb{Z}}_K\) of
norm bounded by any constant \(C_1\). We saw this for quadratic fields:
if \(N(J) = m\), which by Lagrange's theorem yields
\(1+1+\cdots 1 = m \equiv 0 \operatorname{mod}J\). So \(m\in J\) and
thus \(J\divides m\). But by unique factorization, any ideal only has
finitely many divisors, so there are only finitely many possible values
of \(m\).

Note also that
\({\left\lvert {N( \alpha )} \right\rvert} = N(\left\langle{ \alpha }\right\rangle ) < C_1\),
so there are only finitely many such ideals. So given an infinite
sequence of \(\alpha\)s, there are only finitely many different
principal ideals they generate. So by the pigeonhole principle, we can
pass to a subsequence such that \(\left\langle{ a }\right\rangle\) is
constant. Note that \(F\) can't be constant on this sequence, since
\(F(\alpha) \to \infty\), so we can choose \(\alpha_1, \alpha_2\) such
that
\begin{align*}
\left\langle{ \alpha_1 }\right\rangle= \left\langle{ \alpha_2 }\right\rangle \quad\text{but}\quad F(\alpha_1) \neq F(\alpha_2)  
.\end{align*}
If this happens, then
\(\alpha_1/ \alpha_2 \in {\mathbb{Z}}_K^{\times}\), and
\begin{align*}
F(\alpha_1/ \alpha_2) = F( \alpha_1)- F(\alpha_2) \neq 0
.\end{align*}
So we have \(F({\mathbb{Z}}_K^{\times})\neq 0\), since we've found a
unit that maps to a nonzero element.

\end{proof}

\hypertarget{ch.-20-continued-friday-may-21}{%
\section{Ch. 20 Continued (Friday, May
21)}\label{ch.-20-continued-friday-may-21}}

\begin{remark}

Goal of Ch.18: prove that units theorem where the rank is left
unspecified, and we proved it for \(g\) the rank of a certain lattice.
For Ch.20, we want to show \(g\) is the right number! Recall the log
function:
\begin{align*}
\operatorname{Log}: K^{\times}\to {\mathbb{R}}^{r_1 + r_2}
\end{align*}
where we take the \(\log{\left\lvert {{-}} \right\rvert}\) of the real
embeddings and \(2\log{\left\lvert {{-}} \right\rvert}\) of one of each
pair of complex embeddings. We know
\(\operatorname{Log}{\mathbb{Z}}_K^{\times}\subseteq {\mathbb{R}}^{r_1 + r_2}\)
is a lattice of rank \(g \leq r_1 + r_2 - 1\), and we want to show
equality. We showed that everything follows if we establish the key
lemma: if \(c_1, \cdots, c_{r_1 + r_2 - 1} \in {\mathbb{R}}\) are not
all zero, defin
\begin{align*}
F(\alpha) \coloneqq{\left[ {c_1, \cdots, c_{r_1 + r_2 - 1}, 0} \right]} \cdot \operatorname{Log}(\alpha)
.\end{align*}
We want to show there is a sequence of \(\alpha\in {\mathbb{Z}}_K\) such
that \({\left\lvert {F(\alpha)} \right\rvert} \to \infty\) with
\({\left\lvert {N( \alpha)} \right\rvert} \leq C_1\) bounded, where
\(C_1\) doesn't depend on \(\alpha\). We want to make the dot product
above large while keeping the norms small while making linear
combinations of \(\log{\left\lvert {\sigma_i( \alpha)} \right\rvert}\)
large. The only embeddings that show up are the ones labeled up to
\(r_1 + r_2 - 1\), since the last component is zero, and we'll show that
we can essentially control the image of \(\alpha\) under these
embeddings while keeping its norm bounded.

Recall the Minkowski embedding realizes \(K\) as a subspace of Euclidean
space:
\begin{align*}
\iota: K\to {\mathbb{R}}^{r_1} \oplus {\mathbb{C}}^{r_2}
,\end{align*}
where the first components are the first \(r_1\) real embeddings and the
latter are the complex embeddings. Identifying the target with
\({\mathbb{R}}^n\), we said last time that
\(\iota({\mathbb{Z}}_K) \subseteq {\mathbb{R}}^n\) is a full rank
lattice, so of rank \(n\), which thus has a finite nonzero covolume.

We'll use this with Minkowski's convex body theorem: suppose \(R\) is a
region in \({\mathbb{R}}^d\) which is centrally symmetric and convex,
and \(\Lambda\) is a full rank lattice in \({\mathbb{R}}^d\). Then if
\(\operatorname{vol}(R)> w^d \operatorname{covol}(\Lambda)\), \(R\)
contains a nonzero point \(x\in \Lambda\). How we'll use these: apply s
the convex body theorem on \(\Lambda = \iota({\mathbb{Z}}_K)\), and use
this to locate the types of \(\alpha\) we want.

\end{remark}

\begin{remark}

Given
\(\lambda_1, \cdots, \lambda_{r_1 + r_2} \in {\mathbb{R}}^{\geq 0}\),
denote this by the vector \(\mathbf{\lambda}\). Define
\begin{align*}
R_{\mathbf{\lambda}}\coloneqq\left\{{ {\left[ {x_1, \cdots, x_{r_1 + r_2}} \right]} \in {\mathbb{R}}^{r_1} \oplus {\mathbb{C}}^{r_2} {~\mathrel{\Big|}~}{\left\lvert {x_i} \right\rvert} \leq \lambda_i\,\,\forall i  }\right\} \subseteq {\mathbb{R}}^{r_1} \oplus {\mathbb{C}}^{r_2} \cong {\mathbb{R}}^{n}
.\end{align*}
This is clearly centrally symmetric: negating anything lands in the
region since we're taking absolute values. It's less clearly convex, but
this follows from the triangle inequality. More intuitively, it's a
product of discs and boxes, each of which are convex. One can compute
its volume as
\begin{align*}
\operatorname{vol}R_{\mathbf{\lambda}} = \prod_{i\leq r_1} 2\lambda_i \prod_{j\leq r_2} \pi \lambda_{r_1 + j}^2
= 2^{r_1} \pi^{r_2} \prod_{i\leq r_1} \lambda_i \prod_{j\leq r_2} \lambda_{r_1 + j}^2
,\end{align*}
since the first \(r_1\) components form boxes and the last form discs.

As a convention, we'll fill out \(r_1+r_2\) tuples to \(r_1 + 2r_2\)
tuples in the following way: for \(1 \leq i\leq r_2\), set

\begin{itemize}
\tightlist
\item
  \(\sigma_{r_1 + r_2 + i} = \mkern 1.5mu\overline{\mkern-1.5mu\sigma_{r_1 + i}\mkern-1.5mu}\mkern 1.5mu\),
  and
\item
  \(\lambda_{r_1 + r_2 + i} = \lambda_{r_1 + i}\).
\end{itemize}

This introduces the remaining embeddings from each conjugate pair and
assigns it the same \(\lambda_i\). We can now write
\begin{align*}
\operatorname{vol}R_{\mathbf{\lambda}} = 2^{r_1}\pi^{r_2} \prod_{i\leq n} \lambda_i
,\end{align*}
since each \(\lambda\) now appears twice in the product.

\end{remark}

\begin{proof}[of key lemma]

Rather than working with arbitrary tuples, we'll only consider
\({ {\lambda}_1, {\lambda}_2, \cdots, {\lambda}_{r_1 + r_2}} \in {\mathbb{R}}^{\geq 0}\)
contained such that
\begin{align*}
\operatorname{vol}R_{\mathbf{\lambda}} = 2 \qty{ 2^n \operatorname{covol}\iota({\mathbb{Z}}_K) }
,\end{align*}
noting that the parenthesized quantity is exactly twice what's needed in
Minkowski's theorem. This allows one to choose the first \(r_1 + r_2-1\)
freely, and the last will be determined uniquely. By Minkowski's
theorem, there is an \(\alpha\in {\mathbb{Z}}_K\) nonzero with
\(\iota( \alpha)\in R_{\mathbf{\alpha}}\). Hence
\begin{align*}
{\left\lvert { \sigma_i( \alpha)} \right\rvert} \leq \lambda_i && \forall i=1,2,\cdots, n
.\end{align*}
Consequently,
\begin{align*}
{\left\lvert {N \alpha} \right\rvert} 
&= \prod_{i} {\left\lvert {\sigma_i( \alpha)} \right\rvert}\leq \prod \lambda_i \\
&= \frac{ \operatorname{vol}R_{\mathbf{\lambda}}} {2^{r_1} \pi^{r_2} } \\
&= C_1
.\end{align*}
Notice that
\begin{align*}
1 \leq {\left\lvert {N \alpha} \right\rvert} 
&= \prod_i {\left\lvert { \sigma_i( \alpha)} \right\rvert} \\
&= {\left\lvert { \sigma_j( \alpha)} \right\rvert} \prod_{i\neq j} {\left\lvert { \sigma_i( \alpha)} \right\rvert} \\
&\leq {\left\lvert { \sigma_j ( \alpha)} \right\rvert} \prod_{i\neq g} \lambda_i \\
&\leq { {\left\lvert { \sigma_j } \right\rvert} \over \lambda_j } \prod_i \lambda_i \\
&\leq { {\left\lvert { \sigma_j } \right\rvert} \over \lambda_j } C_1 
.\end{align*}

Rearranging yields
\begin{align*}
{\left\lvert { \sigma_j ( \alpha)} \right\rvert} \geq C_1^{-1}\lambda_j
.\end{align*}
Thus for every \(1\leq j \leq n\), we have an inequality
\begin{align*}
C_1\int \lambda_j \leq {\left\lvert { \sigma_j( \alpha)} \right\rvert} \leq \lambda_j
.\end{align*}
Taking logarithms yields
\begin{align*}
\log C_1^{-1}\leq \log {\left\lvert { \sigma_j( \alpha)} \right\rvert} - \log \lambda_j \leq 0
.\end{align*}

Recalling the definition of \(\operatorname{Log}( \alpha)\), and the
above inequality says we know
\(\log{\left\lvert { \sigma_j( \alpha)} \right\rvert}\) up to a constant
-- it's about \(\log \lambda_j\). So setting
\begin{align*}
\mathbf{L} \coloneqq
{\left[ { 
\log \lambda_1, \cdots, \lambda_{r_1}, 2\log \lambda_{r_1 + 1}, 2\log \lambda_{r_1 + r_2}
} \right]}
,\end{align*}
we have \(\mathbf{L} \approx \operatorname{Log}\alpha\), where
\(\approx\) means each component is off by at most a constant. So
\({\left\lvert { \operatorname{Log}\alpha - \mathbf{L}} \right\rvert} \leq C_2\),
some other constant. We want to understand
\begin{align*} 
F(\alpha) \coloneqq{\left[ {c_1, \cdots, c_{r_1 + r_2 - 1}, 0} \right]} \cdot \mathbf{\operatorname{Log}}(\alpha)
\approx  
F(\alpha) \coloneqq{\left[ {c_1, \cdots, c_{r_1 + r_2 - 1}, 0} \right]} \cdot \mathbf{L}
.\end{align*}
How can we estimate the error? By Cauchy-Schwarz, the difference is
bounded by
\begin{align*}
{\left\lVert {
{\left[ {c_1, \cdots, c_{r_1 + r_2 - 1}, 0} \right]} 
} \right\rVert}
\cdot 
{\left\lVert {
\operatorname{Log}\alpha - \mathbf{L}
} \right\rVert}
\leq 
{\left\lVert {
{\left[ {c_1, \cdots, c_{r_1 + r_2 - 1}, 0} \right]} 
} \right\rVert}
C_2
\coloneqq
C_3
.\end{align*}

We've show that for any choice of
\(\lambda_1, \cdots, \lambda_{r_1 + r_2}\) satisfying the volume
condition, one can find \(\alpha\) where
\({\left\lvert {N \alpha} \right\rvert}\) is bounded and all of the
above inequalities hold. In particular,
\(F( \alpha) \approx \mathbf{c} \cdot \mathbf{L}\). We claim we can now
choose \(\alpha\)s to make
\({\left\lvert {F( \alpha)} \right\rvert} \to \infty\). Every choice of
\(\lambda_i\)s yield an \(\alpha\), and the \(c_i\) in \(\mathbf{c}\)
are fixed, so \(\mathbf{c} \cdot \mathbf{L}\) is just a function of the
\(\lambda_i\). So to finish the prove, we need to show we can pick
\(\lambda_i\) satisfying the volume condition but making
\(\mathbf{c} \cdot \mathbf{L}\) as large as we want. This
correspondingly makes \(F(\alpha)\) large, since they differ by at most
a constant.

Here's why we can do this: we assumed not all \(c_i\) were zero, so fix
an \(i\) where \(c_i\neq 0\). Choose \(\lambda_j\) with \(\lambda_i\)
very large, and \(\lambda_{j'} =1\) for \(j=1, \cdots, r_1 + r_2 - 1\)
not including \(i\). Then choose \(\lambda_{r_1 + r_2}\) as determined
by the volume condition. The claim is that we're done. The dot product
has a zeroth \(r_1 + r_2\) component, so the dot product doesn't see
\(\sigma_{r_1 + r_2}\). Recall \(\mathbf{L}\) is a function of the
\(\lambda_i\), since the whole point was that we removed the dependence
on \(\alpha\), and is given by
\begin{align*}
\mathbf{L} = {\left[ {
\log \lambda_1, \cdots, \lambda_{r_1}, 2\log \lambda_{r_1 + 1}, \cdots, 2\log \lambda_{r_1 + r_2}
} \right]}
.\end{align*}
What survives the dot product? We're taking \(\log(1) = 0\) in the
\(j\neq i\) components, and in the \(i\)th component we have at most
\(2\log \lambda_i \cdot c_i\), where the \(\lambda_i\) is a large
number. So we can make \(\mathbf{c} \cdot \mathbf{L}\) as large as we
want.

\end{proof}

\hypertarget{ch.-21-applications-of-minkowskis-theorem}{%
\section{Ch. 21: Applications of Minkowski's
Theorem}\label{ch.-21-applications-of-minkowskis-theorem}}

\begin{proposition}[?]

Recall the Minkowski embedding,
\(\iota: K\to {\mathbb{R}}^{r_1} \oplus {\mathbb{C}}^{r_2}\), where we
concatenate all of the real embeddings followed by the complex
embeddings. View the target as \({\mathbb{R}}^n\), then there is a
formula:
\begin{align*}
\operatorname{covol}\iota({\mathbb{Z}}_K) = 2^{-r_2} \sqrt{ {\left\lvert { {\Delta}_K } \right\rvert}}
.\end{align*}

\end{proposition}

\begin{proof}[?]

Pick an integral basis
\({ {\omega}_1, {\omega}_2, \cdots, {\omega}_{n}}\) for \(K\), then
\(\iota(\omega_1), \cdots, \iota(\omega_n)\) span
\(\iota({\mathbb{Z}}_K)\) over \({\mathbb{Z}}\). It suffices to show
that if \(A\) is the \(n\times n\) matrix with \(i\)th column
\(\iota(\omega_i)\), then
\({\left\lvert {\operatorname{det}A} \right\rvert}\) is the right-hand
side in the formula. If this is true, the \(\iota(w_i)\) are linearly
independent over \({\mathbb{R}}\), otherwise the determinant would be
\({\mathbb{R}}\), so they're a basis for the lattice
\(\iota({\mathbb{Z}}_K)\), and this determinant is precisely the formula
for the covolume of a lattice. We claim the matrix looks like the
following.

\begin{align*}
\begin{bmatrix}
\sigma_1(\omega_1) & \cdots & \sigma_1(\omega_1) \\
\vdots & \cdots & \vdots \\
\sigma_{r_1}(\omega_1) & \cdots & \sigma_{r_1}(\omega_1) \\
\Re \sigma_{r_1+1}(\omega_1) & \cdots & \Re \sigma_{r_1+1}(\omega_1) \\
\Im \sigma_{r_1+1}(\omega_1) & \cdots & \Im \sigma_{r_1+1}(\omega_1) \\
\vdots & \cdots & \vdots \\
\Re \sigma_{r_1+r_2}(\omega_1) & \cdots & \Re \sigma_{r_1+r_2}(\omega_1) \\
\Im \sigma_{r_1+r_2}(\omega_1) & \cdots & \Im \sigma_{r_1+r_2}(\omega_1) 
\end{bmatrix}
.\end{align*}

Let \(D_{ij} \coloneqq{\left[ { \sigma_i(\omega_j) } \right]}\), then by
tracking row operations that transform \(D\) to \(A\),
\(\operatorname{det}(A) = (-i/2)^{r_2} \operatorname{det}(D)\). So
\begin{align*}
{\left\lvert {\operatorname{det}A} \right\rvert} = 2^{-r_2} {\left\lvert {\operatorname{det}D} \right\rvert}
.\end{align*}
Since \(\operatorname{det}(D)^2 = {\Delta}_K\), then taking square roots
yields
\({\left\lvert {\operatorname{det}D} \right\rvert} = \sqrt{ {\left\lvert { {\Delta}_K} \right\rvert} }\).

\end{proof}

\begin{remark}

We now now the covolume of this lattice, we'd like to do this not just
for \({\mathbb{Z}}_K\) but also its ideals.

\end{remark}

\begin{lemma}[?]

If \(I{~\trianglelefteq~}{\mathbb{Z}}_K\), then \(I\) is a free abelian
group of rank \(n\).

\end{lemma}

\begin{proof}[?]

Take any nonzero \(\alpha\in I\), then
\(\alpha {\mathbb{Z}}_K \subseteq I \subseteq {\mathbb{Z}}_K\). The
outer terms are free abelian of rank \(n\), so \(I\) must be as well.

\end{proof}

\begin{proposition}[?]

If \(I{~\trianglelefteq~}{\mathbb{Z}}_K\) is nonzero, then \(\iota(I)\)
is a full rank lattice in \({\mathbb{R}}^n\) and
\begin{align*}
\operatorname{covol}\iota(I) = 2^{r_2} \sqrt{{\left\lvert {{\Delta}_K} \right\rvert}} N(I)
.\end{align*}

\end{proposition}

\hypertarget{ch.-21-applications-of-minkowskis-theorem-friday-may-21}{%
\section{Ch. 21: Applications of Minkowski's Theorem (Friday, May
21)}\label{ch.-21-applications-of-minkowskis-theorem-friday-may-21}}

\begin{remark}

Recall that we're considering the Minkowski embedding
\(\iota: K\to{\mathbb{R}}^n\). We saw that
\(\iota({\mathbb{Z}}_K)\subseteq {\mathbb{R}}^n\) is a full rank lattice
whose covolume is the following:
\begin{align*}
\operatorname{covol}\iota {\mathbb{Z}}_K = 2^{r_2} \sqrt{ {\left\lvert { {\Delta}_K} \right\rvert}}
.\end{align*}
We wanted to make this work for arbitrary ideals
\(I{~\trianglelefteq~}{\mathbb{Z}}_K\), and the proposition was that
\begin{align*}
\operatorname{covol}\iota I = 2^{r_2} \sqrt{ {\left\lvert { {\Delta}_K} \right\rvert}} N(I)
.\end{align*}

\end{remark}

\begin{proof}[?]

We know \(I\) is free abelian of rank \(n\), so choose a
\({\mathbb{Z}}{\hbox{-}}\)basis
\({ {\omega}_1, {\omega}_2, \cdots, {\omega}_{n}}\) for
\({\mathbb{Z}}_K\) and
\({ {\theta}_1, {\theta}_2, \cdots, {\theta}_{n}}\) for \(I\). There is
some \(A\in \operatorname{Mat}(n\times n; {\mathbb{Z}})\) such that
\begin{align*}
{\left[ {{ {\theta}_1, {\theta}_2, \cdots, {\theta}_{n}}} \right]} = {\left[ {{ {\omega}_1, {\omega}_2, \cdots, {\omega}_{n}}} \right]}A
.\end{align*}
The index-determinant theorem tells us that
\(\# {\mathbb{Z}}_K/I = {\left\lvert {\operatorname{det}A} \right\rvert}\),
and this is equal to \(N(I)\). Now apply \(\iota\) to both sides above
to obtain \(n\times n\) matrix with the \(\iota(\theta_i)\) and
\(\iota(\omega_i)\) as column vectors:
\begin{align*}
{\left[ {{ { \iota \theta^t }_1, { \iota \theta^t }_2, \cdots, { \iota \theta^t }_{n}} } \right]} 
= 
{\left[ {{ {\iota \omega^r}_1, {\iota \omega^r}_2, \cdots, {\iota \omega^r}_{n}}} \right]}A
.\end{align*}
Taking \({\left\lvert {\operatorname{det}{{-}}} \right\rvert}\) on both
sides, we get
\begin{align*}
\operatorname{covol}\iota(I) 
= \operatorname{covol}\iota({\mathbb{Z}}_K) {\left\lvert {\operatorname{det}A} \right\rvert} 
= \operatorname{covol}\iota({\mathbb{Z}}_K) N(I)
.\end{align*}

\end{proof}

\hypertarget{minkowskis-class-group-bound}{%
\subsection{Minkowski's Class Group
Bound}\label{minkowskis-class-group-bound}}

\begin{remark}

We've proved that \(\# { \operatorname{Cl}} ({\mathbb{Z}}_K) < \infty\)
for a general number field, which is in the book and uses Dirichlet's
approximation theorem and goes similarly to how it did for quadratic
fields. That proof isn't so good if you want to concretely compute the
class group: what you'd like would be an upper bound on the smallest
ideal in any ideal class. This would allow computing all ideals up to
that bound and discerning the structure based on these finitely many
ideals.

\end{remark}

\begin{proposition}[Minkowski's Bound]

Every ideal class contains a representative \(I\) with
\(N(I) \leq M_K\), where
\begin{align*}
M_K \coloneqq\qty{4\over \pi}^{r_2} {n! \over n^n} \sqrt{{\left\lvert {{\Delta}_K} \right\rvert}}
.\end{align*}

\end{proposition}

\begin{remark}

The precise form of this bound isn't important yet. We can prove using
this bound that some specific number fields have class number 1. How do
we prove this?

\end{remark}

\begin{lemma}[?]

Let \(B>0\), then the following are equivalent:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Every ideal class contains a representative of norm at most \(B\),
\item
  Every nonzero ideal \(I{~\trianglelefteq~}{\mathbb{Z}}_K\) contains a
  nonzero \(\alpha\) with
  \({\left\lvert {N \alpha} \right\rvert} \leq N N(I)\).
\end{enumerate}

\end{lemma}

\begin{proof}[?]

\(2\implies 1\): done for quadratic fields.

\(1\implies 2\): Start with \(I{~\trianglelefteq~}{\mathbb{Z}}_K\)
nonzero. Pick \(J \in [I]^{-1}\) with \(N(J) \leq B\) using (1). Then
\(IJ\) is principal, so write
\(IJ = \left\langle{ \alpha }\right\rangle\). This shows that
\(I \divides \left\langle{ \alpha }\right\rangle\), so \(\alpha\in I\)
since to divide is to contain. Moreover
\begin{align*}
{\left\lvert {N ( \alpha)} \right\rvert} 
&= N( \left\langle{ \alpha }\right\rangle ) \\
&= N(I) N(J) \\
&\leq B N(I)
.\end{align*}

\end{proof}

\begin{remark}

So we've reduced the problem, and it suffices to prove (2) in the above
lemma taking \(B = M_K\) from above. Idea: we'll introduce a region
\(R \subseteq {\mathbb{R}}^n\) which is centrally symmetric, convex, and
\(\operatorname{vol}(R) > 2^n \operatorname{covol}\iota(I)\) big enough.
Then we'll be guaranteed an \(\alpha \in I\) nonzero with
\(\iota( \alpha)\in R\), coming from Minkowski's theorem.

What does this tell us? The components of \(\iota( \alpha)\) are the
images of \(\alpha\) under embeddings \(\sigma_i\). If we know these
images, we can recover the norm as the product, so one ought to be to
rig the region \(R\) in order to control the size of \(\alpha\). So
we'll choose \(R\) such that
\begin{align*}
{\left\lvert { N( \alpha)} \right\rvert} \leq M_K N(I)
.\end{align*}

\end{remark}

\begin{question}

What kinds of regions \(R\) correspond to
\({\left\lvert {N( \alpha)} \right\rvert} \leq X\) for an arbitrary
\(X\)?

\end{question}

\begin{answer}

It's precisely the following region:
\begin{align*}
\left\{{ {\left[ {x} \right]}{r_1 + r_2} \in {\mathbb{R}}^{r_1} \oplus {\mathbb{C}}^{r_2} {~\mathrel{\Big|}~}\prod_{i\leq r_1} {\left\lvert {x_i} \right\rvert} \prod_{j\leq r_2} {\left\lvert {x_{r_1 + j}} \right\rvert}^2 \leq X }\right\} 
.\end{align*}
Note that this is centrally symmetric, but the convexity may be a
problem.

\end{answer}

\begin{example}[?]

Let \(K\) be a real quadratic field, so \(r_1 = 2, r_2 = 0\). Then the
region above is
\(R_0 = \left\{{ (x_1, x_2) {~\mathrel{\Big|}~}{\left\lvert {x_1 x_2} \right\rvert} \leq X }\right\}\),
which looks like the following:

\begin{figure}
\centering
\includegraphics{figures/image_2021-05-22-17-53-50.png}
\caption{image\_2021-05-22-17-53-50}
\end{figure}

The solution: choose a convex region inside \(R_0\) and apply
Minkowski's theorem, noting that this maintains the bound. For example,
one can take
\begin{align*}
R = \left\{{ (x_1, x_2) {~\mathrel{\Big|}~}{\left\lvert {x} \right\rvert}_1 + {\left\lvert {x} \right\rvert}_2 \leq 2x^{1/2} }\right\} 
,\end{align*}
which is a diamond lying inside \(R_0\) due to the AM-GM inequality.

\end{example}

\begin{proposition}[AM-GM Inequality]

Let \({ {t}_1, {t}_2, \cdots, {t}_{m}} \geq 0\), then
\begin{align*}
{1\over m} \sum_{i\leq m} t_i \geq \qty{ \prod_{i\leq m} t_i }^{1/m}
.\end{align*}

\end{proposition}

\begin{remark}

So we'll take the region
\begin{align*}
R \coloneqq\left\{{ {\left[ {{ {x}_1, {x}_2, \cdots, {x}_{r_1 + r_2}}} \right]} \in {\mathbb{R}}^{r_1} \oplus {\mathbb{C}}^{r_2} {~\mathrel{\Big|}~}
{1\over n} \qty{ \sum_{i\leq r} {\left\lvert {x_i} \right\rvert} + 2\sum_{r_1 + 1 \leq j \leq r_1 + r_2 } {\left\lvert {x_j} \right\rvert} } \leq X^{1/n}
}\right\} 
.\end{align*}
Then by AM-GM, \(R \subseteq R_0\) and \(n= r_1 + r_2\). This is still
centrally symmetric, and now convex using the triangle inequality. When
is \(\operatorname{vol}R > 2^n \operatorname{covol}\iota(I)\)? We'll
need to compute the volume of \(R\), which is an involved exercises in
multivariable calculus. This is done in the book, and it turns out that
\begin{align*}
\operatorname{vol}(R) = 2^{r_1} \qty{\pi \over 2}^{r_2} {n^n \over n!} X
.\end{align*}
Recall that
\begin{align*}
2^n \operatorname{covol}\iota(I) = 2^n 2^{-r_2} \sqrt{{\left\lvert {{\Delta}_K} \right\rvert}}
,\end{align*}
and solving this linear equality for \(X\) yields \(X > M_K N(I)\) as
defined before.

Now apply Minkowski's theorem: for any \(X > M_K N(I)\) there is a
nonzero \(\alpha\in I\), there is a nonzero \(\alpha\in I\) with
\(\iota( \alpha)\in R\), and hence
\({\left\lvert {N( \alpha)} \right\rvert} \leq X\). Note that the
inequalities don't quite match up as-is, since we can't take
\(X = M_K N(I)\). Does this imply that we can find an \(\alpha\) with
\({\left\lvert {N( \alpha)} \right\rvert} \leq X\)? The answer is yes,
because we can choose \(X\) with
\({\left\lfloor X \right\rfloor} = {\left\lfloor M_K M(I) \right\rfloor}\).
Then
\({\left\lvert {N( \alpha)} \right\rvert} \leq X \implies {\left\lvert {N( \alpha)} \right\rvert} \leq {\left\lfloor X \right\rfloor} = {\left\lfloor M_K N(I) \right\rfloor}\),
since the left-hand side is an integer.

\end{remark}

\hypertarget{example-showing-number-fields-are-pids-using-dedekind-kummer-and-the-minkowski-bound}{%
\subsection{Example: Showing Number Fields are PIDs using
Dedekind-Kummer and the Minkowski
Bound}\label{example-showing-number-fields-are-pids-using-dedekind-kummer-and-the-minkowski-bound}}

\begin{example}[?]

Let \(K = {\mathbb{Q}}( \sqrt[3]{3})\). Then \(n=3\), and
\(r_1 = 1, r_2 = 1\) since embeddings permute the roots of \(x^3-3\),
which has exactly one real root. We need the discriminant, so we need an
integral basis, in which case it helps to know \({\mathbb{Z}}_K\). By a
homework problem, if \(K = {\mathbb{Q}}(\sqrt[3]{d})\) with \(d\)
squarefree and \(d\not\equiv \pm 1\operatorname{mod}9\), then
\({\mathbb{Z}}_K\) is what you guess it'd be! So here
\({\mathbb{Z}}_K = {\mathbb{Z}}[\sqrt[3]{3}]\), and thus
\begin{align*}
{\Delta}_K = \Delta(1, 3^{1/3}, 3^{2/3})
= {\Delta}(x^2 - 3) = -3^5
.\end{align*}

Thus
\begin{align*}
M_K = \qty{4\over \pi} {3! \over 3^3} \sqrt{3^5} \approx 4.42
,\end{align*}
and thus we know every ideal class contains a representative of norm at
most 4, using that \(N(I)\) is an integer. Any such \(I\) is a product
of ideals of norms \(2,3,4\). Prime ideals of norm 2 or 4 lie above the
prime 2, and of norm 3 lie above 3. If we can show that every prime
ideal with \(N(I) = 2,3,4\) is principal, then \(I\) will be a product
of principal ideals and thus principal. Then since every class contains
such an \(I\), \({ \operatorname{Cl}} ({\mathbb{Z}}_K)\) is trivial. We
look at all of the primes above 2 and 3 using Dedekind-Kummer, which
says the factorization of \(\left\langle{ p }\right\rangle\) mirrors the
factorization of \(x^2 - 3 \operatorname{mod}p\).

Note that if \({\mathbb{Z}}(\sqrt[3]{3}) \neq {\mathbb{Z}}_K\), there'd
be some exceptional primes \(p\) to worry about, but since these are
equal here this is literally true for all \(p\).

So we factor polynomials. First mod 2:
\begin{align*}
x^2 - 3 = x^3 - 1 = (x-1) (x^2 + x + 1)
,\end{align*}

where the second term is an irreducible quadratic with no roots in
\({\mathbb{F}}_2\). This yields
\begin{align*}
\left\langle{ 2 }\right\rangle= \left\langle{ 2, \sqrt[3]{3} - 1 }\right\rangle \left\langle{ 2, \qty{\sqrt[3]{3} }^2 + \sqrt[3]{3} + 1 }\right\rangle   
.\end{align*}
Next mod 2:
\begin{align*}
x^3 -3 = x^3
,\end{align*}
which yields
\begin{align*}
\left\langle{ 3 }\right\rangle= \left\langle{ 3, \sqrt[3]{3} }\right\rangle^3  
.\end{align*}

Although we have explicit factorizations, but it may not obvious whether
or not they're principal. It's easy to see that
\(\left\langle{ 3, 3^{1/3} }\right\rangle\) is principal since
\(3^{1/3} \divides 3\) and the first generator is redundant For
\(\left\langle{2}\right\rangle\), it's less clear, but
\begin{align*}
(3^{1/3} - 1) (3^{2/3} + 3^{1/3} + 1) = 2
,\end{align*}
and so the 2s are redundant generators in both terms. So every ideal
above 2 and 3 is principal, so \(I\) is principal. Every ideal class
contains such an \(I\), so \({ \operatorname{Cl}} ({\mathbb{Z}}_K) = 0\)
and \({\mathbb{Z}}_K\) is a PID.

\end{example}

\begin{remark}

Asking whether or not this ring is a PID is an undergraduate-level
question, but it's not clear how one would determine this without the
theory developed in this class.

\end{remark}

\begin{example}[?]

Let \(K = {\mathbb{Q}}(\theta)\) where \(\theta\) is a root of
\(f(x) = x^5-x^3 + 1\). This is irreducible over \({\mathbb{Q}}\) and
has one real root. Then \(n=5, r_1 = 1, r_2 = 2\), but we need the
discriminant to apply the Minkowski bound. If we could prove
\({\mathbb{Z}}_K = {\mathbb{Z}}(\theta)\), \({\Delta}_K = {\Delta}(f)\).
If it were squarefree, it'd correspond to an integral basis, so we
compute
\begin{align*}
{\Delta}(f) = {\Delta}(1, \theta, \theta^2, \theta^3, \theta^4) =3017
,\end{align*}
using determinant formulas in the book. This is squarefree, so the power
basis is an integral basis, so \({\mathbb{Z}}_K = {\mathbb{Z}}[\theta]\)
and \({\Delta}_K = 3017\). Then
\begin{align*}
M_k = \qty{4\over \pi}^2 {5! \over 5^5} \sqrt{3017} \approx 3.41
.\end{align*}
So every ideal class is represented by an ideal of norm at most \(3\).

\begin{itemize}
\tightlist
\item
  Norm 1: Unit ideal and thus principal,
\item
  Norm 2: Primes above 2
\item
  Norm 3: Primes above 3
\end{itemize}

We'll apply Dedekind-Kummer. First mod 2: \(f(x)\) is irreducible mod 2,
so \(\left\langle{2}\right\rangle\) is inert and prime. So there are no
ideals of norm 2, since it would have to factor as a product of primes
above 2, but the only possible factor is 2 which has norm \(2^5 =32\).

Mod 3: \(f\) is no longer irreducible, but factors as \(f=pq\) with
\(\deg p = 2,\deg q = 3\). So there are no ideals of norm 3, since all
prime ideals above 3 would have to have norms of \(3^2\) corresponding
to \(p\) or \(3^3\) corresponding to \(q\). Thus
\({ \operatorname{Cl}} ({\mathbb{Z}}_K) = 0\), since every \(I\) is
represented by an ideal of norm 1 and hence is principal.

\end{example}

\begin{remark}

Next up we'll talk about a lower bound for
\({\left\lvert {{\Delta}_K} \right\rvert}\) Obviously \(M_K \geq 1\),
since the theorem states that every ideal class has a representative of
norm at most \(M_K\), and \(M_K< 1\) wouldn't make sense. Using the
formula, rearranging yields a bound
\begin{align*}
{\left\lvert {{\Delta}_K} \right\rvert} \geq \qty{n^n \over n!}^2 \qty{\pi \over 4}^{2r_2}
.\end{align*}

\end{remark}

\hypertarget{ch.-21-consequences-of-minkowskis-bound-saturday-may-22}{%
\section{Ch. 21: Consequences of Minkowski's Bound (Saturday, May
22)}\label{ch.-21-consequences-of-minkowskis-bound-saturday-may-22}}

\begin{remark}

We were discussing a lower bound on
\({\left\lvert {{\Delta}_K} \right\rvert}\). We have Minkowski's class
number bound: every ideal class has a representative of norm at most
\(M_K\), where
\begin{align*}
M_K \coloneqq\qty{4\over \pi}^{r_2} \qty{n! \over n^n}\sqrt{{\left\lvert {{\Delta}_K} \right\rvert}}
.\end{align*}
A consequence of the theorem is \(M_K \geq 1\), and rearranging yields a
bound
\begin{align*}
{\left\lvert {{\Delta}_K} \right\rvert} \geq \qty{n^n \over n!}^2 \qty{\pi/4}^{2r)2}
.\end{align*}
What is this bound really telling us? Let's bound the right-hand side
from below. Note that \(\pi/4<1\), and since this is raised to a power,
this might make things smaller. The worst case, i.e.~the smallest it
could be, is when \(2r_2\) is as large as possible, so using that
\(r_1 + r_2 = n\) we have
\begin{align*}
{\left\lvert {{\Delta}_K} \right\rvert} \geq \qty{n^n\over n!}^2 \qty{\pi/4}^n \coloneqq B_n
.\end{align*}
How does \(B_n\) grow? We could use Stirling's formula, but we'll take a
crude bound by looking at ratios:
\begin{align*}
{B_{n+1} \over B_n} 
&= (1 + {1\over n})^{2n} (\pi/4)\\
&= \qty{ 1 + {2n \choose 1} {1\over n} + \cdots }(\pi/4) \\
&\geq 3\pi/4
.\end{align*}
Noting that \(B_2 = \pi^2/4\), so by induction
\begin{align*}
B_n \geq (3\pi/4)^{n-2} (\pi^2/4)
.\end{align*}

\end{remark}

\begin{remark}

Some consequences:

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \({\left\lvert {{\Delta}_K} \right\rvert} > 1\) for all number fields
  \(K\neq {\mathbb{Q}}\).
\item
  \({\left\lvert {{\Delta}_K} \right\rvert}\to\infty\) as
  \([K:{\mathbb{Q}}]\to\infty\).
\end{enumerate}

The following says why (a) is important:

\end{remark}

\begin{theorem}[Dedekind]

\(p\) ramifies in \({\mathbb{Z}}_K \iff p \divides {\Delta}_K\).

\end{theorem}

\begin{proof}[?]

Omitted, see book.

\end{proof}

\begin{remark}

So by (a), every number field \(K\geq {\mathbb{Q}}\) there is at least
one ramified prime \(p\).

\end{remark}

\begin{remark}

Note that if \({\left\lvert {{\Delta}_K} \right\rvert} = 1\), then
\(K={\mathbb{Q}}\), i.e.~there is only one such number field. What about
for any fixed number \(n\)? The next theorem says that there are only
finitely many number fields occurring below a prescribed bound:

\end{remark}

\begin{theorem}[Hermite's theorem]

For every \(X>0\), there are finitely many number fields \(K\) with
\({\left\lvert {{\Delta}_K} \right\rvert} \leq X\).

\end{theorem}

\begin{remark}

Since \({\left\lvert {{\Delta}_K} \right\rvert}\to\infty\) as
\([K:{\mathbb{Q}}]\to\infty\), it suffices to prove this theorem with a
fixed \(n\coloneqq[K: {\mathbb{Q}}]\). We'll make a simplifying
assumption that \(r_1 = n\) -- this doesn't simplify the proof so much,
but rather simplifies the notation. The full proof in the book is not so
different.

\end{remark}

\begin{remark}

Define a region
\begin{align*}
R \coloneqq\left\{{{\left[ {{ {x}_1, {x}_2, \cdots, {x}_{n}}} \right]} \in {\mathbb{R}}^n {~\mathrel{\Big|}~}{\left\lvert {x_{i\leq n-1}} \right\rvert} \leq {1 \over 2},\,\, {\left\lvert {x_n} \right\rvert} \leq T }\right\} \subseteq {\mathbb{R}}^n
,\end{align*}
where we'll specify \(T\) in a moment.

Note that

\begin{itemize}
\tightlist
\item
  \(R\) is centrally symmetric
\item
  \(R\) is convex, by the triangle inequality,
\item
  The volume is easily computable:
  \(\operatorname{vol}R = 1\cdot 1\cdot \ldots \cdot 1 \cdot 2T = 2T\).
\end{itemize}

Choose \(T = 2^n \sqrt{X}\). Suppose \(K\) is totally real with
\({\left\lvert { {\Delta}_K } \right\rvert} \leq X\). Recall that
\begin{align*}
\operatorname{covol}\iota {\mathbb{Z}}_K = 2^{r_2} \sqrt{{\left\lvert {{\Delta}_K} \right\rvert}} 
\leq \sqrt{X}
.\end{align*}

Then
\begin{align*}
\operatorname{vol}R = 2^{n+1 } \sqrt{X} > 2^n \operatorname{covol}\iota {\mathbb{Z}}_K
,\end{align*}
so Minkowski's theorem can be applied: there is a nonzero
\(\alpha\in {\mathbb{Z}}_K\) with \(\iota( \alpha) \in R\). Then
\({\left\lvert {\sigma_i( \alpha)} \right\rvert} \leq 1/2\) for
\(i\leq n-1\) and
\({\left\lvert { \sigma_n ( \alpha)} \right\rvert} \leq T\). The claim
is that there are only finitely many such \(\alpha\), since it was a
root of a polynomial in a finite set \(P_{n, T}\). The claim is now that
\(K = {\mathbb{Q}}( \alpha)\), so this \(\alpha\) uniquely determines
\(K\). Since there were finitely many \(\alpha\), there can only be
finitely many such \(K\).

The claim is that the size of \(\sigma_n( \alpha)\) has to be big, say
at least 1. We have control over the product, since
\begin{align*}
1\leq {\left\lvert {N \alpha} \right\rvert} 
= \prod {\left\lvert { \sigma_i( \alpha)} \right\rvert} 
\leq {1\over 2^{n-1}} {\left\lvert { \sigma_n( \alpha)} \right\rvert}
,\end{align*}
since the first \(n-1\) terms contribute at most \(1/2\) each. So
\({\left\lvert {\sigma_n( \alpha)} \right\rvert} \geq 2^{n-1} \geq 1\).
Suppose now that \({\mathbb{Q}}(\alpha) < K\) is a proper subfield, so
\([K : {\mathbb{Q}}( \alpha)] = d > 1\). Then every embedding
\({\mathbb{Q}}(\alpha) \hookrightarrow{\mathbb{C}}\) extends to \(d\)
embeddings \(K\hookrightarrow{\mathbb{C}}\), but this means that for any
element \(x \in {\mathbb{Q}}(\alpha)\), the images
\(\sigma_1(x),\cdots, \sigma_n(x)\) would have the same element repeated
\(d\) times. But we know that \(\sigma_n( \alpha)\) is different from
all of the other \(\sigma_i( \alpha)\), so this is a contradiction.

\end{remark}

\begin{remark}

Idea of proof: making sure some image of \(\alpha\) under one embedding
is different than all of the other images.

\end{remark}

\begin{remark}

Some remarks on modern research! Let \(N_n(X)\) be the set of number
fields of degree \(n\) where
\({\left\lvert {{\Delta}_K} \right\rvert} \leq X\).

\end{remark}

\begin{conjecture}

For each fixed \(n\),
\begin{align*}
\lim_{X\to \infty} {N_n(X) \over X} = \delta_n > 0
,\end{align*}
where \(\delta_n\) is some particular constant.

Some known results:

\begin{itemize}
\tightlist
\item
  \(n=2\): known to Gauss, since this is more or less equivalent to
  counting squarefree numbers.
\item
  \(n=3\): much harder, Davenport-Heilbronn.
\item
  \(n=4, 5\): Bhargava, part of what resulted in his 2014 Fields medal.
\end{itemize}

\end{conjecture}

\begin{remark}

One could restrict this problem, e.g.~by prescribing a particular Galois
group. See Mahler's conjecture.

\end{remark}

\hypertarget{chapter-xyz-relative-extensions-galois-theory-prime-splitting}{%
\section{Chapter XYZ: Relative Extensions, Galois Theory, Prime
Splitting}\label{chapter-xyz-relative-extensions-galois-theory-prime-splitting}}

\begin{remark}

Up until now: we've compared extensions over \({\mathbb{Q}}\):

\begin{center}
\begin{tikzcd}
    K && {{\mathbb{Z}}_K} \\
    \\
    {\mathbb{Q}}&& {\mathbb{Z}}
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJLIl0sWzAsMiwiXFxRUSJdLFsyLDAsIlxcWlpfSyJdLFsyLDIsIlxcWloiXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

We'll next talk about extensions between number fields:

\begin{center}
\begin{tikzcd}
    L && {{\mathbb{Z}}_L} \\
    \\
    K && {{\mathbb{Z}}_K}
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJMIl0sWzAsMiwiSyJdLFsyLDAsIlxcWlpfTCJdLFsyLDIsIlxcWlpfSyJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link
to Diagram}
\end{quote}

We can ask the same sorts of questions about prime ideals factoring.
Note that if \(I = \prod P_i^{e_i}\) with each
\(P_i \in \operatorname{Spec}{\mathbb{Z}}_K\) , then extending to
\({\mathbb{Z}}_L\) yields
\begin{align*}
I{\mathbb{Z}}_L = \prod (P_i {\mathbb{Z}}_L)^{e_i}
.\end{align*}
So we want to understand the following: given a prime ideal \(P\) of
\({\mathbb{Z}}_K\), how does \(P{\mathbb{Z}}_L\) factor?

\end{remark}

\begin{definition}[lies above]

Let \(K\leq L\) be number fields, and suppose
\(Q\in \operatorname{Spec}{\mathbb{Z}}_L, P\in \operatorname{Spec}{\mathbb{Z}}_K\).
Then we say \(Q\) \textbf{lies above} \(P\) if \(Q \supseteq P\), or
equivalently \(Q \divides P{\mathbb{Z}}_L\).

\end{definition}

\begin{proposition}[?]

Every nonzero \(Q\in \operatorname{Spec}{\mathbb{Z}}_L\) lies above a
unique nonzero \(P \in \operatorname{Spec}{\mathbb{Z}}_K\).

\end{proposition}

\begin{proof}[?]

Consider
\(P \coloneqq Q \cap{\mathbb{Z}}_K \in \operatorname{Spec}{\mathbb{Z}}_K\).
This is nonzero because taking the norm of any element of \(Q\) yields a
nonzero integer still in the ideal. Then \(Q\) lies above \(P\) by
definition. Why is this unique? Suppose \(Q\) lies about \(P'\), we'll
show \(P = P'\). Since \(Q\) lies above \(P'\), \(Q\supseteq P'\). But
\(P'{~\trianglelefteq~}{\mathbb{Z}}_K\), so
\(Q\cap{\mathbb{Z}}_K \supseteq P'\), so \(P \supseteq P'\). We know
\(P'\) is maximal, since \({\mathbb{Z}}_K/P'\) is a finite domain and
thus a field, so \(P = P'\).

\end{proof}

\begin{definition}[Ramification and residue degrees]

Let \(P\in \operatorname{Spec}{\mathbb{Z}}_K\), then write
\begin{align*}
P{\mathbb{Z}}_L = \prod Q_i^{e_i}
\end{align*}
with \(Q_i \in \operatorname{Spec}{\mathbb{Z}}_L\). Then the \(Q_i\) are
the prime ideals of \({\mathbb{Z}}_L\) above \(P\). The exponent \(e_i\)
is called the \textbf{ramification degree}, usually denoted
\(e(Q_i/P)\). We have the following picture:

\begin{center}
\begin{tikzcd}
    L && {{\mathbb{Z}}_L} && Q \\
    \\
    K && {{\mathbb{Z}}_K} && P
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=1-3, to=3-3]
    \arrow[no head, from=1-5, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiSyJdLFsyLDAsIlxcWlpfTCJdLFsyLDIsIlxcWlpfSyJdLFs0LDIsIlAiXSxbNCwwLCJRIl0sWzAsMSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDMsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNSw0LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link
to Diagram}
\end{quote}

The inclusion \({\mathbb{Z}}_K \hookrightarrow{\mathbb{Z}}_L\) induces a
ring morphism \({\mathbb{Z}}_K \to {\mathbb{Z}}_L/Q\), where the kernel
is \({\mathbb{Z}}_K \cap Q = P\). Thus there is an injection
\({\mathbb{Z}}_K/P \hookrightarrow{\mathbb{Z}}_L/Q\), which is an
inclusion of finite fields. So we'll define
\begin{align*}
f(Q/P) \coloneqq[{\mathbb{Z}}_L/Q : {\mathbb{Z}}_K/P]
\end{align*}
to be the \textbf{residue degree} of \(Q/P\). Note that
\begin{align*}
\# {\mathbb{Z}}_L/{\mathbb{Q}}= \qty{\# {\mathbb{Z}}_K/P}^{f(Q/P)}
\quad\text{i.e.}\quad 
N_L(Q) = N_K(P)^{f(Q/P)}
.\end{align*}

\end{definition}

\begin{theorem}[efg theorem]

If \(P \in \operatorname{Spec}{\mathbb{Z}}_K\) is nonzero with
\(P{\mathbb{Z}}_L = \prod Q_i^{e_i}\), then
\begin{align*}
\sum_i e(Q_i/P) f(Q_i/P) = [L: K]
.\end{align*}

\end{theorem}

\begin{lemma}[?]

If \(I\in \operatorname{Spec}{\mathbb{Z}}_K\) is nonzero, then extend to
\(L\) to get \(I{\mathbb{Z}}_L\). Then
\begin{align*}
N_L (I{\mathbb{Z}}_L) = \qty{ N_K(I)}^{[L : K]}
.\end{align*}

\end{lemma}

\begin{proof}[?]

Omitted. Idea of why it's true: the norm of an ideal is supposed to be a
``product of conjugates'', although naive conjugates of an ideal might
not remain an ideal in the field one starts with. So interpret norms as
products of images under all embeddings into \({\mathbb{C}}\). But then
just interpret \([L: K]\) is the number of lifts of embeddings
\(K\hookrightarrow{\mathbb{C}}\) to \(L\hookrightarrow{\mathbb{C}}\).

\end{proof}

\begin{proof}[of efg theorem]

Take norms in \(L\), then
\begin{align*}
N( P{\mathbb{Z}}_L) 
= \prod N(Q_i)^{e_i} \\
= N_K(P)^{e_1 \sum f(Q_i/P)}
.\end{align*}
On the other hand, the left-hand side is \(N_K(P)^{[L: K]}\), so the
exponents must be equal.

\end{proof}

\begin{remark}

What are the specific prime ideals involved in the factorization,
i.e.~is there a generalization of Dedekind-Kummer here?

\end{remark}

\begin{proposition}[Generalized Dedekind-Kummer theorem]

Write \(L = K(\theta)\) for some
\(\theta \in L \cap\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu= {\mathbb{Z}}_L\).
Let \(m(x)\) be the minimal polynomial of \(\theta\) over \(K\), so
\(m\in {\mathbb{Z}}_K[x]\). Let
\(P\in \operatorname{Spec}{\mathbb{Z}}_K\) lying above
\(p\in \operatorname{Spec}{\mathbb{Z}}\). Then as long as
\(p\notdivides [{\mathbb{Z}}_L: {\mathbb{Z}}_K[\theta]]\), then the
factorization of \(P{\mathbb{Z}}_L\) mirrors the factorization of \(m\)
over the residue field \({\mathbb{Z}}_K/P\).

\end{proposition}

\begin{proof}[?]

Omitted, see ``Number Rings'' by Marcus. Paul strongly recommends!

\end{proof}

\hypertarget{ch.-xyz-april-6}{%
\section{Ch. XYZ: April 6}\label{ch.-xyz-april-6}}

\hypertarget{multiplicativity-in-towers}{%
\subsection{Multiplicativity in
Towers}\label{multiplicativity-in-towers}}

\begin{remark}

An important topic not in the book: relative extensions of number
fields, vs absolute extensions over \({\mathbb{Q}}\).

\begin{center}
\begin{tikzcd}
    L && {{\mathbb{Z}}_?} && Q \\
    \\
    K && {{\mathbb{Z}}_K} && {P = Q \cap{\mathbb{Z}}_K}
    \arrow[no head, from=1-5, to=3-5]
    \arrow[no head, from=1-3, to=3-3]
    \arrow[no head, from=1-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiSyJdLFsyLDAsIlxcWlpfPyJdLFsyLDIsIlxcWlpfSyJdLFs0LDAsIlEiXSxbNCwyLCJQID0gUSBcXGludGVyc2VjdCBcXFpaX0siXSxbNCw1LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=}{Link
to Diagram}
\end{quote}

Note that if \(Q\supseteq P\), then \(Q\) will divide the extension of
\(P\) to \({\mathbb{Z}}_L\), if \(Q\) will show up in the prime
factorization of \(P{\mathbb{Z}}_L\). We defined \(e(Q/P)\) to be the
exponent of \(Q\) in the factorization of \(P{\mathbb{Z}}_L\), and
\(f(Q/P)\) to be the degree of the field extension \({\mathbb{Z}}_L/Q\)
over \({\mathbb{Z}}_K/P\).

\end{remark}

\begin{remark}

Note that ``lying above'' is transitive, and the following situation
makes sense:

\begin{center}
\begin{tikzcd}
    M && {P''} \\
    \\
    L && {P'} \\
    \\
    K && P
    \arrow[no head, from=3-1, to=5-1]
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=1-3, to=3-3]
    \arrow[no head, from=3-3, to=5-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwyLCJMIl0sWzAsNCwiSyJdLFswLDAsIk0iXSxbMiw0LCJQIl0sWzIsMiwiUCciXSxbMiwwLCJQJyciXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzIsMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs1LDQsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCwzLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link
to Diagram}
\end{quote}

How are the various \(e\) and \(f\) related? It turns out they work
similarly to degrees of field extensions:

\end{remark}

\begin{theorem}[Multiplicativity in towers]

\begin{align*}
e(P''/P) &= e(P''/P') \cdot e(P'/P) \\
f(P''/P) &= f(P''/P') \cdot f(P'/P) 
.\end{align*}

\end{theorem}

\begin{proof}[?]

Note that \(f\) is the degree of a field extension where we take
\({\mathbb{Z}}_K/P \hookrightarrow{\mathbb{Z}}_L\) and consider the
induced quotient map to \({\mathbb{Z}}_L/Q\). By composing inclusions we
have a commutative diagram:

\begin{center}
\begin{tikzcd}
    {{\mathbb{Z}}_K/P} && {{\mathbb{Z}}_L/P'} && {{\mathbb{Z}}_M/P''}
    \arrow[hook, from=1-1, to=1-3]
    \arrow[hook, from=1-3, to=1-5]
    \arrow[curve={height=30pt}, hook, from=1-1, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwwLCJcXFpaX0svUCJdLFsyLDAsIlxcWlpfTC9QJyJdLFs0LDAsIlxcWlpfTS9QJyciXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMiwiIiwyLHsiY3VydmUiOjUsInN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
to Diagram}
\end{quote}

So for \(f\), this reduces to the multiplicativity of degrees in towers
of field extensions.

For \(e\), write \(P{\mathbb{Z}}_L = P'^{e(P'/P)} I\) where
\(I\in \operatorname{Id}({\mathbb{Z}}_L)\) is nonzero and
\(P'\notdivides I\). Similarly we can write
\(P' {\mathbb{Z}}_< = P''^{e(P'/P')} J\) where
\(J\in \operatorname{Id}({\mathbb{Z}}_L)\) and \(P''\notdivides J\).
Using general facts about ideal extensions and substituting yields
\begin{align*}
P{\mathbb{Z}}_M 
&= (P{\mathbb{Z}}_L){\mathbb{Z}}_M \\
&= (P'{\mathbb{Z}}_M)^{e(P'/P)} I{\mathbb{Z}}_M \\
&= P''^{e(P'/P) e(P'/P')} J^{e(P'/P)} (I{\mathbb{Z}}_M) \\
.\end{align*}

We know that \(P''\notdivides J\), so it doesn't divide any power of
\(J\) either. We can check that \(P''\notdivides I{\mathbb{Z}}_M\):
otherwise, if it did then \(P''\supseteq I{\mathbb{Z}}_M\) and
\begin{align*}
P'' \cap{\mathbb{Z}}_L \supseteq (I{\mathbb{Z}}_M) \cap{\mathbb{Z}}_L \supset I
\end{align*}
by intersecting both sides with \({\mathbb{Z}}_L\). But we know
\(P'' \cap{\mathbb{Z}}_L = P'\) since \(P''\) was above \(P'\), and
\(P'\notdivides I\) and thus \(P'\not\supseteq I\).

\end{proof}

\begin{definition}[Splitting completely]

Suppose \(L/K\) is an extension of number fields and say
\(P\supseteq {\mathbb{Z}}_K\) is a nonzero prime ideal. Then \(P\)
\textbf{splits completely} in \(L\) (or \({\mathbb{Z}}_L\)) if
\(e(G/P) = f(G/P) = 1\) for all \(G\) above \(P\) in \({\mathbb{Z}}_L\).
Equivalently, there is a factorization
\(P{\mathbb{Z}}_L = \prod_{i\leq n} Q_i\) with the \(Q_i\) distinct and
\(n \coloneqq[L: K]\).

\end{definition}

\begin{proposition}[?]

Let \(M/L/K\) be a tower of number fields and
\(P\in \operatorname{Spec}{\mathbb{Z}}_K\). Then if \(P\) splits
completely in \(M\), it splits completely in \(L\).

\end{proposition}

\begin{slogan}

Splitting completely in an extension implies splitting completely in
every intermediate extension.

\end{slogan}

\begin{proof}[?]

Suppose \(P\) splits completely in \(M\), we then want to show that for
any \(P'\) in \({\mathbb{Z}}_L\) above \(P\), then \(e=f=1\). We know
\(P'\) has some prime factor, so choose any \(P''\) in
\({\mathbb{Z}}_M\) above \(P'\):

\begin{center}
\begin{tikzcd}
    M && {\exists P''} \\
    \\
    L && {P'} \\
    \\
    K && P
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=3-1, to=5-1]
    \arrow[no head, from=5-3, to=3-3]
    \arrow[no head, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCw0LCJLIl0sWzAsMiwiTCJdLFswLDAsIk0iXSxbMiw0LCJQIl0sWzIsMiwiUCciXSxbMiwwLCJcXGV4aXN0cyBQJyciXSxbMiwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzEsMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFszLDQsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNSw0LCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link
to Diagram}
\end{quote}

Since \(P''\) is above \(P\) and \(e(P''/P) = f(P''/P) =1\), we can use
multiplicativity in towers:
\begin{align*}
e(P'/P) \divides e(P''/P) = 1 &\implies e(P'/P) = 1 \\
f(P'/P) \divides f(P''/P) = 1 &\implies f(P'/P) = 1 \\
.\end{align*}

\end{proof}

\begin{definition}[?]

If \(L/K\) is an extension of number fields and
\(P\in \operatorname{Spec}{\mathbb{Z}}_K\) is nonzero, then \(P\)
\textbf{ramifies} in \(L\) (or \({\mathbb{Z}}_L\)) if \(e(Q/P)>1\) for
some \(Q\in \operatorname{Spec}{\mathbb{Z}}_L\) above \(P\).

\end{definition}

\begin{proposition}[?]

For \(M/L/K\) a tower of number fields and
\(P\in \operatorname{Spec}{\mathbb{Z}}_K\) is nonzero and unramified in
\(M\), then \(P\) is unramified in \(L\).

\end{proposition}

\begin{proof}[?]

Same as the last proof.

\end{proof}

\hypertarget{galois-theory-and-prime-decomposition}{%
\subsection{Galois Theory and Prime
Decomposition}\label{galois-theory-and-prime-decomposition}}

\begin{remark}

This will lead into defining the Frobenius element and the fundamental
theorem of algebraic number theory: Chebotarev Density. Some
setup/notation:

\begin{itemize}
\tightlist
\item
  \(L/K\) will be a Galois extension of number of fields
\item
  \(P \in \operatorname{Spec}{\mathbb{Z}}_K, Q\in \operatorname{Spec}{\mathbb{Z}}_L\)
  will be nonzero prime ideals. We may or may not require \(Q\) to lie
  above \(P\).
\end{itemize}

\end{remark}

\begin{remark}

\envlist

\begin{itemize}
\tightlist
\item
  Any \(\sigma \in G(L/K)\) is an automorphism of \(L\) fixing \(K\),
  which restricts to an automorphism of \({\mathbb{Z}}_L\) since it
  preserves algebraic integers.

  \begin{itemize}
  \tightlist
  \item
    It also preserves prime ideals.
  \end{itemize}
\item
  Suppose \(Q\) lies above \(P\), then
  \(P{\mathbb{Z}}_L = Q^{e(Q/p)} \cdots\). \(\sigma\) fixes \(K\)
  pointwise, so applying it to both sides yields
  \(P{\mathbb{Z}}_L = \sigma(Q)^{e(Q/P)}\cdots\) where \(\sigma(Q)\)
  shows up with the exact same power since it preserves distinctness of
  ideals.

  \begin{itemize}
  \tightlist
  \item
    In particular, \(\sigma(G)\) lies above \(P\), so \(G(L/K)\) acts on
    the set of primes above \(P\), and it turns out (very importantly)
    to be transitive.
  \end{itemize}
\end{itemize}

\end{remark}

\begin{theorem}[?]

The action of \(G(L/K)\) on primes above \(P\) is transitive.

\end{theorem}

\begin{proof}[?]

Let \(Q_1, Q_2\) be primes above \(P\) and suppose toward a
contradiction that \({\mathrm{Orb}}(Q_1)\) does not contain \(Q_2\). We
may choose \(\alpha\in {\mathbb{Z}}_L\) such that

\begin{itemize}
\tightlist
\item
  \(\alpha \equiv 0 \operatorname{mod}Q_2\)
\item
  \(\alpha \equiv 1 \operatorname{mod}Q\) for all \(Q\) of the form
  \(Q = \sigma(Q_1)\), noting that none are \(Q_2\)
\end{itemize}

This system can be solved by the CRT, since the \(Q_i\) are pairwise
comaximal, since nonzero primes of a number ring are maximal. These
congruences say \(\alpha\in Q_2\) but not any other \(Q\), since
\(0\equiv 1\) only in the unit ideal. Define
\(\beta = \prod_{\sigma \in G(L/K)} \sigma( \alpha) \in {\mathbb{Z}}_K\),
and observe that \(\beta\not\in Q_1\). If it were in \(Q_1\), one
\(\sigma(\alpha) \in Q_1\) for some \(\sigma \in G(L/K)\) which would
force \(\alpha = \sigma \sigma^{-1}( \alpha) \in \sigma^{-1}(Q_1)\),
which contradicts the choice of \(\alpha\) since \(\sigma^{-1}(Q_1)\) is
one of the \(Q\)s appearing in the second congruence. On the other hand
we have \(\beta\in Q_2\) since the identity is an element of \(G(L/K)\)
and \(\alpha\in Q_2\) and \(\beta\) is a multiple of \(\alpha\). We know
\(\beta\in {\mathbb{Z}}_K\), so
\(\beta \in Q_2 \cap{\mathbb{Z}}_K = P\). But this is fishy because
\(Q_1, Q_2\) both lie above \(P\) and both must contain \(P\):

\begin{center}
\begin{tikzcd}
    {Q_1} && {Q_2} \\
    \\
    & P
    \arrow[no head, from=1-1, to=3-2]
    \arrow[no head, from=1-3, to=3-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMSwyLCJQIl0sWzAsMCwiUV8xIl0sWzIsMCwiUV8yIl0sWzEsMCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDAsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=}{Link
to Diagram}
\end{quote}

Since \(\beta\) is in every ideal above \(P\), in particular
\(\beta \in Q_1\) since \(Q_1 \supseteq P\), and this is a
contradiction.

\end{proof}

\begin{proposition}[?]

If \(Q, Q'\) both lie above \(P\), then \(e(Q/P) = e(Q'/P)\) and
\(f(Q/P) = f(Q'/P)\).

\end{proposition}

\begin{proof}[?]

Pick \(\sigma \in G(L/K)\) with \(\sigma(Q) = Q'\). Factor
\(P{\mathbb{Z}}_L = Q^{e(Q/P)} J\) where \(J\) is a product of primes
not equal to \(Q\). Applying \(\sigma\) yields
\(P{\mathbb{Z}}_L = Q'^{e(Q/P)} J'\) where \(J'\) is a product of primes
not equal to \(Q'\). This factors \(P{\mathbb{Z}}_L\) into primes, and
by uniqueness of prime factorization this exponent has to be \(e(Q'/P)\)
by definition.

For the \(f\)s, note that \(\sigma\) induces a ring morphism between the
residue fields:
\begin{align*}
\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu: {\mathbb{Z}}_K/Q &\to {\mathbb{Z}}_L/Q' \\
\alpha \operatorname{mod}Q &\mapsto \sigma(\alpha) \operatorname{mod}Q'
.\end{align*}
This is well-defined since \(\sigma(Q) = Q'\), and is an isomorphism
since the inverse comes from \(\sigma ^{-1}\). This will imply that the
\(f\)s are the same: we're looking at the degree of these extensions
over \({\mathbb{Z}}_K/P\). An element of \({\mathbb{Z}}_L/Q\) also
belonging to \({\mathbb{Z}}_K/P\) (as a subfield) has the form
\(\alpha \operatorname{mod}Q\) where \(\alpha\in {\mathbb{Z}}_K\), and
under \(\sigma\) this is sent to
\(\sigma(\alpha)\operatorname{mod}Q' = \alpha \operatorname{mod}Q'\),
which is an element of the copy of
\({\mathbb{Z}}_K/P\hookrightarrow{\mathbb{Z}}_L/Q'\). So \(\sigma\)
identifies the copies of \({\mathbb{Z}}_K/P\) in either side, and
\begin{align*}
[{\mathbb{Z}}_L/Q : {\mathbb{Z}}_K/P] = [{\mathbb{Z}}_L/Q: {\mathbb{Z}}_K/P]
.\end{align*}

\end{proof}

\begin{theorem}[efg theorem for Galois extensions]

Let \(L/K\) be a Galois extension of number fields and
\(P\in \operatorname{Spec}{\mathbb{Z}}_K\) be nonzero. Let \(e,f\) be
the common \(e, f\) for all \(Q\) above \(P\), and let \(g\) be the
number of distinct \(Q\) above \(P\). Then
\begin{align*}
efg = n \coloneqq[L:K]
.\end{align*}

\end{theorem}

\begin{proof}[?]

Using the previous efg theorem, \(\sum_{i=1}^g e_i f_i = n\), but by the
previous proposition, all of the \(e_i\) are the same and all of the
\(f_i\) are the same.

\end{proof}

\begin{example}[?]

Let \(L = {\mathbb{Q}}(\zeta_5)\) over \(K= {\mathbb{Q}}\), which is
Galois with
\(G(L/K) \cong ({\mathbb{Z}}/5)^{\times}\cong {\mathbb{Z}}_4\). We know
\({\mathbb{Z}}_L = {\mathbb{Z}}[\zeta_5]\), which actually happens for
any \(n\), and \(\min_{\zeta_5}(x) = x^4 + x^3 + x^2 + x + 1\). Since
\({\mathbb{Z}}_L\) is a monogenic, we can apply Dedekind-Kummer.

\begin{itemize}
\item
  Factoring \(2{\mathbb{Z}}_L = P\) yields a single factor, so \(g=1\).
  Since \(e\) is the common exponent, \(e=1\), so \(f=4\). If you factor
  \(\min_{\zeta_5}\) mod 2, in order to get a single prime D-K says this
  must be irreducible and \(f_i\) are the degrees of the irreducible
  factors.
\item
  \(19{\mathbb{Z}}_L = P_1 P_2\), so \(g=2, e=1\), and so \(f=2\).
\item
  \(11{\mathbb{Z}}_L = P_1 P_2 P_3 P_4\) yields \(g=4, e=1, f=1\).
\item
  \(5{\mathbb{Z}}_L = P_1^4\) so \(g=1,e=4,f=1\).
\end{itemize}

Can the combination \((e,f,g) = (2,2,1)\) occur? This would require a
prime factoring as \(P^2\), but the answer is know. In fact \(e>1\) only
happens for \(5\) since this corresponds (for all applicable primes
\(p\), which is all primes since \({\mathbb{Z}}_L\) is monogenic here)
to the polynomial having a repeated factor mod \(p\). This would require
\({\Delta}(\min_{\zeta_5}(x)) \equiv 0 \operatorname{mod}p\), but you
can show that \(5\) is the only prime that divides this discriminant. So
\(5\) is the only prime that could ramify, so \(e=2\) never happens.

In general, for cyclotomic extensions, \(p\) is always totally ramified
and \(e=p-1\).

\end{example}

\hypertarget{decomposition-inertia-frobenius}{%
\subsection{Decomposition, Inertia,
Frobenius}\label{decomposition-inertia-frobenius}}

\begin{definition}[Decomposition]

Let \(L/K\) be a Galois extension of number fields and let
\(Q \in \operatorname{Spec}{\mathbb{Z}}_L\) be nonzero. Then \(Q\) lies
above a unique prime \(P \in \operatorname{Spec}{\mathbb{Z}}_K\), where
\(P = Q \cap{\mathbb{Z}}_K\). Define the \textbf{decomposition group of
\(Q\)} is defined as
\begin{align*}
D(Q) = D(Q/P) = \left\{{ \sigma \in G(L/K) {~\mathrel{\Big|}~}\sigma(Q) = Q}\right\} 
.\end{align*}

\end{definition}

\begin{definition}[?]

Better notation: define \({\mathbb{F}}_Q \coloneqq{\mathbb{Z}}_L/Q\) and
\({\mathbb{F}}_P = {\mathbb{Z}}_K/P\), so
\({\mathbb{F}}_P \subseteq {\mathbb{F}}_Q\).

\end{definition}

\begin{remark}

We know that the Galois group will take \(Q\) to some prime above \(P\),
these are the ones that take \(Q\) to itself. These are the
automorphisms that make sense ``modulo \(Q\)'': there is a group
morphism
\begin{align*}
{ \text{red} }_{Q/P}: D(G/P) &\to G({\mathbb{F}}_Q/{\mathbb{F}}_P) \\
\sigma &\mapsto \mkern 1.5mu\overline{\mkern-1.5mu\sigma \mkern-1.5mu}\mkern 1.5mu
.\end{align*}
where we reduce \(\sigma\) mod \(Q\), so
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu(\alpha \operatorname{mod}Q) \coloneqq\sigma(\alpha) \operatorname{mod}Q\).
One can check

\begin{itemize}
\item
  For \(\sigma \in D(Q/P)\),
  \(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu\)
  is a well-defined automorphism of \({\mathbb{F}}_Q\) fixing
  \({\mathbb{F}}_P\), since the elements in \({\mathbb{F}}_P\) are of
  the form \(\alpha \operatorname{mod}Q\) where
  \(\alpha\in {\mathbb{Z}}_K\) and \(\sigma\) fixes \(K\). This
  crucially uses that \(\sigma\) fixes \(Q\), otherwise it won't be
  well-defined.
\item
  \({ \text{red} }_{Q/P}\) defines a group morphism.
\end{itemize}

We'll see later that this is in fact a surjective morphism. So all
automorphisms in the Galois group \(G({\mathbb{F}}_Q/{\mathbb{F}}_P)\)
are reductions mod \(Q\) of automorphisms in the decomposition group,
which are the only ones that make sense to reduce mod \(Q\) anyway!

\end{remark}

\hypertarget{ch.-xyz-galois-theory-and-prime-decomposition-april-13}{%
\section{Ch. XYZ: Galois Theory and Prime Decomposition (April
13)}\label{ch.-xyz-galois-theory-and-prime-decomposition-april-13}}

\begin{remark}

Two recommended resources:

\begin{itemize}
\tightlist
\item
  Samuel, Algebraic Theory of Numbers (Dover)
\item
  Matt Baker's
  \href{https://people.math.gatech.edu/~mbaker/pdf/ANTBook.pdf}{online
  notes on Algebraic number theory}.
\end{itemize}

\end{remark}

\begin{remark}

Setup:

\begin{itemize}
\tightlist
\item
  \(L/K\) a Galois extension of number fields.
\item
  \(P \in \operatorname{mSpec}K\)
\item
  \(Q\in \operatorname{mSpec}L\)
\end{itemize}

We saw that \(G(L/K)\) acts on the prime ideals above \(P\), making all
of the \(e,f\) values the same, and thus the decomposition into prime
ideals in this type of extension is simpler than in a general extension.
We'll look at decomposition and inertia groups today.

Recall that if \(Q\) lies above \(P\), then the \textbf{decomposition
group} \(D(Q/P)\) is the set of \(\sigma\in G(L/K)\) such that
\(\sigma(Q) = Q\). Note that \(P\) is redundant in this notation, since
\(Q \cap{\mathbb{Z}}_K = P\). Also recall that
\({\mathbb{F}}_Q \coloneqq{\mathbb{Z}}_L/Q\) is the residue field
associated to \(Q\) and \({\mathbb{Z}}_P \coloneqq{\mathbb{Z}}_L/P\) and
we view \({\mathbb{F}}_P \subseteq {\mathbb{F}}_Q\). Such an extension
of finite fields is always Galois, and the Galois groups turn out to be
related to the decomposition groups. For \(\sigma \in D(Q/P)\) we
defined
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu: {\mathbb{F}}_Q \to {\mathbb{F}}_Q\)
where
\(\alpha \operatorname{mod}Q \mapsto \sigma(\alpha) \operatorname{mod}Q\).
It's easy to check this is

\begin{itemize}
\item
  Well-defined, precisely by the definition of \(D(Q/P)\),
\item
  An automorphism of \({\mathbb{F}}_Q\): it must be surjective since
  everything in \({\mathbb{Z}}_L\) is in the image of \(\sigma\) since
  \(\sigma\) was an automorphism of \({\mathbb{Z}}_L\) to begin with.
  But then it's a surjective morphism from a finite field to itself,
  hence an automorphism.
\item
  \(\sigma\) fixed \({\mathbb{F}}_P\) pointwise Now consider applying
  \(\sigma\) to an element of \({\mathbb{F}}_P\), which are of the form
  \(\alpha \operatorname{mod}Q\) where \(\alpha\) comes from
  \({\mathbb{Z}}_K\). But \(\sigma\) fixed \({\mathbb{Z}}_K\) pointwise,
  so \(\sigma\) fixed \({\mathbb{F}}_P\) pointwise.
\end{itemize}

So each \(\sigma\in D(Q/P)\) yields a
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma \mkern-1.5mu}\mkern 1.5mu\in G({\mathbb{F}}_Q/{\mathbb{F}}_P)\)
and we get a group morphism
\begin{align*}
{ \text{red} }_{Q/P}: D(Q/P) &&\to G({\mathbb{F}}_Q/{\mathbb{F}}_P) \\
\sigma &\mapsto \mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu
.\end{align*}
The following is the deeper and more important fact about this morphism,
which requires a technical proof:

\end{remark}

\begin{theorem}[?]

\({ \text{red} }_{Q/P}\) is surjective.

\end{theorem}

\begin{proof}[?]

We can assume \({\mathbb{F}}_P\) is a proper subfield of
\({\mathbb{F}}_Q\), since the result follows immediately otherwise. By
the primitive element theorem, since \(|FF_Q/{\mathbb{F}}_P\) is
separable we can write
\({\mathbb{F}}_Q = {\mathbb{F}}_P(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu)\)
where \(\alpha\in {\mathbb{Z}}_L\) and
\(\mkern 1.5mu\overline{\mkern-1.5mu{-}\mkern-1.5mu}\mkern 1.5mu\)
denotes reducing mod \(Q\). Note \(\alpha\not\in Q\), since this would
mean
\(\mkern 1.5mu\overline{\mkern-1.5mu\alpha \mkern-1.5mu}\mkern 1.5mu= 0\)
and thus \({\mathbb{F}}_Q = {\mathbb{F}}_P\). On the other hand, we can
assume \(\alpha\in Q'\) for all other \(Q'\neq Q\) above \(P\). Why?
Sketch: the equation
\({\mathbb{F}}_Q = {\mathbb{F}}_P(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu)\)
only depends on \(\alpha\operatorname{mod}Q\), so if \(\alpha\) is not
in some \(Q'\), just adjust \(\alpha\) modulo \(Q'\) without affecting
its class mod \(Q\) to get a new \(\alpha\) in \(Q'\). So if not,
replace \(\alpha\) with \(\alpha'\) satisfying
\begin{align*}
\alpha' \cong \alpha \operatorname{mod}Q \\
\alpha'\cong 0 \operatorname{mod}Q' && \forall Q'\neq Q \text{ above }P
.\end{align*}

Look at the minimal polynomial
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x)\) of
\(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu\)
over \({\mathbb{F}}_P\), some monic polynomial in \({\mathbb{F}}_P[x]\),
where \({\mathbb{F}}_P\) is the reduction mod \(P\) of elements in
\({\mathbb{Z}}_K\). So we can think of
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x)\) as
some \(m(x) \in {\mathbb{Z}}_K[x]\) whose coefficients have been reduced
mod \(Q\). We'll show that for each root \(r\) of
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x)\) in
\({\mathbb{F}}_Q\), there is some \(\sigma\in D(Q/P)\) such that
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu) = r\)
This is enough, since any automorphism of \({\mathbb{F}}_Q\) fixing
\({\mathbb{F}}_P\) is determined by the image of
\(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu\),
which has to go to some other root of
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu (x)\). If
we can show this statement, this means that
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu\)
has to hit every automorphism in \(G({\mathbb{F}}_Q/{\mathbb{F}}_P)\).

Define \(g(x) = \prod_{\sigma\in G(L/K)} (x- \sigma( \alpha))\). Where
does \(g\) live? \(\alpha\) came from \(L\), so
\(\sigma( \alpha) \in L\), and multiplying over all \(\sigma\) puts the
coefficients in \(K\). Even better, since \(\alpha\in {\mathbb{Z}}_L\)
is algebraic, this will have \({\mathbb{Z}}_K\) coefficients, so
\(g\in {\mathbb{Z}}_K[x]\). Now reduce mod \(Q\) to get
\(\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu(x) \in {\mathbb{F}}_P[X]\),
and moreover
\(\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu) = \mkern 1.5mu\overline{\mkern-1.5mug(\alpha)\mkern-1.5mu}\mkern 1.5mu = 0\)
in \({\mathbb{F}}_Q\), since \(g(\alpha) = 0\) by definition since one
\(\sigma\) in the product is the identity. Thus we know
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x) \divides \mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu(x)\)
in \({\mathbb{F}}_P[x]\). Notice that in \({\mathbb{F}}_Q[x]\), if we
take \(g\) and reduce mod \(Q\) we get
\begin{align*}
\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu(x) = \prod_{\sigma \in G(L/K)} (x - \mkern 1.5mu\overline{\mkern-1.5mu\sigma(\alpha)\mkern-1.5mu}\mkern 1.5mu )
.\end{align*}
Since
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x)\)
divides
\(\mkern 1.5mu\overline{\mkern-1.5mug\mkern-1.5mu}\mkern 1.5mu(x)\),
every root of
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu\) has the
form
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma( \alpha)\mkern-1.5mu}\mkern 1.5mu\)
for some \(\sigma\in G(L/K)\). We want this to be
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu)\)
instead to conclude the proof, so take a root of
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu\) in
\({\mathbb{F}}_Q\) and write it as
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma(\alpha)\mkern-1.5mu}\mkern 1.5mu\)
with \(\sigma\in G(L/K)\)

\begin{claim}

\(\sigma \in D(G/P)\) has to be in the decomposition group.

\end{claim}

If this is true, we're done since
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma( \alpha)\mkern-1.5mu}\mkern 1.5mu = \mkern 1.5mu\overline{\mkern-1.5mu\sigma \mkern-1.5mu}\mkern 1.5mu(\mkern 1.5mu\overline{\mkern-1.5mu\alpha\mkern-1.5mu}\mkern 1.5mu)\).

Suppose toward a contradiction that \(\sigma\not\in D(G/P)\), so neither
\(\sigma\) nor \(\sigma^{-1}\) fixes \(Q\), so
\(\sigma^{-1}(Q) \neq Q\). By choice of \(\alpha\), we have
\(\alpha \in Q' \coloneqq\sigma^{-1}(Q)\), so \(\sigma(\alpha)\in Q\).
Then
\(\mkern 1.5mu\overline{\mkern-1.5mu\sigma(\alpha)\mkern-1.5mu}\mkern 1.5mu = 0\),
but \(0\) is not a root of
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x)\)
since it is irreducible where
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x) \neq x\)
since \([{\mathbb{F}}_Q: {\mathbb{F}}_P] \geq 2\) and thus
\(\deg \mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu(x) \geq 2\).

\end{proof}

\begin{remark}

By the first isomorphism theorem, we have
\begin{align*}
G({\mathbb{F}}_Q/{\mathbb{F}}_P) \cong D(Q/P) / \ker { \text{red} }_{Q/P}
.\end{align*}
, where we call the kernel the \textbf{inertia group}:
\begin{align*}
\ker { \text{red} }_{Q/P} = I(Q/P) \coloneqq\left\{{ \sigma \in D(Q/P) {~\mathrel{\Big|}~}\mkern 1.5mu\overline{\mkern-1.5mu\sigma \mkern-1.5mu}\mkern 1.5mu= \operatorname{id}_{{\mathbb{F}}_Q} }\right\}
.\end{align*}
These are the elements \(\sigma\) in the decomposition group such that
\(\sigma(\alpha) \equiv \alpha \operatorname{mod}Q\) for every
\(\alpha\).

Since we have a group action of the Galois group on primes above \(P\),
we can apply Orbit-Stabilizer: we have
\({\operatorname{Stab}}(Q) = D(Q/P)\) and the orbit is all primes above
\(P\), so
\begin{align*}
G(L/K) / D(Q/P) \cong \left\{{\text{Primes above } P}\right\}
.\end{align*}
Taking cardinalities,
\begin{align*}
[L:K] / \# D(Q/P) = g \implies \# D(Q/P) = {[L:K] \over g} = {efg \over g} = ef
.\end{align*}
We also have
\begin{align*}
{ef \over \# I(Q/P)} = \#G({\mathbb{F}}_Q/{\mathbb{F}}_P) = [{\mathbb{F}}_Q: {\mathbb{F}}_P] = f
.\end{align*}
In summary,

\begin{itemize}
\tightlist
\item
  \(\# D(Q/P) = f\),
\item
  \(\# I(Q/P) = e\),
\item
  If \(P\) is unramified, \(e=1\) and \(I(Q/P)\) is trivial and
  \({ \text{red} }_{Q/P}\) is an isomorphism.
\end{itemize}

This map is usually an isomorphism, since there are only finitely many
\(P\) ion \({\mathbb{Z}}_K\) that ramify in \({\mathbb{Z}}_L\).

\end{remark}

\hypertarget{inertia-and-decomposition-fields}{%
\subsection{Inertia and Decomposition
Fields}\label{inertia-and-decomposition-fields}}

\begin{remark}

It will turn out that the fixed fields appearing here have
number-theoretic interpretations.

\end{remark}

\begin{definition}[?]

If \(Q\) lies above \(P\), then define the \textbf{inertia field}
corresponding to \(Q/P\), written \(L^{I(Q/P)}\), and the
\textbf{decomposition field} \(L^{D(Q/P)}\).

\end{definition}

\begin{remark}

Suppose \(L/K\) is Galois, and consider an intermediate field \(L/M/K\).
We can find intermediate primes:

\begin{center}
\begin{tikzcd}
    L && Q \\
    \\
    M && {R = Q \cap{\mathbb{Z}}_M} \\
    \\
    K && P
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=3-1, to=5-1]
    \arrow[no head, from=5-3, to=3-3]
    \arrow[no head, from=3-3, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiTSJdLFswLDQsIksiXSxbMiw0LCJQIl0sWzIsMiwiUiA9IFEgXFxpbnRlcnNlY3QgXFxaWl9NIl0sWzIsMCwiUSJdLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMSwyLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzMsNCwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs0LDUsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=}{Link
to Diagram}
\end{quote}

We can consider \(e, f\) of \(R/P\) in this picture. Going to bigger
extensions then \(M\) makes \(R\) bigger, so \(e\) increases as \(M\)
gets bigger.

\end{remark}

\begin{theorem}[?]

The inertia field \(L^{I(Q/P)}\) is the largest field extension \(M\)
for which \(e(R/P) = 1\). More precisely,
\begin{align*}
e(R/P) = 1 \iff M \subseteq L^{I(Q/P)}
.\end{align*}

\end{theorem}

\begin{remark}

From Galois theory, \(L/M\) is also Galois, so the key to proving this
theorem involves understanding \(D, I\) of \(Q/R\). Noting that
\(G(L/M) \subseteq G(L/K)\), We have
\begin{align*}
D(Q/R) 
&\coloneqq\left\{{\sigma\in G(L/M) {~\mathrel{\Big|}~}\sigma(Q) = Q}\right\} \\
&= \left\{{\sigma\in G(L/K) {~\mathrel{\Big|}~}\sigma(Q) = Q}\right\} \cap G(L/M) \\
&= D(Q/P) \cap G(L/M)
,\end{align*}
so the decomposition groups are related by restriction. Suppose
\(\sigma \in D(Q/R)\), then \({ \text{red} }_{Q/R}(\alpha)\) is an
automorphism of \({\mathbb{F}}_Q\). We can also get an automorphism of
\({\mathbb{F}}_Q\) by taking \({ \text{red} }_{Q/P}(\sigma)\) -- it
turns out these are the same automorphism. Why? Both map
\(\alpha \operatorname{mod}Q\) to
\(\sigma(\alpha) \operatorname{mod}Q\), which doesn't involve \(R\) or
\(P\). We can thus write
\begin{align*}
I(Q/R) 
&= \left\{{ \sigma \in D(Q/R) {~\mathrel{\Big|}~}\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu = \operatorname{id}_{{\mathbb{F}}_Q} }\right\} \\
&= \left\{{ \sigma \in D(Q/P) \cap G(L/M) {~\mathrel{\Big|}~}\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu = \operatorname{id}_{{\mathbb{F}}_Q} }\right\} \\
&= \left\{{ \sigma \in D(Q/P) {~\mathrel{\Big|}~}\mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu = \operatorname{id}_{{\mathbb{F}}_Q} }\right\}\cap G(L/M) \\
&= I(G/P) \cap G(L/M)
.\end{align*}

\end{remark}

\begin{remark}[Upshot]

\begin{align*}
D(Q/R) &= D)Q/P) \cap G(L/M) \\
I(Q/R) &= I)Q/P) \cap G(L/M) \\
.\end{align*}

\end{remark}

\begin{proof}[of theorem]

Let \(L/M/K\) with \(Q/R/P\), we want to show
\(e(R/P) = 1 \iff M \subset L^{I(Q/P)}\):

\begin{center}
\begin{tikzcd}
    L && Q \\
    \\
    M && R \\
    \\
    K && P
    \arrow[no head, from=5-1, to=3-1]
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=5-3, to=3-3]
    \arrow[no head, from=3-3, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiTSJdLFswLDQsIksiXSxbMiwwLCJRIl0sWzIsMiwiUiJdLFsyLDQsIlAiXSxbMiwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMSwiIiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs1LDQsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCwzLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link
to Diagram}
\end{quote}

We'll use multiplicativity of \(e\) in towers. Recall that
\(e(R/P) = e(Q/R) e(R/P)\), so \(e(R/P) = 1\) iff \(e(Q/P) = e(Q/R)\).
Interpreting this as the size of inertia, this happens iff
\(\# I(Q/P) = \# I(Q/R)\), iff \(\#I(Q/P) = \#( I(Q/P) \cap G(L/M) )\).
This happens iff \(I(Q/P) \subseteq G(L/M)\), iff
\(M \subseteq L^{I(Q/P)}\).

\end{proof}

\begin{theorem}[?]

The decomposition field \(L^{D(Q/P)}\) is the largest \(M\) for which
\emph{both} \(e(R/P) = f(R/P) =1\).

\end{theorem}

\begin{proof}[?]

Replace \(e\) in the previous proof with the product \(ef\) and \(I\)
replaced by \(D\), using \(\#D =ef\).

\end{proof}

\begin{remark}

Next time: why these theorems are interesting! How ramification and
splitting completely behaves in fields vs their Galois closures.

\end{remark}

\hypertarget{decomposition-and-inertia-fields-april-15}{%
\section{Decomposition and Inertia Fields (April
15)}\label{decomposition-and-inertia-fields-april-15}}

\hypertarget{ramification-in-composite-fields}{%
\subsection{Ramification in Composite
Fields}\label{ramification-in-composite-fields}}

\begin{remark}

Recall the setup:

\begin{itemize}
\tightlist
\item
  \(L/K\) is a Galois extension of number fields.
\item
  \(P\in \operatorname{Spec}K\) nonzero, and in fact we'll usually mean
  \(P \in \operatorname{mSpec}{\mathbb{Z}}_K\).
\item
  \(Q\in \operatorname{Spec}L\).
\item
  \({\mathbb{F}}_Q \coloneqq{\mathbb{Z}}_L/Q, {\mathbb{F}}_P \coloneqq{\mathbb{Z}}_K/P\)
\end{itemize}

When \(Q\) lies above \(P\), we defined \(D(Q/P)\) as all
\(\sigma\in G(L/K)\) preserving \(Q\). We have a reduction map
\begin{align*}
{ \text{red} }_{Q/P}: D(Q/P) &\to G({\mathbb{F}}_Q / {\mathbb{F}}_P) \\
\sigma &\mapsto \mkern 1.5mu\overline{\mkern-1.5mu\sigma\mkern-1.5mu}\mkern 1.5mu
,\end{align*}
which we saw was a surjective group morphism with kernel
\(I(Q/P) \coloneqq\ker ({ \text{red} }_{Q/P})\). Since \(D, I\) are
subgroups of \(G({\mathbb{F}}_Q/{\mathbb{F}}_P)\), we can consider the
corresponding intermediate fixed fields, the \emph{inertia field} and
\emph{decomposition field}. Using that the Galois correspondence is
inclusion-reversing, we get the following:

\begin{center}
\begin{tikzcd}
    L \\
    \\
    {L^{I(Q/P)}} \\
    \\
    {L^{D(Q/P)}} \\
    \\
    K
    \arrow[no head, from=7-1, to=5-1]
    \arrow[no head, from=5-1, to=3-1]
    \arrow[no head, from=3-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJMIl0sWzAsNiwiSyJdLFswLDIsIkxee0koUS9QKX0iXSxbMCw0LCJMXntEKFEvUCl9Il0sWzEsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFszLDIsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMiwwLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{remark}

Suppose we have a tower \(L/M/K\) with prime ideals \(Q/R/P\) with
\(R\coloneqq Q \cap{\mathbb{Z}}_M\). We showed that with respect to
containment,

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(L^{I(Q/P)}\) is the largest intermediate field in which
  \(e(R/P) = 1\).
\item
  \(L^{D(Q/P)}\) is the largest intermediate field in which
  \(e(R/P) = f(R/P) = 1\).
\end{enumerate}

Note that (b) is a stronger condition, so this matches the containment
in the previous diagram.

\end{remark}

\begin{theorem}[?]

Let \(M/K\) be an extension of number fields, not necessarily Galois.
Let \(L\) be the \textbf{Galois closure} of \(M/K\), the smallest field
extension containing \(M\) which is Galois over \(K\). Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(P\in \operatorname{Spec}K\) is unramified in \(M\) \(\iff\) \(P\) is
  unramified in \(L\).
\item
  \(P\in \operatorname{Spec}K\) splits completely in \(M\) \(\iff\)
  \(P\) splits completely in \(L\).
\end{enumerate}

\end{theorem}

\begin{proof}[?]

We'll just prove (a), and a very similar argument will yield (b).

\(\impliedby\): This is clear: if \(L/M/K\) is a tower and \(P\) is
unramified in \(L\), why is it unramified in \(M\)? This is
multiplicativity of \(e\) in towers. If we have \(R\) in \(M\) with
\(R/P\), we can choose \(Q\) in \(L\) with \(Q/R/P\). If \(e(R/P) > 1\),
we have \(e(Q/P) = e(Q/R) e(R/P) > 1\), forcing \(P\) to ramify in
\(L\). Note that we haven't used that \(L\) is the Galois closure.

\(\implies\): By the primitive element theorem, write \(M = K(\theta)\),
how can we describe the Galois closure \(L\)? Listing
\({ {\theta}_1, {\theta}_2, \cdots, {\theta}_{r}}\) the roots of
\(\min_{\theta}(x)\) over \(K\), then
\(L = K({ {\theta}_1, {\theta}_2, \cdots, {\theta}_{r}})\):

\begin{center}
\begin{tikzcd}
    &&&& L = K({ {\theta}_1, {\theta}_2, \cdots, {\theta}_{r}}) \\
    \\
    {K(\theta_1)} && {K(\theta_2)} && \cdots && {K(\theta_{r-1})} && {K(\theta_{r})} \\
    \\
    &&&& K
    \arrow[no head, from=3-1, to=1-5]
    \arrow[no head, from=3-3, to=1-5]
    \arrow[dashed, no head, from=3-5, to=1-5]
    \arrow[no head, from=3-7, to=1-5]
    \arrow[no head, from=5-5, to=3-1]
    \arrow[no head, from=3-3, to=5-5]
    \arrow[dashed, no head, from=3-5, to=5-5]
    \arrow[no head, from=3-7, to=5-5]
    \arrow[no head, from=3-9, to=5-5]
    \arrow[no head, from=3-9, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=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}{Link
to Diagram}
\end{quote}

By assumption, \(P\) is unramified in \(M\), which is one
\(K(\theta_i)\). There is always an isomorphism between any of the two
\(K(\theta_i)\) which preserves \(K\), so abstractly they're all
equivalent. So \(P\) is unramified in \(K(\theta_i)\) for all \(i\).
Take any \(Q \in \operatorname{Spec}L\) over \(P\). We can consider the
tower involving \(K(\theta_i)\) for any \(i\) and produce a
corresponding prime \(R_i\):

\begin{center}
\begin{tikzcd}
    L && Q \\
    \\
    {K(\theta_i)} && {R_i} \\
    \\
    K && P
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=3-1, to=5-1]
    \arrow[no head, from=1-3, to=3-3]
    \arrow[no head, from=3-3, to=5-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJMIl0sWzAsMiwiSyhcXHRoZXRhX2kpIl0sWzAsNCwiSyJdLFsyLDQsIlAiXSxbMiwwLCJRIl0sWzIsMiwiUl9pIl0sWzAsMSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsxLDIsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCw1LCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzUsMywiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

Since \(P\) is unramified in all \(K(\theta_i)\), we have
\(e(R_i/P) = 1\). Now use the characterization of the inertia field:
\(K(\theta_i) \subset L^{I(Q/P)}\), which is true for every \(i\). So
their composite is also contained in the inertia field, so we have
\begin{align*}
L = K({ {\theta}_1, {\theta}_2, \cdots, {\theta}_{r}}) \subseteq L^{I(Q/P)} \subseteq L
,\end{align*}
yielding equality. Now note that the inertia field was the largest
intermediate field for which \(e=1\), and we've just conclude
\(L = L^{I(Q/P)}\), so \(e(Q/P) = 1\). Since \(Q/P\) was arbitrary,
\(P\) is unramified.

\end{proof}

\begin{theorem}[?]

Let \(M_1/K, M_2/K\) be any two extensions of \(K\). Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(P\) is unramified in \(M_1\) and \(M_2\) \(\iff\) \(P\) is
  unramified in the composite field \(M_1 M_2\).
\item
  \(P\) splits completely in \(M_1\) and \(M_2\) \(\iff\) \(P\) splits
  completely in the composite field \(M_1 M_2\).
\end{enumerate}

\end{theorem}

\begin{proof}[?]

As before, we'll only prove (a).

\(\impliedby\): Again clear by multiplicativity in towers.

\(\implies\): Note that everything involving inertia only makes sense
for Galois extensions, and we haven't assumed that here. So let \(L\) be
the Galois closure of \(M_1M_2/K\), and let \(L_1, L_2\) be the Galois
closures of \(M_1/K, M_2/K\) respectively. By the previous theorem,
\(P\) is unramified in \(L_1, L_2\). How is \(L\) related to \(L_1\) and
\(L_2\)? You can convince yourself that \(L = L_1 L_2\), since the
right-hand side is a Galois extension of \(K\), and it contains
\(M_1 M_2\). We'll forget the \(M_i\) and work with the following
diagram:

\begin{center}
\begin{tikzcd}
    & {} \\
    & L \\
    {L_1} && {L_2} & {} \\
    & K
    \arrow[no head, from=2-2, to=3-1]
    \arrow[no head, from=2-2, to=3-3]
    \arrow[no head, from=3-1, to=4-2]
    \arrow[no head, from=3-3, to=4-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMSwwXSxbMCwyLCJMXzEiXSxbMiwyLCJMXzIiXSxbMywyXSxbMSwxLCJMIl0sWzEsMywiSyJdLFs0LDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbNCwyLCIiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzEsNSwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDUsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=}{Link
to Diagram}
\end{quote}

We have a prime in \(K\) which is unramified in the \(L_i\), and we want
to show it's also unramified in \(L\), which is equivalent to being
unramified in \(M_1 M_2\) (which was our goal). Strat: let
\(Q \in \operatorname{Spec}L\) with \(Q/P\) and let \(R_1, R_2\) be the
corresponding primes in
\(\operatorname{Spec}L_1, \operatorname{Spec}L_2\) between \(Q\) and
\(P\). Since \(P\) is unramified in the \(L_i\), we have
\(e(R_1/P) = e(R_2/P) = 1\). Thus \(L_i \subseteq L^{I(Q/P)}\), and so
is their composite \(L_1 L_2 \subseteq L^{I(Q/P)}\). But
\(L = L_1 L_2\), and so \(e(Q/P) = 1\).

\end{proof}

\hypertarget{frobenius}{%
\subsection{Frobenius}\label{frobenius}}

\begin{remark}

Let \(E/F\) be an extension of \emph{finite} fields, so \(E/F\) is
automatically Galois. Taking \(F \coloneqq{\mathbb{F}}_q\), then \(E\)
is of the form \(E = {\mathbb{F}}_{q^m}\) for some \(m\). Then
\(G(E/F)\) is cyclic of order \(m\), and is generated by
\(x\mapsto x^q\), the \textbf{Frobenius map} on the finite field
\(E/F\). For number fields, we'll want a slightly different notion of
Frobenius.

\end{remark}

\begin{remark}

Let \(L/K\) be a Galois extension of number fields, \(P\) a prime of
\(K\) unramified in \(L\), and \(Q\) a prime of \(L\) above \(P\). Then
\({\mathbb{F}}_Q/{\mathbb{F}}_P\) is an extension of finite fields, so
we know \({ \text{red} }_{Q/P}\) is a surjective group morphism, and the
inertia group has size \(e(Q/P)\), which is 1 in this situation. So this
is an isomorphism between the ``global'' object \(D(Q/P)\) and the
``local'' object \(G({\mathbb{F}}_Q/{\mathbb{F}}_P)\) on residue fields.
We can pull back the generator on the right-hand side to define the
\textbf{Frobenius element}
\begin{align*}
\mathop{\mathrm{Frob}}_{Q/P} \coloneqq{ \text{red} }_{Q/P}^{-1}(x\mapsto x^q)\in D(Q/P) && q\coloneqq\# {\mathbb{Z}}_K/P \coloneqq N(P)
.\end{align*}

\end{remark}

\begin{observation}

Note that \(\mathop{\mathrm{Frob}}_{Q/P}\) is an element of \(G(L/K)\)
where for every
\(\alpha\in {\mathbb{Z}}_L, \mathop{\mathrm{Frob}}_{Q/P}( \alpha) \equiv \alpha^q \operatorname{mod}Q\)
where \(q\coloneqq N(P)\). So \(\mathop{\mathrm{Frob}}_{Q/P}\) acts by
the \(q\)th power map in \({\mathbb{Z}}_L/Q\), and in fact this is a
characterization of the Frobenius element.

\end{observation}

\begin{lemma}[?]

Suppose \(\sigma\in { \mathsf{Gal}} (L/K)\) satisfies
\(\sigma( \alpha) \equiv \alpha^q \operatorname{mod}Q\) for all
\(\alpha\in {\mathbb{Z}}_L\) where \(q \coloneqq N(P)\). Then
\(\sigma = \mathop{\mathrm{Frob}}_{Q/P}\).

\end{lemma}

\begin{remark}

Why this is useful: you can check if something is the Frobenius element
by just checking this congruence.

\end{remark}

\begin{proof}[?]

Take any \(\alpha\in Q\), then
\(\sigma(\alpha) \equiv \alpha^q\operatorname{mod}Q \equiv 0^q \operatorname{mod}Q\)
and so \(\sigma( \alpha) \in Q\). So \(\sigma(Q) \subseteq Q\), but
these are maximal ideals, forcing equality. Then \(\sigma(Q) = Q\)
implies \(\sigma \in D(Q/P)\) by definition, and now applying
\({ \text{red} }_{Q/P}( \sigma) = (x\mapsto x^q)\) is the \(q\)th power
map and
\(\sigma = { \text{red} }_{Q/P}^{-1}(x\mapsto x^q) \coloneqq\mathop{\mathrm{Frob}}_{Q/P}\).

\end{proof}

\begin{proposition}[?]

Suppose \(L/K\) is a Galois extension of number fields with \(L/M/K\)
where we additionally assume that \(M/K\) is Galois. For the tower
\(L/M/K\), take primes \(Q/R/P\) where \(P\) is unramified in \(L\).
Then
\begin{align*}
\mathop{\mathrm{Frob}}_{R/P} = { \left.{{\mathop{\mathrm{Frob}}_{Q/P}}} \right|_{{M}} }
.\end{align*}

\end{proposition}

\begin{remark}

Note that the right-hand side is an automorphism of \(L\) restricted to
\(M\), which is only an automorphism of \(M\) when \(M/K\) is Galois.

\end{remark}

\begin{proof}[?]

Definition chasing and using characterization of Frobenius. It's enough
to show that for all \(\alpha\in {\mathbb{Z}}_M\), we have
\begin{align*}
{ \left.{{\mathop{\mathrm{Frob}}_{Q/P}}} \right|_{{M}} }( \alpha) \equiv \alpha^q \operatorname{mod}R && q = N(P)
.\end{align*}

Since \(\alpha\in {\mathbb{Z}}_M\) and \(M \subseteq L\) so we can think
of \(\alpha\in L\) and it makes sense to compute
\(\mathop{\mathrm{Frob}}_{Q/P}( \alpha) \equiv \alpha^q \operatorname{mod}Q\)
since this is how Frobenius acts upstairs. So we just need to show that
this congruence that holds for \(Q\) also holds for \(R\). Consider the
difference: it's in \(Q\) by the modular condition, and using that
\(M/K\) is Galois, the Frobenius restricts to an automorphism of \(M\)
and thus \(\mathop{\mathrm{Frob}}_{Q/P}( \alpha) \in M\) (and is in fact
still in \({\mathbb{Z}}_M\)). Thus we have a difference of two things in
\({\mathbb{Z}}_M\) and in \(Q\), so
\begin{align*}
\mathop{\mathrm{Frob}}_{Q/P}( \alpha) - \alpha^q \in Q \cap{\mathbb{Z}}_M = R
.\end{align*}

\end{proof}

\begin{remark}

We'll see a nice example later of how to get the law of quadratic
reciprocity from this!

\end{remark}

\hypertarget{frobenius-april-20}{%
\section{Frobenius (April 20)}\label{frobenius-april-20}}

\begin{remark}

Setup and reminders:

\begin{itemize}
\tightlist
\item
  \(L/K\) a Galois extension of number fields,
\item
  \(P \in \operatorname{Spec}K\) unramified in \(L\)
\item
  \(Q\in \operatorname{Spec}L\) above \(P\), so \(e(Q/P) = 1\).
\item
  The Frobenius was defined as
  \({ \text{red} }_{Q/P}^{-1}(x\mapsto x^q)\) for \(q \coloneqq N(P)\).
\item
  The characterization theorem for Frobenius:
  \(\sigma = \mathop{\mathrm{Frob}}_{Q/P}\) iff
  \(\sigma( \alpha) \equiv \alpha^q \operatorname{mod}Q\), so Frobenius
  acts like that \(q\)th power map mod \(Q\).
\item
  When \(L/M/K\) with \(M/K\) Galois with \(Q/R/P\),
  \(\mathop{\mathrm{Frob}}_{R/P} = { \left.{{\mathop{\mathrm{Frob}}_{Q/P}}} \right|_{{M}} }\)
\end{itemize}

\end{remark}

\begin{proposition}[?]

Let \(L/K\) be Galois and let \(Q,Q'\) be two primes lying above \(P\):

\begin{center}
\begin{tikzcd}
    L && Q && {Q'} \\
    \\
    K &&& P
    \arrow[no head, from=1-1, to=3-1]
    \arrow[no head, from=1-3, to=3-4]
    \arrow[no head, from=1-5, to=3-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJMIl0sWzAsMiwiSyJdLFszLDIsIlAiXSxbMiwwLCJRIl0sWzQsMCwiUSciXSxbMCwxLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzMsMiwiIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFs0LDIsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XV0=}{Link
to Diagram}
\end{quote}

Since Galois acts transitively, we can write \(Q' = \sigma(Q)\) for some
\(\sigma \in G(L/K)\). Then
\begin{align*}
\mathop{\mathrm{Frob}}_{Q'/P} = \sigma \circ \mathop{\mathrm{Frob}}_{Q/P} \circ \sigma^{-1}
.\end{align*}

\end{proposition}

\begin{proof}[Sketch]

An exercise, use the characterization theorem. Set \(\tau\) to be the
right-hand side, and check that
\(\tau(\alpha) \equiv \alpha^p \operatorname{mod}Q'\) for all
\(\alpha \in {\mathbb{Z}}_L\) and \(q\coloneqq N(P)\).

\end{proof}

\begin{remark}

What can we do when we are given a \(P\) but not a \(Q\)? There are many
choices of \(Q\), all related by conjugation.

\end{remark}

\begin{definition}[Frobenius Conjugacy Class]

Define the \textbf{Frobenius conjugacy class} as
\begin{align*}
\qty{L/K \over P} = \left\{{ \mathop{\mathrm{Frob}}_{Q/P} {~\mathrel{\Big|}~}Q \text{ lies above } P }\right\} \subseteq G(L/K)
.\end{align*}
This is a conjugacy class of \(G(L/K)\).

\end{definition}

\begin{remark}

Note that this collapses to a single element when
\(G(L/K)\in {\mathsf{Ab}}\)! By abuse of notation, we'll identify
\(\qty{L/K \over P}\) with that common element (despite it being a
singleton set).

\end{remark}

\begin{proposition}[Order of Frobenius]

Every element of \(\qty{L/K \over P}\) has order \(f\), where
\(f = f(Q/P)\) for any \(Q/P\).

\end{proposition}

\begin{proof}[?]

Let \(Q/P\) be a prime over \(P\), then
\(\mathop{\mathrm{Frob}}_{Q/P} \coloneqq{ \text{red} }_{Q/P}^{-1}(x\mapsto x^q)\),
where the \(q\)th power map is the generator of
\(G({\mathbb{F}}_Q/{\mathbb{F}}_P)\), which has size
\([{\mathbb{F}}_Q : {\mathbb{F}}_P] = f\). Since \({ \text{red} }\) was
an isomorphism, we're done.

\end{proof}

\hypertarget{cyclotomic-fields}{%
\subsection{Cyclotomic Fields}\label{cyclotomic-fields}}

\begin{remark}

Fix \(m\). Recall that the \(m\)th cyclotomic field \(K\) is defined by
\begin{align*}
K \coloneqq{\mathbb{Q}}(\zeta_m) && \zeta_m = e^{2\pi i / m}
.\end{align*}
This is the splitting field of \(x^m-1\), and in characteristic zero
this implies \(K/{\mathbb{Q}}\) is Galois.

\end{remark}

\begin{proposition}[?]

\begin{align*}
[K : {\mathbb{Q}}] = \phi(m)
.\end{align*}

\end{proposition}

\begin{definition}[Cyclotomic Polynomials]

Define the \textbf{\(m\)th cyclotomic polynomial} as
\begin{align*}
\Phi_m(x) \coloneqq\prod_{\substack{a\operatorname{mod}m \\ (a, m) = 1}} (x - \zeta_m^a)
.\end{align*}

\end{definition}

\begin{remark}

Note that \(\deg \Phi_m(x) = \phi(m)\), since this is precisely how many
terms show up in the indexing set. We'll show that this is the minimal
polynomial of \(\zeta_m\)

\begin{itemize}
\item
  The coefficients are algebraic integers, since the roots are all roots
  of \(x^m-1\), which is monic in \({\mathbb{Z}}[x]\). Since
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
  is a ring, we have
  \(\Phi_m(x) \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu[x]\).
\item
  If \(\sigma \in G(K/{\mathbb{Q}})\), then
  \(\sigma(\zeta_m) = \zeta_m^b\) for some \(b\) coprime to \(m\). Then
  \begin{align*}
  \sigma(\Phi_m(x)) = \prod_{\substack{a\operatorname{mod}m \\ (a, m) = 1}} (x - \zeta_m^{ab}) = \Phi_m(x)
  ,\end{align*}
  since as \(a\) runs through the number coprime to \(m\), so does
  \(ab\). Thus \(\Phi_m(x) \in {\mathbb{Q}}[x]\), since its coefficients
  are fixed by every element of the Galois group.
\item
  Combining these, the coefficients are in
  \(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu[x] \cap{\mathbb{Q}}[x] = {\mathbb{Z}}[x]\).
\end{itemize}

\end{remark}

\begin{proposition}[?]

\(\Phi_m(x) = m(x) \coloneqq\min_{\zeta_m}(x)\) is the minimal
polynomial of \(\zeta_m\) over \({\mathbb{Q}}\).

\end{proposition}

\begin{proof}[?]

Clearly \(m(x) \divides \Phi_m(x)\) in \({\mathbb{Q}}[x]\) since
\(\Phi_m(x)\) vanishes at \(\zeta_m\). So every root of \(m(x)\) is a
primitive \(m\)th root of unity, so we just need to show that every
primitive \(m\)th root of unity is a root of \(m\), i.e.~we get all of
them. Observe that if \(\zeta\) is \emph{any} \(m\)th root of unity,
then \(\zeta \in {\mathbb{Z}}_K\): \(\zeta\) is an algebraic integer, as
a root of \(x^m-1\), and is in \(K\) since it's a power of \(\zeta_m\).
Also note that \(m(x) \in {\mathbb{Z}}[x]\) since the minimal polynomial
of \emph{any} algebraic integer has rational integer coefficients. So
\(m(\zeta)\in {\mathbb{Z}}_K\) since \({\mathbb{Z}}_K\) is a ring and
\(\zeta\in {\mathbb{Z}}_K\), and \(N(m(\zeta)) \in {\mathbb{Z}}\).

\begin{claim}

There is an \(M\in {\mathbb{Z}}^{\geq 0}\) such that if \(q\) is any
prime with \(q> M\) and \(\zeta\) is any root of \(m(x)\), then
\(\zeta^q\) is still a root of \(m(x)\).

I.e. the roots of \(m(x)\) are closed under taking \(q\)th powers for
large enough \(q\). Moreover, it's enough to pick any
\(M> \max_{\zeta} {\left\lvert {N(m(\zeta))} \right\rvert}\), taking
\(\zeta\) over all \(m\)th roots of 1.

\end{claim}

\begin{proof}[of claim]

Let \(q>M\) be chosen as above and let \(\zeta\) be a root of \(m(x)\).
We know \(\zeta\) is a primitive \(m\)th root of unity and in
\({\mathbb{Z}}_K\). Work modulo \(Q\): since \(0 = m(\zeta)\) we have
\begin{align*}
0 
\equiv m(\zeta)^q \equiv m(\zeta^q)
.\end{align*}
Hence
\begin{align*}
q \divides N(q) \divides N(m(\zeta^q))
,\end{align*}
using that \(N(q) = q^d\) since it's a rational integer where \(d\) is
the degree of the number field. But we have
\(q > {\left\lvert { N(m(\zeta^q))} \right\rvert}\), which can only
happen if the right-hand side is zero. The only element of norm zero in
a number field is zero, so \(m(\zeta^q) = 0\).

\end{proof}

So if \(q>M\) and \(\zeta\) is a root of \(m(x)\), so is \(\zeta^q\).
\(\zeta_m\) is a root of \(m(x)\), thus so us
\(\zeta_m^{q_1}, \zeta_m^{q_1 q_2}, \cdots, \zeta_m^{A}\) for any
\(A \in {\mathbb{Z}}^{\geq 0}\) which can be written as a product of
primes bigger than \(M\).

\begin{exercise}[?]

Show that we can choose any \(a\in {\mathbb{Z}}\) with
\(\gcd(a, m) = 1\), and choose \(A\) as above with
\(A \cong a \operatorname{mod}m\).

\end{exercise}

With this, \(\zeta_m^a = \zeta_m^A\) will be a root of \(m(x)\) and
we're done.

\end{proof}

\begin{remark}

Note that the exercise does follow from Dirichlet's theorem for
arithmetic progressions, but there are easier proofs.

\end{remark}

\hypertarget{galois-theory-of-mathbbqzeta_m-mathbbq-and-the-frobenius}{%
\subsection{\texorpdfstring{Galois Theory of
\({\mathbb{Q}}(\zeta_m) / {\mathbb{Q}}\) and The
Frobenius}{Galois Theory of \{\textbackslash mathbb\{Q\}\}(\textbackslash zeta\_m) / \{\textbackslash mathbb\{Q\}\} and The Frobenius}}\label{galois-theory-of-mathbbqzeta_m-mathbbq-and-the-frobenius}}

\begin{remark}

Let \(\sigma \in G(K/{\mathbb{Q}})\), then we saw that
\(\sigma(\zeta_m) = \zeta_m^a\) for some \(a\) coprime to \(m\). So we
know \(\# G(K/{\mathbb{Q}}) < \phi(m)\), since there are only that many
possibilities for the right-hand side. Since
\([K:{\mathbb{Q}}] = \phi(m)\), for each \(a\) coprime to \(m\) there is
a \(\sigma_a\) with \(\sigma_a(\zeta_m) = \zeta_m^a\), so we can write
\begin{align*}
G(K/{\mathbb{Q}}) = \left\{{ \sigma_a {~\mathrel{\Big|}~}a\in ({\mathbb{Z}}/m)^{\times}}\right\} 
.\end{align*}
Noting that
\(\sigma_a \sigma_b (\zeta_m) = \zeta_m^{ab} = \sigma_{ab}\), so we get
an isomorphism
\begin{align*}
G(K/{\mathbb{Q}}) &\xrightarrow{\sim} ({\mathbb{Z}}/m) ^{\times}\\
\sigma_a &\rightleftharpoons a\operatorname{mod}m
,\end{align*}
so \(K/{\mathbb{Q}}\) is an abelian extension.

\end{remark}

\begin{remark}

Consider \(K \coloneqq{\mathbb{Q}}(\zeta_m)\), then for every rational
prime \(P\) unramified in \(K\), there is a well-defined element in
\(\qty{K/{\mathbb{Q}}\over P} \in G(K/{\mathbb{Q}}) = ({\mathbb{Z}}/m)^{\times}\).
For a given \(P\), which element in \(({\mathbb{Z}}/m)^{\times}\) do you
get?

\end{remark}

\begin{proposition}[?]

If \(p\notdivides m\), then \(p\) is unramified in
\({\mathbb{Q}}(\zeta_m)\).

\end{proposition}

\begin{remark}

The converse is more or less true:
\({\mathbb{Q}}(\zeta_2) = {\mathbb{Q}}\) since \(\zeta_2 = -1\), but 2
is not ramified in \({\mathbb{Q}}\) since nothing is ramified in itself.
So if you avoid \(m\equiv 2 \operatorname{mod}4\), the converse becomes
true. We'll just prove the mentioned direction.

\end{remark}

\begin{lemma}[?]

\begin{align*}
{\mathbb{Z}}_K = {\mathbb{Z}}[\zeta_m]
.\end{align*}

\end{lemma}

\begin{exercise}[?]

Show that \(p\) ramifies in \(K\) iff \({\mathbb{Z}}_K/p{\mathbb{Z}}_K\)
has a nonzero nilpotent element.

\end{exercise}

\begin{proof}[of proposition, cute!]

Suppose \(p\notdivides m\), then it suffices to show that
\({\mathbb{Z}}[\zeta_m] / p{\mathbb{Z}}[\zeta_m]\) has no nonzero
nilpotents. Let \(\alpha\in {\mathbb{Z}}[\zeta_m]\) with
\(\alpha \operatorname{mod}p\) nilpotent in the quotient, we'll show it
must be zero in the quotient. By the lemma, we can write
\(\alpha = \sum_{i=0}^{d-1} a_i \zeta^i\), using that
\({\mathbb{Z}}_K = {\mathbb{Z}}[\zeta]\) where all of the \(a_i\) are in
\({\mathbb{Z}}\), we've set \(\zeta\coloneqq\zeta_m\) and
\(d = \phi(m) = [K : {\mathbb{Q}}]\).

Look mod \(p\), then
\begin{align*}
\alpha^p &\equiv \sum_{i=0}^{d-1} a_i \zeta^{ip} \operatorname{mod}p
\implies \alpha^{p^f} &\equiv \sum_{i=0}^{d-1} a_i \zeta^{ip^f} \operatorname{mod}p && \forall f
.\end{align*}
If \(f\) is large enough, \(\alpha^{p^f} \equiv 0 \operatorname{mod}p\),
since some power of \(\alpha\) is zero. On the other hand,
\(p\notdivides m\), so there are powers of \(p\) that are
\(1\operatorname{mod}m\) that show up regularly, and we can choose \(f\)
so that \(p^f \cong 1 \operatorname{mod}m\), e.g.~by choosing \(f\) to
be any multiple of the order of \(p\) mod \(m\). But then the second
line above reduces to
\(\sum a_i\zeta^{ip^f} \equiv \sum a_i \zeta^{i} = \alpha\) since
\(p^f \equiv 1 \operatorname{mod}m\), so \(\alpha^{p^f} \equiv \alpha\),
but we know \(\alpha^{p^f} \equiv 0\).

\end{proof}

\begin{theorem}[?]

Suppose \(p\notdivides m\), which guarantees \(p\) does not ramify in
\(K\). Then
\begin{align*}
\qty{K/{\mathbb{Q}}\over P} = \sigma_p && \sigma_p(\zeta_m) \coloneqq\zeta_m^p
.\end{align*}

\end{theorem}

\begin{proof}[?]

Choose any \(Q\) lying above \(p\). We'll show that for all
\(\alpha\in {\mathbb{Z}}_K\),
\(\sigma_p( \alpha) = \alpha^p \operatorname{mod}Q\), which by the
characterization theorem will show
\(\sigma_p = \mathop{\mathrm{Frob}}_{Q/P}\), and in the abelian case
\(\qty{K/{\mathbb{Q}}\over p}\) reduces to this single element. We know
that \(Q\supseteq p{\mathbb{Z}}_K\) since \(Q\) is above \(p\), so it's
enough to show \(\sigma_p( \alpha) \equiv \alpha^p \operatorname{mod}p\)
for all \(\alpha\in {\mathbb{Z}}_K\) -- this is because the difference
is a multiple of \(p\) in \({\mathbb{Z}}_K\), which is a subset of
\(Q\). Using \({\mathbb{Z}}_K = {\mathbb{Z}}[\zeta_m]\) to write
\(\alpha = \sum_{i=0}^{d-1} a_i \zeta_m^{i}\) where \(d\) is the degree
of the extension. Now reducing \(\operatorname{mod}p\) and applying the
Freshman's dream yields
\begin{align*}
\sigma_p( \alpha) 
&= \sum_{i=0}^{d-1} a_i \zeta_m^{ip} \\
&\equiv \qty{ \sum_{i=0}^{d-1} a_i \zeta_m^{i}}^p \operatorname{mod}p \\
&\equiv \alpha^p
.\end{align*}

\end{proof}

\begin{remark}

This can be used to determine how all of the primes \(p\notdivides m\)
factor in \({\mathbb{Q}}(\zeta_m)\), i.e.~easily determining the
relevant values of \(f\) and \(g\).

\end{remark}

\hypertarget{chebotarev-density-april-26}{%
\section{Chebotarev Density (April
26)}\label{chebotarev-density-april-26}}

\hypertarget{setup-1}{%
\subsection{Setup}\label{setup-1}}

\begin{remark}

Today: the Frobenius element, cyclotomic extensions of \({\mathbb{Q}}\),
Chebotarev density theorem, and some applications. We'll then look at
applying Frobenius to understanding squares modulo a prime and get a
proof of quadratic reciprocity.

Setup: let \(L/K\) be a Galois extension of number fields,
\(P \in \operatorname{Spec}K\) unramified in \(L\), and
\(Q\in \operatorname{Spec}L\) lying above \(p\).

\begin{itemize}
\item
  There is a unique element
  \(\mathop{\mathrm{Frob}}_{Q / P} \in { \mathsf{Gal}} (L/K)\)
  satisfying
  \begin{align*}
  \mathop{\mathrm{Frob}}_{Q/P}(\alpha) = \alpha^p \operatorname{mod}q
  &&
  p \coloneqq\# {\mathbb{Z}}_K/p
   .\end{align*}
\item
  As \(Q\) ranges over the primes of \(L\) over \(P\), then
  \(\mathop{\mathrm{Frob}}_{Q/P}\) ranges over an entire conjugacy
  class, which we denoted \({ \qty{ L/K \over P } }\). For abelian
  Galois groups, we'll identify this class with its single
  representative.
\end{itemize}

\end{remark}

\begin{remark}

How does this theory play out for cyclotomic extensions? We've seen that
for \(K\coloneqq{\mathbb{Q}}(\zeta_n)\), the extension
\(K/{\mathbb{Q}}\) is Galois of degree
\(\phi(m) = \# ({\mathbb{Z}}/m)^{\times}\), and moreover these are
isomorphic via
\begin{align*}
({\mathbb{Z}}/m)^{\times}&\to { \mathsf{Gal}} (K/{\mathbb{Q}})\\
a\operatorname{mod}m &\mapsto \sigma_a: \zeta_a \mapsto \zeta_m^a
.\end{align*}

\end{remark}

\begin{proposition}[?]

If \(p\) is prime and \(p\notdivides m\), then \(p\) is unramified in
\(K\). We proved this last time, and it follows that
\({ \qty{ K/Q \over P } } \in { \mathsf{Gal}} (K/{\mathbb{Q}}) \cong ({\mathbb{Z}}/m)^{\times}\).
Moreover, we saw that
\begin{align*}
{ \qty{ K/Q \over P } } \equiv p \operatorname{mod}m
.\end{align*}

\end{proposition}

\begin{remark}

Thus for any two primes not dividing \(m\) which become equal modulo
\(m\), then they split in the same way in the cyclotomic field. So the
way a rational prime splits in \({\mathbb{Q}}(\zeta_m)\) is entirely
determined by its residue class \(\operatorname{mod}m\).

\end{remark}

\begin{corollary}[?]

If \(p\) is a prime not dividing \(m\), then \(e,f,g\) depend only on
\(p\operatorname{mod}m\).

\end{corollary}

\begin{proof}[?]

If \(p\) doesn't divide \(m\), \(p\) is unramified and \(e=1\). Recall
that \(f\) is the order of any Frobenius element above \(p\) in the
Galois group, \({ \qty{ K/Q \over P } } = p \operatorname{mod}m\) in
\(({\mathbb{Z}}/m)^{\times}\). Note that \(efg = \phi(m)\) is the degree
of the field, and solving yields \(g = \phi(m)/f\).

\end{proof}

\begin{remark}

In class field theory, there are analogs of this, foreshadowing
something that happens in general for abelian extensions of number
fields. In fact, every such extension is a subfield of the cyclotomic
field.

\end{remark}

\hypertarget{chebotarevs-theorem}{%
\subsection{Chebotarev's Theorem}\label{chebotarevs-theorem}}

\begin{question}

A classical question: let \(a\operatorname{mod}m\) be a coprime residue
class, are there infinitely many (positive) rational primes \(p\) with
\(p\equiv a\operatorname{mod}m\)? For instance, are there infinitely
many primes \(p\) satisfying \(p \equiv 5 \operatorname{mod}11\)?

\end{question}

\begin{remark}

Note that only coprime residue classes are interesting here.
Considering, say, \(p\equiv 5 \operatorname{mod}10\) would only yield
multiples of 5 and thus \(p=5\) is the only solution. In general, for
\(a\operatorname{mod}m\) with \(a, m\) not coprime, there is at most one
solution.

\end{remark}

\begin{answer}

For \(a, m\) coprime, it is a theorem of Dirichlet that there are
infinitely many. This proof uses \(L{\hbox{-}}\)functions and their
analytic behavior.

\end{answer}

\begin{theorem}[Chebotarev's Density Theorem: The Fundamental Theorem of Algebroanalytic Number Theory]

Let \(L/K\) be a Galois extension of number fields and \(C\) a conjugacy
class of \({ \mathsf{Gal}} (L/K)\). Note that for every prime \(P\) of
\(K\) that is unramified in \(L\), we can associate the Frobenius
conjugacy class. There are only finitely many primes of \(K\) that
ramify in \(L\), so all but finitely many primes yield a conjugacy class
-- is it this fixed conjugacy class \(C\)?

The theorem is that there are infinitely many primes \(P\) of \(K\) for
which \(P\) is unramified in \(L\) and the Frobenius conjugacy class
satisfies \({ \qty{ L/K \over P } } = C\).

\end{theorem}

\begin{remark}

This is qualitatively stated, but one can make this quantitative --
there is a well-defined proportion of primes for which this is true.
Letting \(P\) range over the primes of \(K\), the \emph{proportion} with
\(\qty{L/K \over P} = C\) will be \(\# C / [L:K]\). Here
\emph{proportion} is defined in the following way:
\begin{align*}
\lim_{x\to\infty} 
\qty{ \# \left\{{ P \in \operatorname{Spec}K {~\mathrel{\Big|}~}N(P) \leq x, \, \qty{L/K \over P} = C}\right\} \hspace{1em} \over \# \left\{{ P\in \operatorname{Spec}K {~\mathrel{\Big|}~}N(P) \leq x}\right\} }
,\end{align*}
which is sometimes called the \emph{natural density} of these primes.
What Chebotarev proved is slightly weaker and involved the
\emph{Dirichlet density} instead.

\end{remark}

\begin{corollary}[?]

We recover Dirichlet's theorem on primes
\(p\equiv a \operatorname{mod}m\) by taking
\(L \coloneqq{\mathbb{Q}}(\zeta_m)\), \(K={\mathbb{Q}}\), and
\(C \coloneqq\left\{{a\operatorname{mod}m}\right\}\). The density
theorem yields infinitely many primes \(P\) with Frobenius equal to
\(C\), but then
\(\qty{L/K\over p} = \left\{{p\operatorname{mod}m}\right\}\), yielding
\(p\equiv a \operatorname{mod}m\). Moreover the proportion of such
primes is \(1/\phi(m)\), the degree of \([L:K]\), so the coprime residue
classes are essentially distributed uniformly.

\end{corollary}

\begin{remark}

This is useful when one needs infinitely many primes of a certain form,
for which one can apply the density theorem to a well-chosen extension
of number fields.

\end{remark}

\begin{question}

Consider primes \(p\) and consider the multiplicative order of
\(2\operatorname{mod}p\), which is a divisor of \(p-1\). How often is
this order even or odd? One might expect this to happen half of the
time, but this is not true. Instead:

\end{question}

\begin{corollary}[of a strong Chebotarev theorem, due to Hasse]

The proportion of \(p\) for which \(2\) has even order
\(\operatorname{mod}p\) is \(17/24\).

\end{corollary}

\begin{remark}

Note that 2 is special, and this proportion changes for
\(3\operatorname{mod}p\) to something around \(2/3\). Interesting
algebraic number theory governs why this is \emph{not} \(17/24\), and
involves how different number fields intersect.

\end{remark}

\hypertarget{residues-operatornamemodp-and-quadratic-reciprocity}{%
\subsection{\texorpdfstring{Residues \(\operatorname{mod}p\) and
Quadratic
Reciprocity}{Residues \textbackslash operatorname\{mod\}p and Quadratic Reciprocity}}\label{residues-operatornamemodp-and-quadratic-reciprocity}}

\begin{remark}

If \(G\) is a cyclic group of even order, there is a unique morphism
\begin{align*}
\phi_G:G &\to \left\{{\pm 1}\right\} \\
g &\mapsto 
\begin{cases}
1 &  g \in G^2
\\
-1 & \text{else}.
\end{cases}
\end{align*}

\end{remark}

\begin{definition}[Legendre Symbol]

Let \(p\) be an odd prime, and let
\(G \coloneqq({\mathbb{Z}}/p)^{\times}\) which is cyclic of order
\(p-1\). For each \(a\in {\mathbb{Z}}\) coprime to \(p\), define the
\textbf{Legendre symbol}
\begin{align*}
{ \qty{ a \over p } } \coloneqq\phi_G(a\operatorname{mod}p) 
=
\begin{cases}
1 &  a = a_0^2\operatorname{mod}p \text{ for some } a_0\in G
\\
-1 & \text{else]}.
\end{cases}
\end{align*}
Note that this is a group morphism, and thus multiplicative and we get
\begin{align*}
{ \qty{ ab \over p } } = { \qty{ a \over p } } { \qty{ b \over p } } && a,b \in {\mathbb{Z}}\text {coprime to} p
.\end{align*}
One can also extend the symbol by defining \({ \qty{ n \over p } } = 0\)
when \(n\) is not coprime to \(p\).

\end{definition}

\begin{question}

Given \(a\), can we characterize the primes \(p\) (not dividing \(a\))
for which \({ \qty{ a \over p } } =1\).

\end{question}

\begin{remark}

Let \(K = {\mathbb{Q}}(\sqrt d)\) where \(d\) is squarefree, then
\(K/{\mathbb{Q}}\) is abelian and the Galois group can be thought of as
\(G \coloneqq{ \mathsf{Gal}} (K/{\mathbb{Q}}) = \left\{{\pm 1}\right\}\).
Each rational prime \(P\) unramified in \(K\) will yields a well-defined
Frobenius conjugacy class in \(K\), and since \(G\) is a abelian this is
a single element -- which element is it?

If \(p\) is an odd prime not dividing \(d\), we know from our study of
quadratic fields that \(p\) is unramified in \(K\).

\end{remark}

\begin{proposition}[?]

Let \(p\) be an odd prime not dividing \(d\), then
\begin{align*}
{ \qty{ K/{\mathbb{Q}} \over p } } = { \qty{ d \over p } }
,\end{align*}
where we've identified the Galois group as \(\left\{{\pm 1}\right\}\).
Thus is \(d\) is a square \(\operatorname{mod}p\), the Frobenius is
trivial, and conversely if \(d\) is not a square then the Frobenius is
nontrivial.

\end{proposition}

\begin{proof}[?]

Since \(p\) is odd not dividing \(d\), \(p\) doesn't ramify and thus
\(e=1\). Since we're in a quadratic field, the only other things that
can happen are being split (completely) or inert. So when does \(p\)
split? This is precisely when \(g=2\), so there are two primes above
\(p\), and we know \(efg=2\) is the degree of the extension, so \(f=1\).
But \(f\) is the order of any Frobenius element, so the order of
\({ \qty{ K/{\mathbb{Q}} \over p } } = 1\), making it the identity. So
\(p\) splits \(\iff { \qty{ K/{\mathbb{Q}} \over p } } = 1\).

On the other hand, when we worked out how primes factor in quadratic
fields, we used Dedekind-Kummer, and if \(p\) is odd not dividing \(d\)
then \({ \qty{ d \over p } } = 1\). Equivalently, \(p\) splitting
involves looking at how \(x^2-d\) factors in \(K\), and that's what
yields this condition. So \(p\) splits
\(\iff { \qty{ d \over p } } = 1\).

\begin{quote}
Note that looking at how arbitrary primes factor involves looking at the
minimal polynomial of \(\tau = \sqrt{d}, {1 + \sqrt d \over 2}\), but we
can throw out the second case here if we're willing to throw out primes
dividing the index
\([{\mathbb{Z}}_K : {\mathbb{Z}}[\sqrt d] ] \in \left\{{1, 2}\right\}\).
So for odd primes, the way \(p\) factors is exactly the way \(x^2-d\)
factors.
\end{quote}

Thus \(p\) splits iff both of these quantities are \(+1\) and doesn't
split iff they're both \(-1\), so they must be equal.

\end{proof}

\begin{proposition}[?]

If \(p\) is an odd prime,
\begin{align*}
{ \qty{ -1 \over p } } = 
\begin{cases}
1 & p\equiv1\operatorname{mod}4 
\\
-1 & \text{else}.
\end{cases}
\end{align*}

\end{proposition}

\begin{proof}[?]

Consider \(K \coloneqq{\mathbb{Q}}(\sqrt{-1})\). If \(p\) is an odd
prime, we can identify
\({ \mathsf{Gal}} (K/{\mathbb{Q}}) = \left\{{\pm 1}\right\}\) to obtain
\begin{align*}
{ \qty{ K/{\mathbb{Q}} \over p } } = { \qty{ -1 \over p } }
.\end{align*}
Note that \(K\) is also the cyclotomic extension
\(K = {\mathbb{Q}}(\zeta_4)\), and for cyclotomic extensions we can
compute the Frobenius elements as
\({ \qty{ K/{\mathbb{Q}} \over p } } \equiv p \operatorname{mod}4\),
viewing
\({ \mathsf{Gal}} (K/{\mathbb{Q}}) = ({\mathbb{Z}}/4)^{\times}\).

From the first perspective, we have
\({ \qty{ K/{\mathbb{Q}} \over p } } = \operatorname{id}\iff { \qty{ -1 \over p } } = 1\),
and from the second perspective this happens
\(\iff p \equiv 1 \operatorname{mod}4\).

\end{proof}

\begin{remark}

This is perhaps overkill, but it's nice that it follows easily from the
general theory! What about \(2\) instead of \(-1\)?

\end{remark}

\begin{proposition}[?]

If \(p\) is an odd prime,
\begin{align*}
{ \qty{ 2 \over p } } = 
\begin{cases}
1 & p\equiv \pm 1\operatorname{mod}8
\\
-1 & \text{else}.
\end{cases}
\end{align*}

\end{proposition}

\begin{remark}

The usual elementary proof uses Gauss' lemma, and there are other proofs
that use clever counting.

\end{remark}

\begin{proof}[?]

We have a tower of extensions
\(K/L/{\mathbb{Q}}\coloneqq{\mathbb{Q}}(\zeta_8) / {\mathbb{Q}}(\sqrt 2) / {\mathbb{Q}}\),
and consider corresponding primes \(Q/R/P\). Since the overall extension
is Galois, by the Galois correspondence we can write
\({\mathbb{Q}}(\sqrt 2) = {\mathbb{Q}}(\zeta_8)^H\) as the fixed field
of some
\(H\leq G\coloneqq{ \mathsf{Gal}} ({\mathbb{Q}}(\zeta_8)/{\mathbb{Q}})\)
of index 2.

We know that \(G = ({\mathbb{Z}}/8)^{\times}\) which has size 4, so
\(H\) has 2 elements: the identity and an order 2 element. All of the
\(g\in G\) are order 2, so which is it? We look for an automorphism of
\(K/{\mathbb{Q}}\) fixing \(L\), and taking \(\sigma\) to be complex
conjugation works since \(\sigma(\zeta_8) = \zeta_8^{-1}\neq \zeta_8\),
so \(\sigma \neq \operatorname{id}\) is nontrivial. So
\(\sigma = -1\operatorname{mod}8\) and we can write
\(H = \left\{{\pm 1\operatorname{mod}8}\right\}\).

Now note that
\begin{align*}
{ \qty{ 2 \over p } } = 1 \iff { \qty{ {\mathbb{Q}}(\sqrt 2)/{\mathbb{Q}} \over p } } = \operatorname{id}
,\end{align*}
since the Legendre and Frobenius symbols are equal here. This happens
\(\iff \mathop{\mathrm{Frob}}_{R/P} = \operatorname{id}\iff { \left.{{\mathop{\mathrm{Frob}}_{Q/P}}} \right|_{{{\mathbb{Q}}(\sqrt 2)}} } = \operatorname{id}\),
using what we've previously proved about how Frobenius behaves in towers
of extensions. This happens
\(\iff { \qty{ {\mathbb{Q}}(\zeta_8)/{\mathbb{Q}} \over P } } \in H \iff p\operatorname{mod}8 \in \left\{{\pm 1 \operatorname{mod}8}\right\}\).

\end{proof}

\begin{remark}

For a general \(a\in {\mathbb{Z}}_{\neq 0}\), we can factor
\(a = (\pm 1)(2^k) \prod p_i\) where the \(p_i\) are odd primes. Using
multiplicativity of the Legendre symbol, since we understand the first
two terms, it only remains to understand \({ \qty{ q \over p } }\) for
\(q\) an odd prime -- this is precisely what quadratic reciprocity gives
us.

\end{remark}

\hypertarget{quadratic-reciprocity-april-29}{%
\section{Quadratic Reciprocity (April
29)}\label{quadratic-reciprocity-april-29}}

\begin{remark}

Last time: quadratic residues mod \(p\), and how all of the usual
content from a first course in number theory follows immediately from
the high-powered machinery of the Frobenius and cyclotomic fields. The
problem we were looking at was the following: given
\(a\in {\mathbb{Z}}_{\neq 0}\), classify all odd primes \(p\) where
\({ \qty{ a \over p } } = 1\).

For \(a=-1\), we saw that this happens iff
\(p \equiv 1 \operatorname{mod}4\), and for \(a=2\),
\(p\equiv \pm 1 \operatorname{mod}8\). The key facts for the latter were
that \({\mathbb{Q}}(i) = {\mathbb{Q}}(\zeta_4)\) and
\({\mathbb{Q}}(\sqrt 2) \subseteq {\mathbb{Q}}(\zeta_8)\). Then because
the Legendre symbol is multiplicative, we can reduce to the case \(a=q\)
is an odd prime.

This was solved by Gauss in the early 19th century, which he called the
``golden theorem'', but is more commonly known now as the law of
quadratic reciprocity.

\end{remark}

\hypertarget{quadratic-reciprocity}{%
\subsection{Quadratic Reciprocity}\label{quadratic-reciprocity}}

\begin{definition}[?]

For an odd prime \(p\), define \(p^* \coloneqq\pm p\) with the sign
chosen such that \(p^* \equiv 1 \operatorname{mod}3\). Explicitly, one
can check that \(p^* = (-1)^{p-1\over 2}p\). The law of
\textbf{quadratic reciprocity} states the following: if \(p, q\) are
distinct odd primes, then
\begin{align*}
{ \qty{ q^* \over p } } = { \qty{ p \over q } }
.\end{align*}

\end{definition}

\begin{remark}

This is not the usual formulation, but it can be recovered easily: note
that
\begin{align*}
{ \qty{ q^* \over p } }
&=
{ \qty{ (-1)^{q-1 \over 2} q \over p } }\\
&=
{ \qty{ -1 \over p } }^{q-1 \over 2} { \qty{ q \over p } } \\
&= 
\qty{ (-1)^{p-1\over 2} }^{q-1 \over 2} { \qty{ q \over p } } \\
\implies 
(-1)^{ {p-1\over 2}{q-1 \over 2} } 
&= { \qty{ p \over q } } { \qty{ q \over p } }
,\end{align*}
where the last step follows from multiplying through by
\({ \qty{ q \over p } }\) and noting that
\({ \qty{ q \over p } }^2 = 1\)

\end{remark}

\begin{proof}[of quadratic reciprocity]

Let \(p\neq q\) be distinct odd primes. Then
\({ \mathsf{Gal}} ({\mathbb{Q}}(\zeta_q) / {\mathbb{Q}})\cong ({\mathbb{Z}}/q)^{\times}\).
Since this is a cyclic group of even order, there is a unique index 2
subgroup, which we can take to be the subgroup of square
\(U({\mathbb{Z}}/q)^2\). By Galois theory, the fixed field
\({\mathbb{Q}}(\zeta_q)^H\) corresponds to a unique quadratic subfield.

\begin{claim}

This subfield is exactly \({\mathbb{Q}}(\sqrt{q^*})\).

\end{claim}

\begin{proof}[?]

Let \(K\coloneqq{\mathbb{Q}}(\zeta_q)^H\), so we have a tower
\({\mathbb{Q}}(\zeta_q)/K/{\mathbb{Q}}\). Suppose \(r\) is a prime that
ramifies in \(K\), then it ramifies in any larger field by
multiplicativity of the \(e\) values. So \(r\) ramifies in
\({\mathbb{Q}}(\zeta_q)\), but the only primes that can possibly ramify
in any \({\mathbb{Q}}(\zeta_m)\) are those that divide \(m\). So
\(r\divides q\), forcing \(r=q\), so the only prime that ramifies is
\(q\).

Write \(K = {\mathbb{Q}}(\sqrt d)\) for \(d\) squarefree, then every
prime dividing \(d\) is ramified in \(K\). So \(d\) is not divisible by
any primes other than \(q\). Thus \(d\) could be \(\pm q\) or \(-1\),
noting that the \(d=1\) case doesn't yield a quadratic extension, and
we'd like to show that \(q^*\) is the only viable option. We can rule
out \(d=-1\), since 2 ramifies in
\({\mathbb{Q}}(\sqrt{-1}) = {\mathbb{Q}}(i)\) -- but the only prime that
can ramify is \(q\), which was an \emph{odd} prime.

We can also rule out which ever value of \(q\) yields
\(q\equiv 3 \operatorname{mod}4\), since again 2 ramifies in this case.
This was a general fact: for \(K={\mathbb{Q}}(\sqrt d)\) with
\(d\equiv 3 \operatorname{mod}4\), then 2 automatically ramifies. So
\(q\equiv 1\operatorname{mod}4\), and \(d = q^*\).

\end{proof}

\begin{remark}

This can also be proved with Gauss sums, but we don't necessarily need
that here.

\end{remark}

Continuing the proof, consider
\({\mathbb{Q}}(\zeta_q)/{\mathbb{Q}}(\sqrt{q^*})/{\mathbb{Q}}\) with
primes \(Q/R/p\) lying over \(p\). We'll show that
\({ \qty{ p \over q } } = 1 \iff { \qty{ q^* \over p } } = 1\), making
them equal. Note that \({ \qty{ p \over q } } = 1 \iff p\) is a square
in \(U({\mathbb{Z}}/q) \iff \mathop{\mathrm{Frob}}_{Q/p} \in H\). This
happens
\(\iff { \left.{{\mathop{\mathrm{Frob}}_{Q/p}}} \right|_{{{\mathbb{Q}}\sqrt{q^*}}} } = \operatorname{id}\),
which we can write as
\(\mathop{\mathrm{Frob}}_{R/P} = \operatorname{id}\), or equivalently
the symbol
\({ \qty{ {\mathbb{Q}}(\sqrt{q^*} /{\mathbb{Q}}) \over p } } = \operatorname{id}\).
Since we can write this as the Legendre symbol
\({ \qty{ q^* \over p } } = 1\) when we identify
\({ \mathsf{Gal}} ({\mathbb{Q}}(\sqrt{d^*})/{\mathbb{Q}}) \cong \left\{{\pm 1}\right\}\),
which is precisely what we wanted.

\end{proof}

\hypertarget{applying-quadratic-reciprocity-recovering-classical-results}{%
\subsection{Applying Quadratic Reciprocity: Recovering Classical
Results}\label{applying-quadratic-reciprocity-recovering-classical-results}}

\begin{example}[of applying quadratic reciprocity]

Let \(a = 6\), then for odd primes \(p\) not dividing \(6\),
\begin{align*}
{ \qty{ 6 \over p } } = 1 \iff p = 1,5,19,23\operatorname{mod}24
.\end{align*}
Note that \(6\divides 24\), and importantly \(\phi(24) = 8\) and this
allowed list is 4 elements, so exactly half are squares.

\end{example}

\begin{theorem}[?]

Let \(a\in {\mathbb{Z}}_{\neq 0}\) with \(a\) not a square, then

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  If \(p, p'\) are distinct odd primes not dividing \(a\) with
  \(p\equiv p' \operatorname{mod}4{\left\lvert {a} \right\rvert}\), then
  \({ \qty{ a \over p } } = { \qty{ a \over p' } }\). So
  \({ \qty{ a \over p } } = 1\) is entirely a function of the
  equivalence class of
  \(p \operatorname{mod}4{\left\lvert {a} \right\rvert}\).
\item
  The number of allowable residue class modulo
  \(4{\left\lvert {a} \right\rvert}\) is precisely
  \({1\over 2}\phi(4{\left\lvert {a} \right\rvert})\).
\end{enumerate}

\end{theorem}

\begin{remark}

For odd \(p\) not dividing \(a\), we have
\({ \qty{ a \over p } } = { \qty{ {\mathbb{Q}}(\sqrt a) / {\mathbb{Q}} \over p } }\)
where we identify
\({ \mathsf{Gal}} ({\mathbb{Q}}(\sqrt a)/{\mathbb{Q}})= \left\{{\pm 1}\right\}\).
We did this for \(a\) squarefree, but we can generally write
\(a = a_0 a_1^2\) with \(a_0\) squarefree, and the claim is that
replacing \(a\) with \(a_0\) doesn't change either side of this
equality. The right-hand side doesn't change since
\({\mathbb{Q}}(\sqrt a) \cong {\mathbb{Q}}(\sqrt{a_0})\), and the
left-hand side doesn't change because
\begin{align*}
{ \qty{ a \over p } } = { \qty{ a_0 \over p } } { \qty{ a_1 \over p } }^2 = { \qty{ a_0 \over p } } \cdot 1 = { \qty{ a_0 \over p } }
.\end{align*}

\end{remark}

\begin{lemma}[?]

The proportion of primes \(p\) with \({ \qty{ a \over p } } = 1\) is
exactly \(1/2\).

\end{lemma}

\begin{proof}[of lemma]

If \(p\) is odd and doesn't divide \(a\), so \(p\) doesn't divide
\(2a\), then
\({ \qty{ a \over p } } = { \qty{ {\mathbb{Q}}(\sqrt a) / {\mathbb{Q}} \over p } }\).
Applying Chebotarev's density to this Frobenius yields that the
proportion of \(p\) for which the right-hand side equals 1 is
\(\#\left\{{\operatorname{id}}\right\}/2 = 1/2\), noting that the
denominator is just the degree of the extension, which is quadratic.

\end{proof}

\begin{exercise}[A fun one]

For what proportion of \(p\) is 2 a cube mod \(p\)?

\end{exercise}

\begin{proof}[of theorem, part 1]

Suppose \(p\equiv p' \operatorname{mod}4{\left\lvert {a} \right\rvert}\)
with \(p\notdivides 2a\). From a homework problem,
\({\mathbb{Q}}(\sqrt a) \subseteq {\mathbb{Q}}(\zeta_{4{\left\lvert {a} \right\rvert}})\).
So we have a tower
\({\mathbb{Q}}(\zeta_{4{\left\lvert {a} \right\rvert}})/{\mathbb{Q}}(\sqrt a)/{\mathbb{Q}}\),
and corresponding towers of primes \(Q/R/p\) and \(Q'/R'/p'\). Since
we're assuming
\(p\equiv p' \operatorname{mod}4{\left\lvert {a} \right\rvert}\), we
have \(\mathop{\mathrm{Frob}}_{Q/p} = \mathop{\mathrm{Frob}}_{Q'/p'}\).
Restricting both to \({\mathbb{Q}}(\sqrt a)\) we get

\begin{itemize}
\tightlist
\item
  \(\mathop{\mathrm{Frob}}_{R/p} = { \qty{ {\mathbb{Q}}(\sqrt a) / {\mathbb{Q}} \over p } } = { \qty{ a \over p } }\),
\item
  \(\mathop{\mathrm{Frob}}_{R'/p'} = { \qty{ {\mathbb{Q}}(\sqrt a) / {\mathbb{Q}} \over p' } } = { \qty{ a \over p' } }\),
\end{itemize}

so the two Legendre symbols are equal.

\end{proof}

\begin{proof}[of theorem, part 2]

List the coprime residue classes
\(\operatorname{mod}4{\left\lvert {a} \right\rvert}\) corresponding to
\({ \qty{ a \over p } } = 1\), say
\(a_1, \cdots, a_k \operatorname{mod}4{\left\lvert {a} \right\rvert}\).
By Dirichlet's theorem on prime progressions (or Chebotarev density on
\({\mathbb{Q}}(\zeta_{4{\left\lvert {a} \right\rvert}})\)), we know that
the proportion of \(p\) with
\(p\equiv a_i \operatorname{mod}4{\left\lvert {a} \right\rvert}\) is
\(1/\phi(4{\left\lvert {a} \right\rvert})\). Adding these up yields
\(k/\phi(4{\left\lvert {a} \right\rvert})\). But we just computed this
proportion alternatively and found it to be \(1/2\), so this forces
\(k = (1/2) \phi(4{\left\lvert {a} \right\rvert})\).

\end{proof}

\begin{exercise}[?]

Think about the theorem of Hasse which stated that the primes \(p\) for
which \(2\) has an even order \(\operatorname{mod}p\) has density
\(17/24\). A way to start: divide up the primes according to the largest
power of 2 that divides \(p-1\). For \(2^1\divides p-1\) but \(2^2\)
does not, these are primes \(p\equiv 3\operatorname{mod}4\), \(a\) has
even order precisely when \(a\) is not a square \(\operatorname{mod}p\),
since otherwise if \(a\) is a square then the order will divide
\(p-1\over 2\), which is odd, and the order will be odd. So one should
think about which primes \(p\equiv 3\operatorname{mod}4\) for which (for
example) \(2\) is not a square \(\operatorname{mod}p\), and these ideas
will apply.

For \(2^2 \divides p-1\) but \(2^3\) does not, such primes are of the
form \(p\equiv 5 \operatorname{mod}8\). When does 2 have even order? Now
one needs 2 not to be a \emph{fourth} power, and one can find the
density of primes for which \(2\) is not a 4th power
\(\operatorname{mod}p\) and \(p\equiv 5\operatorname{mod}8\) using these
ideas, but it's harder.

For \(2^3\), one gets \(p\equiv 9\operatorname{mod}16\) and needs \(2\)
to not be an eighth power.

Doing this for every \(k\) and summing all of the densities will yield
\(17/24\).

\end{exercise}

\addsec{ToDos}
\listoftodos[List of Todos]
\cleardoublepage

% Hook into amsthm environments to list them.
\addsec{Definitions}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]
\cleardoublepage

\addsec{Theorems}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]
\cleardoublepage

\addsec{Exercises}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]
\cleardoublepage

\addsec{Figures}
\listoffigures
\cleardoublepage


\printbibliography[title=Bibliography]


\end{document}
