\input{"preamble.tex"}

\addbibresource{CommutativeAlgebra.bib}

\let\Begin\begin
\let\End\end
\newcommand\wrapenv[1]{#1}

\makeatletter
\def\ScaleWidthIfNeeded{%
 \ifdim\Gin@nat@width>\linewidth
    \linewidth
  \else
    \Gin@nat@width
  \fi
}
\def\ScaleHeightIfNeeded{%
  \ifdim\Gin@nat@height>0.9\textheight
    0.9\textheight
  \else
    \Gin@nat@width
  \fi
}
\makeatother

\setkeys{Gin}{width=\ScaleWidthIfNeeded,height=\ScaleHeightIfNeeded,keepaspectratio}%

\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Commutative Algebra
  }
    \\ {\normalsize Lectures by Daniel Litt. University of Georgia,
Spring 2022} \\
  \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 Commutative Algebra taught by Daniel Litt at
the University of Georgia in Spring 2022. 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:} 2022-05-29 }
\end{flushleft}


\newpage

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

\hypertarget{thursday-january-06}{%
\section{Thursday, January 06}\label{thursday-january-06}}

\begin{quote}
Topics: Localization and completion, Nakayama's lemma, Dedekind domains,
Hilbert's basis theorem, Hilbert's Nullstellensatz, Krull dimension,
depth and Cohen-Macaulay rings, regular local rings.
\end{quote}

\begin{remark}

References:

\begin{itemize}
\item
  Atiyah-MacDonald, \emph{Commutative Algebra}. Be sure to check the
  erratum!
\item
  Chambert-Loir,
  \href{https://webusers.imj-prg.fr/~antoine.chambert-loir/enseignement/2014-15/ga/commalg.pdf}{Mostly
  Commutative Algebra}
\item
  Miles Reid, \emph{Undergraduate Algebraic Geometry}
\item
  Altman-Kleiman,
  \href{https://www.mi.fu-berlin.de/en/math/groups/arithmetic_geometry/teaching/exercises/Altman_-Kleiman---A-term-of-commutative-algebra-_2017_.pdf}{A
  Term of Commutative Algebra}
\end{itemize}

\end{remark}

\begin{example}[?]

Some examples of module morphisms:

\begin{itemize}
\item
  \(\mathop{\mathrm{Hom}}_\mathsf{Ring}({\mathbb{Z}}, S) = \left\{{{\operatorname{pt}}}\right\}\),
  since \(1\to 1\) is necessary.
\item
  \(\mathop{\mathrm{Hom}}_\mathsf{Ring}({\mathbb{Z}}[x], S) \cong S\).
  Why? Since \(1\to 1\) is forced, \(x\mapsto s\) can be sent to any
  \(s\in S\).
\item
  \(\mathop{\mathrm{Hom}}_\mathsf{Ring}\qty{ { {\mathbb{Z}}[x, y]\over \left\langle{y^2-x^3-1}\right\rangle }, S} = \left\{{(a, b)\in S{ {}^{ \scriptscriptstyle\times^{2} } } {~\mathrel{\Big\vert}~}a^2-b^3=1}\right\}\).
\item
  \(\mathop{\mathrm{Hom}}_\mathsf{Ring}(R/I, S) = \left\{{f\in \mathop{\mathrm{Hom}}_\mathsf{Ring}(R, S) {~\mathrel{\Big\vert}~}f(I) = 0}\right\}\)
\end{itemize}

\end{example}

\begin{exercise}[?]

Show that
\(\operatorname{Id}(R/I)\rightleftharpoons\left\{{J\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J \supseteq I}\right\}\)
using
\begin{align*}
\Pi: R &\to R/I \\
x &\mapsto [x] \\
\pi^{-1}(J) &\mapsfrom J
.\end{align*}
Show that \(\pi^{-1}(J)\) is in fact an ideal, construct a proposed
inverse \(\Pi^{-1}\), and show
\(\Pi\circ \Pi^{-1}= \Pi^{-1}\circ \Pi = \operatorname{id}\).

\end{exercise}

\begin{exercise}[?]

Show that \(R\) is a field iff \(R\) is a simple ring iff any ring
morphism \(R\to S\) is injective. For \(3\implies 1\), directly shows
that every nonzero element is a unit.

\end{exercise}

\begin{exercise}[?]

Chapter 1 of A\&M:

\begin{itemize}
\tightlist
\item
  1,8,10,13,15,16,19.
\end{itemize}

\end{exercise}

\hypertarget{thursday-january-13}{%
\section{Thursday, January 13}\label{thursday-january-13}}

\begin{remark}

Last time: fields are simple rings.

\end{remark}

\begin{exercise}[?]

Let \(k\in \mathsf{Field}\) and show that

\begin{itemize}
\tightlist
\item
  If \(f\in R\coloneqq k[x]\) is irreducible then
  \(\left\langle{f}\right\rangle \in \operatorname{Spec}R\).
\item
  \(\left\langle{xy}\right\rangle \in \operatorname{Id}(k[x,y])\) is not
  prime and not maximal.
\item
  There exist nonzero non-prime maximal ideals.
\item
  Show that \(I\in \operatorname{Spec}R \iff A/I\) is an integral
  domain.
\item
  Show that \(I\in \operatorname{mSpec}R \iff A/I\in \mathsf{Field}\).
\end{itemize}

\end{exercise}

\begin{exercise}[?]

Prove that if \(f: R\to S\) is a ring morphism then there is a
well-defined map
\begin{align*}
f^*: \operatorname{Spec}S &\to \operatorname{Spec}R \\
{\mathfrak{p}}&\mapsto f^{-1}({\mathfrak{p}})
,\end{align*}
i.e.~if \({\mathfrak{p}}\) is prime in \(S\) then
\(f^{-1}({\mathfrak{p}})\) is prime in \(R\). Show that this doesn't
generally hold with \(\operatorname{Spec}\) replaced by
\(\operatorname{mSpec}\).

\end{exercise}

\begin{exercise}[?]

Show that defining
\(V(I) \coloneqq\left\{{{\mathfrak{p}}\in \operatorname{Spec}R {~\mathrel{\Big\vert}~}{\mathfrak{p}}\supseteq I}\right\}\)
as closed sets defines a topology on \(\operatorname{Spec}R\). Show that
\(\operatorname{Spec}R\) is Hausdorff iff it's discrete, and \(V\) is
functorial.

\end{exercise}

\begin{exercise}[?]

Describe

\begin{itemize}
\tightlist
\item
  \(\operatorname{Spec}k\) for \(k\in \mathsf{Field}\)
\item
  \(\operatorname{Spec}{\mathbb{Z}}\), show it is not Hausdorff, and
  \(\left\langle{0}\right\rangle\) is a generic point, and the subspace
  topology on its closed points is cofinite.
\item
  \(\operatorname{Spec}R\) for \(R\) a local ring.
\item
  \(\operatorname{Spec}R\) for \(R\) a DVR
\item
  \(\operatorname{Spec}k[x,y]\)
\item
  \(\operatorname{Spec}R\) for \(R = {\mathbb{Z}}[i]\).
\item
  \(\operatorname{Spec}{\mathcal{O}}_K\), the ring of integers of a
  number field \(K\).
\item
  \(\operatorname{Spec}R\) for \(R\) a Dedekind domain
\end{itemize}

\end{exercise}

\begin{exercise}[?]

Show that every \(I\in \operatorname{Id}(R)\) is contained in some
\({\mathfrak{m}}\in \operatorname{mSpec}R\).

\end{exercise}

\begin{exercise}[?]

Show that the following rings are local:

\begin{itemize}
\tightlist
\item
  For \(p\in {\mathbb{Z}}\) prime,
  \(R\coloneqq{\mathbb{Z}} \left[ { \scriptstyle { {S}^{-1}} } \right]\)
  for
  \(S\coloneqq\left\{{\ell \in {\mathbb{Z}}\text{ prime} {~\mathrel{\Big\vert}~}\ell \neq p}\right\}\).
\item
  \(k\in \mathsf{Field}\)
\item
  \(k { \left[ {x} \right] }\) with
  \({\mathfrak{m}}= \left\langle{x}\right\rangle\)
\item
  \(k { \left[ {x, y} \right] }\). What is the maximal ideal?
\end{itemize}

\end{exercise}

\begin{exercise}[?]

For \((R, {\mathfrak{m}})\) a local ring, show that
\({\mathfrak{m}}= R\setminus(R^{\times})\).

\end{exercise}

\begin{remark}

Recall that

\begin{itemize}
\tightlist
\item
  \(\sum I_j\) is the smallest ideal containing all of the \(I_j\).
\item
  \(\displaystyle\bigcap I_j\) is again an ideal
\item
  \(IJ \coloneqq\left\langle{xy {~\mathrel{\Big\vert}~}x\in I, y\in J}\right\rangle\)
  is an ideal
\item
  \(IJ \subseteq I \cap J\).
\item
  \({\sqrt{0_{R}} }\) is the set of nilpotent elements.
\end{itemize}

\end{remark}

\begin{theorem}[?]

Show that \({\sqrt{0_{R}} }\) is the intersection of all prime ideals.

\end{theorem}

\begin{slogan}

Regarding elements \(f\in R\) as functions on \(\operatorname{Spec}R\),
\(f\) nilpotent is like being zero at every point of
\(\operatorname{Spec}R\).

\end{slogan}

\begin{exercise}[?]

Show that \(x\in {J ({R}) } \iff 1-xy\in R^{\times}\) for all
\(y\in R\).

\end{exercise}

\hypertarget{tuesday-january-18}{%
\section{Tuesday, January 18}\label{tuesday-january-18}}

\begin{remark}

A reference for pictures: Mumford's red book. Note the typo in A\&M
problem 10: it is false for the zero ring.

Recall some definitions:

\begin{itemize}
\tightlist
\item
  \(R{\hbox{-}}\)modules, what are some examples?
\item
  Submodules and quotient modules.
\item
  The submodule generated by a subset.
\item
  A morphism of modules and the \(R{\hbox{-}}\)module structure on
  \(\mathop{\mathrm{Hom}}_R(M, N)\), \((rf)(x) \coloneqq r\cdot f(x)\).

  \begin{itemize}
  \tightlist
  \item
    This makes \({\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) into a category
    enriched over itself.
  \end{itemize}
\item
  \(\operatorname{im}f, \ker f, \operatorname{coker}f\).
\item
  \(\prod M_i, \bigoplus M_i\)
\end{itemize}

\end{remark}

\begin{exercise}[Module structure on quotients]

Show that if \(M\leq N \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), then
the following action makes \(M/N\) into an \(R{\hbox{-}}\)module:
\begin{align*}
r\cdot [x] \coloneqq[rx]
,\end{align*}
i.e.~that if \([x] = [y]\) then \(r\cdot [x] = r\cdot [y]\).

\end{exercise}

\begin{exercise}[Universal properties of kernels and cokernels]

Show the universal properties of kernels and cokernels: given
\(f: M\to N\) and \(Q\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\),
\begin{align*}
\mathop{\mathrm{Hom}}_R(Q, \ker f) &= \left\{{ s\in \mathop{\mathrm{Hom}}_R(Q, M) {~\mathrel{\Big\vert}~}f\circ s = 0}\right\} \\
\mathop{\mathrm{Hom}}_R(\operatorname{coker}f, Q) &= \left\{{t\in \mathop{\mathrm{Hom}}_R(N, Q) {~\mathrel{\Big\vert}~}t\circ f = 0 }\right\}
.\end{align*}

\end{exercise}

\begin{exercise}[Direct sum and product coincide for finite index sets]

Show that if \(I\) is a finite indexing set, there is an isomorphism
\begin{align*} 
\bigoplus _{i\in I} M_i  { \, \xrightarrow{\sim}\, }\prod_{i\in I} M_i
, \end{align*}
and that
\begin{align*}
\mathop{\mathrm{Hom}}_R\qty{ T, \prod M_i} &\cong \prod_{i\in I} \mathop{\mathrm{Hom}}_R\qty{T, M_i} \\
\mathop{\mathrm{Hom}}_R\qty{ \bigoplus M_i, T} &\cong \bigoplus \mathop{\mathrm{Hom}}_R\qty{ M_i, T}
.\end{align*}

Show that by Yoneda, this satisfies the universal property.

\end{exercise}

\begin{solution}

Sketch:

\begin{itemize}
\tightlist
\item
  Define projections \(\pi_j: \prod_i M_i \to M_j\).
\item
  Send \(f\in \mathop{\mathrm{Hom}}(T, \prod_i M_i)\) to
  \(\pi_j \circ f\in \prod_i \mathop{\mathrm{Hom}}(T, M_i)\)
\item
  For the other direction, given
  \((f_i) \in \prod_i \mathop{\mathrm{Hom}}(M_i, T)\), send \((f_i)\) to
  \(\sum f_i\).
\end{itemize}

\end{solution}

\begin{exercise}[Modules have products and coproducts]

Show that \(\prod\) is a categorical product and \(\bigoplus\) is a
categorical coproduct. What are the product and coproduct in
\({\mathsf{Top}}\)?

\end{exercise}

\begin{definition}[Colon ideals and annihilators]

Given \(N \leq M \in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), define the
\textbf{colon ideal}
\begin{align*}
(N: M) \coloneqq\left\{{a\in R{~\mathrel{\Big\vert}~}aM \subseteq N}\right\}
\end{align*}
and the \textbf{annihilator} of \(M\)
\begin{align*}
\operatorname{Ann}(M) = (0: M) = \left\{{a\in R {~\mathrel{\Big\vert}~}aM = 0}\right\}
.\end{align*}

\end{definition}

\begin{example}[Annihilators]

Some annihilators:

\begin{itemize}
\tightlist
\item
  \(C_n \in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\), so
  \(\operatorname{Ann}(C_n) = n{\mathbb{Z}}= \left\langle{n}\right\rangle\).
\item
  Again in \({\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\),
  \(\operatorname{Ann}(C_n \oplus C_m) = n{\mathbb{Z}}\cap m{\mathbb{Z}}= \operatorname{lcm}(m, n) {\mathbb{Z}}\).
\end{itemize}

\end{example}

\begin{remark}

An \(R{\hbox{-}}\)module is free iff \(R \cong \bigoplus _{i\in I} R\),
where \(I\) can be infinite but (importantly) we need the direct sum
instead of the direct product. Note that generally \(\prod_{i\in I} R\)
may not be free as an \(R{\hbox{-}}\)module.

A module is finitely generated if there exists a generating set
\(X \coloneqq\left\{{x_i}\right\}_{i\leq n} \subseteq M\) such that any
submodule containing \(X\) is all of \(M\), or equivalently
\(x\in M \implies x = \sum r_i x_i\) for some \(r_i \in R\).

\end{remark}

\begin{exercise}[Finitely generated iff surjective image of a free module]

Show that \(M\) is finitely-generated iff there is a surjective morphism
\(R^n \to M\) for some \(n\in {\mathbb{Z}}_{\geq 0}\).

\end{exercise}

\begin{solution}

Sketch:

\begin{itemize}
\tightlist
\item
  finitely-generated \(\implies\) surjection:

  \begin{itemize}
  \tightlist
  \item
    Take \(e_i = \left\{{0,\cdots, 1,\cdots, 0}\right\} \in R^n\) and
    define \(f(e_i) \coloneqq x_i\)
  \item
    Use the universal property of the direct sum.
  \end{itemize}
\item
  \(\impliedby\):

  \begin{itemize}
  \tightlist
  \item
    Show that the \(f(e_i)\) generate \(M\): by surjectivity,
    \(m = f(x) = f(\sum r_i e_i) = \sum r_i f(e_i)\).
  \end{itemize}
\end{itemize}

\end{solution}

\begin{remark}

Thus every \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is the quotient
of a free module: find a surjection \(f: F\to M\), so
\(\operatorname{im}f = M\), then use that
\(\operatorname{im}f \cong /\ker f\). For example, one can take
\(F \coloneqq\bigoplus _{m\in M} R \to M\).

Recall

\begin{itemize}
\tightlist
\item
  The definition of exact sequences
\item
  \(0\to A \xrightarrow{f} B\) is exact iff \(f\) is injective,
\item
  \(A \xrightarrow{f} B \to 0\) iff \(f\) is surjective,
\item
  \(0\to A \xrightarrow{f} B \to 0\) iff \(f\) is an isomorphism,
\item
  \(0\to A\to B\to C\to 0\) is called a short exact sequence.
\end{itemize}

\end{remark}

\begin{exercise}[Direct sums are exact]

Check that \(({-}) \oplus M\) is exact.

\end{exercise}

\begin{exercise}[Left exactness of hom]

Show that \(\mathop{\mathrm{Hom}}(N, {-})\) and
\(\mathop{\mathrm{Hom}}({-}, N)\) are both left-exact for any
\(N\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\), where
\(0 \to A \xrightarrow{f} B \xrightarrow{g} C\to 0\) is sent to
\(0 \to \mathop{\mathrm{Hom}}(N, A) \xrightarrow{f\circ {-}} \cdots\).
Give an example of when right-exactness fails.

\begin{quote}
Hint: try
\(0\to {\mathbb{Z}}\xrightarrow{\cdot 2} {\mathbb{Z}}\to C_2 \to 0\) and
apply \(\mathop{\mathrm{Hom}}_{\mathbb{Z}}({-}, C_2)\).
\end{quote}

\end{exercise}

\begin{remark}

Recall that
\(\mathop{\mathrm{Hom}}_{{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}}({\mathbb{Z}}, {-}) \cong \operatorname{id}\)
and
\(\mathop{\mathrm{Hom}}_{{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}}(C_n, {-}) \cong ({-})[n]\)
picks out the \(n{\hbox{-}}\)torsion.

\end{remark}

\hypertarget{thursday-january-20}{%
\section{Thursday, January 20}\label{thursday-january-20}}

\begin{remark}

Last time: \(\mathop{\mathrm{Hom}}(N, {-})\) and
\(\mathop{\mathrm{Hom}}({-}, N)\) are left exact. We can explicitly
describe
\begin{align*}
\mathop{\mathrm{Hom}}_A(A/I, M) = \left\{{m\in M {~\mathrel{\Big\vert}~}im = 0 \forall i\in I}\right\}
,\end{align*}
which is the \(I{\hbox{-}}\)torsion in \(M\). Using that
\(0\to I\to A \to A/I \to 0\) is short exact, we'll get a long exact
sequence
\begin{align*}
0 \to \mathop{\mathrm{Hom}}(I, M) \to \mathop{\mathrm{Hom}}(A, M) \xrightarrow{f} \left\{{I{\hbox{-}}\text{torsion in } M}\right\} \to \operatorname{Ext} ^1_A(I, M) \to \cdots
,\end{align*}
where the \(\operatorname{Ext}\) term measures failure of surjectivity
of \(f\).

\end{remark}

\begin{exercise}[?]

Show that \(\mathop{\mathrm{Hom}}(N, {-})\) and
\(\mathop{\mathrm{Hom}}({-}, N)\) are left exact.

\end{exercise}

\begin{remark}

Recall the snake lemma:

\begin{center}
\begin{tikzcd}
    0 & {\ker a} & {\ker b} & {\ker c} \\
    0 & {A_1} & {B_1} & {C_1} & 0 \\
    \\
    0 & {A_2} & {B_2} & {C_2} & 0 \\
    & {\operatorname{coker}a} & {\operatorname{coker}b} & {\operatorname{coker}c} & 0
    \arrow["a", from=2-2, to=4-2]
    \arrow["b", from=2-3, to=4-3]
    \arrow["c", from=2-4, to=4-4]
    \arrow[from=2-2, to=2-3]
    \arrow[from=2-3, to=2-4]
    \arrow[from=4-2, to=4-3]
    \arrow[from=4-3, to=4-4]
    \arrow[from=2-1, to=2-2]
    \arrow[from=2-4, to=2-5]
    \arrow[from=4-1, to=4-2]
    \arrow[from=4-4, to=4-5]
    \arrow[from=1-1, to=1-2]
    \arrow[from=1-2, to=1-3]
    \arrow[from=1-3, to=1-4]
    \arrow[from=5-2, to=5-3]
    \arrow[from=5-3, to=5-4]
    \arrow[from=5-4, to=5-5]
    \arrow[curve={height=12pt}, dashed, from=1-4, to=5-2]
    \arrow[from=1-2, to=2-2]
    \arrow[from=1-3, to=2-3]
    \arrow[from=1-4, to=2-4]
    \arrow[from=4-2, to=5-2]
    \arrow[from=4-3, to=5-3]
    \arrow[from=4-4, to=5-4]
\end{tikzcd}
\end{center}

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

The recipe for the connecting morphism:

\begin{itemize}
\tightlist
\item
  Start with \(x\in \ker(C_1\to C_2)\), choose a preimage in \(B_1\)
\item
  Push to \(B_2\)
\item
  Use exactness to pull to \(A_2\)
\item
  Project along \(A_2\to \operatorname{coker}(A_1\to A_2)\)
\end{itemize}

\end{remark}

\begin{definition}[Finite presentation]

An object \(M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is of
\textbf{finite presentation} iff there is an exact sequence
\begin{align*}
R^m \to R^n\to M \to 0
,\end{align*}
i.e.~there are finitely many generators and finitely many relations.

\end{definition}

\begin{remark}

A nice application of the snake lemma: for finitely presented modules
\(M, N\), one can extend a morphism \(M\to N\) to

\begin{center}
\begin{tikzcd}
    0 && {R^{m_1}} && {R^{n_1}} && M \\
    \\
    0 && {R^{m_2}} && {R^{n_2}} && N
    \arrow["f", from=1-7, to=3-7]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=3-7]
    \arrow[from=1-5, to=3-5]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=3-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMiwwLCJSXnttXzF9Il0sWzQsMCwiUl57bl8xfSJdLFs2LDAsIk0iXSxbMiwyLCJSXnttXzJ9Il0sWzQsMiwiUl57bl8yfSJdLFs2LDIsIk4iXSxbMCwwLCIwIl0sWzAsMiwiMCJdLFsyLDUsImYiXSxbMCwxXSxbMSwyXSxbMyw0XSxbNCw1XSxbMSw0XSxbMCwzXSxbNiwwXSxbNywzXV0=}{Link
to Diagram}
\end{quote}

Since \(R\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is free, the extended
maps can be represented by matrices, which is a significant
simplification. In fact, \(f\) can be recovered uniquely by knowing the
map on generators.

\end{remark}

\begin{exercise}[?]

Prove the snake lemma, and show exactness at all 6 places.

\end{exercise}

\begin{remark}

Recall the universal property of \(M\otimes_R N\) in
\({\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) in terms of bilinear morphisms.

\end{remark}

\begin{exercise}[?]

Prove uniqueness of any object satisfying a universal property using the
Yoneda lemma.

\end{exercise}

\begin{exercise}[?]

Prove using universal properties:

\begin{itemize}
\tightlist
\item
  \(R\otimes_R R \cong R\), using universal properties. Why is this
  unique?
\item
  \(R\otimes_R M \cong M\).
\item
  \((M_1 \oplus M_2) \otimes_R N \cong (M_1 \otimes_R N) \oplus (M_2\otimes_R N)\).
\end{itemize}

\end{exercise}

\hypertarget{tuesday-january-25}{%
\section{Tuesday, January 25}\label{tuesday-january-25}}

\begin{exercise}[?]

Do A\&M:

\begin{itemize}
\tightlist
\item
  2.12, 24, 25
\item
  3.1, 4, 12
\end{itemize}

See website: \url{https://www.daniellitt.com/commutative-algebra}

\end{exercise}

\begin{remark}

Last time:

\begin{itemize}
\tightlist
\item
  Defined \(M\otimes_R N\), need to show it exists.
\item
  Showed \(R\otimes_R M \cong M\)
\item
  Showed sums commute with tensor products
\end{itemize}

\end{remark}

\begin{exercise}[?]

Show that if \(A \xrightarrow{f} B \xrightarrow{g} C\to 0\) is exact and
\(A\otimes_R N, B\otimes_R N\) exist, then \(C\otimes_R N\) also exists.

\begin{quote}
Hint: Use the universal property to produce a map
\(\psi: A\otimes N\to B\otimes N\) and set
\(C\otimes N \coloneqq\operatorname{coker}\psi\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Prove that \(M\otimes_R N\) exists by constructing it.

\end{exercise}

\begin{solution}

Some hints: Construction 1: use
\(0\to \ker f\to R{ {}^{ \scriptscriptstyle\oplus^{I} } } \xrightarrow{f} M\to 0\),
and find \(R{ {}^{ \scriptscriptstyle\oplus^{J} } }\to \ker f\) to
assemble an exact sequence
\begin{align*}
R{ {}^{ \scriptscriptstyle\oplus^{J} }  }\to R{ {}^{ \scriptscriptstyle\oplus^{I} }  }\to M\to 0
.\end{align*}
Now apply \(({-})\otimes_R N\) to exhibit \(M\otimes_R N\) as a cokernel
\(N{ {}^{ \scriptscriptstyle\oplus^{J} } }\to N{ {}^{ \scriptscriptstyle\oplus^{I} } }\to M\otimes_R N\to 0\).
Separately, use the hands-on construction and prove it satisfies the
universal property.

\end{solution}

\begin{exercise}[?]

Prove \(C_2 \otimes_{\mathbb{Z}}C_3 \cong 0\) using construction 1
above.

\end{exercise}

\begin{solution}

Sketch:

Take \({\mathbb{Z}}\xrightarrow{\times 2} {\mathbb{Z}}\to C_2\to 0\),
apply \(({-})\otimes C_3\), and check that
\(\operatorname{coker}(C_3 \xrightarrow{\times 2} C_3) = 0\) since
multiplication by 2 is invertible. Alternatively, use bilinearity:
\begin{align*}
B(x,y) = 3B(x, y) - 2B(x, y) = B(x, 3y) - B(2x, y) = B(x, 0) - B(0, y) = 0
.\end{align*}

\end{solution}

\begin{exercise}[?]

Show that

\begin{itemize}
\tightlist
\item
  \(({-})\otimes_R N\) is right exact
\item
  Tensoring is associative, distributive, commutative
\item
  There is a canonical isomorphism \(R\otimes_R M\to M\) induced by
  \(r\otimes m \mapsto r.m\).
\item
  Morphisms \(f: A\to B\in \mathsf{CRing}\) induce functors
  \(f^\sharp: {\mathsf{A}{\hbox{-}}\mathsf{Mod}} \to {\mathsf{B}{\hbox{-}}\mathsf{Mod}}\).
  Also show that \(M\otimes_A B\) has a \(B{\hbox{-}}\)module structure
  given by \(b_1(m\otimes b_2) \coloneqq m\otimes(b_1 b_2)\).
\item
  For \(N\in{\mathsf{B}{\hbox{-}}\mathsf{Mod}}\), there is an
  isomorphism
  \(\mathop{\mathrm{Hom}}_{A}(M, N) { \, \xrightarrow{\sim}\, }\mathop{\mathrm{Hom}}_{B}(M\otimes_A B, N)\).

  \begin{itemize}
  \tightlist
  \item
    Use
    \(f\mapsto (m\otimes b \mapsto bf(m)) \in \mathrm{Bil}_B(M\times B, N)\),
    with inverse \(Q\mapsto Q({-}, 1)\).
  \end{itemize}
\item
  Show that \(M\otimes_R R/I \cong M/IM\), using
  \(R/I = \operatorname{coker}(I\hookrightarrow R)\) and applying
  \(({-})\otimes M\). Also show that
  \(I\otimes_R M { \, \xrightarrow{\sim}\, }IM\) canonically.
\item
  Show that
  \(k[t]{ {}^{ \scriptscriptstyle\oplus^{2} } } \otimes_{k[t]} {k[t]\over\left\langle{t^2}\right\rangle} \cong \qty{k[t]\over \left\langle{t^2}\right\rangle}{ {}^{ \scriptscriptstyle\oplus^{2} } }\).
\item
  Show that
  \(R/I \otimes_R R/J \cong {R/I \over I \cdot R/J} \cong {R\over I+J}\).
\item
  Tensoring need not be left exact.
\item
  \(C_p \not\in {\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}^\flat\) .
\item
  \(R\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}^\flat\).
\item
  \({\mathsf{R}{\hbox{-}}\mathsf{Mod}}^\flat\) is closed under
  \(\otimes_R\) and \(\oplus\).
\item
  \({\mathbb{Q}}\in{\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}^\flat\)
  but not in
  \({\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}^{\mathrm{free}}\)
\end{itemize}

\end{exercise}

\begin{definition}[$A\dash$algebras]

\({\mathsf{Alg}}_{/ {A}}\) is the coslice category
\(\mathsf{CRing}_{{A/}}\): objects are rings \(B\) equipped with ring
morphisms \(A\to B\), and morphisms are cones under \(A\):

\begin{center}
\begin{tikzcd}
    & A \\
    \\
    B && C
    \arrow[from=1-2, to=3-1]
    \arrow[from=1-2, to=3-3]
    \arrow[from=3-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMSwwLCJBIl0sWzAsMiwiQiJdLFsyLDIsIkMiXSxbMCwxXSxbMCwyXSxbMSwyXV0=}{Link
to Diagram}
\end{quote}

\end{definition}

\begin{example}[?]

Examples of algebras:

\begin{itemize}
\tightlist
\item
  \(k[t] \in{\mathsf{Alg}}_{/ {k}}\)
\item
  \(R\in \mathsf{CRing}\implies R\in{\mathsf{Alg}}_{/ {{\mathbb{Z}}}}\)
\item
  Every \(B\in {\mathsf{Alg}}_{/ {A}}\) is a quotient of some polynomial
  algebra \(A[t_1,\cdots]\) on potentially infinitely many generators.
\end{itemize}

\end{example}

\begin{definition}[Finiteness]

An object \(B\in{\mathsf{Alg}}_{/ {A}}\) is \textbf{finitely generated}
if it is a quotient of some \(A[t_1, \cdots, t_n]\). Equivalently, there
exist \(x_1,\cdots, x_n\in B\) such that any subring containing the
\(x_i\) and the image of \(A\) is all of \(B\).

\(B\) is \textbf{finite} if
\(B\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).

\end{definition}

\begin{example}[?]

Examples:

\begin{itemize}
\tightlist
\item
  \(k[t] \in {\mathsf{Alg}_{/k} }\) is finitely generated but not
  finite.
\item
  \(k[t]/\left\langle{t^2}\right\rangle\in {\mathsf{Alg}_{/k} }\) is
  finitely generated and finite.
\item
  \(({-})\otimes_A ({-}) \in {\mathsf{Fun}}({\mathsf{Alg}}_{/ {A}} { {}^{ \scriptscriptstyle\times^{2} } }, {\mathsf{Alg}}_{/ {A}} )\),
  defined by
  \((b_1 \otimes c_1)(b_2\otimes c_2) \coloneqq b_1 b_2 \otimes c_1 c_2\).
\item
  \(\mathop{\mathrm{Hom}}_{{\mathsf{Alg}}_{/ {A}} }(B\otimes_A C, S) = \left\{{f: B\to S, g: C\to S {~\mathrel{\Big\vert}~}{ \left.{{f}} \right|_{{A}} } = { \left.{{g}} \right|_{{A}} }}\right\}\).
\item
  \(k[t_1] \otimes_k k[t_2] \cong k[t_1, t_2]\) is not isomorphic to
  \(k[t_1]\times k[t_2]\) via
  \((f(t_1), g(t_2)) \mapsto f(t_1) \cdot g(t_2)\), since
  e.g.~\(h(t_1, t_2) \coloneqq t_1 + t_2\) is not in the image of this
  map.
\end{itemize}

\end{example}

\hypertarget{thursday-january-27}{%
\section{Thursday, January 27}\label{thursday-january-27}}

\begin{remark}

Nakayama: called a lemma, but arguably the most important statement in
commutative algebra! Recall that
\({J ({A}) } = \cap_{{\mathfrak{m}}\in \operatorname{mSpec}A} {\mathfrak{m}}\),
and being zero mod every \({\mathfrak{m}}\in \operatorname{mSpec}A\)
means being zero mod \({J ({A}) }\).

\end{remark}

\begin{theorem}[Nakayama's Lemma]

Let \(A \in \mathsf{CRing}\) with \(I \subseteq {J ({I}) }\) and let
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\). Then
\begin{align*}
M=IM \implies M=0
.\end{align*}

\end{theorem}

\begin{example}[?]

This reduces statements about local rings to statements about fields.
Commonly used examples:

\begin{itemize}
\tightlist
\item
  \(I = {\sqrt{0_{R}} }\).
\item
  \(A\in \mathsf{Loc}\mathsf{Ring}\) with \(I = {\mathfrak{m}}_A\) its
  maximal ideal.
\end{itemize}

\end{example}

\begin{corollary}[?]

If \(A\) is local and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\),
\begin{align*}
M/{\mathfrak{m}}_A M = 0 \iff M = 0
.\end{align*}

\end{corollary}

\begin{remark}

\(M\) yields a sheaf (or vector bundle) on \(\operatorname{Spec}A\), so
think of \(m\in M\) as a function on \(\operatorname{Spec}A\) in the
following way: if \(m\in M\),
\begin{align*}
m: {\mathfrak{p}}\mapsto m \operatorname{mod}{\mathfrak{p}}\in M/{\mathfrak{p}}M
.\end{align*}

\includegraphics{figures/2022-01-27_12-57-28.png}

\end{remark}

\begin{proposition}[Equivalent formulation of Nakayama]

If \(n\in N\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) and
\(n\equiv 0 \operatorname{mod}{\mathfrak{m}}\) for all
\({\mathfrak{m}}\in \operatorname{mSpec}A\), then \(M=0\).

\end{proposition}

\begin{exercise}[?]

Show that if \(A \in \mathsf{Loc}\mathsf{Ring}\) and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) with
\(\left\{{x_i}\right\}_{i\leq n} \subseteq M\), then
\begin{align*}
\left\langle{x_1,\cdots, x_n}\right\rangle = M \iff \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux_1\mkern-1.5mu}\mkern 1.5mu, \cdots, \mkern 1.5mu\overline{\mkern-1.5mux_n\mkern-1.5mu}\mkern 1.5mu}\right\rangle = M/{\mathfrak{m}}M, \qquad \mkern 1.5mu\overline{\mkern-1.5mux_i\mkern-1.5mu}\mkern 1.5mu \coloneqq x_i \operatorname{mod}{\mathfrak{m}}
.\end{align*}

\end{exercise}

\begin{solution}

\(\implies\): If
\(\mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu\in M/{\mathfrak{m}}M\),
lift to \(y\in M\) to write \(y = \sum c_i x_i\) and thus
\(\mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu = \sum c_i \mkern 1.5mu\overline{\mkern-1.5mux_i\mkern-1.5mu}\mkern 1.5mu\).

\(\impliedby\): Present \(M\) by
\(A^n \xrightarrow{f} M \to C =\operatorname{coker}f \to 0\) where
\(f(e_i) = x_i\) -- we want to show \(C = 0\). Reduce mod
\({\mathfrak{m}}\) by applying \(({-})\otimes_A A/{\mathfrak{m}}\) to
get
\begin{align*}
(A/{\mathfrak{m}})^n \xrightarrow{\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu} M/{\mathfrak{m}}M \to C/{\mathfrak{m}}C\to 0
.\end{align*}
This is surjective so \(C/{\mathfrak{m}}C = 0\). By Nakayama, \(C=0\).

\end{solution}

\begin{example}[?]

Let the local ring \(R = k { \left[ {t} \right] }\) with
\({\mathfrak{m}}_R = \left\langle{t}\right\rangle\), and let
\(M = R{ {}^{ \scriptscriptstyle\oplus^{3} } }\). Consider
\begin{align*}
{\left[ {1+t^3 + t^5, t+t^2, t^3} \right]}, {\left[ {t^{22}, 1+t^9, t^{10} + t^{10^{10}}} \right]}, {\left[ {1+ t^{10^{10^{10}}}, 1+t^5 + t^9, 1+t^7 + t^{11} } \right]}
,\end{align*}
which reduced \(\operatorname{mod}t\) yields
\begin{align*}
{\left[ {1,0,0} \right]}, {\left[ {0,1,0} \right]}, {\left[ {1,1,1} \right]}
.\end{align*}
So the original elements generate \(M\).

\end{example}

\begin{remark}

Next goal: proving Nakayama. We'll need a version of Cayley-Hamilton.
Recall the definition of multiplicative subsets and localization, along
with its universal property.

\end{remark}

\begin{example}[of multiplicative sets $S$]

Examples:

\begin{itemize}
\tightlist
\item
  For any element \(f\), \(S\coloneqq\left\{{1, f, f^2,\cdots}\right\}\)
\item
  For \(A\) an integral domain,
  \(S \coloneqq A\setminus\left\{{0}\right\}\)
\item
  All nonzero zero divisors.
\end{itemize}

\end{example}

\begin{exercise}[?]

Prove \(S^{-1}A\) exists and is unique for \(A\in \mathsf{CRing}\). Give
an example where \(s'a = sb\) is not sufficient.

\end{exercise}

\begin{remark}

Remarks on localization:

\begin{itemize}
\tightlist
\item
  \({a\over s} = {b\over s'} \iff s'' s' a - s'' s b\) for some
  \(s''\in S\) -- this is needed because it will hold in \(S^{-1}A\),
  since \(s''\in (S^{-1}A)^{\times}\) for any \(s''\in S\) by
  construction.
\item
  \({a\over s}=0\) iff \(a\) is annihilated by an element of \(S\).
\item
  Producing the actual map for the universal property: if \(f:A\to B\)
  sends \(S\) to invertible elements,
\end{itemize}

\begin{center}
\begin{tikzcd}
    A && B & {f(a)f(s)^{-1}} \\
    \\
    {S^{-1}A} \\
    {{a\over s}}
    \arrow["\iota", from=1-1, to=3-1]
    \arrow["f", from=1-1, to=1-3]
    \arrow[from=3-1, to=1-3]
    \arrow[dashed, maps to, from=4-1, to=1-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJBIl0sWzIsMCwiQiJdLFswLDIsIlNcXGludiBBIl0sWzAsMywie2FcXG92ZXIgc30iXSxbMywwLCJmKGEpZihzKVxcaW52Il0sWzAsMiwiXFxpb3RhIl0sWzAsMSwiZiJdLFsyLDFdLFszLDQsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifSwiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to Diagram}
\end{quote}

\begin{itemize}
\tightlist
\item
  \(A_f \coloneqq S^{-1}A\) for
  \(S \coloneqq\left\{{1,f,f^2,\cdots}\right\}\). For
  \(A = {\mathbb{Z}}\) and \(f=p\) a prime,
  \({\mathbb{Z}}_f = {\mathbb{Z}}{ \left[ { \scriptstyle \frac{1}{p} } \right] } \subseteq {\mathbb{Q}}\)
  are fractions whose denominator is a power of \(p\).
\item
  For \(A\) an integral domain and \(S=A\setminus\left\{{0}\right\}\)
  yields \(S^{-1}A = \operatorname{ff}(A)\) the fraction field.
\item
  For \({\mathfrak{p}}\in \operatorname{Spec}A\),
  \(S\coloneqq A\setminus{\mathfrak{p}}\), then
  \(A_{\mathfrak{p}}\coloneqq S^{-1}A\) is localization at a prime ideal
\item
  \({\mathbb{Z}}_{\left\langle{p}\right\rangle} = {\mathbb{Z}}{ \left[ { \scriptstyle \frac{1}{\ell} } \right] }_{\ell\neq p \text{ prime}}\),
  fractions in \({\mathbb{Q}}\) with denominators not divisible by
  \(p\).
\end{itemize}

\end{remark}

\begin{exercise}[?]

Some exercises:

\begin{itemize}
\tightlist
\item
  Use the universal property to show
  \(({\mathbb{Z}}/15{\mathbb{Z}})_5 \cong {\mathbb{Z}}/3{\mathbb{Z}}\).
\item
  Show that for \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\), \(S^{-1}M\)
  exists and is unique.
\item
  Show that \(S^{-1}A \curvearrowright S^{-1}M\).
\item
  Show that \(S^{-1}M = M\otimes_A S^{-1}A\) using the universal
  property. Use
  \begin{align*}
  M = M\otimes_A A \xrightarrow{\operatorname{id}\otimes S^{-1}({-})} M\otimes_A S^{-1}A
  \end{align*}
  where \(m\mapsto m\otimes 1\) For \(f:M\to N\) where \(s\) acts
  invertibly on \(N\), produce a map \(M\times S^{-1}A\to N\) where
  \((m, a/s)\mapsto as^{-1}f(m)\) where \(s^{-1}\) is the inverse of the
  action \(s: N\to N\).
\item
  Show that \(({-})\otimes_A S^{-1}A\) is left exact and thus exact.

  \begin{itemize}
  \tightlist
  \item
    Injectivity: use that \({m\over s}\mapsto 0 \iff s'f(m) = 0\) for
    some \(s'\in S\).
  \end{itemize}
\item
  Show that \(S^{-1}A \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^\flat\).
\end{itemize}

\end{exercise}

\hypertarget{tuesday-february-01}{%
\section{Tuesday, February 01}\label{tuesday-february-01}}

\begin{proposition}[Cayley-Hamilton]

For
\(A\in \mathsf{CRing}, M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\),
\({\mathfrak{a}}\in \operatorname{Id}(A)\), and
\(\phi: M\to {\mathfrak{a}}M \subseteq M\), there exists
\(\left\{{a_i}\right\}_{i\leq n} \subseteq {\mathfrak{a}}\) such that
\begin{align*}
\phi^r + a_1 \phi^{r-1} + \cdots + a_r \operatorname{id}= 0
.\end{align*}

\end{proposition}

\begin{proof}[Cayley-Hamilton: Reductions]

Reduce to showing this for \(M = A^r\) a free module. Use the diagram:

\begin{center}
\begin{tikzcd}
    {e_i} & {A^r} && M && 0 \\
    \\
    & {{\mathfrak{a}}^r} && {{\mathfrak{a}}M} && 0 \\
    & {} && {a_i}
    \arrow[from=3-2, to=3-4]
    \arrow[from=3-4, to=3-6]
    \arrow[from=1-2, to=1-4]
    \arrow[from=1-4, to=1-6]
    \arrow["\phi", from=1-4, to=3-4]
    \arrow["{\tilde f}", dashed, from=1-2, to=3-2]
    \arrow[from=1-1, to=4-4]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOSxbMSwwLCJBXnIiXSxbMywwLCJNIl0sWzUsMCwiMCJdLFszLDIsIlxcbWZhIE0iXSxbNSwyLCIwIl0sWzEsMiwiXFxtZmFeciJdLFsxLDNdLFswLDAsImVfaSJdLFszLDMsImFfaSJdLFs1LDNdLFszLDRdLFswLDFdLFsxLDJdLFsxLDMsIlxccGhpIl0sWzAsNSwiXFx0aWxkZSBmIiwwLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzcsOF1d}{Link
to Diagram}
\end{quote}

Lift by sending \(e_i\) to any element \(a_i \in {\mathfrak{a}}\). Then:
STS \(\tilde f\) satisfies some polynomial, since \(\phi\) will satisfy
the same polynomial and map to zero by commutativity of the square
above. \(\tilde f: A^r\to A^r\) can be written as a matrix \((a_{ij})\).
Now reduce to \({\mathbb{C}}\): consider the map
\begin{align*}
{\mathbb{Z}}[ \left\{{ x_{ij}}\right\}_{i, j \leq r} ] &\to A \\
x_{ij} &\mapsto a_{ij}
.\end{align*}

Forming the matrix \(M = (x_{ij})\) yields a commutative diagram:

\begin{center}
\begin{tikzcd}
    {{\mathbb{Z}}[\left\{{x_{ij}}\right\}_{i,j\leq r}]{ {}^{ \scriptscriptstyle\times^{r} }  }} && {A^r} \\
    \\
    {{\mathbb{Z}}[\left\{{x_{ij}}\right\}{i,j\leq r}]{ {}^{ \scriptscriptstyle\times^{r} }  }} && {A^r}
    \arrow[from=1-1, to=1-3]
    \arrow["{\tilde f = (a_{ij})}", from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow["{M = (x_{ij})}"', from=1-1, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXFpaW1xcdHN7eF97aWp9fV97aSxqXFxsZXEgcn1dXFxjYXJ0cG93ZXJ7cn0iXSxbMCwyLCJcXFpaW1xcdHN7eF97aWp9fXtpLGpcXGxlcSByfV1cXGNhcnRwb3dlcntyfSJdLFsyLDAsIkFeciJdLFsyLDIsIkFeciJdLFswLDJdLFsyLDMsIlxcdGlsZGUgZiA9IChhX3tpan0pIl0sWzEsM10sWzAsMSwiTSA9ICh4X3tpan0pIiwyXV0=}{Link
to Diagram}
\end{quote}

Since \({\mathbb{Z}}[\left\{{ x_{ij} }\right\}]\) is an integral domain,
we can pass to the fraction field
\({\mathbb{Q}}(\left\{{x_{ij}}\right\})\), where by linear algebra \(M\)
has a characteristic polynomial in the \(x_{ij}\). So for some
homogeneous polynomials \(p_i\) in the \(x_{ij}\), we have
\begin{align*}
\tilde f^r + p_1(x_{ij}) \tilde f^{r-1} + \cdots + p_r(x_{ij}) = 0
.\end{align*}
Now choose an arbitrary embedding
\({\mathbb{Q}}(x_{ij}) \hookrightarrow{\mathbb{C}}\), and prove
Cayley-Hamilton here using (e.g.) Jordan normal form.

\end{proof}

\begin{exercise}[Fun, everyone should have a proof!]

Write a careful proof of Cayley-Hamilton for arbitrary fields.

\end{exercise}

\begin{remark}

Useful strategy: reduce to the ``universal matrix''.

\end{remark}

\begin{proof}[Cayley-Hamilton for arbitrary fields, sketch]

Recall that there exists an \emph{adjugate} of any square matrix
satisfying
\(M \operatorname{adj}(M) = \operatorname{adj}(M) M = \operatorname{det}(M) \operatorname{id}\).
Apply this to
\(M \coloneqq tI - \tilde f \in \mathop{\mathrm{End}}_{A[t]}(A(t){ {}^{ \scriptscriptstyle\times^{r} } })\).
Write \(\operatorname{adj}(tI-\tilde f) = \sum B_i t^i\) with
\(B_i \in \operatorname{Mat}_{n\times n}(A)\). We have
\begin{align*}
(tI-\tilde f) \operatorname{adj}(tI - \tilde f) = \operatorname{det}(\tilde f)I
,\end{align*}
but we can't plug in \(\tilde f\) here because we don't know if this
lands in a \emph{commutative} ring, so e.g.~\(t m t^r \neq mt^{r+1}\)
doesn't necessarily hold.

Note that \(B_i\) commutes with \(tI - \tilde f \iff B_i\) commutes with
\(\tilde f\), e.g.~by equating coefficients. Write \(R = Z(\tilde f)\)
for the centralizer, those matrices commuting with \(\tilde f\). Then
\((tI - \tilde f) \in R[t]\), which reduces us to the world of
commutative rings. Then
\({ \left.{{ \operatorname{det}(tI - \tilde f) }} \right|_{{\tilde f}} } = (\tilde f - \tilde f)\cdot g = 0\).

\end{proof}

\begin{exercise}[?]

Show that if \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
and \({\mathfrak{a}}\in \operatorname{Id}(A)\),
\begin{align*}
M = {\mathfrak{a}}M \implies \exists x\cong 1{\mathsf{{\mathfrak{a}}}{\hbox{-}}\mathsf{Mod}}\text{ such that } xM = 0
.\end{align*}

\begin{quote}
Hint: apply Cayley-Hamilton to \(\operatorname{id}_M\).
\end{quote}

\end{exercise}

\begin{theorem}[Nakayama]

For
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}, I \in \operatorname{Id}(A)\)
with \(I \in {J ({A}) }\),
\begin{align*}
M = IM \implies M = 0
.\end{align*}

\end{theorem}

\begin{remark}

Apply the corollary to get \(x\equiv 1 \operatorname{mod}I\) with
\(xM = 0\). But then \(x \equiv 1 \operatorname{mod}{J ({A}) }\), so
\(x\) is a unique and \(xM = M\).

Alternatively, pick a minimal set of generators \(\left\{{x_i}\right\}\)
of \(M\), so \(m = \sum a_i x_i\) with \(a_i\in I\) since \(M=IM\).
Since \(1-a_n\in {J ({A}) }\) and is a unit, so
\begin{align*}
(1-a_n)x_n = \sum_{i\leq n-1} a_i x_i \implies x_n = (1-a_n)^{-1}\sum_{i\leq n-1}a_i x_i
.\end{align*}
\(\contradiction\)

\end{remark}

\begin{remark}

Notes:

\begin{itemize}
\tightlist
\item
  Proved last time:
  \(A\in \mathsf{Loc}\mathsf{Ring}, M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\),
  if \(X\coloneqq\left\{{x_i}\right\} \subseteq M\) with
  \(\left\{{\mkern 1.5mu\overline{\mkern-1.5mux_i\mkern-1.5mu}\mkern 1.5mu}\right\}\)
  generating \(M/{\mathfrak{m}}M\), then \(X\) generates \(M\).
\item
  Suppose \(f\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}(M, N)\) with
  \(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu: M/{\mathfrak{m}}_A M \to N/{\mathfrak{m}}_A N\)
  an isomorphism -- \(f\) is not necessarily an isomorphism.

  \begin{itemize}
  \tightlist
  \item
    Counterexample: \(k[[x]]\to k\) is not an isomorphism in
    \({\mathsf{k}{\hbox{-}}\mathsf{Mod}}\) but reduces mod \(x\) to
    \(k \xrightarrow{\operatorname{id}} k\).
  \end{itemize}
\item
  Show that \(f\) need not be injective, but is always surjective.

  \begin{itemize}
  \tightlist
  \item
    For surjectivity: use \(M \xrightarrow{f} N \to C \to 0\), use that
    \(C/{\mathfrak{m}}C =0\) to conclude \(C=0\) by Nakayama.
  \item
    For injectivity: use \(0\to K\to M \xrightarrow{f} N \to 0\) and try
    to show \(K=0\). Not true, take
    \(0\to \left\langle{x}\right\rangle \to k[[x]] \to k\to 0\) and
    apply \(({-})\otimes_{k[[x]]} k\) to get
    \(k \xrightarrow{0} k \xrightarrow{\operatorname{id}} k\to 0\).
  \end{itemize}
\item
  The special SES \(0\to M \to M \oplus N \to N \to 0\) has a
  left-section \(s: M \oplus N\to M\); applying \(({-})\otimes_A S\) or
  \(\mathop{\mathrm{Hom}}_A(S, {-})\) actually produces a SES, since
  this induces left sections on the resulting sequences.
\item
  Prove that free modules are projective.
\item
  Prove that divisible abelian groups are injective in
  \({\mathsf{{\mathbb{Z}}}{\hbox{-}}\mathsf{Mod}}\).

  \begin{itemize}
  \tightlist
  \item
    Prove that \({\mathbb{Q}}\) and \({\mathbb{Q}}/{\mathbb{Z}}\) are
    injective.
  \end{itemize}
\item
  Show that a SES splits iff it admits a right section or a left
  section.
\item
  Show that \(0\to I\to M\to P\to 0\) splits if either \(P\) is
  projective or \(I\) is injective.
\item
  Show that
  \(({-})\otimes_A S, \mathop{\mathrm{Hom}}_A({-}, S), \mathop{\mathrm{Hom}}_A(S, {-})\)
  are exact on SES's \(0\to A\to B \to P\to 0\) with \(P\) projective.
\end{itemize}

\end{remark}

\hypertarget{thursday-february-03}{%
\section{Thursday, February 03}\label{thursday-february-03}}

\begin{remark}

Exercises:

\begin{itemize}
\tightlist
\item
  Find a way to remember which \(\hom\) is covariant vs contravariant.
\item
  Show \(P\) is projective \(\iff \mathop{\mathrm{Hom}}(P, {-})\) is
  exact, \(I\) is injective \(\iff \mathop{\mathrm{Hom}}({-}, I)\) is
  exact (sends injections to surjections).
\item
  Find a non-free projective module.
\item
  Show that \(P\) is projective iff \(P\) is a direct summand of a free
  module. Use that
  \(0\to K\to A{ {}^{ \scriptscriptstyle\oplus^{I} } } \to P\to 0\) and
  lift \(P \xrightarrow{\operatorname{id}_P} P\) to get
  \(A{ {}^{ \scriptscriptstyle\oplus^{I} } } = K \oplus P\). For the
  other direction:
\end{itemize}

\begin{center}
\begin{tikzcd}
    &&&&& N \\
    {A{ {}^{ \scriptscriptstyle\oplus^{m} }  }} && {P\oplus K} && P & M \\
    &&&&& 0
    \arrow[two heads, from=1-6, to=2-6]
    \arrow[from=2-6, to=3-6]
    \arrow["{\exists \text{ by freeness}}", curve={height=-18pt}, dotted, from=2-1, to=1-6]
    \arrow["\cong"', Rightarrow, no head, from=2-1, to=2-3]
    \arrow["{{\operatorname{pr}}_1}"', from=2-3, to=2-5]
    \arrow[from=2-5, to=2-6]
    \arrow["{\exists \iota_1}"', curve={height=12pt}, dotted, from=2-5, to=2-3]
    \arrow["{\therefore \exists}", dashed, from=2-5, to=1-6]
\end{tikzcd}
\end{center}

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

\begin{itemize}
\tightlist
\item
  Show that in \({\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\), \(I\) is
  injective iff divisible.

  \begin{itemize}
  \tightlist
  \item
    For one direction, show \(ni' = i\) using the following diagram:
  \end{itemize}
\end{itemize}

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,92;green,92;blue,214}{1} & {\mathbb{Z}}&& {\mathbb{Z}}& \textcolor{rgb,255:red,92;green,92;blue,214}{1} \\
    \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{i} & I & \textcolor{rgb,255:red,92;green,92;blue,214}{i'}
    \arrow["{\times n}", from=1-2, to=1-4]
    \arrow[from=1-2, to=3-2]
    \arrow["\exists", dashed, from=1-4, to=3-2]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=1-1, to=3-1]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, maps to, from=1-5, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbMSwwLCJcXFpaIl0sWzEsMiwiSSJdLFswLDIsImkiLFsyNDAsNjAsNjAsMV1dLFswLDAsIjEiLFsyNDAsNjAsNjAsMV1dLFszLDAsIlxcWloiXSxbMiwyLCJpJyIsWzI0MCw2MCw2MCwxXV0sWzQsMCwiMSIsWzI0MCw2MCw2MCwxXV0sWzAsNCwiXFx0aW1lcyBuIl0sWzAsMV0sWzQsMSwiXFxleGlzdHMiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMywyLCIiLDAseyJjb2xvdXIiOlsyNDAsNjAsNjBdLCJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV0sWzYsNSwiIiwwLHsiY29sb3VyIjpbMjQwLDYwLDYwXSwic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoibWFwcyB0byJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\begin{itemize}
\tightlist
\item
  For the other direction, produce a map:
\end{itemize}

\begin{center}
\begin{tikzcd}
    0 && X && Y \\
    \\
    && I
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-3, to=3-3]
    \arrow["{\exists ?}"{description}, dashed, from=1-5, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCIwIl0sWzIsMCwiWCJdLFs0LDAsIlkiXSxbMiwyLCJJIl0sWzAsMV0sWzEsMl0sWzEsM10sWzIsMywiXFxleGlzdHMgPyIsMSx7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\begin{itemize}
\item
  Use Zorn's lemma on pairs \((Y, f)\) where
  \((Y, f) \leq (Y', g) \iff Y \subseteq Y'\) and
  \({ \left.{{g}} \right|_{{Y}} } = f\). Show every chain has an upper
  bound by setting \(Y_\infty = \displaystyle\bigcup Y_i\) and
  \(f_\infty = \cup f_i\), thus producing
  \(Y_{\mathrm{max}}, f_{\mathrm{max}}\) Take a SES
  \(0 \to \left\langle{y}\right\rangle \to \left\langle{y, Y_{\mathrm{max}}}\right\rangle \to \left\langle{y, Y_{\mathrm{max}}}\right\rangle/\left\langle{y}\right\rangle\to 0\)
  and map the last term to \(I\).
\item
  Recall that projective resolutions are complexes
  \({ {P}_{\scriptscriptstyle \bullet}} = \cdots P_1\to P_0\to 0\) with
  \(H_{i>0}{ {P}_{\scriptscriptstyle \bullet}} = 0\) and
  \(H_0{ {P}_{\scriptscriptstyle \bullet}} = M\), equivalently an exact
  complex \(\cdots\to P_1\to P_0\to M\to 0\).
\item
  Find a projective resolution of
  \(C_2\in{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\) and
  \(k[t]/\left\langle{t^2}\right\rangle\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\).
  Why must the latter necessarily be infinite length?
\item
  Compute
  \(\operatorname{Tor}_*^{\mathbb{Z}}(C_2, C_2) \in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\)
  and
  \(\operatorname{Tor}_*^{\mathbb{Z}}({\mathbb{Z}}{ {}^{ \scriptscriptstyle\oplus^{2} } }, C_2)\).
\item
  Show \(\operatorname{Tor}_*^{R}(k, k) = \bigoplus_{i\geq 0} k\) for
  \(R = k[t]/\left\langle{t^2}\right\rangle\)? Use the resolution
  \(P_i = k[t]/\left\langle{t^2}\right\rangle\) with maps
  \(({-})\times t\), using that \(t\curvearrowright k\) by zero.
\item
  Show that if \(f\simeq g\), then \(f-g\) induces the zero map in
  homology. Use \(d_{i+1} s_i + s_{i-1} d_i: H_i(C)\to H_i(D)\), pick
  \(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\in C_i\)
  with
  \(d_i \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu = 0\)
  and check
  \((d_{i+1} s_i + s_{i-1} d_{i})\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\in \operatorname{im}d_{i+1}\).
\end{itemize}

\end{remark}

\hypertarget{tuesday-february-08}{%
\section{Tuesday, February 08}\label{tuesday-february-08}}

\begin{exercise}[?]

Show that chain-homotopic maps induce the same map in homology.

\end{exercise}

\begin{corollary}[?]

\({ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)\)
for \(M, N\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) is well-defined,
since the identity \(M \xrightarrow{\operatorname{id}_M} M\) induces an
isomorphism on
\({ {P}_{\scriptscriptstyle \bullet}} \otimes N \to { {P}_{\scriptscriptstyle \bullet}} '\otimes N\)
for any two projective resolution
\({ {P}_{\scriptscriptstyle \bullet}} , { {P}_{\scriptscriptstyle \bullet}} ' \rightrightarrows M\).

\end{corollary}

\begin{proposition}[?]

If \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(M, N)\), then there is an
induced morphism
\(\tilde f\in \mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}({ {P}_{\scriptscriptstyle \bullet}} ^M, { {P}_{\scriptscriptstyle \bullet}} ^N)\)
between resolutions
\({ {P}_{\scriptscriptstyle \bullet}} ^M\rightrightarrows M, { {P}_{\scriptscriptstyle \bullet}} ^N\rightrightarrows N\),
where \(\tilde f\) is unique up to homotopy.

\end{proposition}

\begin{proof}[Hint]

For existence, use projectivity to lift through surjections onto
kernels. For uniqueness, create \(s_0\) such that \(d_1 s_0 = d_0-g_0\)
and check that \(\operatorname{im}f_0 - g_0 \subseteq \ker d_0\) after
lifting through \(P_1^N\to \ker d_0\):

\begin{center}
\begin{tikzcd}
    \vdots && \vdots \\
    {P_1^M} && {P_1^N} \\
    \\
    {P_0^M} && {P_0^N} \\
    \\
    M && N && {\ker d_0} \\
    &&&&& 0 \\
    0 && 0 &&&& {}
    \arrow[from=6-1, to=8-1]
    \arrow[from=6-3, to=8-3]
    \arrow["f", from=6-1, to=6-3]
    \arrow["{d_0}", from=4-1, to=6-1]
    \arrow["{d_0'}", from=4-3, to=6-3]
    \arrow["{f_0, g_0}", from=4-1, to=4-3]
    \arrow[two heads, from=4-3, to=6-5]
    \arrow[from=6-5, to=7-6]
    \arrow["\exists"{pos=0.3}, shift right=2, curve={height=12pt}, dashed, from=4-1, to=6-5]
    \arrow[from=2-1, to=4-1]
    \arrow[from=2-3, to=4-3]
    \arrow["{\exists s}", dashed, from=4-1, to=2-3]
\end{tikzcd}
\end{center}

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

\end{proof}

\begin{exercise}[?]

Show that a SES of modules induces a SES of chain complexes between
their projective resolutions.

Hint: use the following diagram.

\begin{center}
\begin{tikzcd}
    && \vdots && \vdots && \vdots \\
    {} && {P_{21}} && \vdots && {P_{23}} \\
    \\
    0 && {P_{11}} && {P_{11} \oplus P_{13}} && {P_{13}} && 0 \\
    \\
    0 && A && B && C && 0
    \arrow[from=6-1, to=6-3]
    \arrow[from=6-3, to=6-5]
    \arrow[from=6-5, to=6-7]
    \arrow[from=6-7, to=6-9]
    \arrow[from=4-7, to=4-9]
    \arrow[from=4-1, to=4-3]
    \arrow[from=2-3, to=4-3]
    \arrow[from=4-3, to=6-3]
    \arrow[from=4-7, to=6-7]
    \arrow[from=2-7, to=4-7]
    \arrow["\exists", dashed, hook, from=4-3, to=4-5]
    \arrow["\exists", dashed, two heads, from=4-5, to=4-7]
    \arrow["\exists", dashed, from=4-7, to=6-5]
    \arrow["{\therefore {\exists}}"', dotted, from=4-5, to=6-5]
\end{tikzcd}
\end{center}

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

\end{exercise}

\begin{exercise}[?]

Show that a SES of chain complexes induces a LES in their homology.
There are about eight conditions one needs to check here (as in the
snake lemma, of which this is a special case for two-term complexes).

\end{exercise}

\begin{exercise}[?]

Show that given \(0\to M_1 \to M_2\to M_3\to 0\), taking projective
resolutions and applying \(({-})\otimes N\) yields a SES
\begin{align*}
0\to { {P}_{\scriptscriptstyle \bullet}} ^1\otimes N \to { {P}_{\scriptscriptstyle \bullet}} ^2\otimes N \to { {P}_{\scriptscriptstyle \bullet}} ^3\otimes N\to 0
,\end{align*}
so there is an induced LES in
\({ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}\).

\end{exercise}

\begin{exercise}[?]

Show

\begin{itemize}
\tightlist
\item
  \(\operatorname{Tor}_0^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) = M\otimes_R N\)
\item
  \(\tau_{\geq 1} { {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) = 0\)
  if either \(M\) or \(N\) is projective.
\item
  \({ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^R(M, N)\) is
  uniquely determined by these properties.
\end{itemize}

\end{exercise}

\begin{remark}

Hint:
\begin{align*}
\operatorname{Tor}_0^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)
&= H_0({ {P}_{\scriptscriptstyle \bullet}} ^M \otimes_R N) \\
&= \operatorname{coker}\qty{P_1\otimes N \to P_0\otimes_R N}\\
&= \operatorname{coker}(P_1\to P_0) \otimes_R N \\
&= M\otimes_R N
.\end{align*}

For vanishing, use that projective implies flat and exact complexes have
zero higher homology. Note that if \(M\) is projective, it is its own
projective resolution.

For uniqueness, induct on \(i\): write
\(0\to K \to R{ {}^{ \scriptscriptstyle\oplus^{J} } }\to M\to 0\), use
that free implies projective, and consider the LES:

\begin{center}
\begin{tikzcd}
    \cdots && 0 && {\operatorname{Tor}_1^R(M, N) = \ker f} \\
    \\
    {\operatorname{Tor}_1(K, N)} && {\operatorname{Tor}_1(R{ {}^{ \scriptscriptstyle\oplus^{I} }  }, N) = 0} && {\operatorname{Tor}_1^R(M, N) = \ker f} \\
    \\
    {K\otimes_R N} && {R{ {}^{ \scriptscriptstyle\oplus^{I} }  }\otimes_R N} && {M\otimes_R N} \\
    &&&& 0
    \arrow[from=5-5, to=6-5]
    \arrow[from=5-3, to=5-5]
    \arrow["f", from=5-1, to=5-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=5-1]
    \arrow[from=1-5, to=3-1]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTAsWzQsNSwiMCJdLFs0LDQsIk1cXHRlbnNvcl9SIE4iXSxbMiw0LCJSXFxzdW1wb3dlcntJfVxcdGVuc29yX1IgTiJdLFswLDQsIktcXHRlbnNvcl9SIE4iXSxbNCwyLCJcXFRvcl8xXlIoTSwgTikgPSBcXGtlciBmIl0sWzIsMiwiXFxUb3JfMShSXFxzdW1wb3dlcntJfSwgTikgPSAwIl0sWzAsMiwiXFxUb3JfMShLLCBOKSJdLFs0LDAsIlxcVG9yXzFeUihNLCBOKSA9IFxca2VyIGYiXSxbMiwwLCIwIl0sWzAsMCwiXFxjZG90cyJdLFsxLDBdLFsyLDFdLFszLDIsImYiXSxbNiw1XSxbNSw0XSxbNCwzXSxbNyw2XSxbOCw3XV0=}{Link
to Diagram}
\end{quote}

Now induct up using the isomorphisms in the LES.

\end{remark}

\begin{exercise}[?]

Show that
\begin{align*}
{ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) \cong {{ {H}_{\scriptscriptstyle \bullet}} }({ {P}_{\scriptscriptstyle \bullet}} ^M \otimes_R N) \cong {{ {H}_{\scriptscriptstyle \bullet}} }(M\otimes_R { {P}_{\scriptscriptstyle \bullet}} ^N )\cong { { \operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(N, M)
.\end{align*}

\end{exercise}

\hypertarget{thursday-february-10}{%
\section{Thursday, February 10}\label{thursday-february-10}}

\begin{exercise}[?]

Let \(R= k[x,y]\) and
\(M = k \cong k[x,y]/\left\langle{x, y}\right\rangle\), and compute
\({ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(k, k)\).

\end{exercise}

\begin{solution}

Hint: use the following resolution

\begin{center}
\begin{tikzcd}
    && 1 && {(y,-x)} && 1 && 1 \\
    0 && {k[x,y]} && {k[x,y]{ {}^{ \scriptstyle\otimes_{R}^{2} }  }} && {k[x,y]} && k && 0 \\
    && {} && {e_1} && x \\
    &&&& {e_2} && y
    \arrow[from=2-1, to=2-3]
    \arrow[from=2-3, to=2-5]
    \arrow[from=2-5, to=2-7]
    \arrow[from=2-7, to=2-9]
    \arrow[from=2-9, to=2-11]
    \arrow[maps to, from=1-3, to=1-5]
    \arrow[maps to, from=3-5, to=3-7]
    \arrow[maps to, from=4-5, to=4-7]
    \arrow[maps to, from=1-7, to=1-9]
\end{tikzcd}
\end{center}

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

\end{solution}

\begin{exercise}[?]

Compute
\({ {\operatorname{Tor}}_{\scriptscriptstyle \bullet}} ^{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(k, k)\)
for \(R=k[x_1, \cdots, x_{n}]\) and
\(M = k = k[x_1, \cdots, x_{n}]/\left\langle{x_1,\cdots, x_n}\right\rangle\).

\end{exercise}

\begin{remark}

\(\operatorname{Tor}\) measures failure of injectivity of tensoring
against a module \(M\),
\({ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}}\) measures
failure of surjectivity when mapping against \(M\).

\end{remark}

\begin{exercise}[?]

Show that for \(R\in\mathsf{CRing}\), every
\(M\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) admits an injective
resolution.

Hint: it suffices to show any \(M\) injects into an injective object.
Use the following diagram:

\begin{center}
\begin{tikzcd}
    0 && M && {I^0} && {I^0/M} && 0 \\
    \\
    &&&& {I'} \\
    \\
    &&&& {\operatorname{coker}f} && {I^2} & \cdots \\
    \\
    &&&& 0
    \arrow[from=1-1, to=1-3]
    \arrow["\exists", color={rgb,255:red,92;green,92;blue,214}, hook, from=1-3, to=1-5]
    \arrow[two heads, from=1-5, to=1-7]
    \arrow["\exists", color={rgb,255:red,92;green,92;blue,214}, hook, from=1-5, to=3-5]
    \arrow[two heads, from=3-5, to=5-5]
    \arrow["\exists", color={rgb,255:red,92;green,92;blue,214}, hook, from=5-5, to=5-7]
    \arrow[from=5-5, to=7-5]
    \arrow["{\therefore \exists}", dashed, from=1-3, to=3-5]
    \arrow[from=1-7, to=1-9]
    \arrow["{\therefore \exists}", dashed, from=3-5, to=5-7]
\end{tikzcd}
\end{center}

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

First show this for \(R={\mathbb{Z}}\) using that
\({\mathbb{Q}}{ {}^{ \scriptscriptstyle\oplus^{J} } }/K\) is divisible
and thus injective:

\begin{center}
\begin{tikzcd}
    && 0 & 0 && 0 \\
    \\
    && 0 & K && K & 0 \\
    \\
    && 0 & {{\mathbb{Z}}{ {}^{ \scriptscriptstyle\oplus^{J} }  }} && {{\mathbb{Q}}{ {}^{ \scriptscriptstyle\oplus^{J} }  }} & {\operatorname{coker}g} \\
    \\
    0 && \textcolor{rgb,255:red,214;green,92;blue,92}{\ker f} & M && {{\mathbb{Q}}{ {}^{ \scriptscriptstyle\oplus^{J} }  }/K} & {\operatorname{coker}f} \\
    \\
    &&& 0 && 0 & 0
    \arrow[from=1-4, to=3-4]
    \arrow[from=1-6, to=3-6]
    \arrow[hook, from=3-4, to=5-4]
    \arrow[hook, from=3-6, to=5-6]
    \arrow[two heads, from=5-4, to=7-4]
    \arrow[two heads, from=5-6, to=7-6]
    \arrow[from=7-4, to=9-4]
    \arrow[from=7-6, to=9-6]
    \arrow[hook, two heads, from=3-4, to=3-6]
    \arrow["g", hook, from=5-4, to=5-6]
    \arrow[from=7-1, to=7-3]
    \arrow[from=7-3, to=7-4]
    \arrow["f", from=7-4, to=7-6]
    \arrow[from=3-3, to=5-3]
    \arrow[from=5-3, to=7-3]
    \arrow[from=3-7, to=5-7]
    \arrow[from=5-7, to=7-7]
    \arrow[dotted, from=7-3, to=3-7]
    \arrow[from=5-3, to=5-4]
    \arrow[from=5-6, to=5-7]
    \arrow[from=7-6, to=7-7]
    \arrow[from=3-3, to=3-4]
    \arrow[from=3-6, to=3-7]
    \arrow[from=7-7, to=9-7]
    \arrow[from=1-3, to=3-3]
\end{tikzcd}
\end{center}

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

Now reduce to \(R={\mathbb{Z}}\) using that \(M\hookrightarrow D\) for
\(D\) a divisible abelian group, and show
\(M\hookrightarrow I \coloneqq\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(R, D)\in{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\).
Form the map as the composition:

\begin{center}
\begin{tikzcd}
    && 0 && M && D \\
    \\
    0 && M && {\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, M)} && {\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, D)} \\
    && m && {{\operatorname{mult}}_m: r\mapsto rm} \\
    &&&& f && {i\circ f}
    \arrow[""{name=0, anchor=center, inner sep=0}, "i", hook, from=1-5, to=1-7]
    \arrow["\exists", dashed, from=3-3, to=3-5]
    \arrow[""{name=1, anchor=center, inner sep=0}, "{i^*}", hook, from=3-5, to=3-7]
    \arrow[maps to, from=4-3, to=4-5]
    \arrow[maps to, from=5-5, to=5-7]
    \arrow[from=1-3, to=1-5]
    \arrow[dashed, from=3-1, to=3-3]
    \arrow["\iota", curve={height=-24pt}, dotted, from=3-3, to=3-7]
    \arrow["{\mathop{\mathrm{Hom}}_{\mathbb{Z}}(R, {-})}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1]
\end{tikzcd}
\end{center}

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

Then show that
\(I\coloneqq\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(R, D)\)
is injective using the universal property:

\begin{center}
\begin{tikzcd}
    0 & X & Y \\
    \\
    {\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(R, D)} \\
    \\
    {\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(Y, \mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(R, D))} && {\mathop{\mathrm{Hom}}_{\mathbb{Z}}(Y, D)} \\
    f && {y\mapsto f(y)(1)} \\
    {y\mapsto(r\mapsto g(ry))} && g
    \arrow[from=1-1, to=1-2]
    \arrow[from=1-2, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow["{?}"', dashed, from=1-2, to=3-1]
    \arrow[tail reversed, from=5-1, to=5-3]
    \arrow[maps to, from=6-1, to=6-3]
    \arrow[maps to, from=7-3, to=7-1]
\end{tikzcd}
\end{center}

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

\end{exercise}

\begin{exercise}[?]

Show \({\mathbb{Z}}\in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\) has an
injective resolution.

\end{exercise}

\begin{solution}

\begin{align*}
0 \to {\mathbb{Z}}\hookrightarrow{\mathbb{Q}}\twoheadrightarrow{\mathbb{Q}}/{\mathbb{Z}}\to 0
.\end{align*}

\end{solution}

\begin{exercise}[?]

Show
\({ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_3, {\mathbb{Z}}) \cong C_3[1]\).

\begin{quote}
Hint: apply
\(\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_3, {-})\)
to the above resolution and use
\(\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_3, {\mathbb{Q}}/{\mathbb{Z}}) \cong C_3\)
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that if \(f\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}(M, N)\) then
there is an induced chain map
\(\tilde f\in \mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}( { {I}^{\scriptscriptstyle \bullet}} _M, { {I}^{\scriptscriptstyle \bullet}} _N)\)
which is unique up to homotopy. Conclude that
\({ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)\)
is independent of injective resolution.

\begin{quote}
Hint: take \(f=\operatorname{id}_M\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that if \(0\to A\to B\to C\to 0\) is a SES in
\({\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) with
\({ {I}^{\scriptscriptstyle \bullet}} _A \leftleftarrows A, { {I}^{\scriptscriptstyle \bullet}} _C\leftleftarrows B\),
then there is a complex
\({ {I}^{\scriptscriptstyle \bullet}} _B\leftleftarrows B\) making
\(0\to { {I}^{\scriptscriptstyle \bullet}} _A \to { {I}^{\scriptscriptstyle \bullet}} _B \to { {I}^{\scriptscriptstyle \bullet}} _C \to 0\)
a SES in \(\mathsf{Ch}{\mathsf{R}{\hbox{-}}\mathsf{Mod}}\).

\end{exercise}

\begin{exercise}[?]

Show that a SES in \({\mathsf{R}{\hbox{-}}\mathsf{Mod}}\) induces a LES
in
\({ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}\).
Do this for both homs: start with \(0\to A\to B\to C\to 0\), and produce
a LES for
\({ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}} (M, {-})\)
and
\({ { \operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}({-}, M)\).

\begin{quote}
Hint: for the first case, apply
\(\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, {-})\)
to the SES of chain complexes of injective resolutions, and use that if
\(I_1\) is injective then the SES splits. For the second, use that
\(\mathop{\mathrm{Hom}}({-}, I)\) is exact iff \(I\) is injective and
take an injective resolution of \(M\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that
\({ { \operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)\)
is uniquely characterized by

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\operatorname{Ext} ^0_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) \cong \mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N)\)
\item
  \(\tau_{\geq 1} { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M, N) = 0\)
  if \(M\) is projective or \(N\) is injective.
\item
  The two LESs above exist.
\end{enumerate}

\end{exercise}

\begin{solution}

Hints:

\begin{itemize}
\tightlist
\item
  Resolve \({ {I}^{\scriptscriptstyle \bullet}} _N \leftleftarrows N\),
  apply
  \(\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M,{-})\),
  and identify
  \(\operatorname{Ext} ^0_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}\) as a
  kernel.
\item
  \(N\) is its own injective resolution when \(N\) is injective.
\item
  \(\mathop{\mathrm{Hom}}_{{\mathsf{R}{\hbox{-}}\mathsf{Mod}}}(M,{-})\)
  is exact when \(M\) is projective.
\item
  For uniqueness, use that if \(0\to N\to I \to C\to 0\) with \(I\)
  injective, then the middle terms in the LES vanish to get
  isomorphisms.
\end{itemize}

\end{solution}

\hypertarget{tuesday-february-15}{%
\section{Tuesday, February 15}\label{tuesday-february-15}}

\begin{exercise}[?]

Check that \(\operatorname{Ext} _R^i(M, {-})\) is independent of
injective resolutions, and \(\operatorname{Ext} _R^i({-}, N)\) is
independent of projective resolutions.

\end{exercise}

\begin{exercise}

Check that \({ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}}\) is
determined by

\begin{itemize}
\tightlist
\item
  \(\operatorname{Ext} ^0 = \mathop{\mathrm{Hom}}\)
\item
  \(\operatorname{Ext} ^{i>0}(P, I) = 0\) if \(P\) is projective or
  \(I\) is injective.
\item
  It extends SESs to LESs
\end{itemize}

\end{exercise}

\begin{exercise}[?]

Show
\begin{align*}
 { {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{k[t]/\left\langle{t^2}\right\rangle } (k, k) = \bigoplus_{i\geq 0} k[i] 
.\end{align*}
Use the projective resolution with entries
\(k[t]/\left\langle{t^2}\right\rangle\) with differential
\({\partial}= \cdot t\)

\end{exercise}

\begin{exercise}[?]

Compute
\({ {\operatorname{Ext} }^{\scriptscriptstyle \bullet}} _{k[x_1, \cdots, x_{n}]}(k, k)\)
using the Koszul resolution
\({ {\bigwedge\nolimits}^{\scriptscriptstyle \bullet}} k[x_1, \cdots, x_{n}]\rightrightarrows k\).

\end{exercise}

\begin{remark}

Defining Noetherian rings and modules:

\begin{itemize}
\tightlist
\item
  \(A\in \mathsf{CRing}\) is Noetherian iff every
  \(\operatorname{Id}(A) \subseteq {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).
\item
  \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is Noetherian iff every
  \(N\leq M\) satisfies
  \(N\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
\end{itemize}

\begin{quote}
Thank you Emmy Noether!!
\end{quote}

\end{remark}

\begin{exercise}[?]

Show that TFAE:

\begin{itemize}
\tightlist
\item
  \(R\) is Noetherian
\item
  Every \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) is
  Noetherian.
\end{itemize}

\end{exercise}

\begin{solution}

For \(1\implies 2\), it STS that \(R^n\) is Noetherian. To reduce, use
the diagram

\begin{center}
\begin{tikzcd}
    && {R^n} && M && 0 \\
    \\
    {R^n} && {\pi^{-1}(N)} && N
    \arrow["\exists", two heads, from=3-1, to=3-3]
    \arrow[hook, from=3-3, to=1-3]
    \arrow["\pi", two heads, from=1-3, to=1-5]
    \arrow[from=3-3, to=3-5]
    \arrow[hook, from=3-5, to=1-5]
    \arrow[from=1-5, to=1-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMiwwLCJSXm4iXSxbNCwwLCJNIl0sWzYsMCwiMCJdLFs0LDIsIk4iXSxbMiwyLCJcXHBpXFxpbnYoTikiXSxbMCwyLCJSXm4iXSxbNSw0LCJcXGV4aXN0cyIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFs0LDAsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMSwiXFxwaSIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFs0LDNdLFszLDEsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMl1d}{Link
to Diagram}
\end{quote}

To show \(R^n\) is Noetherian, use induction since we know \(R^1\) is
Noetherian. Use the following diagram, using the snake lemma on
\(s_1, s_3\) to show \(s_2\) is surjective:

\begin{center}
\begin{tikzcd}
    && 0 && {R^{n-1}} && {R^n} && R && 0 \\
    \\
    && 0 && {N \cap R^{n-1}} && N && {N \over N \cap R^{n-1}} && 0 \\
    &&&& {{\mathrm{fg}}\text{ by IH}} && {\therefore {\mathrm{fg}}} && {{\mathrm{fg}}\text{ as a submodule of } R} \\
    \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{0} && \textcolor{rgb,255:red,92;green,92;blue,214}{R^a} && \textcolor{rgb,255:red,92;green,92;blue,214}{R^{a+b}} && \textcolor{rgb,255:red,92;green,92;blue,214}{R^b} && \textcolor{rgb,255:red,92;green,92;blue,214}{0}
    \arrow[hook, from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[two heads, from=1-7, to=1-9]
    \arrow[two heads, from=3-7, to=3-9]
    \arrow[from=3-5, to=3-7]
    \arrow[hook, from=3-3, to=3-5]
    \arrow[from=1-9, to=1-11]
    \arrow[from=3-9, to=3-11]
    \arrow["{s_1}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, two heads, from=6-3, to=3-5]
    \arrow["{s_3}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, two heads, from=6-7, to=3-9]
    \arrow["{\therefore s_2}", color={rgb,255:red,92;green,92;blue,214}, curve={height=-12pt}, dashed, two heads, from=6-5, to=3-7]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=6-1, to=6-3]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, hook, from=6-3, to=6-5]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, two heads, from=6-5, to=6-7]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, from=6-7, to=6-9]
    \arrow[hook, from=3-5, to=1-5]
    \arrow[hook, from=3-7, to=1-7]
    \arrow[hook, from=3-9, to=1-9]
\end{tikzcd}
\end{center}

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

\end{solution}

\begin{exercise}[?]

Show TFAE:

\begin{itemize}
\tightlist
\item
  \(M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is Noetherian.
\item
  The ACC for submodules.
\end{itemize}

\end{exercise}

\begin{solution}

Hint: for \(1\implies 2\), set
\(M_\infty = \displaystyle\bigcup_i M_i \leq M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\).
Write \(M_i = \left\langle{x_1,\cdots, x_n}\right\rangle\) and use that
each \(x_k \in M_{i_k}\) to choose \(N\gg 1\) with \(M_\infty = M_N\).
For \(2\implies 1\), for \(M\) non-Noetherian find \(S\leq M\)
infinitely generated as \(S = \left\langle{x_1,\cdots}\right\rangle\)
and take the chain \(\left\{{S_k}\right\}_{k\geq 0}\) where
\(S_k = \left\langle{x_1,\cdots, x_k}\right\rangle\).

\end{solution}

\begin{exercise}[?]

Show that the following are Noetherian:

\begin{itemize}
\tightlist
\item
  Fields
\item
  PIDs
\item
  \(R/I\) for \(R\) Noetherian and \(I\) arbitrary
\item
  For \(A\) Noetherian, \(A[x]\) (Hilbert's basis theorem).\footnote{A
    very important result! Marks the end of invariant theory
    historically.}

  \begin{itemize}
  \tightlist
  \item
    Similarly \(A{\left[\left[ x \right]\right] }\).
  \end{itemize}
\item
  Localizations of Noetherian rings
\end{itemize}

\end{exercise}

\begin{theorem}[?]

If \(A\) a Noetherian local ring and
\(M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fg}}, \mathop{\mathrm{proj}}}\),
then \(M\) is free.

\end{theorem}

\begin{exercise}[?]

Prove this!

\end{exercise}

\begin{solution}

Hints:

\begin{itemize}
\tightlist
\item
  Surjectivity:

  \begin{itemize}
  \tightlist
  \item
    Choose a basis
    \(M/{\mathfrak{m}}M = \left\langle{\left\{{ \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu}\right\}_{k\leq n} }\right\rangle\)
    and lifts \(x_k\) to \(M\).
  \item
    Take a surjective map \(A^n \to M\) where \(e_i \mapsto x_i\).
  \item
    Take the SES
    \(A^n \xrightarrow{f} M \twoheadrightarrow\operatorname{coker}f\to 0\)
    and apply \(({-})\otimes_A A/{\mathfrak{m}}\); use that
    \((A/{\mathfrak{m}})^n { \, \xrightarrow{\sim}\, }M/{\mathfrak{m}}M\)
    and apply Nakayama.
  \end{itemize}
\item
  Injectivity:

  \begin{itemize}
  \tightlist
  \item
    Write \(0\to K = \ker f \to A^n \xrightarrow{f} M \to 0\) and apply
    \(({-})\otimes_A A/{\mathfrak{m}}\) to get
    \begin{align*}
    \cdots \to \operatorname{Tor}^0(M, A/{\mathfrak{m}}) \to K/{\mathfrak{m}}K \to (A/{\mathfrak{m}})^n  { \, \xrightarrow{\sim}\, }M/{\mathfrak{m}}M\to 0
    ,\end{align*}
    then \(K/{\mathfrak{m}}K = 0\) and apply Nakayama to conclude
    \(K=0\).
  \end{itemize}
\end{itemize}

\end{solution}

\begin{remark}

Try an example: \(k[x]\in {\mathsf{k}{\hbox{-}}\mathsf{Mod}}\), which is
free after reduction mod
\({\mathfrak{m}}= \left\langle{x}\right\rangle\) but not free before
reduction.

\end{remark}

\begin{remark}

Note that this works with projective replaced by flat.

\end{remark}

\begin{remark}

Why care about Noetherian rings? Hilbert studied group actions \(G\) on
modules over (say) polynomial rings and wanted to find the submodules of
\(G{\hbox{-}}\)invariants. There was an industry of writing down
generating sets in order to show existence and finiteness, and the basis
theorem (which is partially effective) showed that this is no longer
necessary -- finite generating sets always exist.

\end{remark}

\begin{theorem}[Hilbert's basis theorem]

If \(A\) is Noetherian then \(A[x]\) is Noetherian.

\end{theorem}

\begin{proof}[?]

Fix \(U \subseteq A[x]\) an ideal, so
\(I = \left\{{f\in I{~\mathrel{\Big\vert}~}f = \sum a_{f, i} x^i}\right\}\).
Write \(J = \left\langle{a_{f, \deg f}}\right\rangle\) be the ideal
generated by all leading coefficients of elements in \(I\). Since \(A\)
is Noetherian, \(J\) is finitely-generated, so write
\(J = \left\langle{a_1,\cdots, a_n}\right\rangle\) and choose
\(\left\{{f_1,\cdots, f_n}\right\}\) so that \(a_i\) is the leading
coefficient of \(f_i\). Note that these exist since
\(J = A\left\{{a_1,\cdots, a_n}\right\}\) (i.e.~these already form an
ideal). Consider \(L\coloneqq I \cap A[x]^{\deg \leq d}\): since \(A\)
is Noetherian,
\(A[x]^{\deg \leq d} \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\),
and \(L\) also forms a finitely generated \(A{\hbox{-}}\)module. Write
the generators as \(L = \left\{{g_1,\cdots, g_m}\right\}\).

\begin{claim}

\begin{align*}
I = I_{fg} \coloneqq\left\langle{f_1,\cdots, f_n, g_1,\cdots, g_m}\right\rangle
.\end{align*}

\end{claim}

\begin{proof}[?]

If not, pick \(f\in I\setminus I_{fg}\) of minimal degree. Then
\(\deg f > d\) by construction of \(L\). Write \(f = \sum b_i x^i\)
where \(b_{\deg f} = \sum c_i a_i\), and check
\(f - \sum c_i f_i x^{\deg f - \deg f_i}\in I\).

\end{proof}

\end{proof}

\begin{remark}

Idea: the \(f_k\) take care of low degree elements, the \(g_k\) knock
things down in degree.

\end{remark}

\begin{corollary}[?]

Any quotient \(k[x_1, \cdots, x_{n}]/I\) is Noetherian, as is
\({\mathbb{Z}}[x_1,\cdots, x_n]/I\).

\end{corollary}

\begin{example}[Non-Noetherian rings]

Some examples:

\begin{itemize}
\tightlist
\item
  \(k[x_1,\cdots, ]\) is not Noetherian: take
  \(I = \left\langle{x_1,\cdots}\right\rangle\).
\item
  \(k[t. t^{1\over 2}, t^{1\over 3}, t^{1\over 4},\cdots]\).
\item
  \({\mathbb{Z}} { \left[ \scriptstyle { \left\{{ 2^{1\over n}}\right\}_{n\geq 0} } \right] }\).
  Note this is countable!
\end{itemize}

\end{example}

\begin{remark}

Next time: other finiteness conditions, integrality, the
Nullstellensatz.

\end{remark}

\hypertarget{thursday-february-17}{%
\section{Thursday, February 17}\label{thursday-february-17}}

\begin{exercise}[?]

Prove the correspondence theorem between \(\operatorname{Id}(A)\) and
\(\operatorname{Id}(S^{-1}A)\).

\begin{center}
\begin{tikzcd}
    A && {S^{-1}A} \\
    \\
    {\iota^{-1}(I)} && I
    \arrow["\iota", from=1-1, to=1-3]
    \arrow[hook, from=3-3, to=1-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-1, to=1-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJBIl0sWzIsMCwiU1xcaW52IEEiXSxbMCwyLCJcXGlvdGFcXGludihJKSJdLFsyLDIsIkkiXSxbMCwxLCJcXGlvdGEiXSxbMywxLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDNdLFsyLDBdXQ==}{Link
to Diagram}
\end{quote}

Use that

\begin{itemize}
\tightlist
\item
  \(\iota^{-1}I\in \operatorname{Id}(A)\)
\item
  \(\iota(\iota^{-1}(I))\) generates \(I\).
\item
  \({a\over s} = {1\over s}{a\over 1}\) and \({a\over 1}\) is in the
  image of the lower map.
\end{itemize}

\end{exercise}

\begin{exercise}[?]

Show that the localization of any Noetherian ring is again Noetherian.

\end{exercise}

\begin{exercise}[?]

Say \(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{ \operatorname{fp} }\)
iff there is an exact sequence \(R^a\to R^b\to M\to 0\). Show that if
\(R\) is Noetherian,
\({\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{ \operatorname{fp} }= {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fg}}}\).

\end{exercise}

\begin{definition}[Integrality]

For \(f\in \mathsf{CRing}(A, B)\) and \(b\in B\) define
\begin{align*}
f_b: A[x] &\to B \\
x&\mapsto b \\
\sum a_i x^j &\mapsto \sum f(a_i) b^i
.\end{align*}
The element \(b\) is \textbf{integral over \(A\)} iff
\(\operatorname{im}f_b \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).
The ring \(B\) is \textbf{integral over \(A\)} iff every \(b\in B\) is
integral over \(A\) as above.

\end{definition}

\begin{exercise}[?]

Show that
\({\mathbb{Z}}{ \left[ { \scriptstyle \frac{1}{2} } \right] } \subseteq {\mathbb{Q}}\)
is not integral.

\end{exercise}

\begin{exercise}[?]

Show TFAE:

\begin{itemize}
\tightlist
\item
  \(b\) is integral over \(A\)
\item
  \(b\) satisfies a monic polynomial over \(A\), so
  \(b^n + \sum_{i = 1}^n a_i b^{n-i} = 0\).
\end{itemize}

\end{exercise}

\begin{solution}

Hints:

\begin{itemize}
\tightlist
\item
  Write
  \(\operatorname{im}f_b = I\coloneqq\left\langle{1,b,\cdots, b^{n-1}}\right\rangle\),
  suppose \(b^n\in I\), and show \(b^{n+1} \in I\) by expanding as a
  sum.
\item
  Set \(M_i = \left\langle{1,b, \cdots, b^i}\right\rangle\) and use that
  these stabilize to conclude \(b^n \in M_{n-1}\) for some \(n\).
\end{itemize}

\end{solution}

\begin{warnings}

\(B\) Noetherian over \(A\) does not imply \(B\) is Noetherian! Consider
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\),
the algebraic integers, which is integral over \({\mathbb{Z}}\). Note
that \(\left\langle{2^{1\over n}}\right\rangle\) is not
finitely-generated.

\end{warnings}

\hypertarget{toward-the-nullstellensatz}{%
\section{Toward the Nullstellensatz}\label{toward-the-nullstellensatz}}

\begin{remark}

Idea: there is a dictionary between \(k[x_1, \cdots, x_{n}]\) and
\({\mathbb{A}}^n_{/ {k}}\) for
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\):

\begin{itemize}
\tightlist
\item
  Points in \({\mathbb{A}}^n\) correspond to maps
  \(k[x_1, \cdots, x_{n}]\to k\) given by \(x_i\mapsto c_i\in k\).
\item
  Rings \(R\) correspond to their prime spectra
  \(\operatorname{Spec}R\).
\item
  Kernels correspond to maximal ideals
\item
  Ideals \(I\in \operatorname{Id}(R)\) correspond to
  \(V(I) =\left\{{{\mathfrak{m}}{~\mathrel{\Big\vert}~}I \subseteq {\mathfrak{m}}}\right\}\)
\item
  Ideals \(I\in \operatorname{Id}(k[x_1, \cdots, x_{n}])\) correspond to
  \(V(I) = \left\{{\mathbf{c}\in k{ {}^{ \scriptscriptstyle\times^{n} } } {~\mathrel{\Big\vert}~}f(\mathbf{c}) = 0 \, \forall f\in I}\right\}\).
\end{itemize}

\end{remark}

\begin{theorem}[Nullstellensatz V1]

Any \({\mathfrak{m}}\in \operatorname{mSpec}k[x_1, \cdots, x_{n}]\) is
given by \(\ker \pi_p\) for some morphism
\begin{align*}
\pi_p: k[x_1, \cdots, x_{n}]\to k \\
x_i &\mapsto c_i
\end{align*}
for some
\(p = {\left[ {c_1,\cdots, c_n} \right]} \in k{ {}^{ \scriptscriptstyle\times^{n} } }\).
Equivalently,
\({\mathfrak{m}}= \left\langle{x_1-c_1,\cdots, x_n - c_n}\right\rangle\).

\end{theorem}

\begin{theorem}[Nullstellensatz]

If \(f\in k[x_1, \cdots, x_{n}]\) satisfies
\({ \left.{{f}} \right|_{{V(I)}} } = 0\) for a fixed
\(I\in \operatorname{Id}(k[x_1, \cdots, x_{n}])\), then \(f^n\in I\) for
some \(n>0\). I.e. there is a bijective correspondence
\(I\rightleftharpoons V(I)\) for radical ideals \(\sqrt{I} = I\).

\end{theorem}

\begin{example}[?]

Necessity of conditions:

\begin{itemize}
\item
  Why
  \(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\)
  is needed: take
  \(\left\langle{x^-2}\right\rangle\in {\mathbb{Q}}[x]\), this is a
  maximal ideal but \(V(I) = 0\).
\item
  Why the radical condition is needed: take
  \(I = \left\langle{x^{10} }\right\rangle\) so
  \(V(I) = \left\{{0}\right\} \subseteq k\), but
  \({ \left.{{x}} \right|_{{V(I)}} } = 0\).
\end{itemize}

\end{example}

\begin{corollary}[?]

For
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\)
and \(R\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\),
\begin{align*}
\operatorname{mSpec}R \rightleftharpoons\mathop{\mathrm{Hom}}_{{\mathsf{Alg}_{/k} }}(R, k)
.\end{align*}

\end{corollary}

\begin{corollary}[?]

\(V(f) = k{ {}^{ \scriptscriptstyle\times^{n} } } = V(0)\) iff
\(f\in {\sqrt{0_{R}} }\).

\end{corollary}

\begin{corollary}[?]

Let \(R\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\), so
\(R = k[x_1, \cdots, x_{n}]/I\) for some \(n\), and let
\(V(I) \subseteq k{ {}^{ \scriptscriptstyle\times^{n} } }\). Given
\(J \subseteq R\) radical, \(V(J) \subseteq V(I)\), and \(f\in R\)
vanishes on \(V(J)\) iff \(f^n\in J\) for some \(n>0\).

\end{corollary}

\begin{theorem}[Maximal idealansatz]

For
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\)
and \(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) with
\({\mathfrak{m}}\in \operatorname{mSpec}A\),
\begin{align*}
A/{\mathfrak{m}}\cong k
.\end{align*}

\end{theorem}

\begin{theorem}[Maximal idealansatz 2]

For \(k\) an arbitrary field and
\(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) with
\({\mathfrak{m}}\in \operatorname{mSpec}A\),
\begin{align*}
A/{\mathfrak{m}}
\text{ is a finite extension of } k
.\end{align*}

\end{theorem}

\begin{theorem}[EEKS / proto Zariski's lemma]

If \(k \subseteq F\) fields with
\(F\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\), then \(F_{/ {k}}\) is a
finite extension of fields.

\end{theorem}

\begin{corollary}[?]

If \(R\in {\mathsf{Alg}_{/{\mathbb{Z}}}}^{\mathrm{fg}}\) and
\({\mathfrak{m}}\in \operatorname{mSpec}R\), then \(R/{\mathfrak{m}}\)
is finite.

\end{corollary}

\begin{proof}[?]

Check
\({\mathfrak{m}}\cap{\mathbb{Z}}= {\mathfrak{p}}\in \operatorname{Spec}{\mathbb{Z}}\)
and \(R/{\mathfrak{m}}\) is a finitely generated extension of
\({\mathbb{F}}_p\) and hence finite.

\end{proof}

\begin{lemma}[Zariski]

If \(A\in \mathsf{CRing}^{ \mathrm{Noeth} }\) and
\(A \subseteq B \subseteq C\) with

\begin{itemize}
\tightlist
\item
  \(C\in {{A}{\hbox{-}}\mathsf{Alg}}^{\mathrm{fg}}\) and
\item
  \(C\in {\mathsf{B}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\),
\end{itemize}

then \(B\in {{A}{\hbox{-}}\mathsf{Alg}}^{\mathrm{fg}}\).

\end{lemma}

\begin{proof}[?]

Sketch:

\begin{itemize}
\tightlist
\item
  Choose \(x_1,\cdots, x_n\) generating \(C\) as an
  \(A{\hbox{-}}\)algebra
\item
  Choose \(y_1,\cdots, y_m\) generating \(C\) as a
  \(B{\hbox{-}}\)module.
\item
  Write \(x_i = \sum_j b_{ij} y_j\) with \(b_{ij}\in B\)
\item
  Write \(x_i x_j = \sum_k b_{ijk} y_k\) with \(b_{ijk}\in B\)
\item
  Let \(B_0 \subseteq B\) be the \(A{\hbox{-}}\)algebra generated by the
  \(b_{ij}\) and \(b_{ijk}\).
\item
  Observe that \(B_0\) is Noetherian by the Hilbert basis theorem since
  it's finitely generated as an \(A{\hbox{-}}\)algebra.
\item
  For any \(c\in C\), write \(c = \sum_i b_i y_i\) since the \(y_i\)
  generate \(C\) as a \(B_0\) module.
\item
  Idea: can rewrite in terms of lower degree monomials?
\item
  Since each \(b_i = \sum a_{I}^i x^I\), we have
  \(c = \sum_i \sum_I a_I^i x^I y_i\), which is a polynomial in
  \(\sum b_{ij} y_j\).
\item
  Then \(y_{i} \cdots y_{i_k} = \sum p_s y_s\) where the \(p_s\) are
  polynomials in the \(b_{ijk}\)?
\item
  Since \(B_0\) is Noetherian and \(B \subseteq C\), \(B\) is finitely
  generated as a \(B_0\) module and thus as a \(B_0\) algebra.
\item
  Since \(A \subseteq B_0 \subseteq B\) and \(B_0\) is
  finitely-generated as an \(A\) algebra and \(B\) is finitely-generated
  as a \(B_0\) algebra, we have that \(B\) is finitely-generated as an
  \(A\) algebra.
\end{itemize}

\end{proof}

\hypertarget{thursday-february-24}{%
\section{Thursday, February 24}\label{thursday-february-24}}

\hypertarget{going-up-going-down}{%
\subsection{Going up, going down}\label{going-up-going-down}}

\begin{theorem}[Going Up I]

Let \(A \subseteq B\in \mathsf{CRing}\) with \(B\) integral over \(A\).
Then for every \(p\in \operatorname{Spec}A\) there is a
\(q\in \operatorname{Spec}B\) with \(p = A \cap q\).

\end{theorem}

\begin{remark}

Idea: \(\operatorname{Spec}B\to \operatorname{Spec}A\) has finite
fibers.

\end{remark}

\begin{lemma}[?]

If \(A \hookrightarrow B\) is an integral extension, then
\(A\in \mathsf{Field}\iff B\in \mathsf{Field}\).

\end{lemma}

\begin{proof}[of lemma]

\(\implies\): Last time.

\(\impliedby\): Given \(x\in A\) we have \(x^{-1}\in A\). Then if
\(f(x) = x^{-n} + \sum_{0\leq k \leq n-1} a_k x^{-k}\), with
\(f(x) = 0\), then
\(x^{n-1}f(x) = 0 \implies x^{-1}= -\sum a_k x^k \in A\).

\end{proof}

\begin{proof}[of Going Up]

Note
\(A \left[ { \scriptstyle { {p}^{-1}} } \right] \hookrightarrow B \left[ { \scriptstyle { {p}^{-1}} } \right]\)
is still integral, and let
\(q'\in \operatorname{mSpec}B \left[ { \scriptstyle { {p}^{-1}} } \right] \neq \emptyset\).
Write \(p\) for the maximal ideal of \(A\), then (claim)
\(A \left[ { \scriptstyle { {p}^{-1}} } \right] \cap q' = p\). STS
\(A \left[ { \scriptstyle { {p}^{-1}} } \right] \cap q'\) is maximal,
since these are local rings, and so it's ETS
\(A \left[ { \scriptstyle { {p}^{-1}} } \right]/(A \left[ { \scriptstyle { {p}^{-1}} } \right] \cap q') \in \mathsf{Field}\).
Since this is integral iff
\(B \left[ { \scriptstyle { {p}^{-1}} } \right]/q'\) is integral, which
it is, by the lemma this is a field too.

\begin{exercise}[?]

Show tat taking the preimage of \(q'\) in \(B\) works.

\end{exercise}

\end{proof}

\begin{theorem}[Going Up II: partial lifts of chains of primes extend]

Given \(A \hookrightarrow B\) integral and
\begin{align*}
p_1 &\subseteq p_2 \subseteq \cdots \subseteq p_n \in \operatorname{Spec}A \\
q_1 &\subseteq q_2 \subseteq \cdots \subseteq q_m \in \operatorname{Spec}B \\
,\end{align*}
where \(p_i = q_i \cap A\) and e.g.~with \(m\leq n\), then there exist
\(q_{m+1}, \cdots, q_n\) with \(q_j \subseteq q_{j+1}\) and
\(q_i \cap A = p_i\).

\end{theorem}

\begin{proof}[?]

Note that it's enough to lift one stage and induct. So given
\(q_1 \subseteq \cdots q_n\), it's ETF \(q_{m+1} \supseteq q_m\) with
\(q_{m+1} \cap A = p_{m+1}\). Strategy:

\begin{itemize}
\tightlist
\item
  Replace \(A\) with \(A/p_n\) and \(B\) with \(B/q_n\).
\item
  Find
  \(\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu \in B/q_m\)
  with
  \(\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu \cap A/p_n\)
  the image of \(p_{n+1}\)
\item
  Use that
  \(\mkern 1.5mu\overline{\mkern-1.5muq\mkern-1.5mu}\mkern 1.5mu\)
  exists by Going Up I.
\end{itemize}

\end{proof}

\begin{remark}

The geometry:
\(A\to B\leadsto \operatorname{Spec}B \xrightarrow{\pi} \operatorname{Spec}A\).
Increasing chains \(p_i\) means
\(p_{i+1}\in { \operatorname{cl}} _{\operatorname{Spec}A} p_i\), and
``going up'' means sequences can be completed with points in closures in
\(\operatorname{Spec}B\) I.e. \(\pi\) is a closed map, i.e.~closed under
specialization (passing to a point in the closure). Idea: covering map,
possibly with ramification or splitting.

\includegraphics{figures/2022-02-24_13-20-20.png}

\end{remark}

\begin{example}[?]

Consider \(k[x]\hookrightarrow k[x,y]/\left\langle{y^2-x}\right\rangle\)
over \(\operatorname{ch}k = 0\) and
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\).
Take \(p=\left\langle{x-2}\right\rangle\) and
\(q = \left\langle{y-\sqrt 2}\right\rangle\) or
\(\left\langle{y+\sqrt 2}\right\rangle\) extend \(p\).

\includegraphics{figures/2022-02-24_13-21-53.png}

Similarly, take \({\mathbb{Z}}\hookrightarrow{\mathbb{Z}}[i]\) and
consider how primes lift (see inert, split, ramified).

\end{example}

\begin{definition}[Integrally closed]

For \(A \subseteq B\), say \(A\) is \textbf{integrally closed in \(B\)}
iff \(A\) contains every element of \(B\) which is integral over \(A\).

\end{definition}

\begin{example}[?]

\(k[x] \subseteq k[x,y]\) is integrally closed, but
\(k[x] \subseteq k[x,y]/\left\langle{y^2-x}\right\rangle\) is not.

\end{example}

\begin{theorem}[Going Down]

For \(A\leq B\in \mathsf{IntDom}\) with \(A\) integrally closed in
\(\operatorname{ff}(A)\) and \(B\) integral over \(A\). If
\begin{align*}
p_1 &\supseteq p_2 \supseteq\cdots \supseteq p_n \in \operatorname{Spec}A \\
q_1 &\supseteq q_2 \supseteq\cdots \supseteq q_m \in \operatorname{Spec}B \\
\end{align*}
with \(q_i \cap A = p_i\), then there exist
\(q_m \supseteq q_{m+1} \supseteq\cdots \supseteq q_{n+1} \supseteq q_n\)
with \(q_i \cap A = p_i\).

\end{theorem}

\begin{proof}[?]

See A\&M, similar to proof of going up.

\end{proof}

\begin{remark}

Idea: closed under \emph{generization} (opposite of specialization,
given \(x\) finding a point \(y\) with \(x\in { \operatorname{cl}} y\)),
so the geometric map is almost open.

\end{remark}

\begin{example}[?]

Being integrally closed corresponds to a variety being \textbf{normal}
and is a smoothness condition.

\end{example}

\begin{exercise}[Challenge]

Use
\(k[x,y]/\left\langle{y^2-x^3}\right\rangle \cong k[x^2, x^3] \hookrightarrow k[x,y]\)
to construct a counterexample to the going down theorem when
\(A \hookrightarrow\operatorname{ff}(A)\) is \emph{not} integrally
closed.

\end{exercise}

\hypertarget{local-properties}{%
\subsection{Local Properties}\label{local-properties}}

\begin{remark}

On local properties: for
\(M\in {\mathsf{R}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fg}}, { \mathrm{Noeth} }}\),
a HW problem shows if \(p\in \operatorname{Spec}A\) and
\(M \left[ { \scriptstyle { {p}^{-1}} } \right] = 0\) then
\(M \left[ { \scriptstyle { {f}^{-1}} } \right] = 0\) for some \(f\).
This says that the property of being zero extends:

\includegraphics{figures/2022-02-24_13-45-57.png}

\end{remark}

\begin{definition}[Local]

A property \(Q\) is \textbf{local} iff given
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\), TFAE:

\begin{itemize}
\tightlist
\item
  \(Q\) holds for \(M\)
\item
  \(Q\) holds for \(M \left[ { \scriptstyle { {f_i}^{-1}} } \right]\)
  for every \(\left\{{f_i}\right\}\) with
  \(\left\langle{\left\{{f_i}\right\}}\right\rangle = \left\langle{1}\right\rangle\).
\end{itemize}

\end{definition}

\begin{slogan}

Local properties can be checked on an open cover of
\(\operatorname{Spec}A\), and \({\mathsf{A}{\hbox{-}}\mathsf{Mod}}\)
corresponds to \({\mathsf{QCoh}}(\operatorname{Spec}A)\).

\end{slogan}

\begin{remark}

One can always take the set \(\left\{{f_i}\right\}\) to be finite since
if such a collection generates the unit ideal, there is some finite sum
\(\sum a_i f_i = 1\). One can also reformulate the second condition as
follows: for each \(p\in \operatorname{Spec}A\), there exists some
\(f_p \not\in p\) such that every
\(M \left[ { \scriptstyle { {f_p}^{-1}} } \right]\) satisfies \(Q\).
\(\implies\): Check that
\(\left\langle{\left\{{f_p}\right\}}\right\rangle = \left\langle{1}\right\rangle\);
if not then there exists some \(m\in \operatorname{mSpec}A\) with
\(\left\{{f_p}\right\} \subseteq m\) which is maximal and hence prime.

\(\impliedby\): The claim is that given \(p\) there exist
\(f_i\not\in p\). If not, \(\left\{{f_i}\right\} \subseteq p\) and
\(1\in p\).

\end{remark}

\begin{corollary}[?]

If \(\left\langle{\left\{{f_i}\right\}}\right\rangle = 1\) then
\(\operatorname{Spec}A \left[ { \scriptstyle { {f_i}^{-1}} } \right] = \operatorname{Spec}A\setminus V(f_i) \subseteq \operatorname{Spec}A\)
is an open cover of \(\operatorname{Spec}A\). Thus
\(\operatorname{Spec}A\) is quasicompact for any ring \(A\).

\end{corollary}

\begin{remark}

Some local properties:

\begin{itemize}
\tightlist
\item
  Being zero
\item
  Being injective/surjective/bijective
\item
  Being finitely generated (and projective)
\item
  Being flat
\end{itemize}

\end{remark}

\hypertarget{tuesday-march-01}{%
\section{Tuesday, March 01}\label{tuesday-march-01}}

\begin{remark}

A property is \emph{local} on \(A\) if it can be checked on an affine
open cover of \(\operatorname{Spec}A\). Note that
e.g.~\({\mathbb{A}}^2_{/ {k}}\) is affine but
\(\operatorname{Spec}k[x,y]/\left\langle{x,y}\right\rangle \cong{\mathbb{A}}^2_{/ {k}} \setminus\left\{{0}\right\}\)
is not an affine open subset.

\end{remark}

\begin{exercise}[?]

Show that the property of being zero is local.

\begin{quote}
Hint: \(M_{f_i} = 0 \implies\) for every \(m\in M\) there is an \(n_i\)
with \(f_i^{n_i} m = 0\), write \(\sum a_i f_i = 1\), and consider
\((\sum a_i f_i)^N m\) for \(N\gg 1\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that being injective/surjective/bijective is local.

\begin{quote}
Hint: localization is exact, so take the SES
\(0\to \ker g \to M \xrightarrow{g} N\to \operatorname{coker}g\to 0\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that being finitely generated is local.

\begin{quote}
Hint: tensor a presentation. For the other direction, take generators
\(M_{f_i} = \left\langle{{x_{i1} \over a_{i1}}, {x_{i2} \over a_{i2}},\cdots, {x_{in} \over a_{in}} }\right\rangle\)
for \(\left\{{f_1,\cdots, f_m}\right\}\), where without loss of
generality \(a_{ij} = 1\), and take \(\left\{{x_{ij}}\right\}\) and
check surjectivity locally.
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that flatness is local, i.e.~if
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^\flat\) then
\(M_f\in {\mathsf{A_f}{\hbox{-}}\mathsf{Mod}}^\flat\).

\begin{quote}
Hint: to show \(M_f\) is flat assuming \(M\) is flat, show that
\(\operatorname{Tor}^{A_f}_i(A_f/a, M_f) = 0\) by taking
\({ {P}^{\scriptscriptstyle \bullet}} \twoheadrightarrow M\) and
\({ {P}^{\scriptscriptstyle \bullet}} _f \twoheadrightarrow M_f\). Then
compute
\begin{align*}
\operatorname{Tor}_1(M_f, A_f/a)
&= H_1( { {P}^{\scriptscriptstyle \bullet}} _f \otimes A_f/a) \\
&= H_1( { {P}^{\scriptscriptstyle \bullet}}  \otimes A/a' \otimes A_f) \\
&= H_1( { {P}^{\scriptscriptstyle \bullet}}  \otimes A/a') \otimes A_f \\
&= 0 \otimes A_f = 0
,\end{align*}
using that localization is exact and thus commutes with taking homology.
In the other direction, show that for
\(0\to A \xrightarrow{f} B\to C\to 0\) leads to
\(A\otimes M \xrightarrow{\tilde f} B\otimes M\), and injectivity can be
checked locally.
\end{quote}

\end{exercise}

\begin{remark}

Define the \textbf{support} of \(M\) as
\(\mathop{\mathrm{supp}}(M) \coloneqq\left\{{p\in \operatorname{Spec}A{~\mathrel{\Big\vert}~}M_p\neq 0}\right\}\),
thought of as points where ``functions on \(A\)'' defined by \(M\) do
not vanish.

\end{remark}

\begin{exercise}[?]

Show that if \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
then \(\mathop{\mathrm{supp}}M = V(\operatorname{Ann}_A(M))\) where
\(\operatorname{Ann}_A(M) \coloneqq\left\{{a\in A{~\mathrel{\Big\vert}~}am=0 \, \forall m\in M}\right\}\).

\end{exercise}

\begin{exercise}[?]

Show that for \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
with \(A\) Noetherian, \(M=0 \iff \mathop{\mathrm{supp}}M = 0\).

\end{exercise}

\begin{remark}

Modules give sheaves over \(\operatorname{Spec}A\), and the following
theorem is a special case of faithfully flat descent:

\end{remark}

\begin{theorem}[Serre]

If \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) and
\(\left\{{f_i}\right\}\) is a finite generating set, then the following
sequence is exact:
\begin{align*}
0 \to M \to \bigoplus_{f_i} M_{f_i} &\to \bigoplus M_{f_i f_i} \\
x\in M_{f_i} &\mapsto {\left[ {\cdots, {x\over f_i f_j}, \cdots, -{x\over f_j f_i}, \cdots } \right]}
.\end{align*}
with the positive sign in the \(i\)th component and the negative in the
\(j\)th.

\end{theorem}

\begin{exercise}[?]

Prove this: check injectivity locally, and use that localization
commutes with direct sums. Note that essentially the same proof goes
through for faithfully flat descent.

\end{exercise}

\begin{theorem}[Classification of flat finitely-generated modules over a Noetherian ring]

If \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\flat, {\mathrm{fg}}}\)
for \(A\) Noetherian, then \(M\) is locally free, i.e.~there exist
\(f_i\) generating the unit ideal with \(M_{f_i}\) free for all \(i\).

\end{theorem}

\begin{proof}[?]

Philosophy: reduce to local case.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  For \(A\) local: finitely-generated flat flat modules over a
  Noetherian \emph{local} ring is free.
\item
  Hence \(M_p\) is free over \(A_p\) for all
  \(p\in \operatorname{Spec}A\).
\item
  Spreading out: by the HW, there exist \(f_i^p\) not in \(p\) such that
  \(M_{f_1^p \cdots f_\ell^p}\) is free and equals \(M_{f^p}\).
\item
  The finite collection \(\left\{{P^*}\right\}\) generated the unit
  ideal.
\end{enumerate}

\end{proof}

\begin{remark}

Next: Artinian, local, complete local rings, DVRs, etc -- the building
blocks of the theory of local rings!

\end{remark}

\hypertarget{thursday-march-03}{%
\section{Thursday, March 03}\label{thursday-march-03}}

\begin{remark}

Next topic: Artin rings. The general way we reduce the study of
arbitrary rings:

\begin{center}
\begin{tikzcd}
    \mathsf{CRing}\\
    && {\text{Checking on covers by }\operatorname{Spec}A} \\
    \mathsf{Loc}\mathsf{CRing}\\
    && {\mathsf{Loc}\mathsf{CRing}^{\text{complete}}} \\
    \mathsf{Art}\mathsf{CRing}\\
    && {} \\
    \mathsf{Field}
    \arrow[squiggly, from=1-1, to=3-1]
    \arrow[squiggly, from=3-1, to=5-1]
    \arrow[squiggly, from=3-1, to=4-3]
    \arrow[squiggly, from=4-3, to=5-1]
    \arrow[squiggly, from=5-1, to=7-1]
    \arrow[squiggly, from=1-1, to=2-3]
    \arrow[squiggly, from=2-3, to=3-1]
\end{tikzcd}
\end{center}

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

\begin{itemize}
\tightlist
\item
  Recall that Artin rings are defined by the DCC condition on ideals.
\item
  The \textbf{length} of a module is the maximal length of a strictly
  increasing filtration.
\end{itemize}

\end{remark}

\begin{exercise}[?]

Show the following:

\begin{itemize}
\item
  \({\mathbb{C}}[t]/\left\langle{t^n}\right\rangle\) is Artin.
\item
  \(C_{p^n}\) is Artin.
\item
  \({\mathbb{Z}}\) is Noetherian but not Artin.
\item
  Any finite product of fields is Artin.
\item
  An Artin domain is a field.
\item
  Prime implies maximal in any Artin ring.

  \begin{quote}
  Hint: quotient by the prime, and use that any element \(a\) satisfies
  \(a^n = ba^{n+1}\) for some \(n\) and \(b\) to produce an inverse.
  \end{quote}
\item
  \(\operatorname{Spec}A = \left\{{{\operatorname{pt}}}\right\}\) for a
  local Artin ring.
\item
  Artin rings \(A\) have finite length in
  \({\mathsf{A}{\hbox{-}}\mathsf{Mod}}\).
\item
  Quotients of Artin rings by their Jacobson radicals are products of
  fields.
\item
  \(\operatorname{Spec}A = \displaystyle\coprod_i \left\{{{\operatorname{pt}}_i}\right\}\)
  is a disjoint union of points, and is Hausdorff.

  \begin{itemize}
  \tightlist
  \item
    The only Noetherian rings with Hausdorff spectra are Artin.
  \end{itemize}
\end{itemize}

\end{exercise}

\begin{theorem}[?]

Artin rings are Noetherian.

\end{theorem}

\begin{proof}[?]

Use that

\begin{itemize}
\tightlist
\item
  Any module has a maximal proper submodule.
\item
  Choose \({\mathfrak{a}}_i \subseteq A\) simple, so
  \(A/{\mathfrak{a}}_i\) is Artinian, to produce an increasing chain
  \(0 \subseteq {\mathfrak{a}}_1 \subseteq {\mathfrak{a}}_2 \cdots\)
  where \({\mathfrak{a}}_i/{\mathfrak{a}}_{i+1}\) is simple for all
  \(i\).
\item
  Enumerate maximal ideals \({\mathfrak{m}}_i\) and produce a chain
  \({\mathfrak{m}}_1 \supseteq{\mathfrak{m}}_2 \supseteq\cdots\) and
  take colon ideals.
\item
  Reduce to this case by showing every chain refines.
\end{itemize}

Steps:

\begin{itemize}
\tightlist
\item
  Make a descending filtration with semisimple associated graded, whose
  filtration is finite.
\item
  Use Jordan-Holder, every such sequence has the same length.
\item
  Refine an arbitrary filtration to one in which the quotients are
  simple.
\end{itemize}

\end{proof}

\begin{corollary}[?]

\({\sharp}\operatorname{mSpec}A < \infty\) for
\(A\in \mathsf{Art}\mathsf{CRing}\).

\end{corollary}

\begin{proof}[?]

Hints:

\begin{itemize}
\tightlist
\item
  If not, take a decreasing chain of
  \({\mathfrak{m}}_1 \supseteq{\mathfrak{m}}_1{\mathfrak{m}}_2 \cdots\),
  stabilize, use that the \({\mathfrak{m}}_i\) are finitely-generated
  and apply Nakayama.
\item
  If \(I\) is defined as the product at the minimal stabilized step,
  \(I \subseteq {J ({A}) }\).
\item
  Without loss of generality, assume \({J ({A}) } = 0\), so \(I=0\)
  since
  \(\operatorname{mSpec}A/{J ({A}) } \cong \operatorname{mSpec}A\).
\item
  \(A/{\mathfrak{m}}_1\cdots {\mathfrak{m}}_n = \prod A/{\mathfrak{m}}_i\)
  is a product of fields
\item
  Corollary: \(n> N\), \({\mathfrak{m}}_N\) will have empty support.
\end{itemize}

\end{proof}

\hypertarget{tuesday-march-15}{%
\section{Tuesday, March 15}\label{tuesday-march-15}}

\hypertarget{artin-rings}{%
\subsection{Artin Rings}\label{artin-rings}}

\begin{remark}

Last time: \(A/{J ({A}) }\) is a product of Artin local rings.

\end{remark}

\begin{exercise}[?]

Show that if \(A\) is Artin local, then \({\mathfrak{m}}_A^n = 0\) for
some \(n\).

\begin{quote}
Hint: use that \(\left\{{{\mathfrak{m}}^k}\right\}_{k\geq 0}\)
stabilizes, so
\({\mathfrak{m}}^n/{\mathfrak{m}}^{n+1} \cong {\mathfrak{m}}^n/{\mathfrak{m}}{\mathfrak{m}}^{n} = 0\)
so \({\mathfrak{m}}^n=0\) since \(A\) is Noetherian.
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that if \(A\) is an Artin local ring that is finitely generated
over an algebraically closed field \(k\). Then \(A\) is a quotient of
\(k[x_1, \cdots, x_{n}]/\left\langle{x_1,\cdots,x_n}\right\rangle^N\)
for some \(n\) and \(N\).

\end{exercise}

\begin{solution}

Hint: use that \({\mathfrak{m}}\) is finitely generated, construct a
surjection \(k[x_1, \cdots, x_{n}]\to A\), and show
\(I_{N} \coloneqq\left\langle{x_1,\cdots, x_n}\right\rangle^N\) is in
the kernel. Also use the corollary of the Nullstellensatz (EEKS) that
finite extensions of \(k\) which are fields are necessarily \(k\)
itself. Apply the snake lemma:

\begin{center}
\begin{tikzcd}
    &&& 0 \\
    0 & {\left\langle{x_1,\cdots, x_n}\right\rangle} & k[x_1, \cdots, x_{n}]& k & 0 \\
    \\
    0 & {\mathfrak{m}}& A & {A/{\mathfrak{m}}\cong k} & 0 \\
    & 0 & \textcolor{rgb,255:red,92;green,92;blue,214}{\therefore 0} & 0
    \arrow[from=2-1, to=2-2]
    \arrow[from=2-2, to=2-3]
    \arrow[from=2-3, to=2-4]
    \arrow[from=2-4, to=2-5]
    \arrow[from=4-1, to=4-2]
    \arrow[from=4-2, to=4-3]
    \arrow[from=4-3, to=4-4]
    \arrow[from=4-4, to=4-5]
    \arrow["\sim", from=2-2, to=4-2]
    \arrow[two heads, from=2-3, to=4-3]
    \arrow["{?}", color={rgb,255:red,92;green,92;blue,214}, from=4-3, to=5-3]
    \arrow["\sim", from=2-4, to=4-4]
    \arrow[from=5-2, to=5-3]
    \arrow[from=5-3, to=5-4]
    \arrow[from=1-4, to=5-2]
\end{tikzcd}
\end{center}

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

This shows surjectivity, and
\(\left\langle{x_1,\cdots, x_n}\right\rangle^N\) being in the kernel
follows from the previous proposition.

\end{solution}

\hypertarget{dvrs}{%
\subsection{DVRs}\label{dvrs}}

\begin{exercise}[?]

Recall that a Noetherian local domain \(A\) is a DVR if
\({\mathfrak{m}}\neq 0\) is principal. Show that
\(\dim_k {\mathfrak{m}}/{\mathfrak{m}}^2 = 1\) for
\(k=A/{\mathfrak{m}}\).

\begin{quote}
Hint: use Nakayama to bound the dimension by 1.
\end{quote}

\end{exercise}

\begin{example}[?]

Examples of DVRs:

\begin{itemize}
\tightlist
\item
  \(k{\left[\left[ t \right]\right] }\)
\item
  \({ {\mathbb{Z}}_{\widehat{p}} }\)
\item
  \({\mathbb{Z}}_{(p)} = {\mathbb{Z}} { \left[ \scriptstyle {\left\{{ q^{-1}{~\mathrel{\Big\vert}~}q\neq p }\right\}} \right] }\).
\end{itemize}

A non-example: \(k{\left[\left[ x,y \right]\right] }\).

\end{example}

\begin{exercise}[?]

Show that if \(A\) is a DVR with uniformizer \(\pi\) and
\(a\in A\setminus\left\{{0}\right\}\), then there is a unique
\(n\in {\mathbb{Z}}_{\geq 0}\) such that \(a = \pi^n a_u\) with \(a_u\)
a unit.

\begin{quote}
Hint: for existence, set
\(N\coloneqq\max \left\{{N{~\mathrel{\Big\vert}~}a\in \left\langle{\pi^N}\right\rangle}\right\}\)
which exists because \(\cap_N \left\langle{\pi^N}\right\rangle = 0\).
Use that \(\pi I = I \implies I=0\) by Nakayama, to write
\(a = \pi^n a_0\), and if \(a_0\) is not a unit then
\(a_0\in\left\langle{\pi}\right\rangle\) contradicting maximality.
\end{quote}

\end{exercise}

\begin{corollary}[?]

If \(A\) is a DVR,
\(\operatorname{Id}(A) = \left\{{\left\langle{0}\right\rangle, \left\langle{\pi^n}\right\rangle }\right\}_{n\geq 0}\)
and
\(\operatorname{Spec}A = \left\{{\left\langle{\pi}\right\rangle, \left\langle{0}\right\rangle}\right\}\).
This follows from writing
\begin{align*}
I = \left\langle{a_1,\cdots, a_N}\right\rangle = \left\langle{\pi^{n_1} b_1,\cdots, \pi^{n_N} b_N }\right\rangle = \left\langle{\pi^m}\right\rangle,\qquad m\coloneqq\min\left\{{n_j}\right\}_{j\leq N}
.\end{align*}

\end{corollary}

\begin{exercise}[?]

Show that DVRs \(A\) biject with fields \(K\) equipped with a valuation
\(v: K\to {\mathbb{Z}}\cup\left\{{\infty}\right\}\) satisfying:

\begin{itemize}
\tightlist
\item
  \(v(a+b) \geq \min( v(a), v(b) )\),
\item
  \(v(ab) = v(a) + v(b)\),
\item
  \(v(a) =\infty \iff a=0\),
\item
  \(v(K^{\times}) \neq {\mathbb{Z}}\).
\end{itemize}

\begin{quote}
Hint: for \(A\in\mathsf{DVR}\), set \(K = \operatorname{ff}(A)\) and
\(v(a/b)\coloneqq v(a)-v(b)\) where \(v(\pi^n a_0) = n\). Given
\((K, v)\), set
\(A = \left\{{x\in K {~\mathrel{\Big\vert}~}v(x)\geq 0}\right\}\) with
\({\mathfrak{m}}= \left\{{v(x) > 0}\right\}\), showing
\({\mathfrak{m}}^c \subseteq A^{\times}\) and \({\mathfrak{m}}\) is
generated by any \(x\) with \(v(x) = 1\)?
\end{quote}

\end{exercise}

\hypertarget{classifying-finitely-generated-modules-over-a-dvr}{%
\subsection{Classifying finitely-generated modules over a
DVR}\label{classifying-finitely-generated-modules-over-a-dvr}}

\begin{remark}

Recall that \(M\in{\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is torsionfree
iff
\(\operatorname{Ann}_A(M) \coloneqq\left\{{a\in A{~\mathrel{\Big\vert}~}am=0}\right\}\)
contains only zero divisors iff \(M \xrightarrow{\times a} M\) is
injective for all nonzero \(a\in A\).

\end{remark}

\begin{exercise}[?]

Show that if \(A\in \mathsf{CRing}^{{ \mathrm{Noeth} }}\) and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) then
\(0\to M_{\operatorname{tors}}\to M\to M'\to 0\) is a SES where \(M'\)
is torsionfree, and moreover there exists some \(a\in A\) such that
\(a M_{\operatorname{tors}}= 0\).

\begin{quote}
Hint: for the latter statement, use that \(M_{\operatorname{tors}}\) is
finitely-generated and take a product of annihilators of generators. For
the former, take \(a\in \operatorname{Ann}_A(\tilde m)\) for some
\(\tilde m\in M'\), lift to \(m\in M\) and show
\(am\in M_{\operatorname{tors}}\) for some \(a\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that if \(A\) is a PID, then
\(M\in{\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is flat iff \(M\) is
torsionfree.

\begin{quote}
Hint: use \(A \xrightarrow{\times a} A\) and apply
\(({-}) \otimes_A M\). For the reverse, show
\(\operatorname{Tor}^1(M, A/I) = 0\) for any ideal in \(A\) and compute
using the projective resolution
\(0\to A \xrightarrow{\times a} A \to A/\left\langle{a}\right\rangle \to 0\).
Note that
\(H_1(M \xrightarrow{\times a} M ) = \ker(M \xrightarrow{\times a} M) = 0\)
since \(M\) is torsionfree.
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that for \(A\) a DVR and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), then \(M\)
torsionfree implies \(M\) is free.

\begin{quote}
Hint: torsionfree \(\implies\) flat \(\implies\) free for
finitely-generated Noetherian local rings.
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that if \(M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
for \(A\) a DVR then \(M \cong M_{\operatorname{tors}}\oplus A^n\) for
some \(n\).

\begin{quote}
Hint: use that the SES involving \(M_{\operatorname{tors}}\to M\)
splits.
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that any finitely-generated torsion module over \(A\) is isomorphic
to \(\bigoplus_i A/\pi^{n_i} A\). Use this to classify all
finitely-generated modules over \(k{\left[\left[ t \right]\right] }\).

\end{exercise}

\hypertarget{tuesday-march-22}{%
\section{Tuesday, March 22}\label{tuesday-march-22}}

\hypertarget{classification-of-finitely-generated-modules-over-a-dedekind-domain}{%
\subsection{Classification of finitely-generated modules over a Dedekind
domain}\label{classification-of-finitely-generated-modules-over-a-dedekind-domain}}

\begin{definition}[Dedekind domain]

A ring \(A\) is a \textbf{Dedekind domain} iff

\begin{itemize}
\tightlist
\item
  \(A\in{ \mathrm{Noeth} }\mathsf{Domain}, \operatorname{krulldim}A = 1\),
\item
  The local rings \(A_p\) are DVRs for all
  \(p\in \operatorname{mSpec}A\).
\end{itemize}

\end{definition}

\begin{theorem}[Structure theorem for Dedekind domains]

If
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}, A\in \mathsf{Dedekind}\mathsf{Domain}\),
then
\begin{align*}
M\underset{{\mathsf{A}{\hbox{-}}\mathsf{Mod}}}{\cong} T \oplus F \coloneqq\qty{ \bigoplus (A/{\mathfrak{p}}_i)^{n_i} } \oplus \qty{ \bigoplus {\mathcal{L}}_i }
\end{align*}
where
\({\mathcal{L}}_i \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{{\mathsf{locfree}}, \operatorname{rank}=1}\)
(and are in particular torsionfree) and the \(A/{\mathfrak{p}}_i\) are
torsion.

\end{theorem}

\begin{exercise}[?]

Show that if \(p\in \operatorname{Spec}A\) then \(M_p \cong A_p^m\)
where \(m = \operatorname{rank}M\).

\begin{quote}
Hint: use that
\(M_p \oplus \operatorname{ff}(A_p) = \operatorname{ff}(A)^n = \operatorname{ff}(A)^n\)
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that for \(A\) a Dedekind domain and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), TFAE:

\begin{itemize}
\tightlist
\item
  \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^\flat\)
\item
  \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}(\mathop{\mathrm{proj}})\)
\item
  \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{{\mathsf{locfree}}}\)
\item
  \(M\) is torsionfree
\end{itemize}

Conclude that the following SES splits:
\begin{align*}
0\to M_{\operatorname{tors}}\to M \to M_{{\operatorname{tors}}{\mathrm{free}}} \to 0
.\end{align*}

\begin{quote}
Hint: the first three already hold for Noetherian domains. To show flat
\(\implies\) torsionfree, that
\(0\to A \xrightarrow{\cdot a} A \to A/\left\langle{a}\right\rangle \to 0\)
for \(a\in \mathsf{Tors}(M)\) and tensor with \(M\). For the converse,
show \(M_p\) is flat for all \(p\) and that \(A_p\) is a DVR -- if
\({a\over s'}{m\over s} = 0 \implies \tilde s' s am = 0\) for some
\(\tilde s\), then \(m\) is torsion.
\end{quote}

\end{exercise}

\begin{remark}

Note that torsionfree \(\implies\) flat fails for most rings!

\end{remark}

\begin{exercise}[?]

Show that if \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
is torsion over a Dedekind domain \(A\) then
\(M \cong \bigoplus _{i} (A/p_i^{n_i})^{m_i}\).

\begin{quote}
Hint: use that \(\mathop{\mathrm{supp}}M\) is finite to produce a map
\(M\to \bigoplus _i M_{p_i}\) and show it is locally an isomorphism
since \((M_{p})_q = M_q\) when \(q\neq p\). Also use that
\(M_p = \bigoplus _j A_p/p^{i_j} \cong \oplus_j A/p^{i_j}\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that for \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
torsionfree over \(A\) a Dedekind domain, there is a (not necessarily
unique) decomposition \(M \cong \bigoplus _i {\mathcal{L}}_i\) with
\({\mathcal{L}}_i\) locally free of rank 1.

\begin{quote}
Induct on rank, where it's ETS there exists an \({\mathcal{L}}\) where
\(M/{\mathcal{L}}_i\) is torsionfree since
\(0\to {\mathcal{L}}\to M\to M/{\mathcal{L}}_i\to 0\) splits (tensor to
fraction field). To find such an \({\mathcal{L}}\), take any \(m\neq 0\)
and take \({\mathcal{L}}\) to be the preimage of
\((M/\left\langle{m}\right\rangle)_{\operatorname{tors}}\) under
\(M \to M/\left\langle{m}\right\rangle\); then
\({\mathcal{M}}/{\mathcal{L}}\) will be torsionfree and is rank 1 since
\({\mathcal{L}}\otimes\operatorname{ff}(A) \cong \left\langle{m}\right\rangle\otimes\operatorname{ff}(A)\).
\end{quote}

\end{exercise}

\hypertarget{classification-of-locally-free-rank-1-modules-over-a-dedekind-domain}{%
\subsection{Classification of locally free rank 1 modules over a
Dedekind
domain}\label{classification-of-locally-free-rank-1-modules-over-a-dedekind-domain}}

\begin{exercise}[?]

Show that if \(I{~\trianglelefteq~}A\) is nonzero for \(A\) a Dedekind
domain, then \(I\) is locally free of rank 1.

\begin{quote}
Hint: show that \(I_p\leq A_p\) is nonzero, so
\(I_p = \left\langle{\pi^n}\right\rangle\) for \(\pi\) a uniformizer of
\(A_p\).
\end{quote}

\end{exercise}

\begin{definition}[Invertible modules]

A module \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is
\textbf{invertible} iff \(M\) is locally free of rank 1. Equivalently,
defining
\(M {}^{ \vee }\coloneqq{\mathsf{A}{\hbox{-}}\mathsf{Mod}}(M, A)\) there
is an evaluation isomorphism
\(M\otimes_A M {}^{ \vee } { \, \xrightarrow{\sim}\, }A\). Define
\begin{align*}
{\operatorname{Pic}}(A) \coloneqq({\mathsf{A}{\hbox{-}}\mathsf{Mod}}, \otimes_A)/\cong
.\end{align*}
Note that for
\(A = {\mathcal{O}}_K, K\in \mathsf{Field}_{/ {{\mathbb{Q}}}}\),
\begin{align*}
{\operatorname{Pic}}(A) \cong { \operatorname{Cl}} (K)
.\end{align*}

\end{definition}

\begin{definition}[(Weil) Divisors]

For \(A\) a Dedekind domain, a \textbf{divisor} is a formal linear
combination
\begin{align*}
D = \sum_{{\mathfrak{p}}\neq 0 \in \operatorname{Spec}A} n_p {\mathfrak{p}}\in \mathsf{Free}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(\operatorname{Spec}A)
.\end{align*}
such that \(n_p = 0\) for almost all \(p\). These form a group
\(\operatorname{Div}(A)\). A divisor \(D\) is \textbf{effective} iff
\(n_p\geq 0\) for all \(p\), yielding a submonoid
\(\operatorname{Div}^+(A)\) of effective divisors.

\end{definition}

\begin{remark}

More generally, this will be a sum over height 1 primes.

\end{remark}

\begin{exercise}[?]

Show that there is a natural bijection
\begin{align*}
\operatorname{Div}^+(A) &\rightleftharpoons\operatorname{Id}(A)\setminus\left\{{0}\right\}\\
\sum n_p p &\rightleftharpoons\prod p^{n_p}
.\end{align*}

\begin{quote}
Hint: if \(\operatorname{krulldim}A = 1\), use that
\(\operatorname{Id}(A)\cong \left\{{I_p {~\trianglelefteq~}A_p}\right\}_{p\in P}\)
where \(I_p = A_p\) for almost all \(p\), and each
\(I_p \cong \left\langle{\pi_p^{n_p}}\right\rangle\).
\end{quote}

\end{exercise}

\begin{exercise}[?]

Show that if \({\mathcal{L}}\) is invertible then
\({\mathcal{L}} {}^{ \vee }\) is invertible.

\begin{quote}
Hint: show \({\mathcal{L}} {}^{ \vee }\) is locally free of rank 1,
using
\({\mathcal{L}}_p {}^{ \vee } { \, \xrightarrow{\sim}\, }{\mathsf{A}{\hbox{-}}\mathsf{Mod}}({\mathcal{L}}, A)_p { \, \xrightarrow{\sim}\, }{\mathsf{A_p}{\hbox{-}}\mathsf{Mod}}({\mathcal{L}}_p, A_p)\)
by post-composing with localization. Then check
\({\mathcal{L}}\otimes{\mathsf{A}{\hbox{-}}\mathsf{Mod}}({\mathcal{L}}, A)\to A\)
where \((n, f) \mapsto f(n)\) is an isomorphism by checking locally
where everything is free.
\end{quote}

\end{exercise}

\begin{proposition}[Picard group SES]

Write \(D\in \operatorname{Div}(A)\) as \(D = D^+ - D^-\) where
\(D^{\pm} \in \operatorname{Div}^+(A)\) are effective. There is a SES

\begin{center}
\begin{tikzcd}
    0 && {\ker f} && {\operatorname{Div}(A)} && {{\operatorname{Pic}}(A)} && 0 \\
    &&&& {D^+ - D^-} && {\qty{\prod_{p} p^{n_p^+} } \otimes\qty{ \prod_p p^{n_p^-}} {}^{ \vee }}
    \arrow["f", two heads, from=1-5, to=1-7]
    \arrow[maps to, from=2-5, to=2-7]
    \arrow[from=1-7, to=1-9]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbNCwwLCJcXERpdihBKSJdLFs2LDAsIlxcUGljKEEpIl0sWzQsMSwiRF4rIC0gRF4tIl0sWzYsMSwiXFxxdHl7XFxwcm9kX3twfSBwXntuX3BeK30gfSBcXHRlbnNvciBcXHF0eXsgXFxwcm9kX3AgcF57bl9wXi19fVxcZHVhbCJdLFs4LDAsIjAiXSxbMiwwLCJcXGtlciBmIl0sWzAsMCwiMCJdLFswLDEsImYiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMiwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV0sWzEsNF0sWzUsMF0sWzYsNV1d}{Link
to Diagram}
\end{quote}

\end{proposition}

\begin{remark}

This is generally far from injective, and will instead biject with
\emph{fractional ideals}:

\end{remark}

\begin{definition}[Fractional ideals]

For \(A \in \mathsf{Dedekind}\), a \textbf{fractional ideal} is a
nonzero \(A{\hbox{-}}\)submodule \({\mathcal{L}}\leq A\) where
\({\mathcal{L}}\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  \(I{~\trianglelefteq~}A\) or
  \(I{~\trianglelefteq~}\operatorname{ff}(A)\) any ideal
\item
  \({1\over a}I{~\trianglelefteq~}\operatorname{ff}(A)\).
\end{itemize}

\end{example}

\begin{exercise}[?]

Show that any fractional ideal is invertible.

\begin{quote}
Hint: use that any such \(I\) is torsionfree since it's a subset of
\(\operatorname{ff}(A)\), and
\({\mathcal{L}}\otimes_A \operatorname{ff}(A) \cong \operatorname{ff}(A)\)
implies rank 1.
\end{quote}

\end{exercise}

\hypertarget{tuesday-march-29}{%
\section{Tuesday, March 29}\label{tuesday-march-29}}

\begin{remark}

Last time: proving the following theorem.

\end{remark}

\begin{theorem}[?]

If \(A\in{ \mathrm{Noeth} }{\mathsf{loc}}\mathsf{Domain}\) with
\(\operatorname{krulldim}A = 1\), then \(A\in\mathsf{DVR}\iff A\) is
integrally closed.

\end{theorem}

\begin{proof}[?]

Last time: DVR \(\implies\) integrally closed.

\(\impliedby\): It suffices to show that the maximal ideal is principal.
Let \(a\in {\mathfrak{m}}\setminus\left\{{0}\right\}\), then
\(A/\left\langle{a}\right\rangle\) is Artinian (which is true here for
any quotient) and thus \({\mathfrak{m}}^k = 0\) in
\(A/\left\langle{a}\right\rangle\) for some \(k\) (chosen minimally), so
\({\mathfrak{m}}^k \in \left\langle{a}\right\rangle\). By minimality,
pick \(b\in {\mathfrak{m}}^{k-1}\setminus\left\langle{a}\right\rangle\)
-- the claim is that \(a/b\) generates \({\mathfrak{m}}\). First,
\(\qty{b\over a}{\mathfrak{m}}{~\trianglelefteq~}A\) since
\(bm\in {\mathfrak{m}}^k \subseteq \left\langle{a}\right\rangle\). It's
ETS \(\qty{b\over a} m\) is not contained in \({\mathfrak{m}}\) -- given
this, \(\qty{b\over a}{\mathfrak{m}}= A\) by maximality and this implies
\({\mathfrak{m}}= {b\over a} A\).

Why this last claim is true: the map
\(x\mapsto {a\over b}x: {\mathfrak{m}}\to {\mathfrak{m}}\) is
annihilated by some polynomial with coefficients in \(A\) by
Cayley-Hamilton, and by integral closedness this implies \(a/b\in A\).
Then \(a/b\in {\mathfrak{m}}\), a contradiction since \(a/b\) is a unit
and \({\mathfrak{m}}\) is proper. \(\contradiction\)

\end{proof}

\begin{theorem}[?]

If \(A\in{ \mathrm{Noeth} }\mathsf{Domain}\) (not necessarily local)
\(\operatorname{krulldim}A = 1\), then
\(A\in\mathsf{Dedekind}\mathsf{Domain}\iff A\) is integrally closed.

\end{theorem}

\begin{remark}

\(A\) Dedekind means all local rings are DVRs, so it's ETS \(A\) is
integrally closed iff \(A \left[ { \scriptstyle { {p}^{-1}} } \right]\)
is integrally closed (since we can apply the last theorem to
\(A \left[ { \scriptstyle { {p}^{-1}} } \right]\)).

\end{remark}

\begin{lemma}[Integral closure commutes with localization]

Let \(A \subseteq B\) be an inclusion of rings and let
\(\operatorname{cl}^{\mathrm{int}} _B A\) be the integral closure of
\(A\) in \(B\). If \(S \subseteq A\) is a multiplicative subset, then
\begin{align*}
 \operatorname{cl}^{\mathrm{int}} _B (A \left[ { \scriptstyle { {S}^{-1}}  } \right]) \cong ( \operatorname{cl}^{\mathrm{int}} _B A)  \left[ { \scriptstyle { {S}^{-1}}  } \right]
.\end{align*}

\end{lemma}

\begin{proof}[of lemma]

\(\supseteq\): Take \(b\in A_{\mathrm{int}}\) and \(s\in S\), we then
WTS \(b/s\) satisfies a monic polynomial over
\(A \left[ { \scriptstyle { {S}^{-1}} } \right]\). If
\(f(b) = b^n + a_1 b^{n-1} + \cdots = 0\), note that \(s^nf(b) = 0\) and
is of the form \(\qty{b\over s}^n + sa_1{b\over s}^{n-1} + \cdots\) is a
polynomial with coefficients in
\(A \left[ { \scriptstyle { {S}^{-1}} } \right]\).

\(\subseteq\): Let \(b/s\) in the LHS, so
\(\qty{b\over s}^n + {a_1\over s_1}\qty{b\over s}^{n-1} + \cdots = 0\).
Multiply through by \(s^n\qty{\prod s_i}^n\) to get
\(\qty{\prod s_i}^n b^n + a_1 s' b^{n-1} \qty{\prod s_i}^{n-1} + \cdots\)
by absorbing some factors into the coefficient \(s'\), so \(b\prod s_i\)
satisfies a monic polynomial.

\end{proof}

\begin{proof}[of theorem]

By the lemma, \(A\) integrally closed
\(\implies A \left[ { \scriptstyle { {p}^{-1}} } \right]\) is integrally
closed because
\((A \left[ { \scriptstyle { {p}^{-1}} } \right])_{\mathrm{int} } = (A_{\mathrm{int}}) \left[ { \scriptstyle { {p}^{-1}} } \right] = A \left[ { \scriptstyle { {p}^{-1}} } \right]\).
Suppose that \(A \left[ { \scriptstyle { {p}^{-1}} } \right]\) is
integrally closed. There is a map \(A\to A_{\mathrm{int}}\) which we
want to show is an isomorphism. Check locally:
\(A \left[ { \scriptstyle { {p}^{-1}} } \right] \to (A_{\mathrm{int}}) \left[ { \scriptstyle { {p}^{-1}} } \right] \to (A \left[ { \scriptstyle { {p}^{-1}} } \right])_{\mathrm{int}} = A{ \scriptsize {}_{ \left[ { \scriptstyle { {p}^{-1}} } \right] } }\).

\end{proof}

\begin{theorem}[Why Dedekind domains arise in nature: Dedekind is closed under certain integral closures]

Let \(A\in\mathsf{Dedekind}\) with \(K = \operatorname{ff}(A)\) and let
\(K'/K\) be a finite separable extension. Then
\(\operatorname{cl}^{\mathrm{int}} _{K'} A\in\mathsf{Dedekind}\) is
Dedekind.

\begin{quote}
Note that in characteristic zero, finite extensions are automatically
separable.
\end{quote}

\end{theorem}

\begin{example}[?]

\envlist

\begin{itemize}
\item
  For \(K/{\mathbb{Q}}\) a finite separable extension, then
  \(\operatorname{cl}^{\mathrm{int}} _K {\mathbb{Z}}\) is Dedekind.
\item
  For \(L/k[t]\) finite separable,
  \(\operatorname{cl}^{\mathrm{int}} _{L} k[t]\) is Dedekind.
\end{itemize}

\end{example}

\begin{lemma}[?]

Let \(A' \coloneqq \operatorname{cl}^{\mathrm{int}} _{K'} A\), then
\(A'\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).

\end{lemma}

\begin{proof}[of theorem, using the lemma]

\envlist

\begin{itemize}
\tightlist
\item
  By the Hilbert basis theorem, \(A'\) is Noetherian
\item
  \(A'\) is a domain since \(A' \subseteq K'\)
\item
  \(A'\) is integrally closed: STS satisfying a monic polynomial over
  \(A'\) implies satisfying a monic polynomial over \(A\). To check if
  \(b\in K'\) is integral, one needs that
  \(A'[b]\in {\mathsf{A'}{\hbox{-}}\mathsf{Mod}}\) is finite, which
  implies \(A'[b] \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is finite.
  This is true because \(A[b] \leq A'[b]\) is a submodule and \(A\) is
  Noetherian, so \(b\) is integral over \(A\).
\item
  \(\operatorname{krulldim}A' = 1\): STS
  \(\operatorname{Spec}A' \subseteq \operatorname{mSpec}A'\). Let \(p\)
  be prime in \(A\) and let
  \(\tilde p \coloneqq p \cap A \in \operatorname{Id}(A)\). There is a
  finitely-generated extension \(A/\tilde p \leq A'/p\), and it's ETS
  \(A/\tilde p\in \mathsf{Field}\) by EEKS (finitely-generated ring
  extensions of fields are fields). ETS
  \(\tilde p \in \operatorname{mSpec}A\) -- consider
  \(\operatorname{Spec}A' \xrightarrow{\pi} \operatorname{Spec}A\), we
  WTS
  \(\pi^{-1}(\left\langle{0}\right\rangle) = \left\{{\left\langle{0}\right\rangle}\right\}\).
  Fact:
  \(\pi^{-1}(\left\langle{0}\right\rangle) = \operatorname{Spec}A' \left[ { \scriptstyle { {A^{\times}}^{-1}} } \right] = \operatorname{Spec}(A' \otimes K) = \operatorname{Spec}(K')\).
  This uses the fact that
  \(A'\otimes_A K { \, \xrightarrow{\sim}\, }K'\), which can be checked
  locally on DVRs.
\end{itemize}

\end{proof}

\begin{proof}[of lemma]

Assume \(\operatorname{ch}K = 0\). We know that \(K'/K\) is finite, so
pick a basis \(\left\{{e_i}\right\}\). We can scale the \(e_i\) by
elements of \(A\) so that they are in \(A'\) -- use that
\(K' = A'\otimes_A R\), so \(e_i = \sum a_i'/s_i\) and one can scale by
the \(s_i\).

There is nondegenerate pairing
\({\left\langle {x},~{y} \right\rangle}\coloneqq\operatorname{Tr}_{K'/K}(xy)\)
where \(\operatorname{Tr}(z)\) is the trace of the \(K{\hbox{-}}\)linear
map \(K' \xrightarrow{\cdot z} K'\). Why this is nondegenerate: use the
separability assumption. Given \(x\in K'\setminus\left\{{0}\right\}\),
produce a \(y\) such that
\({\left\langle {x},~{y} \right\rangle}\neq 0\) by setting \(y=x^{-1}\)
(noting that this won't work in characteristic \(p\)).

Use this to define a dual basis \(\left\{{f_i}\right\}\) such that
\({\left\langle {e_i},~{f_j} \right\rangle} = \delta_{ij}\). We can then
express
\(x = \sum_{1\leq i\leq n} {\left\langle {x},~{f_i} \right\rangle} e_i\),
and it's ETS that if \(x\in A'\) then
\({\left\langle {x},~{f_i} \right\rangle}\in A\). For \(x\in A'\), we
have \(\operatorname{Tr}(x)\in A\), e.g.~because it is a sum of Galois
conjugates (after passing to a Galois extension), or that
\(\operatorname{Tr}(x) = \sum r_i\) the roots of a characteristic
polynomial with coefficients in \(A\). This exhausts the possible
factors of the characteristic polynomial.

\end{proof}

\hypertarget{thursday-march-31}{%
\section{Thursday, March 31}\label{thursday-march-31}}

\begin{remark}

Last time: \(A\in \mathsf{Dedekind}\) and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) implies
\(A\cong \bigoplus A/I_i \oplus \bigoplus_{{\mathcal{L}}_i\in {\operatorname{Pic}}(A)} {\mathcal{L}}_i\).

\end{remark}

\begin{exercise}[?]

Show that this implies the classification of finitely-generated modules
over a PID: \(M \cong \bigoplus A/p_i^{n_i} \oplus A^n\).

\end{exercise}

\begin{solution}

Sketch:

\begin{itemize}
\tightlist
\item
  PIDs are Dedekind, so ETS
  \({\operatorname{Pic}}(A) = \left\{{A}\right\}\).
\item
  Use the SES
  \(K^{\times}\xrightarrow{x\mapsto \sum v_p(x) p} \operatorname{Div}A\to {\operatorname{Pic}}A \to 0\),
  ETS the given map is surjective.
\item
  Reduce: ETS that for \(g\in p\in \operatorname{Spec}A\) with \(p\)
  nonzero, there exists an \(x\in K^{\times}\) such that \(v_p(x) = 1\)
  and \(v_{p'}(x) = 0\) for all \(p'\neq p\).

  \begin{itemize}
  \tightlist
  \item
    If \(p = \left\langle{x}\right\rangle\), this works: \(v_p(x) = 1\)
    since this is the maximal \(k\) such that \(x\in p^k\), but
    \(x\in p^2\implies p=p^2\), contradicting Nakayama.
    \(\contradiction\)
  \item
    If \(v_{p'}(x) > 0\), then \(x\in p'\), \(p \subseteq p'\)
    contradicting maximality of \(p\). \(\contradiction\)
  \end{itemize}
\end{itemize}

\end{solution}

\begin{remark}

What about rings of dimension \(d \geq 2\)? E.g.
\(\dim k[x_1, \cdots, x_{n}]= n\), regarded as functions on
\({\mathbb{A}}^n_{/ {k}}\), noting that we haven't quite defined
dimension yet.

\end{remark}

\begin{conjecture}[Serre-Swan, proved by Quillen]

Finitely-generated locally free modules over \(k[x_1, \cdots, x_{n}]\)
are free.

\end{conjecture}

\hypertarget{toward-dimension-filteredgraded-rings}{%
\subsection{Toward dimension: filtered/graded
rings}\label{toward-dimension-filteredgraded-rings}}

\begin{remark}

It is easy to slightly modify this statement to make such a
classification impossible!

\end{remark}

\begin{question}

How many homogeneous polynomials of degree \(d\) in \(n\) variables are
there? Counting the dimension as a \(k{\hbox{-}}\)vector space is yields
\({n+d\choose d} \sim d^n\).

\end{question}

\begin{remark}

\envlist

Recall:

\begin{itemize}
\tightlist
\item
  An increasing filtration \(\left\{{M_i}\right\}\) of
  \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) satisfies
  \(M_i \subseteq M_{i+1}\) and \(\cup_i M_i = M\) and there exists some
  \(i_0\) with \(M_{i_k} = 0\) for all \(i_k\leq i_0\).
\item
  A module morphism \(M \xrightarrow{f} N\) respects filtrations iff
  \(f(M_i) \subseteq N_i\).
\item
  For \(N\leq M\) there is an induced filtrations
  \(N_i \coloneqq M_i \cap N\).
\item
  For \(M \xrightarrow[]{f} { \mathrel{\mkern-16mu}\rightarrow }\, Q\) a
  \emph{surjective} morphism, there is an induced filtration
  \(Q_i \coloneqq f(M_i)\).
\item
  A \textbf{graded abelian group} is a group
  \(M \cong \bigoplus _{i\in {\mathbb{Z}}} M_i\) with \(M_i = 0\) for
  \(i\ll 0\). Typical example:
  \(k[x_1, \cdots, x_{n}]= \bigoplus _d k[x_1, \cdots, x_{n}]_d\) is
  graded by homogeneous degree \(d\) parts.
\item
  Passing between filtrations and gradings: given a grading
  \(\oplus_i M_i\), set
  \({\operatorname{Fil}}^i M = \bigoplus _{k\leq i} M_i\). Given a
  filtration, take the associated graded \(\bigoplus_i M_i/M_{i-1}\).
\item
  Taking the filtration is innocuous and doesn't change the module, but
  taking the associated graded does. Example: \(M = {\mathbb{Z}}\), take
  the filtration
  \(\left\langle{0}\right\rangle \subseteq \left\langle{2}\right\rangle \subseteq {\mathbb{Z}}\)
  whose associated graded is
  \(\left\langle{2}\right\rangle \bigoplus {\mathbb{Z}}/2{\mathbb{Z}}\cong {\mathbb{Z}}\oplus C_2\),
  which now has torsion.
\item
  A morphism of filtered modules induces a well-defined map on the
  associated graded modules.
\end{itemize}

\end{remark}

\begin{exercise}[?]

Show that if \(f: M\to N\) is a morphism of filtered modules, then
\({\mathsf{gr}\,}f\) is injective/surjective \(\implies f\) is
injective/surjective respectively. Show that the converse is not
necessarily true unless \(f\) is injective/surjective and the
filtrations are induced.

\end{exercise}

\begin{solution}

For injectivity, if \({\mathsf{gr}\,}f\) is injective choose \(i\)
minimally so that \(M_i\) intersects \(\ker f\), and contradict
\(i > -\infty\). By minimality, \(x \not\in M_{i-1}\) so
\(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\neq 0\in {\mathsf{gr}\,}^i(M) = M_{i} / M_{i-1}\),
but
\({\mathsf{gr}\,}(f)(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu) = 0\).

For surjectivity, ETS \(f: M_i\to N_i\) is surjective for all \(i\).
Induct on \(i\), using that \(M_i = N_i = 0\) for \(i \ll 0\). Given
\(x\in N_i\), take
\(\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\in N_i/N_{i-1}\)
and lift to \(y\in M_i\) such that
\({\mathsf{gr}\,}(f)(\mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu) = \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\);
take \(x-f(y)\in N_{i-1}\).

\end{solution}

\begin{remark}

Definitions:

\begin{itemize}
\tightlist
\item
  A \textbf{graded ring} is a ring \(A = \bigoplus_i A_i\) with
  \(A_i A_j \to A_{i+j}\), a filtered ring has a filtration
  \(\left\{{A_i}\right\}\) with \(A_i A_j \to A_{ i+j }\).
\item
  A \textbf{graded module \(M\) over a graded ring \(A\)} satisfies
  \(A_i M_j \subseteq M_{i+j}\)
\item
  A \textbf{filtered module over a filtered ring} satisfies
  \(A_i M_j \subseteq M_{i+j}\).
\end{itemize}

\end{remark}

\begin{exercise}[?]

Show that if \(A\) is filtered then \({\mathsf{gr}\,}A\) is naturally
graded.

\end{exercise}

\begin{solution}

Show
\begin{align*}
{A_i \over A_{i-1}}\cdot {A_j \over A_{j-1}} \to {A_{i+j} \over A_{i+j-1} }
.\end{align*}

\end{solution}

\begin{exercise}

Show that if \(M\) is a filtered module over \(A\) a filtered ring, if
\({\mathsf{gr}\,}(M) \in {\mathsf{{\mathsf{gr}\,}(A)}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
then \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).

\end{exercise}

\begin{solution}

Pick homogeneous generators
\(\left\{{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_i}\right\}\)
for \({\mathsf{gr}\,}M\) and show there is a surjection
\begin{align*}
f: A^r &\to M \\
e_i &\to x_i
.\end{align*}
Reduce to showing \({\mathsf{gr}\,}f\) is surjective.

\end{solution}

\begin{exercise}[?]

Show that if \(A\) is a filtered ring and \({\mathsf{gr}\,}A\) is
Noetherian then \(A\) is Noetherian.

\end{exercise}

\begin{solution}

Use that \(I\in \operatorname{Id}(A)\) is a submodule with an induced
filtration and \({\mathsf{gr}\,}I \subseteq {\mathsf{gr}\,}A\) is
finitely-generated to show that \(I \subseteq A\) is finitely-generated.

\end{solution}

\begin{definition}[Good filtrations]

If \(A \in \mathsf{CRing}\) is filtered and
\(M\in{\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is filtered, then
\({\operatorname{Fil}}M\) is a \textbf{good filtration} if
\({\mathsf{gr}\,}(M) \in {\mathsf{{\mathsf{gr}\,}(A)}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).

\end{definition}

\begin{example}[?]

Letting \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) for
\(A=k[x_1, \cdots, x_{n}]\), if
\(\left\{{{\operatorname{Fil}}^i M}\right\}\) is a good filtration then
\(P(i) \coloneqq\dim_k {\operatorname{Fil}}^i M\) will be a polynomial
for \(i \gg 0\) and one can define \(\dim M = \deg P\). To be justified:

\begin{itemize}
\tightlist
\item
  A good filtration exists,
\item
  \(P\) is asymptotically polynomial,
\item
  \(P\) is independent of good filtration chosen.
\end{itemize}

In the free rank 1 case:
\(\dim k[x_1, \cdots, x_{n}]_d = {d+n\choose d} \sim d^{n}\), so
\(\dim k[x_1, \cdots, x_{n}]_{\leq d} = \sum_{k\leq n} d^k \sim d^n\).

\end{example}

\hypertarget{tuesday-april-05}{%
\section{Tuesday, April 05}\label{tuesday-april-05}}

\hypertarget{hilbert-dimension}{%
\subsection{Hilbert dimension}\label{hilbert-dimension}}

\begin{remark}

Preliminary definitions of dimension:

\begin{itemize}
\tightlist
\item
  For \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), find a
  \emph{good filtration} \(\left\{{{\operatorname{Fil}}_i M}\right\}\)
  where \(\dim_k M_i\) is eventually a polynomial \(p\) in \(i\), so
  define \(\dim M \coloneqq\deg p\).
\item
  For \(A_{/ {k}}\) a finitely-generated domain, define
  \(\dim A \coloneqq\operatorname{trdeg}_k A\) as the transcendence
  degree.
\item
  Krull dimension: \(\dim A = n\) iff the longest chain of prime ideals
  \({\mathfrak{p}}_0 \subsetneq {\mathfrak{p}}_1 \subsetneq \cdots \subsetneq {\mathfrak{p}}_n\)
  (note that \(n\) is the number of \emph{inclusions}!)
\end{itemize}

When they're all defined, they all agree.

\end{remark}

\begin{definition}[Good filtrations]

Let \(A\) be a filtered ring and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\). A filtration
\({ {{\operatorname{Fil}}}_{\scriptscriptstyle \bullet}} M\) is a
\textbf{good filtration} iff

\begin{itemize}
\tightlist
\item
  The pair
  \((M, { {{\operatorname{Fil}}}_{\scriptscriptstyle \bullet}} M)\) is a
  filtered module, i.e.~\(A_i M_j \subseteq M_{i+j}\),
\item
  The more contentful condition:
  \({\mathsf{gr}\,}(M) \in {\mathsf{{\mathsf{gr}\,}(A)}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).
\end{itemize}

\end{definition}

\begin{exercise}[?]

Show that every such \(M\) admits a good filtration.

\end{exercise}

\begin{solution}

Use that \(A^n \xrightarrow{f} M\) and \(A^n\) has a good filtration,
and take its image. Show that \(M_i \coloneqq f(A_i^n)\) is good using
that \({\mathsf{gr}\,}(A)^n \twoheadrightarrow{\mathsf{gr}\,}(M)\),
which generally won't stay surjective but will in this case because
\(M\) receives the induced filtration.

\end{solution}

\begin{theorem}[Artin-Rees lemma (extremely important for any arguments involving filtrations!!)]

Suppose \({\mathsf{gr}\,}A\) is Noetherian and let \((M, M_i)\) be an
\(A{\hbox{-}}\)module with a good filtration and let \(N\leq M\). Then
the induced filtration \(N_i \coloneqq N \cap M_i\) is a good
filtration.

\end{theorem}

\begin{proof}[?]

Since \(N\hookrightarrow M\) and we're taking an induced filtration,
\({\mathsf{gr}\,}(N) \hookrightarrow{\mathsf{gr}\,}(M)\) remains
injective. Since \({\mathsf{gr}\,}(M)\) is finitely-generated over
\({\mathsf{gr}\,}(A)\) which is Noetherian, \({\mathsf{gr}\,}(N)\) is
finitely-generated.

\end{proof}

\begin{theorem}[?]

Let \(A\) be a filtered ring,
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), and let
\({ {F}_{\scriptscriptstyle \bullet}} , { {G}_{\scriptscriptstyle \bullet}}\)
be two good filtrations on \(M\). Then there exists a \(k\) such that
\begin{align*}
F_{i-k} M \subseteq G_i M \subseteq F_{i+k} M
.\end{align*}

\end{theorem}

\begin{remark}

There is a notion of a topology on \(M\) induced by a filtration, and
this theorem says all good filtrations induce the same topology. We'll
need the following to prove this theorem:

\end{remark}

\begin{lemma}[?]

Let \((M, F)\) be a module with a good filtration. Then there exists
some \(n\) and \(i_0\) such that
\begin{align*}
i\geq i_0 \implies F_{i+n} \subseteq A_{i+i_0} F_n
.\end{align*}

\end{lemma}

\begin{proof}[of theorem, assuming lemma]

Choose \(n\) as in the lemma, choose \(m\) such that
\(F_n \subseteq G_{n+m}\), then
\begin{align*}
F_{i+n} \subseteq A_{i+i_0} F_n \subseteq A_{i+ i_0} G_{n+m } \subseteq G_{i+n+m}
.\end{align*}
Now run the same argument on \(G\).

\end{proof}

\begin{proof}[of lemma]

Since \({\mathsf{gr}\,}(M)\) is finitely-generated over
\({\mathsf{gr}\,}(A)\), take a finite set of homogeneous generators
\(m_i\). Choose \(n, i_0\) such that \(n-i_0 < \deg m_i < n\), the claim
is that these work. Induct on \(i\): suppose
\(F_{i+n} \subseteq A_{i+i_0} F_n\) and we WTS this still holds when
\(i\mapsto i+1\). Letting \(m\in F_{i+n+1}\), if \(m\in F_{i+n}\) we're
done. Otherwise
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu\neq 0 \in {\mathsf{gr}\,}_{i+n+1} M\),
so
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu = \sum \mkern 1.5mu\overline{\mkern-1.5muc\mkern-1.5mu}\mkern 1.5mu_i m_i\)
and picking lifts,
\(\mkern 1.5mu\overline{\mkern-1.5mum\mkern-1.5mu}\mkern 1.5mu - \sum c_i m_i \in F_{i+n}\).

\end{proof}

\begin{theorem}[?]

Let \(A \coloneqq k[x_1, \cdots, x_{n}]\),
\((M, M_i)\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) have a
good filtration, and let \(\Phi(i) \coloneqq\dim_k M_i\). Then

\begin{itemize}
\tightlist
\item
  There exists a polynomial \(p\) with \(\deg p \leq n\) such that
  \(\Phi(i) = p(i)\) for \(i\gg 0\).
\item
  The degree and leading coefficient of \(p\) are independent of the
  choice of good filtration
\end{itemize}

We then define
\begin{align*}
\dim_A M = \deg p
.\end{align*}

\end{theorem}

\begin{exercise}[Method of finite differences]

Let \(\Phi: {\mathbb{Z}}\to {\mathbb{Z}}\) such that
\(\Phi(i+1) - \Phi(i)\) is eventually polynomial of degree
\(d' \leq n-1\). Show that \(\Phi\) is eventually polynomial of some
bounded degree \(d\leq m\).

\end{exercise}

\begin{solution}

Write
\(\Phi(i) = C + \sum_{i\geq 1} \Phi(i) - \Phi(i-1) = C + \sum_{i\geq 1} q(i)\)
for some polynomial \(q\) with \(\deg q \leq n-1\). So it's ETS
\(\sum_{0\leq n\leq i}n^{a}\) polynomial in \(i\) for \(a\geq 0\).

\end{solution}

\begin{proof}[of theorem, part 1]

Proceed by induction on the number of variables. Since
\(\dim_k(M_i) = \sum_{i} \dim_k({\mathsf{gr}\,}_i M)\) where
\({\mathsf{gr}\,}_i M = M_i/M_{i+1}\), it's ETS
\(\dim M_i - \dim M_{i-1} = \dim {\mathsf{gr}\,}_i M\). An awesome
maneuver: take a regrading to get a LES
\begin{align*}
0 \to K\coloneqq\ker f \to {\mathsf{gr}\,}M \xrightarrow{f} \Sigma {\mathsf{gr}\,}M \to C\coloneqq\operatorname{coker}f\to 0
.\end{align*}

Note that \(K, C\in k[x_1,\cdots, x_{n-1}]\) has fewer variables, and by
alternating additivity in exact sequences,
\begin{align*}
\dim {\mathsf{gr}\,}_i K - \dim {\mathsf{gr}\,}_i M + \dim {\mathsf{gr}\,}_{i+1} M - \dim {\mathsf{gr}\,}_i C = 0
.\end{align*}
Thus
\begin{align*}
\dim {\mathsf{gr}\,}_{i+1} M - \dim {\mathsf{gr}\,}_{i} M = \dim {\mathsf{gr}\,}_i C - \dim {\mathsf{gr}\,}_i K
,\end{align*}
and since the RHS involves good filtrations, these are eventually
polynomial, thus so is the LHS.

\end{proof}

\begin{proof}[of theorem, part 2]

Let \(F, G\) be good filtrations with
\(F_{i-k} \subseteq G_{i} \subseteq F_{i+k}\) with
\(\Phi_F(i) = \dim F_i, \Phi_G(i) = \dim G_i\). Then
\begin{align*}
\Phi_F(i-k) \leq \Phi_G(i) \leq \Phi_G(i+k)
,\end{align*}
and if \(\Phi_G, \Phi_G\) have different degrees or the same degrees but
different leading terms, it would violate this inequality for large
\(i\).

\end{proof}

\hypertarget{properties-of-dimension}{%
\subsection{Properties of dimension}\label{properties-of-dimension}}

\begin{exercise}[?]

Let \(M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).

\begin{itemize}
\tightlist
\item
  Show that if \(N \leq M\) then \(\dim N \leq \dim M\).
\item
  Show that if \(M \twoheadrightarrow N\) then \(\dim N\leq \dim M\).
\end{itemize}

\end{exercise}

\begin{solution}

\envlist

\begin{itemize}
\tightlist
\item
  Restrict the good filtration on \(M\), this is good by Artin-Rees.
\item
  Take an induced good filtration. Use that quotients of
  finitely-generated are finitely-generated and
  \({\mathsf{gr}\,}(A)^n \twoheadrightarrow{\mathsf{gr}\,}(M) \twoheadrightarrow{\mathsf{gr}\,}(N)\).
\end{itemize}

\end{solution}

\begin{exercise}[?]

Show that if \(0\to M_1\to M_2\to M_3\to 0\) is a SES in
\({\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) for \(A=k[x_1, \cdots, x_{n}]\),
then
\begin{align*}
\dim M_2 = \max( \dim M_1, \dim M_3)
.\end{align*}

\end{exercise}

\begin{solution}

Choose good filtrations \(F_i, M_i, G_i\) on \(M_1, M_2, M_3\)
respectively, so \(\dim M_i = \dim F_i + \dim G_i\). The RHS involves
polynomials in \(i\) with non-negative leading terms, and their sum has
the max of the 2 degrees and again has a nonnegative leading term.

\end{solution}

\begin{exercise}[?]

Show that if \(A=k[x_1, \cdots, x_{n}]\) then \(\dim_A A = n\) and
\(\dim_A A^n = n\).

\end{exercise}

\hypertarget{thursday-april-07}{%
\section{Thursday, April 07}\label{thursday-april-07}}

\hypertarget{hilbert-dimension-of-finitely-generated-algebras-over-a-field}{%
\subsection{Hilbert dimension of finitely generated algebras over a
field}\label{hilbert-dimension-of-finitely-generated-algebras-over-a-field}}

\begin{remark}

Prove:

\begin{itemize}
\tightlist
\item
  \(f: M\hookrightarrow M\) implies
  \(\dim \operatorname{coker}f < \dim M\).
\item
  If \(A = k[x_1, \cdots, x_{n}]\) and
  \(f\in A\setminus\left\{{0}\right\}\) then \(\dim A/f < \dim A\).
\item
  If
  \(A\in {\mathsf{k[x_1, \cdots, x_{n}]}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
  and \(f\in A\) is a nonzero divisor, then \(\dim A/F < \dim A\).
\end{itemize}

\end{remark}

\begin{definition}[Dimension]

If \(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\), choosing a surjection
\(k[x_1, \cdots, x_{n}]\twoheadrightarrow A\), define
\(\dim A \coloneqq\dim_{k[x_1, \cdots, x_{n}]} A\).

\end{definition}

\begin{exercise}[?]

Show this is well-defined. It follows from the following:

\end{exercise}

\begin{theorem}[?]

Let
\(M\in {\mathsf{k[x_1,\cdots, x_n, y_1,\cdots, x_n]}{\hbox{-}}\mathsf{Mod}}\)
with
\(M\in {\mathsf{k[x_1, \cdots, x_{n}]}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).
Then
\begin{align*}
\dim_{k[x_1, \cdots, x_{n}]} M = \dim_{k[x_1,\cdots, x_n, y_1,\cdots, y_n] }
.\end{align*}

\end{theorem}

\begin{solution}

Why the theorem implies the exercise:

\begin{center}
\begin{tikzcd}
    && {} \\
    \textcolor{rgb,255:red,214;green,92;blue,92}{k[x_1,\cdots, x_n, y_1, \cdots, y_m]} && \textcolor{rgb,255:red,214;green,92;blue,92}{k[y_1,\cdots, y_m]} \\
    \\
    \textcolor{rgb,255:red,92;green,92;blue,214}{k[x_1,\cdots, x_n]} && A \\
    \\
    {} && {}
    \arrow["g", color={rgb,255:red,214;green,92;blue,92}, two heads, from=2-1, to=2-3]
    \arrow[color={rgb,255:red,214;green,92;blue,92}, two heads, from=2-3, to=4-3]
    \arrow[color={rgb,255:red,92;green,92;blue,214}, two heads, from=4-1, to=4-3]
    \arrow["{\exists: x_i\mapsto x_i, y_i\mapsto g^{-1}(y_i)}"', dashed, from=2-1, to=4-1]
\end{tikzcd}
\end{center}

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

This can be filled in to a commutative diagram, so
\begin{align*}
\dim_{k[x_1, \cdots, x_{n}]} A = \dim_{k[x_1,\cdots, x_n, y_1,\cdots, y_m]} A = \dim_{k[y_1, \cdots, y_m]} A
.\end{align*}

\end{solution}

\begin{proof}[of theorem]

It's ETS this when \(m=1\), so suppose
\(M\in {\mathsf{ k[x_1, \cdots, x_{n}][y] }{\hbox{-}}\mathsf{Mod}}\)
with
\(M\in {\mathsf{k[x_1, \cdots, x_{n}]}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).
Take the surjection \(k[x_1, \cdots, x_{n}]^r\twoheadrightarrow M\)
where \(e_i\mapsto m_i\) for generators \(m_i\), and let \(M_i\) be the
induced filtration. Similarly take
\(k[x_1,\cdots, x_n, y]^r \twoheadrightarrow M\) with \(e_i\mapsto m_i\)
and let \(\tilde M_i\) be the induced filtration. Since
\(M_i \subseteq \tilde M_i\),
\(\dim_{k[x_1, \cdots, x_{n}]} M \leq \dim_{k[x_1, \cdots, x_{n}][y]} M\).

The claim is that one can choose \(k\) such that \(yM_0 \subseteq M_k\)
and \(\tilde M_i \subseteq M_{ik}\). Why:
\(\tilde M_i = \left\{{m\in M {~\mathrel{\Big\vert}~}m \text{ is in the image of } (f_1,\cdots, f_r) \text{ of degree at most } i}\right\}\).
So write \(f_j = \sum_{n} f_{j, n}(x_1,\cdots, x_n)y^n\), then
\(\tilde M_i \subseteq \mathop{\mathrm{span}}(M_i + yM_{i-1} + y^2 M_{i-2} + \cdots)\).

Why this finishes the proof:
\(\dim_{k[x_1, \cdots, x_{n}]} M = \dim_{k[x_1, \cdots, x_{n}][y]} M\),
so

\begin{itemize}
\tightlist
\item
  \(\Phi(i) = \dim M_i\)
\item
  \(\tilde \Phi(i) = \dim \tilde M_i\)
\item
  \(\Phi(i) \leq \tilde \Phi(i)\leq \Phi(ik)\),
\end{itemize}

which forces \(\deg \Phi = \deg \tilde \Phi\).

\end{proof}

\begin{definition}[Dimensions of modules]

If \(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), choose
\(k[x_1, \cdots, x_{n}]\twoheadrightarrow A\) to write
\(\dim_A M \coloneqq\dim_{k[x_1, \cdots, x_{n}]} M\).

\end{definition}

\begin{remark}

Upshot: we have a notion of dimension for
\(M\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) when \(k\in \mathsf{Field}\)
which doesn't depend on choices.

\end{remark}

\begin{exercise}

Show the following properties:

\begin{itemize}
\item
  \(\dim k[x_1, \cdots, x_{n}]= n\).
\item
  If \(A,B\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) and \(A\to B\) with
  \(B\in{\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), then
  \(\dim_A B = \dim B\) (this just involves adding variables to
  \(k[x_1, \cdots, x_{n}]\)).
\item
  If \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) then
  \(\dim_A M\leq \dim A\). Hint: \(A^n\twoheadrightarrow M\).
\item
  If \(A,B\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) and \(A\to B\) with
  \(B\in{\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), then
  \(\dim B \leq \dim A\).
\item
  If \(A,B\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) with
  \(A\hookrightarrow B\) then \(\dim A < \dim B\). Hint: consider

  \begin{center}
  \begin{tikzcd}
    k[x_1, \cdots, x_{n}]&& A \\
    \\
    {k[x_1, \cdots, x_{n}][y_1,\cdots, y_m]} && B
    \arrow[hook, from=1-3, to=3-3]
    \arrow[two heads, from=3-1, to=3-3]
    \arrow[two heads, from=1-1, to=1-3]
    \arrow[hook, from=1-1, to=3-1]
  \end{tikzcd}
  \end{center}

  \begin{quote}
  \href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXGt4biJdLFsyLDAsIkEiXSxbMCwyLCJcXGt4blt5XzEsXFxjZG90cywgeV9tXSJdLFsyLDIsIkIiXSxbMSwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDMsIiIsMix7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFswLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6ImVwaSJ9fX1dLFswLDIsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
  to Diagram}
  \end{quote}

  Take the induced filtrations \(A_i, B_i\) induced by degree on
  \(k[x_1, \cdots, x_{n}]\), then \(\iota(A_i) \subseteq B_i\).
\item
  If \(I {~\trianglelefteq~}A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\)
  then \(\dim A/I \leq \dim A\) (geometrically this is passing to a
  closed subspace).
\item
  If \(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) is a domain with
  \(I{~\trianglelefteq~}A\) then \(\dim A/I < \dim A\). Hint: it's ETS
  \(\dim A/f < \dim A\), so use the SES
  \(0\to A \xrightarrow{f} A \to A/f \to 0\).
\item
  If \(A\) is a domain and \(f\in A\setminus\left\{{0}\right\}\) then
  \(\dim A/f = \dim A - 1\).
\end{itemize}

\end{exercise}

\begin{example}[?]

Using these properties:

\begin{itemize}
\tightlist
\item
  \(A = k[x,y]\) satisfies \(\dim A = 2\), while \(B=k[x,y]/y = k[x]\)
  satisfies \(\dim B = 1\).
\item
  For \(A = k[x,y]/xy\) and \(B = A/y = k[x]\), both are dimension 1.
  Geometrically, \(A\) is the union of the \(x\) and \(y\) axes, and
  \(B\) sets \(y=0\) which results in the \(x\) axis.
\item
  If \(f: A\to B\) yields
  \(B\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), the induced
  map \(f^*: \operatorname{Spec}B\to \operatorname{Spec}A\) has finite
  fibers.

  \begin{itemize}
  \tightlist
  \item
    If additional \(f\) is an injection, the dimensions are equal.
  \end{itemize}
\end{itemize}

\begin{quote}
Injection means dense image, and dense image + finite fibers preserves
dimension.
\end{quote}

\end{example}

\hypertarget{other-notions-of-dimension-transcendence-degree}{%
\subsection{Other notions of dimension: transcendence
degree}\label{other-notions-of-dimension-transcendence-degree}}

\begin{remark}

Setup: let

\begin{itemize}
\tightlist
\item
  \(k \subseteq K\) be an inclusion of fields (with no finiteness
  conditions)
\item
  \(\alpha \in K\) is \textbf{integral over \(k\)} iff \(f(\alpha) = 0\)
  for some monic \(f\in k[x]\).
\item
  \(K\) is \textbf{algebraic over \(k\)} iff every \(\alpha\in K\) is
  integral over \(k\).
\item
  \(K\) is \textbf{transcendental over \(k\)} if it is not algebraic.
\end{itemize}

\end{remark}

\begin{example}[?]

Some examples:

\begin{itemize}
\tightlist
\item
  \({\mathbb{Q}}(\sqrt 5)\) is algebraic over \({\mathbb{Q}}\).
\item
  \({\mathbb{Q}}(t)\) is transcendental over \({\mathbb{Q}}\).
\end{itemize}

\end{example}

\begin{definition}[Algebraically independent elements]

A subset \(\left\{{x_a}\right\} \subseteq K\) is \textbf{algebraically
independent over \(k\)} iff there does not exist a nonzero polynomial
\(p\in k[t_1,\cdots, t_n]\) with \(p(\cdots, x_a, \cdots) = 0\). Such a
subset is \textbf{maximal} if it is not strictly contained in another
such subset.

\end{definition}

\begin{definition}[Transcendence degree]

If \(K/k\) admits a maximal finite algebraically independent subset
\(S\), define \(\operatorname{trdeg}_k K = {\sharp}S\).

\end{definition}

\begin{claim}

This is well-defined. To be justified next time!

\end{claim}

\begin{definition}[?]

If \(A\) is a finitely generated domain over \(k\), then
\(\dim A = \operatorname{trdeg}_k \operatorname{ff}(A)\).

\end{definition}

\hypertarget{thursday-april-14}{%
\section{Thursday, April 14}\label{thursday-april-14}}

\begin{remark}

Recall that if \(A\in {\mathsf{Alg}_{/k} }\), the algebraic
differentials are constructed as
\begin{align*}
\Omega_{A_{/ {k}} } = \mathsf{Free}\left\{{dA{~\mathrel{\Big\vert}~}a\in A}\right\}/ \left\langle{d(ab) = ad(b) + d(a) b}\right\rangle
.\end{align*}

Note that
\(\mathop{\mathrm{Hom}}_{{\mathsf{A}{\hbox{-}}\mathsf{Mod}}}( \Omega_{A_{/ {k}} }, M) \cong \mathop{\mathrm{Der}}_{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}}(A, M)\),
where derivations are importantly not \(A{\hbox{-}}\)linear! This is
meant to emulate a cotangent bundle.

There is a SES
\begin{align*}
0\to I\to A\otimes_k A \xrightarrow{m} A\to 0
.\end{align*}

\end{remark}

\begin{example}[?]

For \(A=k[x]\), check that \(x,y\mapsto x\) and
\(I = \left\langle{x-y}\right\rangle\), and moreover
\(\Omega_{k[x]_{/ {k}} } \cong I/I^2 = \left\langle{x-y}\right\rangle/\left\langle{x^2+xy+y^2}\right\rangle\).

\end{example}

\begin{slogan}

The cotangent bundle is the conormal bundles of the diagonal:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Spring/CommutativeAlgebra/sections/figures}{2022-04-14_12-53.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{slogan}

\begin{exercise}[?]

Show that \(\Omega_{A_{/ {k}} } \cong I/I^2\).

\end{exercise}

\begin{solution}

Hints: a map \(\Omega_{A_{/ {k}} } \xrightarrow{f} I/I^2\) is equivalent
to a derivation \(A\to I/I^2\). Show that if
\(\psi(a) \coloneqq a\otimes 1 - 1\otimes a\in I\), then
\(m(\psi(a)) = 0\) and \(\psi\) is a derivation,
i.e.~\(\psi(ab) - a\psi(b) - \psi(b)a \in I^2\). Check
\begin{align*}
ab \otimes 1 -1 \otimes ab - a(b \otimes 1 - 1 \otimes b) - b(a \otimes 1 - 1 \otimes a)
&= ab \otimes 1 - 1 \otimes ab - ab \otimes 1 + a \otimes b - ba \otimes 1 + b \otimes a \\
&= -1 \otimes ab + a \otimes b - ab \otimes 1 + b \otimes a \\
&= -(a \otimes 1 - 1 \otimes a)(b \otimes 1 - 1 \otimes b) \in I^2
.\end{align*}
For an inverse, take
\begin{align*}
g: I/I^2 &\to \Omega_{A_{/ {k}} } \\
\sum a_i \otimes b_i &\mapsto \sum b_i d(a_i)
\end{align*}
where \(\sum a_i b_i = 0\) and check

\begin{itemize}
\tightlist
\item
  \(g(I)^2=0\)
\item
  \(f \circ g = \operatorname{id}, g \circ f = \operatorname{id}\).
\end{itemize}

\end{solution}

\begin{exercise}[?]

Show that \((1 \circ x - x \circ 1) \to d1 + 1dx\) where
\(x-y = -\,dx\).

\end{exercise}

\begin{exercise}[General algebra]

Show that \(M\to N\to L\to 0\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is
exact iff \(0\to [L, S]_A \to [N, S]_A \to [M, S]_A\) is exact for all
\(S\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\).

\end{exercise}

\begin{solution}

Hint: \(\impliedby\) is the nontrivial direction. Toward a contradiction
take \(S\coloneqq\operatorname{coker}(A\to L)\).

\end{solution}

\begin{exercise}[?]

Show that if \(A,B\in {\mathsf{Alg}_{/k} }\) and
\(0\to I\to A \xrightarrow{f} B\to 0\) is exact then there is an exact
sequence

\begin{center}
\begin{tikzcd}
    {I/I^2} && {\Omega_{A/_{/ {k}} }} && {\Omega_{B/_{/ {k}} }} && 0 && {\in {\mathsf{B}{\hbox{-}}\mathsf{Mod}}} \\
    x && dx \\
    && da && {d(f(a))}
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[maps to, from=2-1, to=2-3]
    \arrow[maps to, from=3-3, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOSxbMCwwLCJJL0leMiJdLFsyLDAsIlxcT21lZ2Ffe0EvXFxzbGljZSBrfSJdLFs0LDAsIlxcT21lZ2Ffe0IvXFxzbGljZSBrfSJdLFs2LDAsIjAiXSxbOCwwLCJcXGluIFxcbW9kc3tBfSJdLFswLDEsIngiXSxbMiwxLCJkeCJdLFsyLDIsImRhIl0sWzQsMiwiZChmKGEpKSJdLFswLDFdLFsxLDJdLFsyLDNdLFs1LDYsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XSxbNyw4LCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d}{Link
to Diagram}
\end{quote}

Idea: \(\operatorname{Spec}B\hookrightarrow\operatorname{Spec}A\) is
like an embedded submanifold, and \(I/I^2\) is the conormal bundle.

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Spring/CommutativeAlgebra/sections/figures}{2022-04-14_13-09.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{exercise}

\begin{solution}

Identify

\begin{center}
\begin{tikzcd}
    0 && {[\Omega_{B_{/ {k}} }, S]_B} && {[\Omega_{A_{/ {k}} }\otimes_A B, S]_B \cong [\Omega_{A_{/ {k}} }, S]_A} && {[I/I^2, S]_B} \\
    \\
    0 && {\mathop{\mathrm{Der}}_k[B,S]} && {\mathop{\mathrm{Der}}_k[A,S]} && {[I/I^2, S]_B}
    \arrow["\cong", from=1-3, to=3-3]
    \arrow["\cong", from=1-5, to=3-5]
    \arrow["\cong", from=1-7, to=3-7]
    \arrow[from=3-3, to=3-5]
    \arrow[from=3-5, to=3-7]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow["{?}"{description}, from=1-1, to=1-3]
    \arrow["\therefore", from=3-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwwLCIwIl0sWzIsMCwiW1xcT21lZ2Ffe0JcXHNsaWNlIGt9LCBTXV9CIl0sWzQsMCwiW1xcT21lZ2Ffe0FcXHNsaWNlIGt9XFx0ZW5zb3JfQSBCLCBTXV9CIFxcY29uZyBbXFxPbWVnYV97QVxcc2xpY2Uga30sIFNdX0EiXSxbNiwwLCJbSS9JXjIsIFNdX0IiXSxbMiwyLCJcXERlcl9rW0IsU10iXSxbNCwyLCJcXERlcl9rW0EsU10iXSxbNiwyLCJbSS9JXjIsIFNdX0IiXSxbMCwyLCIwIl0sWzEsNCwiXFxjb25nIl0sWzIsNSwiXFxjb25nIl0sWzMsNiwiXFxjb25nIl0sWzQsNV0sWzUsNl0sWzEsMl0sWzIsM10sWzAsMSwiPyIsMV0sWzcsNCwiXFx0aGVyZWZvcmUiXV0=}{Link
to Diagram}
\end{quote}

Check:

\begin{itemize}
\tightlist
\item
  \(\psi \in \mathop{\mathrm{Der}}_k(B, S)\) with \(\psi \circ f\)
  surjective implies \(\psi = 0\) when \(f\) is surjective.
\item
  \(\psi \in \mathop{\mathrm{Der}}_k(A, S)\) with \(\rho\psi{I} = 0\)
  implies
  \(\psi\in \operatorname{im}(\mathop{\mathrm{Der}}_K(B, S) \to \mathop{\mathrm{Der}}_k(A, S))\)
\end{itemize}

\end{solution}

\begin{definition}[?]

For \(f\in \mathsf{CRing}(A, B)\), \(\Omega_{B/A}\) is the unique object
in \({\mathsf{B}{\hbox{-}}\mathsf{Mod}}\) such that
\begin{align*}
[\Omega_{B/A}, M] = \mathop{\mathrm{Der}}_A(B, M)
.\end{align*}
Explicitly,
\begin{align*}
\Omega_{B/A} = \mathsf{Free}\left\{{b\in B}\right\}/\left\langle{d(b_1 b_2) = b_1 d(b_2) + d(b_1) b_2, da {~\mathrel{\Big\vert}~}a\in A, b_i\in B }\right\rangle
.\end{align*}

\end{definition}

\begin{exercise}[?]

Show that for \(A,B\in {\mathsf{Alg}_{/k} }\), there is a SES
\begin{align*}
\Omega_{A_{/ {k}} }\otimes_{A} B \to \Omega_{B_{/ {k}} } \to \Omega_{B/A} \to 0 \qquad \in {\mathsf{B}{\hbox{-}}\mathsf{Mod}}
.\end{align*}

\end{exercise}

\begin{remark}

If \(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) and
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\)
and \({\mathfrak{m}}\in \operatorname{mSpec}A\), then
\(0\to {\mathfrak{m}}\to A\to k \to 0\) (by EEKS) and there is a SES
\begin{align*}
0\to {\mathfrak{m}}/{\mathfrak{m}}^2\to \Omega_{A_{/ {k}} } \otimes_A k\to \Omega_{k_{/ {k}} } = 0\to 0
.\end{align*}

\end{remark}

\begin{definition}[Tangent/cotangent spaces]

Define
\({\mathbf{T}} {}^{ \vee }A \coloneqq{\mathfrak{m}}/{\mathfrak{m}}^2\)
and
\({\mathbf{T}}A \coloneqq({\mathfrak{m}}/{\mathfrak{m}}^2) {}^{ \vee }\coloneqq[{\mathfrak{m}}/{\mathfrak{m}}^2, k]_{k}\).

\end{definition}

\begin{exercise}[?]

Let \(A = k[x]\) and \({\mathfrak{m}}=\left\langle{x}\right\rangle\),
then for \(f\in {\mathfrak{m}}\) define
\(\tau\in {\mathbf{T}} {}^{ \vee }A\), so
\(\tau(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu)\in k\),
by \(\tau = {\frac{\partial }{\partial x}\,}\). Then
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\to f'(0)\).

Similarly for \(A=k[x_1, \cdots, x_{n}]\) and
\({\mathfrak{m}}= \left\langle{x_1,\cdots, x_n}\right\rangle\), check
\({\mathbf{T}}_0 = \mathop{\mathrm{span}}_{\mathbb{C}}\left\langle{{\frac{\partial }{\partial x_1}\,}, \cdots, {\frac{\partial }{\partial x_n}\,}}\right\rangle\).

\end{exercise}

\begin{theorem}[?]

For
\(k= { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\)
and \(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) a domain,

\begin{itemize}
\item
  For any \(m\in \operatorname{mSpec}A\),
  \begin{align*}
  \dim_k \Omega_{A_{/ {k}} }\otimes_k A/m = \dim_k m/m^2 \geq \dim A
  .\end{align*}
\item
  There exists a nonempty open \(U \subseteq \operatorname{Spec}A\) such
  that for some \(m\in U\) this is an equality, so
  \begin{align*}
  \dim_k m/m^2 = \dim_{\operatorname{ff}(A)} \operatorname{ff}(A) \otimes_A \Omega_{A_{/ {k}} }
  .\end{align*}
\end{itemize}

\end{theorem}

\begin{remark}

What goes into a proof:

\begin{itemize}
\tightlist
\item
  Find \(k[x_1, \cdots, x_{n}]\hookrightarrow A\) with \(n=\dim A\)
  making
  \(A\in {\mathsf{k[x_1, \cdots, x_{n}]}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).
\item
  Check \(\Omega_{A/k[x_1, \cdots, x_{n}]}\) is torsion.
\end{itemize}

\end{remark}

\begin{definition}[Smooth points]

A point \(m\in \operatorname{mSpec}A\) is \textbf{smooth} if
\begin{align*}
\operatorname{rank}{\mathbf{T}} {}^{ \vee }A = \dim A
.\end{align*}

\end{definition}

\begin{example}[?]

A non-example: \(\operatorname{krulldim}A = 0\) for \(A=k[x]/x^2\),
since \(A\) is Artin. Check
\(\Omega_{A_{/ {k}} } = A\,dx/\left\langle{d(x)^2=0}\right\rangle \cong {k[x]/x^2 \,dx\over 2x\,dx}\),
which is \(k\,dx\) if \(\operatorname{ch}k \neq 2\) and
\(k[x]/x^2 \,dx\) if \(\operatorname{ch}k = 2\).

\end{example}

\begin{example}[?]

For \(A= k[x,y]/\left\langle{y^2-x^3}\right\rangle\) with
\(\operatorname{ch}k = 0\), check
\begin{align*}
\Omega_{A_{/ {k}} } = {A\,dx\oplus A\,dy\over d(y^2-x^3)} = {A\,dx\oplus A\,dy\over dy\,dy- 3x^2\,dx}
.\end{align*}

Given \(m\in \operatorname{mSpec}A\),
\begin{align*}
\dim_{A/m} \Omega_{A/k} \otimes_A A/m = 
\begin{cases}
1 & m\not\in \left\langle{x,y}\right\rangle 
\\
2 & m\in \left\langle{x,y}\right\rangle .
\end{cases}
\end{align*}

Note that this records the nodal singularity:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2022/Spring/CommutativeAlgebra/sections/figures}{2022-04-14_13-55.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{example}

\begin{example}[?]

Over \(k = {\mathbb{F}}_p { \left( {t} \right) }\) and
\(A = {\mathbb{F}}_p { \left( {t^{1\over p}} \right) } = k[x]/\left\langle{x^p-t}\right\rangle\),
check \(\dim A = 0\) and
\begin{align*}
\Omega_{A_{/ {k}} } = {A\,dx\over d(x^p - t)} = {A\,dx\over d(x^p)} = A \,dx
.\end{align*}

So the algebraic differentials detect when a field fails to be
separable.

\end{example}

\hypertarget{tuesday-april-19}{%
\section{Tuesday, April 19}\label{tuesday-april-19}}

\hypertarget{completion}{%
\subsection{Completion}\label{completion}}

\begin{remark}

Recall that
\(\cocolim_{i\in I} M_i\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) is the
universal \(A{\hbox{-}}\)module living above all of the \(M_i\), such
that for any other \(N\) above the \(M_i\) there is a morphism
\(N\to \cocolim_i M_i\). This can be realized as
\(\cocolim M_i = \left\{{ (m_i)_{i\in I} \in \prod_{i\in I} M_i {~\mathrel{\Big\vert}~}\phi(m_i) = m_{i+1}\, \forall i }\right\}\).

\begin{quote}
Note that the inverse limit has a mapping \textbf{in} property, as does
\(\prod\). Limits can be constructed out of equalizers and products:
\begin{align*}
\cocolim F \cong \mathop{\mathrm{=}}\qty{ \prod_{X \in \mathsf{C}} F(X) \to \prod_{f,Y\in \mathsf{C}(X, Y) } F(Y) }
.\end{align*}
\end{quote}

\end{remark}

\begin{exercise}[?]

Show that this satisfies the correct universal property.

\end{exercise}

\begin{exercise}[?]

Show the following:
\begin{align*}
\cocolim_i k[t]/\left\langle{t^i}\right\rangle & \cong k {\left[\left[ t \right]\right]  } \\
\cocolim(\cdots \to {\mathbb{Z}}/p^2{\mathbb{Z}}\to {\mathbb{Z}}/p{\mathbb{Z}}\to 0) &\cong { {\mathbb{Z}}_{\widehat{p}} }\\
\cocolim_n {\mathbb{Z}}/n!{\mathbb{Z}}&\cong \widehat{{\mathbb{Z}}} = \prod_p { {\mathbb{Z}}_{\widehat{p}} }
.\end{align*}

\end{exercise}

\begin{exercise}[?]

Show that the functor \(\cocolim_i({-})\) is left-exact and
\(\colim_i({-})\) is exact, i.e.~if
\(0\to (M_i)\to (N_i) \to (L_i)\to 0\) is a SES of inverse systems, then
\(0\to \cocolim_i M_i \to \cocolim_i N_i \to\cocolim_i L_i\).

\end{exercise}

\begin{solution}

Hint: for injectivity, use the product definition to realize
\(f: \cocolim_i M_i \to \cocolim_i N_i\) as
\(f=\prod_i f_i: M_i\to N_i\).

\end{solution}

\begin{remark}

Show that \(\tau_{\geq 2}{\mathbb{R}}\cocolim_i({-}) = 0\), so there is
always a 6-term exact sequence.

\end{remark}

\begin{theorem}[Mittag-Leffler, sufficient conditions for $\lim^1$ vanishing]

If \(M_i \to M_{i+1}\) is surjective, then \(\lim^1 = 0\) and
\(\cocolim N_i \twoheadrightarrow\cocolim L_i\).

\end{theorem}

\begin{exercise}[?]

Prove this!

Hints:

\begin{itemize}
\tightlist
\item
  Given \((n_i)\in \cocolim N_i\), we want to produce \((l_i)\) with
  \((g_i(n_i)) = \ell_i\).
\item
  Lift \(\ell_0\) to \(n_0\) and induct on \(i\):
\item
  Choose an approximate lift \(\tilde n_i\) of \(\ell_i\).
\item
  Use commutativity of the diagrams of inverse systems to correct this
  choice, using the surjectivity assumption.
\end{itemize}

\end{exercise}

\begin{definition}[Completion]

Recall that given a sequence of submodules \(M_i\) of a module \(M\), so
\(M \geq M_1 \geq M_2 \geq \cdots\), one can form an inverse system
\(\cdots \to M/M_2\to M/M_1\to 0\). The \textbf{completion} of \(M\) is
\(\cocolim_i M/M_i\).

\end{definition}

\begin{remark}

Topological interpretation: define a subset \(U \subseteq M\) iff for
each \(m\in U\) there is some \(M_i\) such that \(m+ M_i \subseteq U\);
equivalently \(\left\{{M_i}\right\}\) forms a basis of neighborhoods of
zero.

\end{remark}

\begin{exercise}[?]

Show that \(\cocolim M/M_i\) is Hausdorff iff
\(\cap_i M_i = \left\{{0}\right\}\) (i.e.~\(\widehat{M}\) is separated).

\end{exercise}

\begin{solution}

Hints:

\(\implies\): If \(m\neq 0\in \cap M_i\), show every open containing
zero contains \(m\).

\(\impliedby\): Pick \(m_1\neq m_2\) and find \(M_i\) with
\((m_1 + M_i) \cap(m_2 + M_i) = \emptyset\).

\end{solution}

\begin{remark}

Call a sequence \((m_i)\) \textbf{Cauchy} iff \(m_n - m_{n'}\in M_i\)
for all \(n, n' > N_i\). Note that one can define a metric this way and
take the Cauchy completion, defined as \(\widehat{M}\), which is
canonically isomorphic to \(\widehat{M}\).

\end{remark}

\begin{exercise}[?]

Produce the isomorphism from Cauchy sequences modulo equivalence to
\(\widehat{M}\).

\end{exercise}

\begin{solution}

For \(\widehat{M} \to M/M_i\), take \(N_i \gg 0\) and project (i.e.~send
the sequence to its limit). For \(\cocolim M/M_i \to \widehat{M}\), send
\((m_i)\mapsto (\tilde m_i)\) by choosing arbitrary lifts.

\end{solution}

\begin{exercise}[?]

Show that if \(0\to M_1\to M_2\to M_3\to 0\) and \(M_2\) admits a
filtration, then there is a SES of the completions at \emph{induced}
filtrations
\(0\to \widehat{M}_1\to \widehat{M}_2\to \widehat{M}_3\to 0\).

\end{exercise}

\begin{solution}

Start with the SES
\begin{align*}
0 \to {M_1\over M_1 \cap{\operatorname{Fil}}_i M_2} \to {M_2\over {\operatorname{Fil}}_i M_2} \to {M_3\over f({\operatorname{Fil}}_i M_2)} \to 0
,\end{align*}
and letting \(i\) vary yields a SES of inverse systems. By
Mittag-Leffler, the limits are exact.

\end{solution}

\begin{exercise}[?]

Show that completion is idempotent, so
\(\widehat{\widehat{M}} \cong \widehat{M}\).

\end{exercise}

\begin{definition}[adic filtration]

For
\(A\in \mathsf{CRing}, M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}, \alpha\in \operatorname{Id}(A)\),
there is an \(\alpha{\hbox{-}}\)adic filtration
\({\operatorname{Fil}}_i M \coloneqq\alpha^i M\), and the corresponding
completion \(\widehat{M}\) is the \textbf{\(\alpha{\hbox{-}}\)adic
completion}.

\end{definition}

\begin{example}[?]

Some examples:

\begin{itemize}
\tightlist
\item
  \(k[t], \left\langle{t}\right\rangle\)
\item
  \({\mathbb{Z}}, \left\langle{p}\right\rangle\)
\item
  \(k[x,y], \left\langle{x,y}\right\rangle\)
\end{itemize}

\end{example}

\begin{slogan}

Completions are inverse limits of Artin rings and are local, and thus
mediate between the two.

\end{slogan}

\begin{exercise}[?]

Show that if \(A\in \mathsf{CRing}, m\in \operatorname{mSpec}A\), then
\(\widehat{A}\) completed with respect to \(m\) is a local ring.

\end{exercise}

\hypertarget{thursday-april-21}{%
\section{Thursday, April 21}\label{thursday-april-21}}

\hypertarget{artin-rees}{%
\subsection{Artin Rees}\label{artin-rees}}

\begin{exercise}[?]

Show that if \(A\in\mathsf{CRing}\) and \(m\in \operatorname{mSpec}A\)
then \(A{ {}_{ \widehat{m} } }\) is local.

\end{exercise}

\begin{solution}

Show that
\(x\in A{ {}_{ \widehat{m} } }\setminus\ker(A{ {}_{ \widehat{m} } } \xrightarrow{\pi} A/m)\)
is invertible. If \(\pi(x) = 1\) then \(1-x\in \ker\pi\) and
\(y = {1\over 1-(1-x)} = \sum_k (1-x)^k\) is an inverse.

\end{solution}

\begin{remark}

Note that \(\alpha{\hbox{-}}\)adic completion is not generally exact,
but \emph{is} exact in most cases of interest, e.g.~for
\(M\in{\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\). There is a
derived functor used in e.g.~Bhatt-Scholze.

\end{remark}

\begin{theorem}[Artin-Rees]

For \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) for
\(A\in { \mathrm{Noeth} }\mathsf{CRing}\) with \(M'\leq M\) and
\(\alpha\in \operatorname{Id}(A)\), the \(\alpha{\hbox{-}}\)adic
topology on \(M'\) coincides with the topology on \(M'\) induced by the
\(\alpha{\hbox{-}}\)adic topology on \(M\).

\end{theorem}

\begin{corollary}[?]

Taking the \(\alpha{\hbox{-}}\)adic completion for finitely-generated
modules on Noetherian rings is exact.

\end{corollary}

\begin{proof}[?]

This coincides with the induced topology, and we showed that taking the
\emph{induced} completions is exact.

\end{proof}

\begin{definition}[?]

For \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\) and
\({\operatorname{Fil}}M\) a descending filtration,

\begin{itemize}
\tightlist
\item
  \({\operatorname{Fil}}M\) is \textbf{compatible} with \(\alpha\) if
  \(\alpha {\operatorname{Fil}}_i M \subseteq {\operatorname{Fil}}_{i+1} M\).

  \begin{itemize}
  \tightlist
  \item
    Equivalently,
    \({\mathsf{gr}\,}_i M\in {\mathsf{{\mathsf{gr}\,}A}{\hbox{-}}\mathsf{Mod}}\).
  \end{itemize}
\item
  \({\operatorname{Fil}}M\) is \textbf{\(\alpha{\hbox{-}}\)good} if for
  \(i\gg 0\) this is an equality.

  \begin{itemize}
  \tightlist
  \item
    Equivalently,
    \({\mathsf{gr}\,}_i M\in {\mathsf{{\mathsf{gr}\,}A}{\hbox{-}}\mathsf{Mod}}^{{\mathrm{fg}}}\).
  \end{itemize}
\end{itemize}

\end{definition}

\begin{observation}

Recall that for Hilbert dimension, we took a good filtration and used
the eventual degree. If
\({\operatorname{Fil}}M, {\operatorname{Fil}}' M\) are two
\(\alpha{\hbox{-}}\)good filtrations,
\begin{align*}
{\operatorname{Fil}}'_{i-k} M \subseteq {\operatorname{Fil}}_i M \subseteq {\operatorname{Fil}}'_{i-k} M
.\end{align*}
As a corollary, any two \(\alpha{\hbox{-}}\)good filtrations induce the
same topology, since this yields a containment of basis elements.

\end{observation}

\begin{theorem}[Artin-Rees reprised]

If \(M, M', A, \alpha\) as in the original statement above,
\(M' \cap\alpha^i M\) is \(\alpha{\hbox{-}}\)good.

\end{theorem}

\begin{remark}

Why this implies the previous version: the induced filtration is
\(\alpha{\hbox{-}}\)good and induces the same topology as the
\(\alpha{\hbox{-}}\)adic filtration.

\end{remark}

\begin{definition}[Rees algebra/blowup algebra]

For \(\alpha\in \operatorname{Id}(A)\), define the \textbf{Rees algebra}
\begin{align*}
{\operatorname{Rees}}A \coloneqq A \oplus \alpha[1] \oplus \alpha^2[2] \oplus \cdots
.\end{align*}

\end{definition}

\begin{exercise}[?]

Show that if \(A\) is Noetherian then \({\operatorname{Rees}}A\) is
Noetherian.

\end{exercise}

\begin{solution}

Show that if \(\left\{{a_i}\right\}\) generate \(\alpha\), there is a
surjection
\(A[x_1,\cdots, x_n] \twoheadrightarrow({\operatorname{Rees}}A)[1]\)
where \(x_i\mapsto a_i\) and apply the Hilbert basis theorem.

\end{solution}

\begin{definition}[Rees modules]

Let \((M, {\operatorname{Fil}})\) be an \(\alpha{\hbox{-}}\)compatible
filtered \(A{\hbox{-}}\)module. Define
\begin{align*}
{\operatorname{Rees}}M \coloneqq M \oplus ({\operatorname{Fil}}_1 M)[1] \oplus ({\operatorname{Fil}}_2 M)[2] \oplus \cdots \qquad \in {\mathsf{{\operatorname{Rees}}A}{\hbox{-}}\mathsf{Mod}}
.\end{align*}

\end{definition}

\begin{proposition}[?]

Let \(A\in{ \mathrm{Noeth} }\mathsf{CRing}\), then TFAE:

\begin{itemize}
\tightlist
\item
  \(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) with an
  \(\alpha{\hbox{-}}\)good filtration,
\item
  \({\operatorname{Rees}}M \in {\mathsf{{\operatorname{Rees}}A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\).
\end{itemize}

\end{proposition}

\begin{proof}[?]

\(2\implies 1\): restrict generators to degree zero, then use that
finitely generated implies that generators are in bounded degree to get
\(\alpha{\hbox{-}}\)goodness.

\(1\implies 1\): find \(i_0\) such that \(\alpha M_i = M_{i+1}\) for
\(i\geq i_0\), so \({\operatorname{Rees}}M\) is generated by
\(\tau_{\leq i_0}{\operatorname{Rees}}M\). Each \(M_i\) is finitely
generated since it's a submodule of a finitely-generated module over a
Noetherian ring.

\end{proof}

\begin{proof}[of Artin-Rees reprised]

Let \({\operatorname{Rees}}M'\) be the induced filtration, then
\({\operatorname{Rees}}M\) is \(\alpha{\hbox{-}}\)good and
finitely-generated over \(A^*\) and we want to show
\({\operatorname{Rees}}M' \subseteq {\operatorname{Rees}}M\). Conclude
using that \({\operatorname{Rees}}A\) is Noetherian.

\end{proof}

\hypertarget{injections-into-completions}{%
\subsection{Injections into
completions}\label{injections-into-completions}}

\begin{corollary}[?]

If \(A\in \mathsf{Loc}{ \mathrm{Noeth} }\mathsf{CRing}\) with
\(m\in \operatorname{mSpec}A\) and
\(M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\),
\begin{align*}
M \hookrightarrow M{ {}_{ \widehat{m} } }
.\end{align*}

Note that this is not true for arbitrary modules!

\end{corollary}

\begin{proof}[?]

Let \(N\) be the kernel, use:

\begin{center}
\begin{tikzcd}
    N && M \\
    \\
    {\widehat{N}} && {\widehat{M}} \\
    \\
    {N/mN}
    \arrow["0", from=1-1, to=3-1]
    \arrow["0", from=1-1, to=3-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=3-1, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow["0"', curve={height=30pt}, dashed, from=1-1, to=5-1]
    \arrow[from=3-1, to=5-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJOIl0sWzAsMiwiXFxoYXQgTiJdLFsyLDAsIk0iXSxbMiwyLCJcXGhhdCBNIl0sWzAsNCwiTi9tTiJdLFswLDEsIjAiXSxbMCwzLCIwIl0sWzIsM10sWzEsM10sWzAsMl0sWzAsNCwiMCIsMix7ImN1cnZlIjo1LCJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMSw0XV0=}{Link
to Diagram}
\end{quote}

So \(N/mn = 0 \implies N=0\) by Nakayama.

\end{proof}

\begin{example}[of when injectivity fails]

Let \(A = k[t]\) and \(M = k[t]/\left\langle{t-1}\right\rangle\) with
\(m = \left\langle{t}\right\rangle\). Then \(M/t^i M = 0\) for all
\(i\), so \(M{ {}_{ \widehat{m} } } = 0\).

\end{example}

\begin{exercise}[?]

Show that for
\(A\in { \mathrm{Noeth} }\mathsf{CRing}, M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}, \alpha\in \operatorname{Id}(A)\),
\begin{align*}
\widehat{A} \otimes_A M  { \, \xrightarrow{\sim}\, }\widehat{M}
.\end{align*}

\end{exercise}

\begin{solution}

Hint: \((a_i)_{i\in I}\otimes m \mapsto (a_i m)_{i\in I}\). Now carry
out a diagram chase and use the 5 lemma (or snake lemma) on the
following diagram:

\begin{center}
\begin{tikzcd}
    0 & {M'} & {A^n} & M & 0 \\
    \\
    {} & {M'\otimes_A \widehat{A}} & {\widehat{A}^n} & {M\otimes\widehat{A}^n} & 0 \\
    \\
    0 & {\widehat{M'}} & {\widehat{A}^n} & {\widehat{M}} & 0
    \arrow[from=1-1, to=1-2]
    \arrow[from=1-2, to=1-3]
    \arrow[from=1-3, to=1-4]
    \arrow[from=1-4, to=1-5]
    \arrow[from=5-1, to=5-2]
    \arrow[from=5-2, to=5-3]
    \arrow[from=5-3, to=5-4]
    \arrow[from=5-4, to=5-5]
    \arrow[from=3-2, to=3-3]
    \arrow[from=3-3, to=3-4]
    \arrow["\therefore", two heads, from=3-2, to=5-2]
    \arrow[from=3-3, to=5-3]
    \arrow["\therefore", two heads, from=3-4, to=5-4]
    \arrow["{({-})\otimes_A \widehat{A}}"{description}, squiggly, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

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

\end{solution}

\begin{exercise}[?]

Show that if \(A\in { \mathrm{Noeth} }\mathsf{CRing}\) and
\(\alpha\in \operatorname{Id}(A)\) then
\(\widehat{A}\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^\flat\).

\end{exercise}

\begin{solution}

Hint: take \(0\to M_1\to M_2\to M_3\to 0\), tensor and use the
proposition, then it STS
\(0\to \widehat{M}_1\to \widehat{M}_2\to \widehat{M}_3\to 0\) is exact.
It suffices to check flatness for
\(M_i\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) by a
previous HW exercise.

\end{solution}

\hypertarget{motivations}{%
\subsection{Motivations}\label{motivations}}

\begin{theorem}[Cohen structure theorem]

If
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu \in \mathsf{Field}\)
and \(A\in { \mathrm{Noeth} }{\mathsf{Alg}_{/k} }\) is \emph{regular}
which is local and complete with respect to its maximal ideal, then
\(A { \, \xrightarrow{\sim}\, }k{\left[\left[ x_1,\cdots, x_n \right]\right] }\).

\end{theorem}

\begin{example}[?]

A non-regular example: take a complete local ring like
\(k{\left[\left[ x, y \right]\right] }/\left\langle{xy}\right\rangle\)
and localize to zoom in on \(\mathbf{0}\in {\mathbb{A}}^2_{/ {k}}\).

An important example in algebraic curves: if \(A_{/ {k}}\) for \(A\)
Dedekind and \(m\in \operatorname{mSpec}A\), the completion is
\(\widehat{A} \cong k{\left[\left[ x \right]\right] }\).

\end{example}

\begin{slogan}

Here \emph{regular} means smooth, so
\({\mathbf{T}}_m {}^{ \vee }= \dim A\).

\end{slogan}

\begin{remark}

A general pattern for studying rings:

\begin{itemize}
\tightlist
\item
  Start with a ring \(A\).
\item
  Localize to achieve smoothness.
\item
  Complete to reduce to questions about power series.
\end{itemize}

\end{remark}

\begin{remark}

What would show up in a 2nd course on commutative algebra: singularity
theory. See Grothendieck duality, Cohen--Macaulay rings.

\end{remark}

\hypertarget{thursday-april-28}{%
\section{Thursday, April 28}\label{thursday-april-28}}

\hypertarget{dimension-theory}{%
\subsection{Dimension theory}\label{dimension-theory}}

\begin{remark}

Given an arbitrary grid, can you tile it with \(2\times 1\) dominoes?
What dominoes corresponding to Young's diagrams for partitions
\(\lambda = (2, 1)\)?

A principled way of approaching such problems: consider labeling the
grid with monomials:

\begin{center}
\begin{tikzcd}
    \vdots \\
    {y^2} & {xy^2} & {x^2y^2} \\
    y & xy & {x^2y} \\
    1 & x & {x^2} & \cdots
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTEsWzAsMywiMSJdLFsxLDMsIngiXSxbMCwyLCJ5Il0sWzAsMSwieV4yIl0sWzEsMiwieHkiXSxbMiwzLCJ4XjIiXSxbMSwxLCJ4eV4yIl0sWzIsMiwieF4yeSJdLFsyLDEsInheMnleMiJdLFswLDAsIlxcdmRvdHMiXSxbMywzLCJcXGNkb3RzIl1d}{Link
to Diagram}
\end{quote}

Now labeling the \(2\times 1\) tile with \(1,x\) and the \(1\times 2\)
tiles with \(1,y\). Note that there is a polynomial
\(f(x, y) = 1 + x + y + x^2 + xy + y^2+\cdots\) associated to the grid;
if there admits a tiling then
\(f\in \left\langle{1+x, 1+y}\right\rangle \subseteq k[x,y]\). One can
then check that
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\in k[x,y]/\left\langle{1+x,1+y}\right\rangle = k\)
satisfies
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu(-1,-1) = 2\),
so
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu\neq 0\).
Note that
\(\mkern 1.5mu\overline{\mkern-1.5muf\mkern-1.5mu}\mkern 1.5mu = 0\) is
a necessary but not sufficient condition.

\end{remark}

\begin{remark}

Last few topics: toward the Cohen structure theorem. Setup:
\(A\in { \mathrm{Noeth} }\mathsf{Loc}\mathsf{Ring}\) and let
\({\mathfrak{m}}\in \operatorname{mSpec}A\), e.g.~

\begin{itemize}
\tightlist
\item
  \(A = k{\left[\left[ x_1,\cdots, x_n \right]\right] }\),
\item
  \(A = (k[x_1, \cdots, x_{n}]/I)_{{\mathfrak{p}}}\)
\item
  \(A= { {\mathbb{Z}}_{\widehat{p}} }{\left[\left[ x_1,\cdots, x_n \right]\right] }/I\).
\end{itemize}

We don't have a good dimension theory for these, since they aren't
finitely generated algebras. Some approaches:

\begin{itemize}
\tightlist
\item
  Hilbert dimension,
\item
  Krull dimension,
\item
  Generator dimension.
\end{itemize}

\end{remark}

\begin{definition}[Hilbert dimension]

Setup:

\begin{itemize}
\tightlist
\item
  Let \(M\in{\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\) and
  choose an \({\mathfrak{m}}{\hbox{-}}\)good filtration
  \({\operatorname{Fil}}_i M\) on \(M\), so
  \({\mathfrak{m}}M_i \subseteq M_{i+1}\) for \(i\gg 0\). Equivalently,
  \({\operatorname{Rees}}M\) is finitely-generated over
  \({\operatorname{Rees}}A\).
\item
  Let \(\Phi(i) = \mathop{\mathrm{len}}(M/{\operatorname{Fil}}_i M)\),
  which is eventually polynomial.
\item
  Define \(\dim_A M = \deg \Phi\).
\end{itemize}

This defines a dimension for any finitely generated module.

\end{definition}

\begin{remark}

There is a naturally good filtration:
\({\operatorname{Fil}}_i M \coloneqq{\mathfrak{m}}^i M\), then
\begin{align*}
\Phi(i) = \mathop{\mathrm{len}}(M/{\operatorname{Fil}}_i M) = \sum_{0\leq j\leq i-1} \dim_{A/{\mathfrak{m}}} {\mathsf{gr}\,}_j M
\end{align*}
where
\({\mathsf{gr}\,}_j M = {\mathfrak{m}}^j M/ {\mathfrak{m}}^{j+1} M\).

\end{remark}

\begin{lemma}[?]

\envlist

\begin{itemize}
\tightlist
\item
  Given a SES \(A\to B\to C\),
  \(\operatorname{hilbdim}B = \max(\operatorname{hilbdim}A, \operatorname{hilbdim}C)\).
\item
  For \(\Phi: M\hookrightarrow M\),
  \(\operatorname{hilbdim}M / m\phi \leq \operatorname{hilbdim}M - 1\)
\end{itemize}

\end{lemma}

\begin{example}[?]

Check
\(\operatorname{hilbdim}k{\left[\left[ x_1,\cdots, x_n \right]\right] } = n\)
using \({\mathfrak{m}}= \left\langle{x_1,\cdots, x_n}\right\rangle\);
then
\(\Phi(i) = \mathop{\mathrm{len}}k{\left[\left[ x_1, \cdots, x_n \right]\right] }/{\mathfrak{m}}^i = \sum_{0\leq j\leq i-1}{n+j\choose j}\)
by counting monomials.

\end{example}

\begin{exercise}[?]

Show
\(\operatorname{hilbdim}{ {\mathbb{Z}}_{\widehat{p}} }{\left[\left[ x_1,\cdots, x_n \right]\right] } = n+1\)
using \({\mathfrak{m}}= \left\langle{p, x_1,\cdots, x_n}\right\rangle\).

\end{exercise}

\begin{exercise}[?]

Find
\(\operatorname{krulldim}k{\left[\left[ x_1,\cdots, x_n \right]\right] }\)
by finding a maximal chain of prime ideals.

\end{exercise}

\begin{definition}[Generator dimension]

Define \(\operatorname{gendim}A\) to be the minimal \(d\) such that
there exist
\(x_1,\cdots, x_d\in {\mathfrak{m}}= \sqrt{\left\langle{x_1,\cdots, x_d}\right\rangle}\).

\end{definition}

\begin{theorem}[?]

If \(A\in{\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) then
\(\dim A = \max_{m\in \operatorname{mSpec}A} \dim A_m\).

\end{theorem}

\begin{exercise}[Dimension $\neq$ minimal number of generators]

Show that for
\(A = k{\left[\left[ x_1, x_2 \right]\right] }/\left\langle{x_2^2}\right\rangle\)
that \(\left\langle{x_1,x_2}\right\rangle\) is a minimal set of
generators for the maximal ideal but \(\operatorname{gendim}A = 1\)
since \(m =\sqrt{x_i}\).

\end{exercise}

\begin{exercise}[?]

Show if \(A\in \mathsf{Loc}{ \mathrm{Noeth} }\mathsf{CRing}\) with
\({\mathfrak{m}}\in \operatorname{mSpec}A\), then
\begin{align*}
\dim A \leq\dim_{A/{\mathfrak{m}}} {\mathfrak{m}}/{\mathfrak{m}}^2
.\end{align*}

\end{exercise}

\begin{solution}

Hint for using generator dimension: pick a basis
\(\left\{{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_i}\right\}_{1\leq i\leq n}\)
of \({\mathfrak{m}}/{\mathfrak{m}}^2\). Lift to \(\left\{{x_i}\right\}\)
and take
\({\mathfrak{m}}\coloneqq\left\langle{x_1,\cdots, x_n}\right\rangle\)
and conclude by Nakayama.

Hint for using Hilbert dimension: show
\(\operatorname{hilbdim}A = \operatorname{hilbdim}\oplus m^i/m^{i+1}\),
since the LHS is \(\deg(i\mapsto \mathop{\mathrm{len}}A/m_i)\) and the
RHS is \(\deg(i \mapsto \sum_{0\leq j\leq i-1} \dim_k m^j/m^{j+1})\).
Then show there is an inequality
\(\operatorname{hilbdim}\bigoplus_i m^i/m^{i+1} \leq \dim_k m/m^2\) by
showing there is an important multiplication map
\(\operatorname{Sym}m/m^2 \twoheadrightarrow\oplus_i m^i/m^{i+1}\).
Conclude using that the LHS is isomorphic to a power series rings
\(k{\left[\left[ \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_1,\cdots, \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu_n \right]\right] }\)
which is dimension \(n\).

\end{solution}

\begin{remark}

\(\oplus m^i/m^{i+1}\) is the tangent cone and
\(\operatorname{Sym}m/m^2\) are functions on the tangent space, and the
tangent cone is a subvariety of the tangent space.

\end{remark}

\begin{remark}

Let
\(M = k{\left[\left[ x,y \right]\right] }/\left\langle{x,y}\right\rangle\),
\(m = \left\langle{x,y}\right\rangle\), and
\(m/m^2 = \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu}\right\rangle\).
Then \(\operatorname{Spec}\operatorname{Sym}m/m^2\) is a curve (?) and
\(\oplus_i m^i/m^{i+1} = k{\left[\left[ x,y \right]\right] }/\left\langle{xy}\right\rangle\).

\end{remark}

\begin{exercise}[?]

For
\(k{\left[\left[ x,y \right]\right] }/\left\langle{y^2-x^3}\right\rangle\),
\(m/m^2 = \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu}\right\rangle\),
\(m^2/m^3 = \left\langle{\mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu^2, \mkern 1.5mu\overline{\mkern-1.5mux\mkern-1.5mu}\mkern 1.5mu\mkern 1.5mu\overline{\mkern-1.5muy\mkern-1.5mu}\mkern 1.5mu}\right\rangle\),
so
\(\oplus_i k{\left[\left[ x,y \right]\right] }/\left\langle{y^2}\right\rangle\).
This yields a thickened line along \(y=0\):

\includegraphics{figures/2022-04-28_13-36-33.png}

\end{exercise}

\begin{slogan}

The dimension of a variety is at most the dimension of its tangent
spaces.

\end{slogan}

\hypertarget{smoothness-and-regularity}{%
\subsection{Smoothness and regularity}\label{smoothness-and-regularity}}

\begin{remark}

Idea: regularity is ``almost smooth''.

\end{remark}

\begin{definition}[Regularity]

A ring \((A, m)\in { \mathrm{Noeth} }\mathsf{Loc}\mathsf{CRing}\) is
\textbf{regular} iff the natural multiplication map
\(\operatorname{Sym}m/m^2 \twoheadrightarrow\oplus_i m^i/m^{i+1}\) is an
isomorphism. For arbitrary \(A\in { \mathrm{Noeth} }\mathsf{CRing}\)
(not necessarily local), \(A\) is \emph{regular} iff \(A_m\) is regular
for all \(m\in \operatorname{mSpec}A\).

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  \(k[x_1, \cdots, x_{n}]\) is regular.
\item
  \(k{\left[\left[ x_1,\cdots, x_n \right]\right] }\) is regular.
\end{itemize}

\end{example}

\begin{exercise}[?]

Any field is regular local, but regularity is not preserved under
extensions. Take
\({\mathbb{F}}_p { \left( {t^{1\over p}} \right) } \supseteq{\mathbb{F}}_p { \left( {t} \right) }\)
and show
\begin{align*}
A \coloneqq{\mathbb{F}}_p { \left( {t^{1\over p}} \right) }  \otimes_{{\mathbb{F}}_p { \left( {t} \right) } } {\mathbb{F}}_p { \left( {t^{1\over p}} \right) }  \cong {\mathbb{F}}_p { \left( {t^{1\over p}} \right) }  { \left[ \scriptstyle {s} \right] } /\left\langle{s^p}\right\rangle
,\end{align*}
which is not reduced. Use that
\({\mathbb{F}}_p(t^{1\over p}) = {\mathbb{F}}_p(t)[x]/\left\langle{x^p-t}\right\rangle\),
so
\begin{align*}
A = {\mathbb{F}}_p(t)[x,y]/\left\langle{x^p-t, y^p-t}\right\rangle \\
= {\mathbb{F}}_p(t^{1\over p})[y] / \left\langle{y^p - (t^{1\over p})^p }\right\rangle \\
= {\mathbb{F}}_p(t^{1\over p})[y]/\left\langle{y - t^{1\over p}}\right\rangle^p
.\end{align*}

\end{exercise}

\begin{definition}[Smoothness]

If \(A\in {\mathsf{Alg}_{/k} }^{\mathrm{fg}}\) and
\(\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu = \operatorname{cl}^{\mathrm{alg}} k\),
then \(A\) is \textbf{smooth} iff
\(A\otimes_k { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\)
is regular.

\end{definition}

\begin{exercise}[?]

Show that a ring \(A\in { \mathrm{Noeth} }\mathsf{Loc}\mathsf{CRing}\)
is regular iff \(\dim A = \dim _{A/m} m/m^2\).

\end{exercise}

\begin{solution}

Hint: use that regularity implies equality implies that the map
\(\pi: \operatorname{Sym}m/m^2\to \oplus m^i/m^{i+1}\) is an
isomorphism. Assume this is not an equality, let \(f\in \ker \pi\), then
\(\dim A = \dim \oplus m^i / m ^{i+1} \leq \dim \operatorname{Sym}m/m^2/f < \dim \operatorname{Sym}m/m^2 = \dim m/m^2\),
a contradiction.

\end{solution}

\begin{exercise}[?]

Show that regular local rings are domains.

\end{exercise}

\begin{solution}

Pick nonzero zero divisors, \(a,b\) with \(ab=0\). Then
\(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu, \mkern 1.5mu\overline{\mkern-1.5mub\mkern-1.5mu}\mkern 1.5mu \in m^i/m^{i+1}\)
with
\(\mkern 1.5mu\overline{\mkern-1.5mua\mkern-1.5mu}\mkern 1.5mu \mkern 1.5mu\overline{\mkern-1.5mub\mkern-1.5mu}\mkern 1.5mu = 0\),
a contradiction.

\end{solution}

\hypertarget{problem-set-1}{%
\section{Problem Set 1}\label{problem-set-1}}

AM Ch. 1, 1, 8, 10, 13, 15, 16, and 19.

\hypertarget{problem-set-2}{%
\section{Problem Set 2}\label{problem-set-2}}

\begin{problem}[AM 2.1]

Show that
\begin{align*}
m{\mathbb{Z}}+ n {\mathbb{Z}}= 1 \implies C_m \otimes_{\mathbb{Z}}C_n = 0
,\end{align*}
and more generally
\begin{align*}
C_m\otimes_{\mathbb{Z}}C_n \cong C_d, \qquad d = \gcd(m, n)
.\end{align*}

\end{problem}

\begin{solution}

To fix notation, set
\(C_m \coloneqq\left\langle{x {~\mathrel{\Big\vert}~}x^m}\right\rangle = \left\{{1=x^0, x, x^2,\cdots, x^{m-1}}\right\}\),
written multiplicatively. The \(n\)th power map \(x\mapsto x^n\) induces
a SES
\begin{align*}
0 \to {\mathbb{Z}}\xrightarrow{({-})^n} {\mathbb{Z}}\to C_n \to 0 \quad \in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}
.\end{align*}

Apply the right-exact functor \(({-})\otimes_{\mathbb{Z}}C_m\) and use
that \({\mathbb{Z}}\otimes_{\mathbb{Z}}({-}) \simeq\operatorname{id}\)
to obtain
\begin{align*}
\cdots \to C_m \xrightarrow{({-})^n} C_m \to \operatorname{coker}(({-})^n) \cong C_n\otimes_{\mathbb{Z}}C_m \to 0
,\end{align*}
so it suffices to show surjectivity, that every element in \(C_m\) has
an \(n\)th root -- i.e that if \(y\in C_m\) then \(y=z^n\) for some
\(z\in C_m\). This immediately reduces to finding \(n\)th roots of the
generator \(x\), since if \(y=z^n \in C_m\), writing \(y=x^k\) for some
\(k\), we have
\begin{align*}
y=x^k = z^n \implies z = x^{k\over n} = (x^{1\over n})^k
,\end{align*}
and thus \(z\) can be expressed as a power of an \(n\)th root of \(x\).
That such a root can always be found follows from Bezout's identity:
since \(m, n\) are coprime, there are solutions \((a, b)\) to
\(1 = am + bn\), so
\begin{align*}
x = x^1 = x^{am + bn} = x^{am} x^{bn} = (x^b)^n
,\end{align*}
using that \(({-})^m\) annihilates every element in \(C_m\), making
\(x^b\) an \(n\)th root of \(x\).

More generally, using the same resolution and tensoring with any
\(A\in {\mathbb{Z}{\hbox{-}}\mathsf{Mod}}\) yields
\begin{align*}
\cdots \to A \xrightarrow{({-})^n} A \to \operatorname{coker}( ({-})^n ) \cong {A\over nA} \to 0
,\end{align*}
the submodule of \(n{\hbox{-}}\)divisible elements, and take \(A = C_m\)
to get
\begin{align*}
{A\over nA} = {C_m \over n C_m} \cong {{\mathbb{Z}}\over m{\mathbb{Z}}+ n {\mathbb{Z}}}\cong {{\mathbb{Z}}\over d{\mathbb{Z}}} \cong C_d
.\end{align*}

\end{solution}

\begin{remark}

Note that similarly applying
\(\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_n, {-})\)
yields
\begin{align*}
\mathop{\mathrm{Hom}}_{{\mathbb{Z}{\hbox{-}}\mathsf{Mod}}}(C_n, C_m) \cong \ker( ({-})^n ) \cong C_d
.\end{align*}

\end{remark}

\begin{problem}[AM 2.2]

Let
\(A\in \mathsf{CRing}, {\mathfrak{a}}\in \operatorname{Id}(A), M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\).
Show that
\((A / {\mathfrak{a}}) \otimes_{A} M \cong M / {\mathfrak{a}}M \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}\).

\begin{quote}
Tensor the exact sequence
\(0 \rightarrow \mathfrak{a} \rightarrow A \rightarrow A / \mathfrak{a} \rightarrow 0\)
with \(M\).
\end{quote}

\end{problem}

\begin{solution}

Applying the hint yields the following:

\begin{center}
\begin{tikzcd}
    0 && {\mathfrak{a}}&& A && {A/{\mathfrak{a}}} && 0 \\
    \\
    \cdots && {{\mathfrak{a}}\otimes_A M} && {A\otimes_A M \cong M} && {A/{\mathfrak{a}}\otimes_A M} && 0
    \arrow[from=1-1, to=1-3]
    \arrow[""{name=0, anchor=center, inner sep=0}, "\iota", hook, from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-7, to=1-9]
    \arrow[from=3-1, to=3-3]
    \arrow[""{name=1, anchor=center, inner sep=0}, "{\iota_*}", hook, from=3-3, to=3-5]
    \arrow[from=3-5, to=3-7]
    \arrow[from=3-7, to=3-9]
    \arrow["{({-})\otimes_A M}", shorten <=9pt, shorten >=9pt, Rightarrow, from=0, to=1]
\end{tikzcd}
\end{center}

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

Thus it suffices to show
\(\operatorname{im}\iota_* \cong {\mathfrak{a}}M\). This is clear since
\({\mathfrak{a}}\hookrightarrow A\) is an inclusion, and the natural map
\(A\otimes_A M { \, \xrightarrow{\sim}\, }M\) is given by
\((a, m) \mapsto am\).

\end{solution}

\begin{remark}

Note that there is a map
\begin{align*}
f: {\mathfrak{a}}\times M &\to {\mathfrak{a}}M \\
(a, m) &\mapsto am
,\end{align*}
which is clearly surjective and bilinear, lifting to map out of the
tensor product by the universal property. However, it is \emph{not}
always an isomorphism, and it being an isomorphism for all ideals is
equivalent to \(M\) being flat as an \(A{\hbox{-}}\)module. In other
words,
\begin{align*}
{\mathfrak{a}}\otimes_A M \cong {\mathfrak{a}}M \quad \forall {\mathfrak{a}}\in\operatorname{Id}(A) \iff M\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^\flat
.\end{align*}

An easy counterexample:

\begin{itemize}
\tightlist
\item
  \(A = k[{\varepsilon}]/\left\langle{{\varepsilon}^2}\right\rangle\)
\item
  \({\mathfrak{a}}= \left\langle{{\varepsilon}}\right\rangle \in \operatorname{Id}(A)\)
\item
  \(M = {\mathfrak{a}}\)
\end{itemize}

Then
\({\mathfrak{a}}{ {}^{ \scriptstyle\otimes_{A}^{2} } } \to {\mathfrak{a}}^2\)
is not injective. Note that \({\mathfrak{a}}^2=0\) in \(A\), so it STS
\({\mathfrak{a}}{ {}^{ \scriptstyle\otimes_{A}^{2} } }\neq 0\). The
claim is that
\begin{align*}
{\mathfrak{a}}{ {}^{ \scriptstyle\otimes_{A}^{2} }  } \cong {\mathfrak{a}}{ {}^{ \scriptstyle\otimes_{A/{\mathfrak{a}}}^{2} }  } \cong {\mathfrak{a}}{ {}^{ \scriptstyle\otimes_{k}^{2} }  }
,\end{align*}
which is the tensor product of two 1-dimensional \(k{\hbox{-}}\)vector
spaces, and is thus 1-dimensional over \(k\).

\end{remark}

\begin{problem}[AM 2.9]

Let
\begin{align*}
0\to A \xrightarrow{d_1}  B \xrightarrow{d_2}  C \to 0 \in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}
\end{align*}
with \(A, C\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\), and
show \(B\in {\mathsf{A}{\hbox{-}}\mathsf{Mod}}^{\mathrm{fg}}\)
(i.e.~\(B\) is finitely generated as an \(A{\hbox{-}}\)module).

\end{problem}

\begin{solution}

Let \({\mathcal{C}}, {\mathcal{A}}\) be generators for \(C\) and \(A\)
respectively, and consider
\begin{align*}
{\mathcal{B}}\coloneqq\left\{{ d_1(a) {~\mathrel{\Big\vert}~}a\in {\mathcal{A}}}\right\}\cup\left\{{\tilde c \in d_2^{-1}(c) {~\mathrel{\Big\vert}~}c\in {\mathcal{C}}}\right\}
,\end{align*}
where the \(\tilde c\) are arbitrarily chosen lifts of the generators
\(c\in {\mathcal{C}}\). Then \({\mathcal{B}}\) is a finite set, and the
claim is that it generates \(B\) as an \(A{\hbox{-}}\)module.

\end{solution}

\begin{problem}[AM 2.11]

Let \(A\) be a ring \(\neq 0\).

\begin{itemize}
\tightlist
\item
  Show that \(A^{m} \cong A^{n} \Rightarrow m=n\).
\item
  If \(\phi: A^{m} \rightarrow A^{n}\) is surjective, then
  \(m \geq n .\)
\item
  If \(\phi: A^{m} \rightarrow A^{n}\) is injective, is it always the
  case that \(m \leq n\)?
\end{itemize}

\begin{quote}
Hint: Let \(m\) be a maximal ideal of \(A\) and let
\(\phi: A^{m} \rightarrow A^{n}\) be an isomorphism. Then
\(1 \otimes \phi:(A / m) \otimes A^{m} \rightarrow(A / m) \otimes A^{n}\)
is an isomorphism between vector spaces of dimensions \(m\) and \(n\)
over the field \(k=A / m .\) Hence \(m=n .\) (Cf. Chapter 3, Exercise
15.)
\end{quote}

\end{problem}

\begin{problem}[AM 2.14]

A partially ordered set \(I\) is said to be a directed set if for each
pair \(i, j\) in \(I\) there exists \(k \in I\) such that \(i \leq k\)
and \(j \leq k\).

Let \(A\) be a ring, let \(I\) be a directed set and let
\(\left(M_{i}\right)_{i \in I}\) be a family of \(A\)-modules indexed by
\(\cdot I\). For each pair \(i, j\) in \(I\) such that \(i \leq j\), let
\(\mu_{i j}: M_{i} \rightarrow M\), be an \(A\)-homomorphism, and
suppose that the following axioms are satisfied:

\begin{enumerate}
\def\labelenumi{(\arabic{enumi})}
\tightlist
\item
  \(\mu_{l t}\) is the identity mapping of \(M_{i}\), for all
  \(i \in I\);
\item
  \(\mu_{i k}=\mu_{j k} \circ \mu_{l y}\) whenever \(i \leq j \leq k\).
\end{enumerate}

Then the modules \(M_{1}\) and homomorphisms \(\mu_{i f}\) are said to
form a direct system \(\mathbf{M}=\left(M_{t}, \mu_{t j}\right)\) over
the directed set \(I\).

We shall construct an \(A\)-module \(M\) called the direct limit of the
direct system M. Let \(C\) be the direct sum of the \(M_{1}\), and
identify each module \(M_{i}\) with its canonical image in \(C\). Let
\(D\) be the submodule of \(C\) generated by all elements of the form
\(x_{i}-\mu_{t j}\left(x_{i}\right)\) where \(i \leq j\) and
\(x_{i} \in M_{i}\). Let \(M=C / D\), let \(\mu: C \rightarrow M\) be
the projection and let \(\mu_{i}\) be the restriction of \(\mu\) to
\(M_{i}\).

The module \(M\), or more correctly the pair consisting of \(M\) and the
family of homomorphisms \(\mu_{i}: M_{1} \rightarrow M\), is called the
direct limit of the direct system \(M\), and is written
\(\colim_i M_{i}\). From the construction it is clear that
\(\mu_{t}=\mu_{j} \circ \mu_{i f}\) whenever \(i \leq j\).

\end{problem}

\begin{problem}[AM 2.16]

Show that the direct limit is characterized (up to isomorphism) by the
following property. Let \(N\) be an \(A\)-module and for each
\(i \in I\) let \(\alpha_{i}: M_{t} \rightarrow N\) be an \(A-\) module
homomorphism such that \(\alpha_{t}=\alpha_{j} \circ \mu_{t s}\)
whenever \(i \leq j\). Then there exists a unique homomorphism
\(\alpha: M \rightarrow N\) such that
\(\alpha_{1}=\alpha \circ \mu_{t}\) for all \(i \in I\).

\end{problem}

\begin{problem}[AM 2.20]

Keeping the same notation as in Exercise 14 , let \(N\) be any
\(A\)-module. Then \(\left(M_{1} \otimes N, \mu_{i j} \otimes 1\right)\)
is a direct system; let \(P=\colim_i \left(M_{i} \otimes N\right)\) be
its direct limit.

For each \(i \in I\) we have a homomorphism
\begin{align*}
\mu_{i} \otimes 1: M_{i} \otimes N \rightarrow M \otimes N
,\end{align*}
hence by Exercise 16 a homomorphism \(\psi: P \rightarrow M \otimes N\).
Show that \(\psi\) is an isomorphism, so that
\begin{align*}
\colim_i \left(M_{i} \otimes N\right) \cong\left(\colim_{i} M_{i}\right) \otimes N .
\end{align*}

\begin{quote}
Hint: For each \(i \in I\), let
\begin{align*}
g_{i}: M_{t} \times N \rightarrow M_{t} \otimes N
\end{align*}
be the canonical bilinear mapping. Passing to the limit we obtain a
mapping \(g: M \times N \rightarrow P\). Show that \(g\) is
\(A\)-bilinear and hence define a homomorphism
\(\phi: M \otimes N \rightarrow P\). Verify that \(\phi \circ \psi\) and
\(\psi \circ \phi\) are identity mappings.
\end{quote}

\end{problem}

\hypertarget{problem-set-3}{%
\section{Problem Set 3}\label{problem-set-3}}

\begin{problem}[AM 3.1]

Let \(S\) be a multiplicatively closed subset of a ring \(A\), and let
\(M\) be a finitely generated \(A\)-module. Prove that \(S^{-1} M=0\) if
and only if there exists \(s \in S\) such that \(s M=0\).

\end{problem}

\begin{problem}[AM 3.4]

Let \(f: A \rightarrow B\) be a homomorphism of rings and let \(S\) be a
multiplicatively closed subset of \(A\). Let \(T=f(S)\). Show that
\(S^{-1} B\) and \(T^{-1} B\) are isomorphic as \(S^{-1} A\)-modules.

\end{problem}

\begin{problem}[AM 3.12]

Let \(A\) be an integral domain and \(M\) an \(A\)-module. An element
\(x \in M\) is a torsion element of \(M\) if Ann \((x) \neq 0\), that is
if \(x\) is killed by some non-zero element of \(A\). Show that the
torsion elements of \(M\) form a submodule of \(M\). This submodule is
called the torsion submodule of \(M\) and is denoted by \(T(M)\). If
\(T(M)=0\), the module \(M\) is said to be torsion-free. Show that

\begin{enumerate}
\def\labelenumi{\roman{enumi})}
\item
  If \(M\) is any \(A\)-module, then \(M / T(M)\) is torsion-free.
\item
  If \(f: M \rightarrow N\) is a module homomorphism, then
  \(f(T(M)) \subseteq T(N)\).
\item
  If
  \(0 \rightarrow M^{\prime} \rightarrow M \rightarrow M^{\prime \prime}\)
  is an exact sequence, then the sequence
  \(0 \rightarrow T\left(M^{\prime}\right)\)
  \(\rightarrow T(M) \rightarrow T\left(M^{\prime \prime}\right)\) is
  exact.
\item
  If \(M\) is any \(A\)-module, then \(T(M)\) is the kernel of the
  mapping \(x \mapsto 1 \otimes x\) of \(M\) into \(K \otimes_{1} M\),
  where \(K\) is the field of fractions of \(A\).
\end{enumerate}

\begin{quote}
For iv), show that \(K\) may be regarded as the direct limit of its
submodules \(A \xi(\xi \in K)\); using Chapter 1, Exercise 15 and
Exercise 20, show that if \(1 \otimes x=0\) in \(K \otimes M\) then
\(1 \otimes x=0\) in \(A \xi \otimes M\) for some \(\xi \neq 0\). Deduce
that \(\xi^{-1} x=0\).
\end{quote}

\end{problem}

\begin{problem}[Ex. 1]

Show that \(\mathbb{Q}\) is a flat \(\mathbb{Z}\)-module which is not
free

\end{problem}

\begin{problem}[Ex. 2]

Prove that if \(B, C\) are \(A\) algebras, the tensor product algebra
\(B \otimes_{A} C\) has the following universal property: an algebra
homomorphism \(B \otimes_{A} C \rightarrow S\) is the same as a pair of
algebra homomorphisms \(B \rightarrow S, C \rightarrow S\).

\end{problem}

\hypertarget{problem-set-4}{%
\section{Problem Set 4}\label{problem-set-4}}

\begin{problem}[?]

Let \(f: M \rightarrow N\) be a map of modules over a local ring \(A\),
with \(N\) finitely generated. Show that if the induced map
\(M / \mathfrak{m} M \rightarrow N / \mathfrak{m} N\) is surjective, the
same is true for \(f\). Does a similar statement hold for injectivity?

\end{problem}

\begin{problem}[?]

Let \(M\) be a finitely-generated module over a ring \(A\), and let
\(p\) be a prime ideal of \(A\). Suppose \(M_{p}=0\). Show that there
exists a finite set \(x_{1}, \cdots, x_{n}\) of elements of
\(A \backslash \mathfrak{p}\) such that the localization of \(M\) at the
multiplicative set generated by the \(x_{i}\) is zero.

\end{problem}

\begin{problem}[?]

Let \(M\) be a finitely-generated module over a Noetherian ring \(A\)
(that means any submodule of a finitely-generated module is finitely
generated), and let \(p\) be a prime ideal of \(A\). Suppose \(M_{p}\)
is free. Show that there exists a finite set \(x_{1}, \cdots, x_{n}\) of
elements of \(A \backslash p\) such that the localization of \(M\) at
the multiplicative set generated by the \(x_{i}\) is free.

\end{problem}

\begin{problem}[?]

Give an example of a module \(M\) over a ring \(A\) such that \(M_{p}\)
is free for each prime ideal p of \(A\), but \(M\) itself is not free.
Such modules are called locally free.

\end{problem}

\begin{problem}[?]

Give an example of a flat module which is not projective.

\end{problem}

\begin{problem}[?]

Write a careful proof of the Cayley-Hamilton theorem over an arbitrary
field.

\end{problem}

\hypertarget{problem-set-5}{%
\section{Problem Set 5}\label{problem-set-5}}

HW 5 (due Feb 17): AM Chapter 2, exercises 24, 25, 26

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Let \(M\) be an \(A\)-module and let
  \(F_{\bullet}: \cdots \rightarrow F_{2} \rightarrow F_{1} \rightarrow F_{0}\)
  be a flat resolution of \(M\), i.e.~a chain complex with each
  \(F_{i}\) flat, and such that \(H_{i}\left(F_{\bullet}\right)=0\) for
  \(i>0\) and \(H_{0}\left(F_{\bullet}\right)=M\). Show that for any
  \(A\)-module
  \(N, H_{i}\left(F_{\bullet} \otimes_{A} N\right)=\operatorname{Tor}_{i}^{A}(M, N)\).
\item
  Let \(M\) be a finitely-generated flat module over a Noetherian local
  ring \(A\). Show that \(M\) is free.
\item
  Carefully check that Ext is well-defined, i.e.~independent of the
  choice of injective resolution in the definition.
\item
  Compute
  \(\operatorname{Ext}_{\mathbb{Z}}^{i}(\mathbb{Z} / p \mathbb{Z}, \mathbb{Z})\)
  for each prime \(p\).
\end{enumerate}

\hypertarget{problem-set-6}{%
\section{Problem Set 6}\label{problem-set-6}}

HW 6 (due Feb 24): AM Chapter 5, exercises 1, 2, 10, 12, 14, and Chapter
6, exercises 2 and 5

\hypertarget{problem-set-7}{%
\section{Problem Set 7}\label{problem-set-7}}

HW 7 (due Thursday, March 17, after Spring Break):

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Let \(M\) be an \(A\)-module, and \(f \in A\). Construct an
  isomorphism between \(M_{f}\) and
  \(\underset{\longrightarrow}{\rightarrow} M \stackrel{\cdot f}{\rightarrow} M \stackrel{\cdot f}{\rightarrow} M \stackrel{\cdot f}{\rightarrow} \cdots\).
\item
  Construct a module \(M\) over a ring \(A\) such that for each prime
  ideal \(\mathfrak{p}\) of \(A, M_{\mathfrak{p}}\) is finitely
  generated, but \(M\) is not finitely-generated.
\item
  Let \(M\) be a finitely-generated module over a Noetherian ring. Show
  that \(M=0\) if and only if the support of \(M\) is empty.
\item
  Suppose \(\operatorname{Spec}(A)=V_{1} \sqcup V_{2}\), where
  \(V_{1}, V_{2}\) are clopen disjoint subsets. Show that there exists a
  direct sum decomposition \(A=A_{1} \oplus A_{2}\) such that the
  natural quotient maps \(A \rightarrow A_{i}\) induce isomorphisms
  \(\operatorname{Spec}\left(A_{i}\right) \rightarrow V_{i}\) for
  \(i=1,2\).
\item
  Show that exactness of a long exact sequence is a local property.
\item
  Let \(A\) be a Noetherian local domain with residue field \(k\) and
  fraction field \(K\), and \(M\) a finitely-generated \(A\)-module.
  Show that the following are equivalent:
\item
  \(M\) is free
\item
  \(\operatorname{dim}_{k} M \otimes_{A} k=\operatorname{dim}_{K} M \otimes_{A} K\).
\item
  Let \(A\) be a Noetherian ring and \(M\) finitely-generated. Show that
  the following are equivalent:
\item
  \(M\) is locally free
\item
  \(M\) is projective
\item
  \(M\) is flat.
\item
  Show that any Artinian ring is Noetherian
\item
  Show that if \(A\) is a Noetherian ring such that
  \(\operatorname{Spec}(A)\) is Hausdorff, then \(A\) is Artinian.
\end{enumerate}

\hypertarget{problem-set-8}{%
\section{Problem Set 8}\label{problem-set-8}}

HW 8 (due March 31)

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Give an example (with proof) of a rank one locally free module over a
  Dedekind domain that is not free.
\item
  Give an example (with proof) of a Noetherian domain of Krull dimension
  one which is not a Dedekind domain.
\item
  Let \(M\) be a finitely generated module over a Dedekind domain \(A\).
  Show that \(M\) has a projective resolution of length two. Conclude
  that Tor \({ }_{i}^{A}(M,-)\) and \(\operatorname{Ext}_{A}^{i}(M,-)\)
  equal zero for \(i>1\).
\item
  Let \(M, N\) be a finitely generated modules over a Dedekind domain
  \(A\). Show that \(\operatorname{Tor}_{1}^{A}(M, N)\) is a torsion
  \(A\)-module. Can you identify its support?
\item
  Give an example of a Dedekind domain with uncountable Picard group.
\item
  Let \(k\) be a field. Show that the Picard group of \(k[t]\) is
  trivial, i.e.~any rank one locally free sheaf over \(k[t]\) is free.
\item
  Give an example of a domain with a maximal non-zero ideal
  \(\mathfrak{m}\) such that \(\mathfrak{m}^{2}=\mathfrak{m}\).
\end{enumerate}

\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



\newpage
\printbibliography[title=Bibliography]


\end{document}