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\title{
\rule{\linewidth}{1pt} \\
\textbf{
Algebraic Curves
}
\\ {\normalsize University of Georgia, Fall 2020} \\
\rule{\linewidth}{2pt}
}
\titlehead{
\begin{center}
\includegraphics[width=\linewidth,height=0.45\textheight,keepaspectratio]{figures/cover.png}
\end{center}
\begin{minipage}{.35\linewidth}
\begin{flushleft}
\vspace{2em}
{\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in Algebraic Curves taught by Pete Clark at the
University of Georgia in Fall 2020. As such, any errors or inaccuracies
are almost certainly my own. } } \\
\end{flushleft}
\end{minipage}
\hfill
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\end{minipage}
}
\begin{document}
\date{}
\author{D. Zack Garza}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
\textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2021-01-06 }
\end{flushleft}
\newpage
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\addsec{Table of Contents}
\tableofcontents
\newpage
\hypertarget{references}{%
\section{References}\label{references}}
\hypertarget{references-1}{%
\subsection{References}\label{references-1}}
\begin{itemize}
\tightlist
\item
Stichtenoth\autocite{stichtenoth_2009}.
\end{itemize}
\hypertarget{lecture-1-field-theory-preliminaries}{%
\section{Lecture 1: Field Theory
Preliminaries}\label{lecture-1-field-theory-preliminaries}}
The main theorems in this course, in order of importance:
\begin{itemize}
\tightlist
\item
The Riemann-Roch Theorem
\item
The Riemann-Hurwitz Formula
\end{itemize}
\hypertarget{finite-generation-of-fields}{%
\subsection{Finite Generation of
Fields}\label{finite-generation-of-fields}}
See Chapter 11 of Field Theory notes.
\hypertarget{notion-1}{%
\subsubsection{Notion 1}\label{notion-1}}
\begin{definition}[Finitely Generated Field Extension]
A field extension \(\ell/k\) is \emph{finitely generated} if there
exists a finite set \(x_1, \cdots, x_n \in \ell\) such that
\(\ell = k(x_1, \cdots, x_n)\) and \(\ell\) is the smallest field
extension of \(k\).
Concretely, every element of \(\ell\) is a quotient of the form
\({p(x_1, \cdots, x_n) \over q(x_1, \cdots, x_n)}\) with
\(p, q\in k[x_1, \cdots, x_n]\).
\end{definition}
There are three different notions of finite generation for fields, the
above is the weakest.
\hypertarget{notion-2}{%
\subsubsection{Notion 2}\label{notion-2}}
The second is being finitely generated as an algebra:
\begin{definition}[Finitely Generated Algebras]
For \(R\subset S\) finitely generated algebras, \(S\) is finitely
generated over \(R\) if every element of \(S\) is a polynomial in
\(x_1, \cdots, x_n\), with coefficients in \(R\),
i.e.~\(S = R[x_1, \cdots, x_n]\).
\end{definition}
Note that this implies the previous definition, since anything that is a
polynomial is also a quotient of polynomials.
\hypertarget{notion-3}{%
\subsubsection{Notion 3}\label{notion-3}}
The final notion: \(\ell/k\) is finite (finite degree) if \(\ell\) is
finitely generated as a \(k{\hbox{-}}\)module, i.e.~a finite-dimensional
\(k{\hbox{-}}\)vector space.
\begin{definition}[Rational Function Field]
A \emph{rational function field} is
\(k(t_1, \cdots, t_n) \coloneqq ff \qty{ k[t_1, \cdots, t_n]}\).
\end{definition}
Note that we can make a similar definition for infinitely many
generators by taking a direct limit (here: union), and in fact every
element will only involve finitely many generators.
\begin{exercise}
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show \(k(t) / k\) is finitely generated by notion (3) but not by (2).
\item
Show that \(k[t]/k\) is (2) but not (1).\footnote{Note \(k[t]\) is not
a field.}
\item
Show that it is not possible for a \textbf{field} extension to satisfy
(2) but not (1).\footnote{Hint: Zariski's lemma.}
\item
Show that if \(\ell/k\) is finitely generated by (3) and algebraic,
then it satisfies (1).
\end{enumerate}
\end{exercise}
\begin{theorem}[Field Theory Notes 11.19]
If \(L/K/F\) are field extensions, then \(L/F\) is finitely generated
\(\iff\) \(K/F\) and \(L/K\) are finitely generated.\footnote{See
Artin-Tate Lemma, this doesn't necessarily hold for general rings.}
\end{theorem}
\begin{definition}[Algebraically Independent]
For \(\ell/k\), a subset \(\left\{{x_i}\right\}\subset \ell\) is
\emph{algebraically independent} over \(k\) if no finite subset
satisfies a nonzero polynomial with \(k\) coefficients.
In this case, \(k[\left\{{x_i}\right\}] / k\) is \emph{purely
transcendental} as a rational function field.
\end{definition}
\begin{theorem}[Existence of transcendence bases]
For \(\ell/k\) a field extension,
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
There exists a subset \(\left\{{x_i}\right\}\subset \ell\)
algebraically independent over \(k\) such that
\(\ell/k(\left\{{x_i}\right\})\) is algebraic.
\item
If \(\left\{{y_t}\right\}\) is another set of algebraically
independent elements such that \(\ell/k(\left\{{y_t}\right\})\) is
algebraic, then
\({\left\lvert {\left\{{x_i}\right\}} \right\rvert} = {\left\lvert {\left\{{y_t}\right\}} \right\rvert}\).
\end{enumerate}
\end{theorem}
Thus every field extension is algebraic over a purely transcendental
extension. A subset as above is called a \emph{transcendence basis}, and
every 2 such bases have the same cardinality.
We have a notion of generation (similar to ``spanning''), independence,
and bases, so there are analogies to linear algebra (e.g.~every vector
space has a basis, any two have the same cardinality).\footnote{There is
a common generalization: matroids.}
The following notion will be analagous to that of dimension in linear
algebra:
\begin{definition}[Transcendence Degree]
The \emph{transcendence degree} of \(\ell/k\) is the cardinality of any
transcendence basis.
\end{definition}
\begin{theorem}[Transcendence Degree is Additive in Towers]
If \(L/K/F\) are fields then
\(\operatorname{trdeg}(L/F) = \operatorname{trdeg}(K/F) + \operatorname{trdeg}(L/K)\).
\end{theorem}
\begin{theorem}[Bounds on Transcendence Degree]
Let \(K/k\) be finitely degenerated, so \(K = k(x_1, \cdots, x_n)\).
Then \(\operatorname{trdeg}(K/k) \leq n\), with equality iff \(K/k\) is
purely transcendental.
\end{theorem}
\begin{proof}
Suppose \(K\) is monogenic, i.e.~generated by one element. Then
\(\operatorname{trdeg}(F(x)/F) = \indic{x/F\text{ is transcendental}}\).
So the degree increases when a transcendental element is added, and
doesn't change when \(x\) is algebraic.
By additivity in towers, we take
\(k \hookrightarrow k(x_1) \hookrightarrow k(x_1, x_2) \hookrightarrow\cdots \hookrightarrow k(x_1, \cdots, x)n) = K\)
to obtain a chain of length \(n\). The transcendence degree is thus the
number of indices \(i\) such that \(x_i\) is transcendental over
\(k(x_1, \cdots, x_{i-1})\).\footnote{This is similar to checking if a
vector is in the span of a collection of previous vectors.}
\end{proof}
\begin{definition}[Function fields in $d$ variables]
For \(d\in{\mathbb{Z}}^{\geq 0}\), an extension \(K/k\) is \emph{a
function field in \(d\) variables} (i.e.~of dimension \(d\)) if \(K/k\)
is finitely generated of transcendence degree \(d\).
\end{definition}
\begin{remark}
The study of such fields is birational geometry over the ground field
\(k\). \(k={\mathbb{C}}\) is of modern interest, things get more
difficult in other fields. The case of \(d=1\) is much easier: the
function field will itself be the geometric object and everything will
built from that. Our main tool will be \textbf{valuation theory}, where
valuations will correspond to points on the curve.
\end{remark}
\hypertarget{case-study-the-luxfcroth-problem.}{%
\subsection{Case Study: The Lüroth
Problem.}\label{case-study-the-luxfcroth-problem.}}
\begin{question}
For which fields \(k\) and \(d\in {\mathbb{Z}}^{\geq 0}\) is it true
that if \(k \subset \ell \subset k(t_1, \cdots, t_d)\) with
\(k(t_1 ,\cdots, t_d)/\ell\) finite then \(\ell\) is purely
transcendental?
\end{question}
\begin{answer}
It's complicated, and depends on \(d\) and \(k\). We have the following
partial results.
\end{answer}
\begin{theorem}[Lüroth]
True for \(d=1\): For any \(k\subset \ell \subset k(t)\),
\(\ell = k(x)\).
\end{theorem}
\begin{theorem}[Castelnuovo]
Also true for \(d=2, k={\mathbb{C}}\).
\end{theorem}
\begin{theorem}[Zariski]
No if \(d= 2\),
\(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\), and
\(k\) is positive characteristic. Also no if
\(d=2, k\neq \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
in characteristic zero.
\end{theorem}
\begin{theorem}[Clemens-Griffiths]
No if \(d\geq 3\) and \(k= {\mathbb{C}}\).
\end{theorem}
\begin{remark}
Note that unirational need not imply rational for varieties.
\end{remark}
\begin{exercise}
Let \(k\) be a field, \(G\) a finite group with \(G\hookrightarrow S_n\)
the Cayley embedding. Then \(S_n\) acts by permutation of variables on
\(k(t_1, \cdots, t_n)\), thus so does \(G\). Set
\(\ell \coloneqq k(t_1, \cdots, t_n)^G\) the fixed field, then by
Artin's observation in Galois theory: if you have a finite field acting
effectively by automorphisms on a field then taking the fixed field
yields a galois extension with automorphism group \(G\).
So \(\operatorname{Aut}(k(t_1, \cdots, t_n)/ \ell) = G\).A
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
Suppose \(k={\mathbb{Q}}\), and show that an affirmative answer to the
Lüroth problem implies an affirmative answer to the inverse galois
problem for \({\mathbb{Q}}\).
\end{enumerate}
\begin{quote}
Hint: works for any field for which Hilbert's Irreducibility Theorem
holds.
\end{quote}
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{1}
\item
\(\ell /{\mathbb{Q}}\) need not be a rational function field, explore
the literature on this: first example due to Swan with
\({\left\lvert {G} \right\rvert} = 47\).
\item
Can still give many positive examples using the Shepherd-Todd Theorem.
\end{enumerate}
\end{exercise}
\todo[inline]{What's a global field?}
\hypertarget{integrals-closures-and-constant-fields}{%
\subsection{Integrals Closures and Constant
Fields}\label{integrals-closures-and-constant-fields}}
\begin{definition}[Integral Closure and Field of Constants]
For \(K/k\) a field extension, set \(\kappa(K)\) to be the algebraic
closure of \(k\) in \(K\), i.e.~special case of \emph{integral closure}.
If \(K/k\) is finitely generated, then \(\kappa(K)/k\) is finite degree.
Here \(\kappa(K)\) is called the \emph{field of constants}, and \(K\) is
also a function field over \(\kappa(K)\).
\end{definition}
\begin{remark}
In practice, we don't want \(\kappa(K)\) to be a proper extension of
\(k\). If this isn't the case, we replace considering \(K/k\) by
\(K/\kappa(K)\). If \(K/k\) is finitely generated, then
\begin{center}
\begin{tikzcd}
k \arrow[rr, "\text{finite}", hook] & & \kappa(K) \arrow[rr, "\text{finitely generated}", hook] & & K
\end{tikzcd}
\end{center}
Where we use the fact that from above, \(\kappa(K)/k\) is finitely
generated and algebraic and thus finite, and by a previous theorem, if
\(K/k\) is transcendental then \(K/\kappa(K)\) is as well, and thus
finitely generated. Thus if you have a function field over \(k\), you
can replace \(k\) by \(\kappa(K)\) and regard \(K\) as a function field
over \(\kappa(K)\) instead.
\end{remark}
\hypertarget{lecture-1-discussion-and-review}{%
\section{Lecture 1: Discussion and
Review}\label{lecture-1-discussion-and-review}}
\hypertarget{valuations}{%
\subsection{Valuations}\label{valuations}}
\begin{itemize}
\tightlist
\item
Transcendence bases
\item
Lüroth problem
\end{itemize}
For \(K/k\) a one variable function field, if we want a curve \(C/k\),
what are the points? We'll use \emph{valuations}, see NT 2.1. See also
completions, residue fields. If \(R \subset K\) a field, \(R\) is a
\emph{valuation ring} of \(K\) if for all \(x\in K^{\times}\), at least
one of \(x, x^{-1} \in R\).
\begin{example}
The valuation rings of \({\mathbb{Q}}\) are
\({\mathbb{Z}}_{(p)}\coloneqq{\mathbb{Z}}[\left\{{{1\over \ell} {~\mathrel{\Big|}~}\ell\neq p}\right\}]\)
for all primes \(p\).
\end{example}
See also \emph{Krull valuations}, which take values in some totally
ordered commutative group.
\begin{exercise}
Show that a valuation ring is a local ring, i.e.~it has a unique maximal
ideal.
\end{exercise}
\begin{example}
Where does the log come from?
There is a \(p{\hbox{-}}\)adic valuation:
\begin{align*}
v: {\mathbb{Q}}&\to {\mathbb{Z}}_{(p)} \\
{a\over b} = p^n {u \over v} &\mapsto n
.\end{align*}
Then we recover
\begin{align*}
{\mathbb{Z}}_{(p)} = \left\{{x\in {\mathbb{Q}}^{\times}{~\mathrel{\Big|}~}v_p(x) \geq 0}\right\} \cup\left\{{0}\right\} \\
{\mathfrak{m}}_{(p)} = \left\{{x\in {\mathbb{Q}}^{\times}{~\mathrel{\Big|}~}v_p(x) > 0}\right\} \cup\left\{{0}\right\} \\
.\end{align*}
There is a \(p{\hbox{-}}\)adic norm
\begin{align*}
{\left\lvert {{\,\cdot\,}} \right\rvert}_p: {\mathbb{Q}}&\to {\mathbb{R}}\\
0 & \mapsto 0 \\
x &\mapsto p^{-n} = p^{-v_p(x)}
.\end{align*}
Then we get an ultrametric function, a non-archimedean function
\begin{align*}
d_p: {\mathbb{Q}}^2 \to {\mathbb{R}}\\
(x, y) &\mapsto {\left\lvert {x- y} \right\rvert}_p
.\end{align*}
We then recover
\(v_p(x) = -\log_p {\left\lvert {x} \right\rvert}_p\).\footnote{See NT 1
notes for more details on valuations.}
\end{example}
\hypertarget{places}{%
\subsection{Places}\label{places}}
For \(A\subset K\) a subring of a field, we'll be interested in the
place
\(\tilde \Sigma = \left\{{\text{Valuation rings } R_v \text{ of } K}\right\} {~\mathrel{\Big|}~}A \subset R_v \subsetneq K\).
Thus the valuation takes non-negative values on all elements of \(K\).
Can equip this with a topology (the ``Zariski'' topology, not the usual
one). This is always quasicompact, and called the \emph{Zariski-Riemann
space}. Can determine a sheaf of rings to make this a locally ringed
space.
We can define an equivalence of valuations and define the set of
\emph{places}
\begin{align*}
\Sigma(K/k) \coloneqq\left\{{\text{Nontrivial valuations } v\in K {~\mathrel{\Big|}~}v(x) \geq 0\, \forall x\in k^{\times}}\right\}
,\end{align*}
which will be the points on the curve. Here the Zariski topology will be
the cofinite topology (which is not Hausdorff). Scheme-theoretically,
this is exactly the set of closed points on the curve.
\begin{definition}[Generic Points]
A point \(p\in X\) a topological space is a \textbf{generic point} iff
its closure in \(X\) is all of \(X\).
\end{definition}
\begin{remark}
Note we will have unique models for curves, but this won't be the case
for surfaces: blowing up a point will yield a birational but
inequivalent surface.
\end{remark}
\hypertarget{divisors}{%
\subsection{Divisors}\label{divisors}}
\begin{definition}[Divisor Group]
The \emph{divisor group} of \(K\) is the free
\({\mathbb{Z}}{\hbox{-}}\)module on \(\Sigma(K/k)\)
\end{definition}
\begin{remark}
This comes with a degree map
\begin{align*}
\deg: \operatorname{Div}(K) \to {\mathbb{Z}}
\end{align*}
which need not be surjective.
\end{remark}
\begin{definition}[Principal Divisors]
Consider the map
\begin{align*}
\phi_d: K^{\times}&\to \operatorname{Div}(K) \\
f &\mapsto (f)
.\end{align*}
Then we define \(\operatorname{im}\phi_d\) as the subgroup of
\textbf{principal divisors}.
\end{definition}
\begin{definition}[Class Group]
Define the \textbf{class group} of \(K\) as
\begin{align*}
\operatorname{cl}(K) \coloneqq\left\{{\text{Divisors}}\right\} / \left\{{\text{Principal divisors}}\right\} \coloneqq\operatorname{Div}(K) / \operatorname{im}\phi_d
.\end{align*}
\end{definition}
We can define the \textbf{class group} as divisors modulo principle
divisors
\(\operatorname{cl}(K) = \operatorname{Div}(K) / \operatorname{im}(K^{\times})\)
and the Riemann-Roch space \(\mathcal{L}(D)\). The Riemann-Roch theorem
will then be a statement about \(\dim \mathcal{L}(D)\).
\hypertarget{lecture-2-field-theory-preliminaries}{%
\section{Lecture 2: Field Theory
Preliminaries}\label{lecture-2-field-theory-preliminaries}}
\hypertarget{base-extension}{%
\subsection{Base Extension}\label{base-extension}}
Given some object \(A/k\) and \(k\hookrightarrow\ell\) is a field
extension, we would like some extended object \(A/\ell\).
\begin{example}
An \emph{affine variety} \(V/k\) is given by finitely many polynomials
in \(p_i \in k[t_1, \cdots, t_n]\), and base extension comes from the
map \(k[t_1, \cdots, t_n] \hookrightarrow\ell[t_1, \cdots, t_n]\). More
algebraically, we have the affine coordinate ring over \(k\) given by
\(k[V] = k[t_1,\cdots, t_n]/\left\langle{p_i}\right\rangle\), the ring
of polynomial functions on the zero locus corresponding to this variety.
We can similarly replace \(k\) be \(\ell\) in this definition. Here we
can observe that \(\ell[V] \cong k[V] \otimes_k \ell\).
\end{example}
In general we have a map
\begin{align*}
{\,\cdot\,}\otimes_k \ell & \\
\left\{{k{\hbox{-}}\text{vector spaces}}\right\} &\to \left\{{\ell{\hbox{-}}\text{vector spaces}}\right\} \\
\left\{{k{\hbox{-}}\text{algebras}}\right\} &\to\left\{{\ell{\hbox{-}}\text{algebras}}\right\}
.\end{align*}
This will be an exact functor on the category
\(k{\hbox{-}}\text{Vect}\), i.e.~\(\ell\) is a flat module. Here
everything is free, and free \(\implies\) flat, so things work out
nicely.
What about for function fields? Since \(k\) is a \(k{\hbox{-}}\)algebra,
we can consider \(k\otimes_k \ell\), however this need not be a field.
Note that tensor products of fields come up very often, but don't seem
to be explicitly covered in classes! We will broach this subject here.
\begin{exercise}
If \(\ell/k\) is algebraic and \(\ell\otimes_k \ell\) is a domain, the
\(\ell = k\).
\end{exercise}
\begin{remark}
In other words, this is rarely a domain. A hint: start with the
monogenic case, and also reduce to the case where the extension is not
just algebraic but finite.
\end{remark}
\begin{remark}
Tensor products of field extensions are still interesting: if \(\ell/k\)
is finite, it is galois \(\iff\)
\(\ell \otimes_k \ell \cong \ell^{[\ell: k]}\). So its dimension as an
\(\ell{\hbox{-}}\)algebra is equal to the degree of \(\ell/k\), so it
splits as a product of copies of \(\ell\).
We'd like the tensor product of a field to be a field, or at least a
domain where we can take the fraction field and get a field. This hints
that we should not be tensoring algebraic extensions, but rather
transcendental ones.
\end{remark}
\begin{exercise}
For \(\ell/k\) a field extension,
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show \(k(t) \otimes_k \ell\) is a domain with fraction field
\(\ell(t)\).
\item
Show it is a field \(\iff\) \(\ell/k\) is algebraic.
\end{enumerate}
\end{exercise}
\begin{proposition}[FT 12.7, 12.8]
Let \(k_1, k_2 / k\) are field extensions, and suppose
\(k_1 \otimes_k k_2\) is a domain. Then this is a field \(\iff\) at
least one of \(k_1/k\) or \(k_2/k\) is algebraic.\footnote{Reminder: for
\(\ell/k\) and \(\alpha\in \ell\) algebraic over \(k\), then
\(k(\alpha) = k[\alpha]\).}
\end{proposition}
So we'll concentrate on when \(K \otimes_k \ell\) is a domain.
\hypertarget{when-extensions-preserve-being-a-domain}{%
\subsection{When Extensions Preserve Being a
Domain}\label{when-extensions-preserve-being-a-domain}}
\begin{question}
What's the condition on a function field \(K/k\) that guarantees this,
i.e.~when extending scalars from \(k\) to \(\ell\) still yields a
domain?
\end{question}
\begin{definition}[Base Change]
If this remains a domain, we'll take the fraction field and call it the
\textbf{base change}.
\end{definition}
\begin{exercise}
If \(K/k\) is finitely generated (i.e.~a function field) and
\(K\otimes_k \ell\) is a domain, then \(ff(K\otimes_k \ell)/\ell\) is
finitely generated.
\end{exercise}
\begin{remark}
The point here is that if taking a function field and extending scalars
still results in a domain, we'll call the result a function field as
well. Most of all, we want to base change to the algebraic closure.
We'll have issues if the constant field is not just \(k\) itself:
\end{remark}
\begin{lemma}
If
\(K\otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
is a domain, then the constant field \(\kappa(K) = k\).
\end{lemma}
\begin{proof}
Use the fact that \({\,\cdot\,}\otimes_k V\) is exact. We then get an
injection
\begin{center}
\begin{tikzcd}
\kappa(K) \otimes_k \kappa(K) \ar[rr, hookrightarrow]\ar[rd, hookrightarrow] & &
K \otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu \\
& \kappa(K) \otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\ar[ru, hookrightarrow] &
\end{tikzcd}
\end{center}
Here we use the injections
\(\kappa(K) \hookrightarrow\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
and \(\kappa(K) \hookrightarrow K\).
We now have an injection of \(k{\hbox{-}}\)algebras, and subrings of
domains are domains. So apply the first exercise: the only way this can
happen is if \(\kappa(K) = k\).
\end{proof}
\begin{exercise}[the simplest possible case]
Describe \({\mathbb{C}}(t) \otimes_{\mathbb{R}}{\mathbb{C}}\), tensored
as \({\mathbb{R}}{\hbox{-}}\)algebras.
\end{exercise}
\begin{remark}
Won't be a domain by the lemma, and will instead be some
\({\mathbb{C}}(t){\hbox{-}}\)algebra of dimension 2.
\end{remark}
\hypertarget{good-base-change-for-function-fields}{%
\subsection{Good Base Change For Function
Fields}\label{good-base-change-for-function-fields}}
In order to have a good base change for our function fields, we want to
constant extension to be trivial, i.e.~\(\kappa(K) = k\). This requires
that the ground field be algebraically closed.
In this case, you might expect that extending scalars to the algebraic
closure would yield a field again. This is true in characteristic zero,
but false in positive characteristic.
\begin{question}[a more precise one]
If \(\kappa(K) = k\), must
\(K\otimes_K \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
be a field?
\end{question}
If that's true and we're in positive characteristic, recalling the for
an algebraic extension this being a field is equivalent to it being a
domain. But if that's a domain, the tensor product of every algebraic
extension must be a domain, which is why this is an important case.
If so, then \(K\otimes_k k^{1\over p}\) is a field, where
\(k^{1\over p} \coloneqq k\qty{\left\{{x^{1\over p} {~\mathrel{\Big|}~}x\in k }\right\}}\)
is obtained by adjoining all \(p\)th roots of all elements. This is a
purely inseparable extension. The latter condition (this tensor product
being a field) is one of several equivalent conditions for a field to be
separable.\footnote{Note that the frobenius maps
\(k^{1\over p} \twoheadrightarrow k\), so this is sort of like
inverting this map.}
\begin{remark}
Recall that \(K/k\) is transcendental, and there is an extended notion
of separability for non-algebraic extensions. Another equivalent
condition is that every finitely generated subextension is separably
generated, i.e.~it admits a transcendence basis \(\left\{{x_i}\right\}\)
such that \(k\hookrightarrow k(\left\{{x_i}\right\}) \hookrightarrow F\)
where \(F/k(\left\{{x_i}\right\})\) is algebraic and separable. Such a
transcendence basis is called a \emph{separating transcendence basis}.
Since we're only looking at finitely generated extensions, we wont' have
to worry much about the difference between separable and separably
generated.
\end{remark}
\begin{question}
What's the point? There's an extra technical condition to ensure the
base change is a field: the function field being separable over the
ground field. Is this necessarily the case if \(\kappa(K) = k\)?
\end{question}
\begin{answer}
No, for fairly technical reasons. \lmargnote{\danger}
\end{answer}
\begin{example}\label{technical_example}
Set \(k = {\mathbb{F}}_p(a, b)\) a rational function field in two
variables as the ground field. Set
\begin{align*}
A \coloneqq k[x, y]/ \left\langle{ax^p + b-y^b}\right\rangle
.\end{align*}
Then \(A\) is a domain, so set \(k = ff(A)\).
\begin{claim}
\(\kappa(K) = k\), so \(k\) is algebraically closed in this extension,
but \(K/k\) is \emph{not} separable.
\end{claim}
How to show: extending scalars to \(k^{1\over p}\) does not yield a
domain.
Let
\(\alpha, \beta \in \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
such that \(\alpha^p = a, \beta^b = b\), so
\begin{align*}
ax^p + b-y^b = (\alpha x + \beta - y)^p
,\end{align*}
which implies \(K \otimes_k k^{1\over p}\) is not a domain: \(k[x, y]\)
is a UFD, so the quotient of a polynomial is a domain iff the polynomial
is irreducible. However, the \(p\)th power map is a homomorphism, and
this exhibits the image of the defining polynomial as something
non-irreducible.
\end{example}
\begin{remark}
Note that \(f(x, y) = ax^p + b - y^p\) is the curve in this situation.
The one variable function field is defined by quotienting out a function
in two variables and taking the function field. Every 1-variable
function field can be obtained in this way. Therefore this polynomial is
irreducible, but becomes reducible over the algebraic closure. So we'd
like the polynomial to be irreducible over both.
\end{remark}
\begin{remark}
This is pretty technical, but we won't have to worry if
\(k = k^{1\over p}\). Equivalently, frobenius is surjective on \(k\),
i.e.~\(k\) is a perfect field.
If \(k\) is not perfect, it can happen (famous paper of Tate) making an
inseparable base extension can decrease the genus of the curve.
\end{remark}
Recall that the perfect fields are given by:
\begin{itemize}
\tightlist
\item
Anything characteristic zero, every reducible polynomial is separable.
\item
Any algebraically closed field
\item
Finite fields (frobenius is always injective)
\end{itemize}
Imperfect fields include:
\begin{itemize}
\tightlist
\item
Function fields in characteristic \(p\)
\item
Complete discretely valued fields \(k((t))\) in characteristic \(p\)
\footnote{This is a good time to review valuations and uniformizing
elements from NTII.}
\end{itemize}
\begin{theorem}[FT 12.20: Regular Field Extensions]
For field extensions \(K/k\), TFAE
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
\(\kappa(K) = k\) and \(K/k\) is separable
\item
\(K\otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
is a domain, or equivalently a field
\item
For all field extensions \(\ell/k\), \(K\otimes_k \ell\) is a domain.
\end{enumerate}
\end{theorem}
\begin{remark}
Note that this allows making not just an algebraic base change, but a
totally arbitrary one.
\end{remark}
\begin{definition}[?]
A field extension satisfying these conditions is called
\textbf{regular}.
\end{definition}
\begin{remark}
Regular corresponds to ``nonsingular'' in this neck of the woods. The
implication \(2\implies 3\) is the interesting one. To prove it, reduces
to showing that if
\(k= \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\) and
\(R_i\) are domains that are finitely generated as
\(k{\hbox{-}}\)algebras, then \(R_1 \otimes_k R_2\) is also a domain.
This doesn't always happen,
e.g.~\({\mathbb{Q}}(\sqrt{2}) \otimes_{\mathbb{Q}}{\mathbb{Q}}(\sqrt{2})\)
is not a domain. Really need algebraically closed.
This is a result in affine algebraic geometry. An algebra that is a
domain and finitely generated over a field is an \emph{affine algebraic
variety}, more precisely it is integral. The tensor product on the
coordinate ring side corresponds to taking the product of varieties.
Thus the fact here is that a product of integral varieties remains
integral, as long as you're over an algebraically closed field. Proof
uses Hilbert's Nullstellensatz.
\end{remark}
\begin{exercise}
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that \(k(t) / k\) is regular. \footnote{I.e.
\(k(t)\otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
is a domain.}
\item
Show every purely transcendental extension is regular.
\item
Show that for a field \(k\), every extension is regular \(\iff\)
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\).
\item
Show \(K/k\) is regular \(\iff\) every finitely generated subextension
is regular.
\end{enumerate}
\end{exercise}
\hypertarget{example-of-a-non-regular-family-of-function-fields}{%
\subsection{Example of a Non-Regular Family of Function
Fields}\label{example-of-a-non-regular-family-of-function-fields}}
Choose an elliptic curve \(E/{\mathbb{Q}}(t)\) with
\(j{\hbox{-}}\)invariant \(t\). For \(N\in {\mathbb{Z}}^{+}\), define
\(\tilde K_N \coloneqq{\mathbb{Q}}(t)(E[N])\) the \(N{\hbox{-}}\)torsion
field of \(E\). Then \(\tilde K_N/{\mathbb{Q}}(t)\) is a finite galois
extension with galois group isomorphic to the image of the modular
galois representation \footnote{See Cornell-Silverman-Stevens covering
the proof of FLT, modular curves from the function field perspective.}
\begin{align*}
\rho_N: g({\mathbb{Q}}(t)) \to \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}}) \pmod N
.\end{align*}
\begin{proposition}[Some Facts]
\(\rho_N\) is surjective, and
\begin{align*}
{\operatorname{Aut}}(\tilde K_N / {\mathbb{Q}}(t)) \cong \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}})
.\end{align*}
\(\det \rho_N = \chi_N \pmod N\), the cyclotomic character, and
therefore \(\chi_N\) restricted to \(g(\tilde K_N)\) is trivial, so
\(\tilde K_N \supset {\mathbb{Q}}(\zeta_N)\). For \(N\geq 3\),
\({\mathbb{Q}}(\zeta_N) \supsetneq {\mathbb{Q}}\), so
\(\tilde K_N / {\mathbb{Q}}(t)\) is a non-regular function field.
\end{proposition}
\begin{remark}
Actually \(\tilde K_N\) depends on the choice of \(E\): difference
choices of nonisomorphic curves with the same \(j{\hbox{-}}\)invariant
differ by a quadratic twist and the \(\rho_N\) differ by a quadratic
character on \(g({\mathbb{Q}}(t))\). Importantly, this changes the
kernel, and thus the field.
\end{remark}
To fix this, we look at the \emph{reduced galois representation}, the
following composition:
\begin{align*}
\mkern 1.5mu\overline{\mkern-1.5mu\rho\mkern-1.5mu}\mkern 1.5mu_N: g({\mathbb{Q}}(t)) \to \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}}) \twoheadrightarrow\operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}})/\left\{{\pm I}\right\}
.\end{align*}
We obtain a field theory diagram
\begin{center}
\begin{tikzcd}
\ar[dd, bend right, "{\operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}})}"'] \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu_N \ar[d, bend left, "\left\{{\pm I}\right\}"] &\\
K_N \ar[d, bend left, "{\operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}})/\left\{{\pm I}\right\}}"] &\\
{\mathbb{Q}}(t)&
\end{tikzcd}
\end{center}
So if you just take the field fixed by \(\pm I\), you get \(K_N\). In
this case, the reduced galois representation depends only on the
\(j{\hbox{-}}\)invariant, and not on the model chosen. So the function
field \(K_N/{\mathbb{Q}}(t)\) is the ``canonical'' choice.
\begin{question}
Does this make \(K_N/{\mathbb{Q}}(t)\) regular?
\end{question}
\begin{answer}
No, \(\rho_N(g(K_N)) = \left\{{\pm I}\right\}\) and \(\det(\pm I) = 1\),
so we still have \(K_N \supset {\mathbb{Q}}(\zeta_N)\).
\end{answer}
In this course, we'll identify algebraic curves over \(k\) and
one-variable function fields \(K/k\). The function field \(K_N\)
corresponds to an algebraic curve \(X(N)/{\mathbb{Q}}\) that is
``nicer'' over \({\mathbb{Q}}(\zeta_N)\). In fact, see Rohrlich:
\(\kappa(K_N) = {\mathbb{Q}}(\zeta_N)\). Our curves will have points
(equal to valuations) which will have degrees. If the constant subfield
is not just \(k\), this prevents degree 1 points on the curve. By Galois
theory, for every subgroup
\(H \subseteq \operatorname{GL}(2, {\mathbb{Z}}/N{\mathbb{Z}}) / \left\{{\pm I}\right\}\),
we'll get a function field \({\mathbb{Q}}(H) \coloneqq H_N^H\).gg In
this case, \({\mathbb{Q}}(H)/{\mathbb{Q}}\) is regular \(\iff\)
\(\det(H) = ({\mathbb{Z}}/N{\mathbb{Z}})^{\times}\).
Later we'll understand the residues at points as the residue fields of
some DVRs, then the residue field will always contain the field of
constants.
\hypertarget{lecture-3-last-of-preliminaries}{%
\section{Lecture 3: Last of
Preliminaries}\label{lecture-3-last-of-preliminaries}}
Today we'll be wrapping up the last of the preliminaries. Upcoming:
one-variable function fields and their valuation rings.
\hypertarget{polynomials-defining-regular-function-fields}{%
\subsection{Polynomials Defining Regular Function
Fields}\label{polynomials-defining-regular-function-fields}}
\begin{question}
Where's the curve in all of this?
\end{question}
\begin{answer}
This will come from an equation like \(f(x, y) = 0\).
\end{answer}
\begin{exercise}
Let \(R_1, R_2\) be \(k{\hbox{-}}\)algebras that are also domains with
fraction fields \(K_i\). Show \(R_1 \otimes_k R_2\) is a domain \(\iff\)
\(K_1 \otimes_k K_2\) is a domain.\footnote{Hint: use a denominator
clearing argument.}
\end{exercise}
\hypertarget{geometric-irreducibility}{%
\subsection{Geometric Irreducibility}\label{geometric-irreducibility}}
\begin{definition}[Geometrically Irreducible Polynomial]
A polynomial of positive degree \(f\in k[t_1, \cdots, t_n]\) is
\textbf{geometrically irreducible} if
\(f\in \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu[t_1, \cdots, t_n]\)
is irreducible as a polynomial.
\end{definition}
\begin{remark}
If \(n=1\) then \(f\) is geometrically irreducible \(\iff\) \(f\) is
linear, i.e.~of degree 1. Let \(f\) be irreducible, then since
polynomial rings are UFDs then \(\left\langle{f}\right\rangle\) is a
prime ideal (irreducibles generate principal ideals) and
\(k[t_1, \cdots, t_n]/\left\langle{f}\right\rangle\) is a domain. Let
\(K_f\) be the fraction field.
\end{remark}
\begin{exercise}[an easy one]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Above for \(1\leq i \leq n\) let \(x_i\) be the image of \(t_i\) in
\(K_f\). Show that \(K_f = k(x_1, \cdots, x_n)\).
\item
Show that if \(K/k\) is generated by \(x_1, \cdots, x_n\), then it is
the fraction field of \(k[t_1, \cdots, t_n] /{\mathfrak{p}}\) for some
prime ideal \({\mathfrak{p}}\) (equivalently, a height 1 ideal).
\end{enumerate}
\end{exercise}
\begin{proposition}[?]
Suppose that \(f\) is geometrically irreducible.
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
The function field \(K/k\) is regular.
\item
For all \(\ell/k\), \(f\in \ell[t_1, \cdots, t_n]\) is irreducible.
\end{enumerate}
\end{proposition}
\begin{definition}[Absolutely Irreducible Polynomial]
In this case we say \(f\) is \textbf{absolutely irreducible} as a
synonym for geometrically irreducible.
\end{definition}
\begin{proof}
By definition of geometric irreducibility,
\(\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu[t_1, \cdots, t_n]/\left\langle{f}\right\rangle = k[t_1, \cdots, t_n]/\left\langle{f}\right\rangle \otimes_k \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
is a domain. The exercise shows that \(K_f \otimes_k k\) is a domain, so
\(K_f\) is regular. It follows that for all \(\ell/k\),
\(K_f \otimes_k \ell\) is a domain, so
\(\ell[t_1, \cdots, t_n]/\left\langle{f}\right\rangle\) is a domain.
\end{proof}
\begin{slogan}
Geometrically irreducible polynomials are good sources of regular
function fields.
\end{slogan}
\begin{exercise}
Let \(k\) be a field, \(d\in {\mathbb{Z}}^+\) such that \(4\nmid d\) and
\(p(x) \in k[x]\) be positive degree. Factor
\(p(x) = \prod_{i=1}^r (x-a_i)^{\ell_i}\) in
\(\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu[x]\).
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Suppose that for some \(i\), \(d\nmid\ell_i\). Show that
\(f(x, y) \coloneqq y^d - p(x) \in k[x, y]\) is geometrically
irreducible. Conclude that
\(K_f \coloneqq ff\qty{k[x, y] / \left\langle{y^d - p(x)}\right\rangle}\)
is a regular one-variable function field over \(k\), and thus elliptic
curves yield regular function fields.\footnote{Referred to as
\emph{hyperelliptic} or \emph{superelliptic} function fields. Hint:
use FT 9.21 or Lang's Algebra.}
\item
What happens when \(4\divides d\)?
\end{enumerate}
\end{exercise}
\begin{exercise}[Nice, Recommended]
Assume \(k\) is a field, if necessary assuming
\(\operatorname{ch}(k) \neq 2\).
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Let \(f(x, y) = x^2 - y^2 -1\) and show \(K_f\) is is rational:
\(K_f = k(z)\).
\item
Let \(f(x, y) = x^2 + y^2 - 1\). Show that \(K_f\) is again rational.
\item
Let \(k = {\mathbb{C}}\) and \(f(x, y) = x^2 + y^2 + 1\), \(K_f\) is
rational.
\item
Let \(k= {\mathbb{R}}\). For \(f(x ,y) = x^2 + y^2 + 1\), is \(K_f\)
rational?\footnote{This is an example of a non-rational genus zero
function field.}
\end{enumerate}
\end{exercise}
\begin{question}
Can we always construct regular function fields using geometrically
irreducible polynomials?
\end{question}
\begin{answer}
In several variables, no, since not every variety is birational to a
hypersurface. In one variable, yes, as the following theorem shows:
\end{answer}
\hypertarget{our-function-fields-are-geometrically-irreducible}{%
\subsection{Our Function Fields are Geometrically
Irreducible}\label{our-function-fields-are-geometrically-irreducible}}
\begin{theorem}[Regular Function Fields in One Variable are Geometrically Irreducible]
Let \(K/k\) be a one variable function field (finitely generated,
transcendence degree one). Then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
If \(K/k\) is separable, then \(K = k(x, y)\) for some \(x, y\in K\).
\item
If \(K/k\) is regular (separable + constant subfield is \(k\), so
stronger) then \(K \cong K_f\) for a geometrically irreducible
\(f\in k[x ,y]\).
\end{enumerate}
\end{theorem}
Recall separable implies there exists a separating transcendence basis.
\begin{proof}[of a]
This means there exists a primitive element \(x\in K\) such that
\(K/k(x)\) is finite and separable. By the Primitive Element Corollary
(FT 7.2), there exist a \(y\in K\) such that \(K = k(x, y)\).
\end{proof}
\begin{proof}[of b]
Omitted for now, slightly technical.
\end{proof}
\begin{remark}
Importance of last result: a regular function field on one variable
corresponds to a nice geometrically irreducible polynomial \(f\).
\end{remark}
\begin{remark}
Note that the plane curve module may not be smooth, and in fact usually
is not possible. I.e. \(k[x ,y]/\left\langle{f}\right\rangle\) is a
one-dimensional noetherian domain, which need not be integrally closed.
\end{remark}
\begin{question}
Can every one variable function field be 2-generated?
\end{question}
\begin{answer}
Yes, as long as the ground field is perfect. In positive characteristic,
the suspicion is no: there exists finite inseparable extensions
\(\ell/k\) that need arbitrarily many generators. However, what if
\(K/k\) has constant field \(k\) but is not separable? Riemann-Roch may
have something to say about this.
\end{answer}
\begin{example}
\hyperref[technical_example]{Example from earlier lecture:}
\begin{align*}
ax^p + b - y^b
\end{align*}
\end{example}
\begin{remark}
We can find examples of nice function fields by taking irreducible
polynomials in two variables. This will define a one-variable function
field. If the polynomial is geometrical reducible, this produces regular
function fields.
\end{remark}
\hypertarget{lecture-4-chapter-1-one-variable-function-fields-and-their-valuations}{%
\section{Lecture 4: Chapter 1, One Variable Function Fields and Their
Valuations}\label{lecture-4-chapter-1-one-variable-function-fields-and-their-valuations}}
Since we have the field-theoretic preliminaries out of the way, we now
start studying one-variable function fields in earnest. The main
technique that we use to extract the geometry will be the theory of
valuations. These may be familiar from NTII, but we will cover them in
more generality here.
\hypertarget{valuation-rings-and-krull-valuations}{%
\subsection{Valuation Rings and Krull
Valuations}\label{valuation-rings-and-krull-valuations}}
Recall that NTII approach to valuations:
\begin{definition}[Valuation]
A \textbf{valuation} on a field \(K\) is a map
\(v:K\to {\mathbb{R}}\cup\left\{{\infty}\right\}\) such that
\(v(K^{\times}) \subset {\mathbb{R}}\), \(v(0) = \infty\), and \(v\) is
of the form \(-\log({\left\lvert {{\,\cdot\,}} \right\rvert})\) where
\({\left\lvert {{\,\cdot\,}} \right\rvert}: K \to [0, \infty)\) is an
\emph{ultrametric norm}.\footnote{In other words,
\(e^{-v({\,\cdot\,})}\) is an ultrametric norm.} Recall that an
\emph{ultrametric norm} satisfies not only the triangle inequality but
the ultrametric triangle inequality, i.e.~\(d(x, z) \leq \max(x, z)\).
\end{definition}
We now take an algebraic approach to this definition, where we'll end up
replacing \({\mathbb{R}}\) with something more general.
\begin{definition}[Valuation Ring]
A subring \(R\) of a field \(K\) is a \textbf{valuation ring} if for all
\(x\in K^{\times}\), at least one of \(x\) or \(x^{-1}\) is in \(R\).
\end{definition}
\begin{remark}
This is a ``largeness'' property. It also implies that
\(K = \operatorname{ff}(R)\).
\end{remark}
\hypertarget{group-of-divisibility}{%
\subsection{Group of Divisibility}\label{group-of-divisibility}}
\begin{definition}[Group of Divisibility]
Given any integral domain \(R\) with fraction field \(K\), the
\textbf{group of divisibility} \(G(R)\) is defined as the
\emph{partially ordered commutative group}\footnote{This means that the
two structures are compatible.}
\begin{align*}
G(R) \coloneqq K^{\times}/ R^{\times}
.\end{align*}
We will write the group law here additively. The ordering is given by
\(x\leq y \iff y/x \in R\).
\end{definition}
\begin{remark}
Note that the way the partial order is written, it's a relation on
\(K^{\times}\), but it is not quite a partial ordering there. It is
reflexive and transitive, but need not be antireflexive: if
\(x/y, y/x\in R\) then \(x,y\) differ by an element of
\(u\in R^{\times}\) so that \(x=uy\). In particular, they need not be
equal. This gives a structure of a \emph{quasiordering}, and if you set
\(x\sim y \iff x\leq y\) and \(y\leq x\), this leads to an equivalence
relation, and modding out by it yields a partial order. Here this is
accomplished by essentially trivializing units.\\
Another way to think of \(G(R)\) is as the nonzero principal fractional
ideals of \(K\), since any two generators of an ideal will differ by a
unit.
\end{remark}
\begin{remark}
Inside this group there is a \emph{positive cone} \(G(R)^+\) of elements
that are ``nonnegative'': since we're in a commutative setting, the zero
element is equal to 1, and the positive cone is given by
\(\left\{{y\geq 0}\right\} = \left\{{y\in R}\right\}\), and is thus
given by the group \(G(R)^+ = (R, \cdot)\).\\
This is very general: if you're studying factorization in integral
domains, many properties are reflected in \(G(R)\). E.g. being a UFD
(the most important factorization property!) implies that \(G(R)\) is a
free commutative group.
\end{remark}
\begin{remark}
In general this is only a \emph{partially} ordered group and not totally
ordered. For example, take \(R = {\mathbb{Z}}\) and \(x=2, y=3\), then
neither of \(2/3, 3/2\) are in \({\mathbb{Z}}\), so \(x\not\leq y\) and
\(y\not\leq x\). On the other hand, if we do have a total order, then
either \(x\) or \(x^{-1}\) is in the ring, which are exactly valuation
subring of a field.
\end{remark}
\begin{claim}
\(R\) is a valuation ring \(\iff\) \(G(R)\) is totally ordered.
\end{claim}
\begin{remark}
Note that \({\mathbb{R}}\) is a totally ordered group.
\end{remark}
\hypertarget{generalized-valuations}{%
\subsection{Generalized Valuations}\label{generalized-valuations}}
This makes \(G(R)\) the ``target group'' of a generalized analytic
valuation. Whenever we have a valuation ring, we have a totally ordered
commutative group, and the valuation \(v: K^{\times}\to G(R)\) is a
quotient map which we can extend to \(K\) by \(v(0) \coloneqq\infty\).
This has some familiar properties:
\begin{itemize}
\tightlist
\item
(VRK1) For all \(x,y\in K^{\times}\),\footnote{This follows from the
fact that the quotient map is a group morphism. Note that the
additive notation makes this more suggestive of what an original
valuation satisfied.}
\end{itemize}
\begin{align*}
v(xy) = v(x) + v(y)
.\end{align*}
\begin{itemize}
\tightlist
\item
(VRK2) For all \(x,y \in K^{\times}\) such that \(x+y\neq 0\),
\begin{align*}
v(x+y) \geq \min(v(x), v(y))
.\end{align*}
\end{itemize}
For ultrametric norms, all triangles are isosceles: is that true for
this type of function? The answer is yes, by the following exercise:
\begin{exercise}[?]
If \(v(x) \neq v(y)\), then \(v(x+y) = \min(v(x), v(y))\).
\end{exercise}
So the properties here are formally identical to the NTII notion of
valuation, with \(({\mathbb{R}}, +, \leq)\) replaced by
\((G(R), +, \leq)\).
\begin{exercise}[?]
Conversely, if \(v: K^{\times}\to G\) is a map into a totally ordered
commutative group satisfying VRK1 and VRK2\footnote{Any such map
satisfying these two properties is a \textbf{Krull valuation}, Krull's
generalization of classical valuations.}, then
\begin{align*}
R_v \coloneqq\left\{{x\in K^{\times}{~\mathrel{\Big|}~}v(x) \geq 0}\right\} \cup\left\{{0}\right\}
\end{align*}
is a valuation ring.\footnote{Note that in a totally ordered group,
either \(v(x) \geq 0\) or \(-v(x) \geq 0\), so we get the property
that either \(x, x^{-1} \in R_v\).} We can thus extract valuation
rings in this situation.
\end{exercise}
\begin{exercise}[?]
A valuation ring is \textbf{local}, i.e.~there is a unique maximal ideal
\begin{align*}
\mathfrak{m}_v \coloneqq\left\{{x\in K^{\times}{~\mathrel{\Big|}~}v(x) > 0}\right\} \cup\left\{{0}\right\}
.\end{align*}
\end{exercise}
\begin{remark}
These two constructions are morally mutually inverse. This doesn't hold
on the nose, since there is extraneous data in the new analytic
valuation. Recall that in NTII we have a notion of equivalence of norms,
and two distinct norms that are equivalent can give rise to the same
valuation. For example, given a valuation, one can scale it by
\(\alpha \in {\mathbb{R}}\), and it's easy to check that this gives the
same valuation. It is possible for the valuation not to surject onto
\({\mathbb{R}}\), but this doesn't happen in practice. The image is
usually infinite cyclic, what we call a \emph{discrete valuation}, and
so one is led to the definition of the \emph{value group} of the
valuation as its image. If you have a notion of equivalence of Krull
valuations, you want to allow for isomorphisms of the value group. The
cleanest notion of equivalence is thus the following:
\end{remark}
\begin{definition}[Equivalence of Krull valuations]
Two Krull valuations on a field \(K\) are \textbf{equivalent} iff their
valuation rings are \emph{equal}.
\end{definition}
\begin{remark}
Going back to NTII, if you have two nonarchimedean norms on a field,
then there are many equivalent conditions for equivalence, and this is
one of them.
\end{remark}
Some general valuation theory:
\begin{itemize}
\item
Every totally ordered commutative group is a group of
divisibility.\footnote{Pete's Commutative Algebra Notes, Ch. 17.10}
\item
A totally ordered group has \textbf{rank 1} if it is nontrivial and
embeds into \({\mathbb{R}}\)
\begin{itemize}
\tightlist
\item
If the value group is trivial, \(R = K\)
\end{itemize}
\item
A Krull valuation of rank at most 1 is the NTII notion of a valuation.
\end{itemize}
\begin{exercise}[?]
For \(n\geq 2\), put the lexicographic order on \({\mathbb{Z}}^n\), and
show this has rank strictly larger than 1. Thus
\({\mathbb{Z}}^n\hookrightarrow{\mathbb{R}}\) as a commutative group,
but not as a totally ordered commutative group.
\end{exercise}
\begin{remark}
In fact, for any ordered group \(G\), one can attach a rank: a cardinal
number \(r(G)\). Here, \(r(({\mathbb{Z}}^n, \text{lex})) = n\). This is
useful when studying \(\operatorname{Spec}(R)\) for \(R\) a DVR.
\end{remark}
A valuation of rank bigger than 1 does not induce a norm on \(K\) in the
metric sense, although this is not so important. A closer notion is
expanding the notion of a metric space by allowing the metric to be
defined on \(X\) as \(d: X\times X \to R\) for some \(R\) more general
than \({\mathbb{R}}\), like a totally ordered group or a nonarchimedean
field. This would yield a class of topological spaces that are
reminiscent of metric spaces.
\hypertarget{regular-or-centered-valuations}{%
\subsection{Regular or Centered
Valuations}\label{regular-or-centered-valuations}}
\begin{definition}[Important: Regular and Centered]
Let \(v:K^{\times}\to (G, +)\) be a Krull valuation and let
\(A \subset K\) be a subring of \(K\). Then \(v\) is
\textbf{\(A{\hbox{-}}\)regular} or \textbf{centered in \(A\)} if \(A\)
is a subset of some valuation ring \(R_v\). In this case,
\(\mathfrak{p} \coloneqq\mathfrak{m}_v \cap A \in \operatorname{Spec}(A)\)
is denoted the \textbf{center of \(v\) in \(A\)}.\footnote{Here
\(\mathfrak{m}_v\) denotes pulling back the maximal ideal along this
morphism of rings.}
\end{definition}
\begin{remark}
The term regularity here arises because we'll want to think of elements
of \(A\) as functions and the valuation as a type of point, then the
notion of being a regular function at a point will carry over. The
center is the subset of \(A\) with strictly positive valuation. Also
recall that pulling back prime ideals yields prime ideals, and maximal
ideals are a special kind of prime ideal, but in general pulling back a
maximal ideal may not result in another maximal ideal. So somehow the
valuation affects every subring on which it is regular.
\end{remark}
\begin{definition}[Key: Zariski-Riemann Space]
For \(A \subset K\), define
\begin{align*}
\Sigma(K/A) &\coloneqq\left\{{\text{valuation rings } A \subset R \subsetneq K {~\mathrel{\Big|}~}K = \operatorname{ff}(R)}\right\} \\
\tilde \Sigma(K/A) &\coloneqq\left\{{\text{valuation rings } A \subset R \subseteq K {~\mathrel{\Big|}~}K = \operatorname{ff}(R)}\right\}
.\end{align*}
The set \(\tilde \Sigma(K/A)\) is the \textbf{Zariski-Riemann space}.
\end{definition}
\begin{remark}
Note that in this definition, we're taking all \(A{\hbox{-}}\)regular
valuation rings \(R\) in \(K\). If someone says \(R\) is a valuation
ring of \(K\), they likely mean that \(K = \operatorname{ff}(R)\). Note
that fields are valuation rings, so otherwise, any subfield of \(K\)
would also be a valuation ring of \(K\). Here, \(K\) itself plays the
role of a generic point. (?) The only difference in these two
definitions is that in the first, the trivial valuation ring is being
excluded.
\end{remark}
\begin{definition}[Key: Places, Points of a Curve]
If \(K/k\) is a one variable function field\footnote{Finitely generated
field extension of transcendence degree one.} , then \(\Sigma(K/k)\)
will be the \textbf{points of the associated algebraic curve} or
\textbf{places}. These can be thought of as valuation rings, or
equivalence classes of Krull valuations, where two valuations are
equivalent if they have the same valuation ring.
\end{definition}
\begin{remark}
In terms of scheme theory, these will be the closed points of our
algebraic curve. We will view elements \(f\in K\) as meromorphic
functions on \(\Sigma(K/k)\).
\end{remark}
\hypertarget{topological-considerations}{%
\subsection{Topological
Considerations}\label{topological-considerations}}
\begin{definition}[Zariski Topology]
The \textbf{Zariski topology} on \(\Sigma(K/A)\) has a sub-base
\begin{align*}
\left\{{U(f) {~\mathrel{\Big|}~}f\in K }\right\} && U(f) \coloneqq\left\{{v\in \tilde \Sigma(K/A) {~\mathrel{\Big|}~}v(f) \geq 0}\right\} = \tilde \Sigma(K/ A[f])
.\end{align*}
and we thus take the minimal topology such that all of these sets are
open. In other words, every open set is a finite intersection and/or
arbitrary unions, including empty intersections/unions. The last term is
precisely the subring generated by \(A\) and \(f\). Thus a base is
\(U(f_1, \cdots, f_n) = \tilde \Sigma(K / A[f_1, \cdots, f_n])\). The
Zariski topology on \(\Sigma(K/A)\) is defined the same way and/or via
the subspace topology, since this removes a single point.
\end{definition}
\begin{remark}
We thus get the subrings of \(K\) that contain \(A\) and are finitely
generated as \(A{\hbox{-}}\)algebras. We'll be specifically looking at
the case where \(A\) is a field and \(K\) is a one variable function
field.
\end{remark}
\begin{theorem}[Zariski]
\(\tilde \Sigma(K/A)\) is quasi-compact.
\end{theorem}
\begin{proof}[?]
See Mazamara (?) in the chapter discussing valuation rings.
\end{proof}
Note that by definition, \(v_n \not\in \Sigma(K/A)\). In
\(\tilde \Sigma(K/A)\), we have a trivial valuation \(v_n\) whose value
group is trivial and valuation ring is \(K\) itself, and \(v_n\) is a
generic point of \(\Sigma(K/A)\): its closure is the entire space. In
other words, it is in every nonempty open subset. Since we have at least
one generic point, and in general there may be many, if
\({\left\lvert {\tilde\Sigma(K/A) > 1} \right\rvert}\) then this is not
a separated (\(T_1\)) space since the point is not closed.\footnote{Note
that in French, separated may be interpreted as Hausdorff, but here we
mean points are closed or equivalently any two distinct points admit
open neighborhoods that do not meet the other point.} Another example
of such a space would be \(\operatorname{Spec}(R)\) for \(R\) a
commutative ring with positive Krull dimension, which will be Kolmogorov
(\(T_0\)) but not separated. Such a spectrum is the underlying
topological space of some affine scheme, and in general, schemes will
have these kinds of properties that are bad (but not \emph{too} bad).
In our case of interest, when \(K/k\) is finitely generated of
transcendence degree one, we'll see that this is the cofinite topology
on an infinite space: the proper closed subsets are precisely the finite
subsets, or equivalently every nonempty open subset has finite
complement. This is far from Hausdorff: the intersection of two open
subsets will still have finite complement, so any two nonempty open
subsets must intersect.\\
It's not generally true that just removing the generic point \(v_n\)
will make the space separated, but in our case, it will be. So if we
restrict to nontrivial valuation rings, then the underlying set will be
infinite and we'll get the cofinite topology. This will be the coarsest
separated topology, i.e.~if you want singletons to be closed, finite
subsets must be closed. If \(k \subset A \subset K\) where \(A\) is a
Dedekind domain with fraction field \(K\), we will see that if we
consider not the \(k{\hbox{-}}\)regular elements but the
\(A{\hbox{-}}\)regular ones, we'll get
\(\Sigma(K/A) = {\operatorname{maxSpec}}(A)\) and both Zariski
topologies are cofinite. Note that in a Dedekind domain, trading in a
prime spectrum for a max spectrum is removing a generic point, so this
matches up. The moral: the topology of \(\Sigma(K/k)\) is not doing
anything interesting and we won't need it much.
\hypertarget{scheme-theory-resolution-of-singularities}{%
\subsection{Scheme Theory, Resolution of
Singularities}\label{scheme-theory-resolution-of-singularities}}
When \(K/k\) instead has transcendence degree \emph{bigger} than 1, then
\(\tilde \Sigma(K/k)\) is much more interesting. If we were doing things
scheme-theoretically, we could try to define a structure sheaf:
attaching a sheaf whose stalks are local commutative rings to make it a
locally ringed space.\footnote{Schemes are a full subcategory of the
much larger category of locally ringed spaces.} Here, the choice of a
ring is straightforward: literally
\(\tilde \Sigma(A, A[f_1, \cdots, f_n])\). There's an exercise that
shows that although defining a sheaf on the entire space is somewhat
annoying, defining it on a basis suffices.
\begin{exercise}[?]
Endow \(\tilde \Sigma(K/k)\) with the structure of a locally ringed
space.
\end{exercise}
\begin{remark}
In dimension 1 (the case we're studying), the corresponding
Zariski-Riemann space will be the scheme associated to the complete
nonsingular model of the curve. So this valuation-theoretic approach
will take you from the function field back to a nice model of the scheme
itself. But note that in larger dimensions, there is no unique complete
nonsingular model -- for example, you can blow any one up to get another
-- so this pattern can't possibly continue to hold. In fact, it's not
clear if we even know if there's \emph{one} such model!
\end{remark}
\begin{remark}
Thus in dimension \(>1\), you get something that is decidedly not a
scheme, but is still relevant to the study of resolution of
singularities for your function field. Where do these come up? Zariski
used \(\Sigma(K/A)\) to prove resolution of singularities \footnote{Resolving
means given \(K/k\), we want to find a complete nonsingular affine
variety whose function field is \(K\).} in characteristic zero and
dimensions 2 and 3 in 1944, although dimension 2 was classical by the
Italian school. Later, Hironaka (1984) got the Fields medal for proving
resolution of singularities for all dimensions in characteristic zero
using an ingenious inductive argument that avoided Zariski-Riemann
spaces entirely. It remarkably doesn't use any new objects/tools, just
uses existing ones in a clever way. So why talk about Zariski-Riemann
spaces at all? In the last 10 years or so, work of Ternkin and Conrad
has revived and generalized them. They study \emph{relative} such
spaces.
\end{remark}
\begin{problem}[Open]
In positive characteristic, resolution of singularities is only known up
to dimension \(\leq 3\).
\end{problem}
\hypertarget{intermediate-rings}{%
\subsection{Intermediate Rings}\label{intermediate-rings}}
The following is an extremely important result from commutative algebra:
\begin{theorem}[CA 17.17]
Let \(A \subset K\) be a subring of a field, then
\begin{align*}
\cap_{v\in \tilde \Sigma(K/A)} R_v
,\end{align*}
the intersection of all valuation subrings of the field, is the integral
closure of \(A\) in \(K\).
\end{theorem}
The proof is mostly a Zorn's lemma type of argument. Note that each
\(R_v\) is generally big, contains \(A\), and
\(\operatorname{ff}(R_v) = K\). Moreover, each valuation ring is
integrally closed, although we haven't proved this yet.
\begin{corollary}[?]
For \(K/k\) function field, \(\cap_{v\in \Sigma(K/k)} R_v = \kappa(K)\),
the constant subfield of \(K\).
\end{corollary}
\begin{proof}[?]
Note that \(\kappa(K)\) is the integral (algebraic) closure of \(k\) in
\(K\). Applying the theorem directly almost works, except the theorem
involves \(\tilde \Sigma\). Can we remove the tilde? Suppose not, this
can only happen if \(\Sigma(K/k) = \emptyset\) and the intersection is
just \(K\) itself, the largest thing in the intersection. But can the
integral closure of \(k\) in \(K\) be \(K\) itself? No, since the
transcendence degree of the function field is positive. So \(K/k\) is
transcendental, while \(\kappa(K) / k\) is an algebraic extension, a
contradiction.
\end{proof}
\begin{remark}
Note that \(\Sigma(K/k)\) is nonempty: there is a nontrivial valuation
ring between \(k\) and \(K\) in great generality, and there are often
many.
\end{remark}
\begin{claim}[Key]
If \(\operatorname{trdeg}(K/k) = 1\), then every \(v\in \Sigma(K/k)\) is
discrete and thus has value group isomorphic to \({\mathbb{Z}}\).
\end{claim}
So despite the fact that we've introduced a more general notion of
higher rank valuations, in dimension 1, every single valuation is
discrete.
\begin{proof}[?]
Let \(v\in \Sigma(K/k)\) be a place, so its a valuation ring with
fraction field \(K\) that is not \(K\), then \(R_v\) is not a field. So
its maximal ideal \(\mathfrak{m}_v\) is nonzero, so choose a nonzero
element \(t\in \mathfrak{m}_v\). Then \(t \in R_v\) and \(R_v\) contains
\(k\), so \(k[t] \subset R_v\). Note that \(k[t]\) is a PID sitting
inside a valuation ring. So restrict this maximal ideal down:
\(\mathfrak{m}_v \cap k[t]\) is a prime ideal of \(k[t]\) containing
\(t\), and thus the center
\(\mathfrak{m}_v \cap k[t] = \left\langle{t}\right\rangle\). This
follows because a prime ideal in the polynomial ring \(k[t]\) which
contains \(t\) is necessarily generated by \(t\), since there's exactly
one such ideal.\\
Now restricting the valuation on \(K\) to \(k(t) \subset K\),
\(K / k(t)\) will be a finite extension (from the first lecture). We
know \(k(t) \subset K\), and we can now check that
\({ \left.{{v}} \right|_{{k(t)}} }\) is the \(t{\hbox{-}}\)adic
valuation \(v_t\). Note that \(\mathfrak{m}_v\) can not contain any
other monic irreducible polynomials, since distinct such polynomials are
coprime. Since we're in a PID, this ideal would contain any linear
combination of them and thus contain 1. So consider the map
\begin{align*}
k[t] \hookrightarrow R_b \to G(R_v) = K^{\times}/ R^{\times}
.\end{align*}
Note that the units of \(k[t]\) map trivially, using the fact that any
element in \(k[t]\) can be written as \(u \prod p_i^{a_i}\) with the
\(p_i\) monic irreducible polynomials. The unit maps to zero, along with
all of the other monic irreducibles, and thus the image is determined
entirely by the image of powers of \(t\). This whole term goes to zero
unless some \(p_i\mapsto t\), in which case it maps to some power of
\(t\).\\
So suppose \(t\mapsto \gamma \neq 0\in G(R)\), which is nonzero because
\(t\) was not a unit (since it was in the maximal ideal). Then the image
is exactly \(\gamma^{\mathbb{N}}\), the non-negative integer powers of
the image of \(t\). But if we know goes on this domain, taking
denominators shows where it goes on the fraction field (of a UFD), so
the image is the cyclic group generated by \(\gamma\), i.e.~the powers
of \(t\) are literally the only valuations we get. So the image of
\(k(t)^{\times}\) in \(G(R_v)\) is \(\gamma^{\mathbb{Z}}\), yielding a
discrete valuation. This proves that the restriction to the rational
function field \(k(t)\) is discrete, and we want to use this to deduce
that the original valuation is discrete.\\
We can now use NTII:\footnote{See NTII, Corollary 1.60: a valuation on a
field whose restriction to a finite index subfield is discrete is
itself discrete.} since \(K/k(t)\) is finite, it follows that \(v\) is
discrete iff \({ \left.{{v}} \right|_{{k(t)}} }\) is discrete, and thus
\(v\) is discrete.
\end{proof}
\hypertarget{valuations-of-every-rank}{%
\subsection{Valuations of Every Rank}\label{valuations-of-every-rank}}
So every place of \(K/k\) is a discrete valuation as long as the
transcendence degree is one, but this is far from the case for degree
\(\geq 2\)! In the following example, we'll have a rational function
field, which makes things easier. You need a theory of extending Krull
valuations, since we'll define a non-rank 1 valuation on the rational
function field. But an arbitrary finitely generated field extension of
degree \(d\) over \(k\) is a finite degree extension of the rational
function field, and valuation theory will tell you that every valuation
downstairs can be extended in full generality to a finite degree field
extension, and the rank will not change.
\begin{exercise}[?]
If \(K/k\) is finitely generated of \(\operatorname{trdeg}\geq 2\), then
\(\Sigma(K/k)\) has valuations of rank \(d\).
\end{exercise}
Note that the Zariski-Riemann space only consists of discrete
valuations, which is a characteristic property of one variable function
fields. So these higher rank valuations may look weird, but when
studying a function field of higher transcendence degree (e.g.~for an
algebraic surface), these occur.
\begin{exercise}[Constructing valuations of arbitrary rank and value group]
Let \(k\) be a field and \(K = k(t_1, \cdots, t_n)\). Set
\(G = {\mathbb{Z}}^n\) with the lex order, so
\(G^{\geq 0} = {\mathbb{N}}^n\).
\begin{itemize}
\item
Show that \(k[t_1, \cdots, t_n] = k[G^{\geq 0}]\), where the RHS is
the associated semigroup ring.
\item
Define \(v: k[G^{\geq 0}]^{\bullet}\to G^{\geq 0}\) by mapping each
polynomial the minimal index of a monomial in its support. For
example,
\begin{align*}
v(a_1 t_1^3 t_2 + a_2 t_1^2 t_2^{10}) = (2, 10)
,\end{align*}
which has support \((3, 1)\) and \((2, 10)\), and we take the min in
the lex order.
\item
Extend \(v\) to \(v: K^{\bullet}\twoheadrightarrow G\) satisfying VRK1
and VRK2. Show that
\(R_v \coloneqq v^{-1}(G^{\geq 0}) \cup\left\{{0}\right\}\) is a
valuation ring with value group \(G\), and in particular, the rank is
\(n\).
\end{itemize}
\end{exercise}
Note that doing this for \(n=1\) reduces to the \(t{\hbox{-}}\)adic
valuation, which just keeps track of the smallest power of \(t\)
appearing. Here you can extend to fraction fields by defining
\(v(x/y) = v(x) - v(y)\). The semigroup ring can't \emph{be} the
valuation ring, since polynomial rings are not local rings, so it's much
bigger. Note also that \({\mathbb{Z}}\) can be replaced with any group
\(G\), since it's never used in anything but a psychological fashion.
\begin{slogan}
There is a huge difference between \(\operatorname{trdeg}= 1\) and
\(\operatorname{trdeg}> 1\), and so we'll only be working with the
former case in this course.
\end{slogan}
\hypertarget{lecture-5-places}{%
\section{Lecture 5: Places}\label{lecture-5-places}}
\begin{definition}[Affine Domain]
An \textbf{affine domain} \(R\) over a field \(k\) is a domain that is
finitely generated as a \(k{\hbox{-}}\)algebra.
\end{definition}
\hypertarget{investigating-the-set-of-places}{%
\subsection{Investigating the Set of
Places}\label{investigating-the-set-of-places}}
We saw an interesting example of a function field in more than one
variable which showed that valuations of rank larger than 1 can arise,
but this does not happen for one variable function fields. That is, for
\(K/k\) of transcendence degree 1, all valuations on \(K\) which are
trivial on \(k\) are discrete. We'll now want to go farther and describe
the places \(\Sigma(K/k)\), which will be the set of points on an
algebraic curve. Scheme-theoretically, this will literally be the set of
closed points on a certain projective curve whose function field is
\(K\). Note that a priori, finding closed points on a curve over an
arbitrary field is hard!
Recall that if \(A\) is a Dedekind domain such that
\(\operatorname{ff}(A) = K\), then for all
\(\mathfrak{p}\in \operatorname{mSpec}(A)\) there exists a discrete
valuation \(v_p\) on \(K\). I.e., every maximal ideal induces a discrete
valuation that is \(A{\hbox{-}}\)regular, so the valuation ring will
contain \(A\). How is this obtained? Take a nonzero \(x\in K^{\times}\),
and take the corresponding principal fractional ideal
\(\left\langle{x}\right\rangle \coloneqq Ax\), which we can factor in a
Dedekind domain as
\(Ax = \prod_{\mathfrak{p} \in \operatorname{mSpec}(A)} \mathfrak{p}^{\alpha_{\mathfrak{p}}}\)
with \(\alpha_{\mathfrak{p}} \in {\mathbb{Z}}\). This looks like an
infinite product, but for any fixed \(x\), only finitely many \(\alpha\)
are nonzero. Note that these \(\alpha\) are exactly what we're looking
for: the \(\mathfrak{p}{\hbox{-}}\)adic evaluation of \(x\) is given
precisely by \(v_{\mathfrak{p}}(x) \coloneqq\alpha_{\mathfrak{p}}\),
where we are using unique factorization of ideals in Dedekind domains.
Thus we have a map
\begin{align*}
v_{{\,\cdot\,}}: \operatorname{mSpec}(A) &\to \Sigma(K/A) \\
\mathfrak{p} &\mapsto v_{\mathfrak{p}}
.\end{align*}
So this sends a maximal ideal to a place that is \(A{\hbox{-}}\)regular,
and it turns out to be a bijection.
\begin{proposition}[?]
The map \(v\) is a bijection, and thus we may write
\begin{align*}
\Sigma(K/A) \cong \operatorname{mSpec}(A)
.\end{align*}
\end{proposition}
\begin{proof}[?]
\begin{claim}
\(v\) is injective.
\end{claim}
If \(\mathfrak{p}_1, \mathfrak{p}_2 \in \operatorname{mSpec}(A)\) are
two different maximal ideals. Then there exists an element
\(x\in \mathfrak{p}_1 \setminus\mathfrak{p}_2\), and so
\(x^{-1} \in A_{\mathfrak{p}_2} \setminus A_{\mathfrak{p}_1}\). This
follows since if \(x\) is not in \(\mathfrak{p}_2\), its
\(\mathfrak{p}_2{\hbox{-}}\)adic valuation is zero, and thus the
\(\mathfrak{p}_2{\hbox{-}}\)adic valuation of \(x^{-1}\) is \(-0 = 0\)
as well. On the other hand, since \(x\in \mathfrak{p}_1\), its
\(\mathfrak{p}_1{\hbox{-}}\)adic valuation is positive and therefore
\(v_{\mathfrak{p}_1}(x^{-1}) < 0\) and \(x^{-1}\) is not in
\(A_{\mathfrak{p}_1}\).
\begin{claim}
\(v\) is surjective.
\end{claim}
Let \(v\in \Sigma(K/A)\), so \(A \subset R_v\), i.e.~take a valuation
whose valuation ring contains \(A\). Note that we're not assuming the
valuation is discrete, this can be a general Krull valuation, but we're
trying to show it's equal to a certain \(p{\hbox{-}}\)adic valuation. As
always with a subring of a valuation ring, we can pull back the maximal
ideal and consider \(\mathfrak{m}_v \cap A \in \operatorname{Spec}(A)\).
We're hoping that this is a maximal ideal, since maximals correspond to
valuations. Since we're in a Dedekind domain, the only prime ideal we
\emph{don't} want this to be is the zero ideal of \(A\), so suppose it
were. Then \(A^{\bullet}\subset R_v^{\times}\), and so
\(K^{\times}\subset R_v^{\times}\). This is because the only element of
the maximal ideal that lies in \(A\) is zero, so every nonzero element
of \(A\) is not in this maximal ideal and is thus a unit. But for any
unit, its inverse is also a unit, yielding the inclusion
\(K^{\times}\subset R_v^{\times}\). The only way this could possibly
happen is if \(R_v = K\), which yields the trivial valuation ring.
However, by definition, in \(\Sigma(K/A)\) we've excluded the trivial
valuation, so this ideal can not be zero.\\
So we can conclude that the pullback
\(\mathfrak{m}_v \cap A \in \operatorname{mSpec}(A)\), and so
\(A_{\mathfrak{p}} \subset R_v\). This is from viewing elements in
\(A_{\mathfrak{p}}\) as quotients of elements in \(A\) whose denominator
have \(\mathfrak{p}{\hbox{-}}\)adic valuation zero. Recall that we want
to show that \(R_v = A_{\mathfrak{p}}\). We know \(R_v \subset K\) is a
proper containment, and we can use the fact that a \emph{discrete}
valuation ring is maximal among all proper subrings of its fraction
field. In other words, for \(R\) a DVR, there is no ring \(R'\) such
that \(R \subset R' \subset \operatorname{ff}(R)\). How do you prove
this? This is similar to an early exercise in commutative algebra, where
we looked at all rings between \({\mathbb{Z}}\) and \({\mathbb{Q}}\),
which generalized to looking at all rings between a PID and its fraction
field, and a DVR is a local PID. So proving this statement is actually
easier.\\
This is enough to show that \(A_{\mathfrak{p}} = R_v\), and this
\(v\sim v_{\mathfrak{p}}\).
\end{proof}
\begin{remark}
What the idea? For a general one variable function field \(K/k\), we'll
produce affine Dedekind domains \(R\) with \(k \subset R \subset K\) and
\(\operatorname{ff}(R) = K\). This will give is subrings of this full
ring of places that are \(\operatorname{mSpec}\) of Dedekind domains.
How many such domains will we need for their union to be the entire set
of places? Just one won't work, since \(\Sigma(K/k)\) is like a complete
or projective object, and a projective variety of dimension 1 can't be
covered by a single affine variety. However, it turns out that you can
always cover it with 2. In fact, if you take any Dedekind domain between
\(k\) and \(\operatorname{ff}(K)\), the set of missing places (the ones
that aren't regular for any of these domains) will be a nonempty finite
set of places. So you can always cover it by finitely many, and two
suffices: as a consequence of the Riemann-Roch theorem, after removing
any nonempty finite set of places, you'll have the
\(\operatorname{mSpec}\) of a canonically associated Dedekind domain.
We'll prove this by starting with the case of \(K = k(t)\).
\end{remark}
\begin{claim}
\begin{align*}
{\left\lvert { \Sigma(k(t)/k) \setminus\operatorname{mSpec}k[t] } \right\rvert} = 1
.\end{align*}
\end{claim}
\begin{question}
Note that \(k \subset k[t] \subset k(t)\) and \(k[t]\) is a Dedekind
domain, so this fits into the above framework, and moreover we know the
maximal ideals of polynomial rings: irreducible monic polynomials.
Taking all of these misses exactly one place.
How do we describe this missing place?
\end{question}
\hypertarget{describing-the-missing-place}{%
\subsection{Describing the Missing
Place}\label{describing-the-missing-place}}
Suppose \(v \in \Sigma(k(t) / k) \setminus\Sigma(k(t) / k[t])\), so the
valuation ring of \(v\) contains \(k\) but does not contain \(k[t]\).
Then the valuation ring can not contain \(t\), and thus \(v(t) < 0\) and
\(v(1/t) = -v(t) > 0\). Since \(k[1/t]\) is a PID, so if the valuation
wasn't \(t dash\)regular, it's \(1/t{\hbox{-}}\)regular by definition.
So \(v\in \Sigma(k(t) / k[1/t])\). Note that \(k[1/t] \cong k[t]\) as
rings. How many valuations on this polynomial ring give positive
valuation to \(1/t\)? Exactly one, since this corresponds to a prime
ideal, namely \(\left\langle{1/t}\right\rangle\), so this unique
valuation is \(v = v_{1 \over t}\), the \(1/t{\hbox{-}}\)adic valuation.
That is, if we write \(f\in k(t)\) as \((1/t)^n a(1/t)/b(1/t)\) with
\(a, b\in k[t]\) polynomials with nonzero constant terms, then
\(v_{1\over t}(f) = n\). Note that this process is the same as the one
used to compute the \(t{\hbox{-}}\)adic valuation \(v_t\).
Recall that a valuation on a domain can be uniquely extended to its
fraction field by setting \(v(x/y) = v(x) - v(y)\).
\begin{exercise}[?]
Define \(v_\infty: k(t)^{\times}\to {\mathbb{Z}}\) by
\(p(t)/q(t) \mapsto \deg q - \deg p\).
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show \(v_ \infty \in \Sigma(k(t) / k[1/t])\).
\item
Show \(v_ \infty \sim v_{1\over t}\) by showing they have the same
valuation ring.
\item
Show that \(v_ \infty = v_{1\over t}\).
\end{enumerate}
\end{exercise}
Note that \(1/t\) is a uniformizer for \(v_ \infty\)
\begin{theorem}[Complete description of places]
\begin{align*}
\Sigma(k(t) / k) = \operatorname{mSpec}k[t] {\coprod}\left\{{v_ \infty}\right\}
.\end{align*}
\end{theorem}
Note that we know the maximal ideals -- the irreducible monic
polynomials -- but it takes some effort to write them down. If \(k\) is
algebraically closed, however, every such polynomial is linear of the
form \(t-\alpha\) for \(\alpha\in k\). In this case,
\(\operatorname{mSpec}k(t) \cong k\), and so
\(\sigma( \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu (t) / \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu) = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu {\coprod}\left\{{\infty}\right\} = {\mathbb{P}}^1(\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu)\).
More generally, the set of places on a rational function field will
yield the scheme-theoretic set of closed points on the projective line
over \(k\), which is more complicated if
\(k\neq \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
since not all closed points are \(k{\hbox{-}}\)rational. Another way to
say this is that if you have a valuation, there is a residue field, and
for any place on a one variable function field the residue field will be
a finite degree extension of \(k\). The degree 1 points will be the
\(k{\hbox{-}}\)rational points, and so \(\Sigma(k(t) / k)\) will always
contain a copy of \(k\) but may have closed points of larger degree,
making things slightly more complicated. This complication is handled
well in both the scheme-theoretic and this valuation-theoretic approach.
\hypertarget{finite-generation-in-towers}{%
\subsection{Finite Generation in
Towers}\label{finite-generation-in-towers}}
The next theorem is a fact from commutative algebra:
\begin{theorem}[?]
Let \(A\) be a domain with \(\operatorname{ff}(A) = K\). Suppose \(A\)
is a finitely generated \(k{\hbox{-}}\)algebra, let \(L/K\) be a finite
degree field extension, and let \(B\) be the integral closure of \(A\)
in \(L\). Then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
\(B\) is finitely generated as an \(A{\hbox{-}}\)module.\footnote{See
CA notes, ``Second Normalization Theorem'', where normalization is a
more geometric synonym for integral closure.}
\end{enumerate}
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{1}
\item
\(B\) is an integrally closed domain with \(\operatorname{ff}(B) = L\)
which is finitely generated as a \(k{\hbox{-}}\)algebra.
\item
\(\dim A = \dim B\)\footnote{Krull dimension, i.e.~the supremum of
lengths of chains of prime ideals.}
\end{enumerate}
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{3}
\tightlist
\item
If \(A\) is Dedekind, so is \(B\).
\end{enumerate}
\end{theorem}
\begin{proof}[?]
See Pete's CA notes sections 18 and 14.
\end{proof}
\begin{remark}
On why these should be true: we have a NTI square
\begin{center}
\begin{tikzcd}
{B} && {L} \\
\\
{A} && {K} \\
\\
{k} && {}
\arrow[from=3-1, to=1-1, hook]
\arrow[from=3-3, to=1-3, hook]
\arrow[from=1-1, to=1-3, hook, "\subset"]
\arrow[from=3-1, to=3-3, hook, "\subset"]
\arrow[from=5-1, to=3-1, hook]
\end{tikzcd}
\end{center}
We have a domain \(A\) with a fraction field \(K\), we take a finite
degree extension \(L/K\), and to complete the square we let \(B\) be the
integral closure of \(A\) in \(L\): the collection of elements in \(L\)
satisfying monic polynomials with coefficients in \(A\).\\
In our case, we're additionally assuming that \(A/k\) is finitely
generated as a \(k{\hbox{-}}\)algebra.
\end{remark}
\begin{remark}
\envlist
On (b): \(B\) being finitely generated as a \(k{\hbox{-}}\)algebra
follows from assuming \(A\) is, and additionally that \(B\) is finitely
generated as an \(A{\hbox{-}}\)module, and finite generation as a module
provides finite generation as an algebra. The result follows from
transitivity of finite generation of algebras.
On (c): This is just a property of integral extensions.
On (d): Use the characterization of being Noetherian, integrally closed,
and Krull dimension 1. The only thing to check is that \(B\) is
Noetherian, which follows from \(B\) being finitely generated as a
\(k{\hbox{-}}\)algebra and applying the Hilbert basis theorem.
\end{remark}
\begin{remark}
Note that we are not assuming that \(L/K\) is separable, which is an
assumption that would simplify things. By the Krull-Akuzuki theorem,
\(B\) will always be a Dedekind domain, but it need not be finitely
generated over \(A\). So the ``stem'' to \(k\) is grounding the
situation: it's not just a Dedekind domain, but rather an \emph{affine}
domain: a domain that is finitely generated over a field. Note that this
is much better than an arbitrary Dedekind domain!
\end{remark}
\hypertarget{regularity-lemma}{%
\subsection{Regularity Lemma}\label{regularity-lemma}}
\begin{proposition}[Regularity Lemma]\label{lemma:regularity}
Suppose that instead of \(K = \operatorname{ff}(A)\), we instead have
\(A \subset K\) an arbitrary subring, and \(L/K\) a finite extension.
Taking the integral closure \(B\) yields another NTI square:
\begin{center}
\begin{tikzcd}
B\ar[r, "\subset", hook] & L \\
A\ar[r, "\text{subring}", hook]\ar[u, hook] & K \ar[u, hook]
\end{tikzcd}
\end{center}
Suppose we have an upstairs valuation \(v\) on \(L\). Then it makes
sense to restrict valuations to subfields, so
\begin{align*}
v\in \Sigma(L/B) \iff { \left.{{v}} \right|_{{K}} } \in \Sigma(K/A)
.\end{align*}
So the original valuation is \(B{\hbox{-}}\)regular iff the restricted
valuation is \(A{\hbox{-}}\)regular.
\end{proposition}
\begin{proof}[?]
\(\impliedby\): Since \(A \subseteq B\), being \(B{\hbox{-}}\)regular
implies being \(A{\hbox{-}}\)regular.\\
\(\implies\): Suppose \(A \subset R_v\) and \(x\in B\), and choose
\(a_0, \cdots, a_{n-1} \in A\) such that
\begin{align*}
p(x) \coloneqq x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0 = 0
.\end{align*}
We can do this precisely because \(B\) is integral over \(A\). So we
have an integral relation for \(x\), and we want to show \(v(x) < 0\)
and derive some contradiction from the fact that \(v(a_i) \geq 0\). Note
that we aren't grounded to the base field here, so this valuation may
not be discrete and is rather some arbitrary Krull valuation.\\
If \(x\not \in R_v\), then \(v(x) < 0\), and we can thus write
\begin{align*}
v(x^n) < \min \left\{{ v(a_j x^j) {~\mathrel{\Big|}~}0\leq j \leq n-1 }\right\} \leq v(p(x))
.\end{align*}
This follows because the first term is \(nv(x)\), and so the next term
can only be \emph{less} negative since \(v(a_j) > 0\). But this is a
contradiction, since we know \(v(x^n) = v(- \sum_{j=0}^{n-1} a_j x^j)\),
and we've exhibited two elements that differ by a unit (\(u=-1\)) which
have different valuations.
\end{proof}
Next, let \(K/k\) be a one variable function, we want to give a nice
description of its places. We already described the places of a
\emph{rational} function field, and we know we can write the former
function fields as finite degree extensions of the latter. Choosing a
transcendental \(t\in K\), to \(K/k(t)\) is a finite extension,
restricting evaluations gives a map
\begin{align*}
r: \Sigma(K/k) \to \Sigma(k(t)/ k)
.\end{align*}
\begin{claim}
This is surjective with finite fibers, so it acts like a branched
covering map.
\end{claim}
This follows from NTI or NTII. The NTI method is taking an extension of
Dedekind domains, taking a prime ideal downstairs, and pushing it
forward to see how it factors upstairs. The NTII method is a field with
a valuation and an extension of the field and you try to figure out how
many ways the downstairs valuation can be extended. If the valuations
are discrete, these are the same problem.
\hypertarget{an-inequality-on-degrees}{%
\subsection{An Inequality on Degrees}\label{an-inequality-on-degrees}}
\begin{theorem}[Degree Inequality (NTII, 1.3)]
Let \(K\) be a field with \(v\) a rank one valuation with valuation ring
\(R\). Let \(L/K\) be a finite extension of degree \(n\). Then the set
of valuations on \(L\) extending \(v\) is finite and nonempty, say
\(\left\{{w_1, \cdots, w_g}\right\}\).
For \(1\leq i \leq g\), define
\begin{align*}
e_i(L/K) &\coloneqq{\left\lvert { w_j (L^{\times}) \over v(K^{\times}) } \right\rvert} && \text{ramification index} \\
f_i(L/K) &\coloneqq[R_{w_i}/\mathfrak{m}_{w_i} : R_v/ \mathfrak{m}_v] && \text{residual degree}
,\end{align*}
so \(e_i, f_i \in {\mathbb{Z}}{>0}\). Then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
We have a useful inequality:
\begin{align*}
\sum_{i=1}^g e_i(L/K) f_i(L/K) \leq [L: K] = n
.\end{align*}
\item
If \(v\) is discrete\footnote{It will be discrete in our case. Note
that this finiteness condition always holds if \(L/K\) is separable.}
and the integral closure \(S\) of \(R\) in \(L\) is finitely generated
as an \(R{\hbox{-}}\)module, then this is an equality.
\end{enumerate}
\end{theorem}
\begin{remark}
Note that a valuation can be extended in at least one way over
\emph{any} field extension, finite or not. For finite extensions,
there's a more precise statement involving completing and taking a
tensor product, then identifying number of valuations with the size of
some \(\operatorname{mSpec}\) over a finite-dimensional algebra over the
field.
NTII shows that \(e_i\) is a finite number by looking at the exponent of
the pushforward. Also note that we view \(\mathfrak{m}_{w_i}\) as an
ideal lying over \(\mathfrak{m}_v\), and there is an inclusion of
residue fields
\(R_v/\mathfrak{m}_v \hookrightarrow R_{w_i} / \mathfrak{m}_{w_i}\)
which is in fact a finite degree field extension.
\end{remark}
\begin{remark}
Part (a) already shows that \(r\) is surjective with fibers of
cardinality at most \([L: K]\), but we want equality. We claim is always
holds when \(K/k\) is a one variable function field and
\(v\in \Sigma(K/k)\). There are examples where the inequality is strict,
however. In our situation, it's not just an arbitrary extension, we have
the aforementioned affine ``grounding'' phenomenon, and all of these
DVRs are going to be localizations of affine Dedekind domains. This is
the key fact: arbitrary extensions of Dedekind domains are nowhere near
as nice as those where the bottom one is finitely generated over a
field.
\end{remark}
\begin{proof}[First step]
We have a discrete valuation \(v\) on \(K\), so let \(t\) be a
uniformizing element\footnote{An element of valuation one.} for \(v\).
Then the argument is that any such uniformizer \(t\) is transcendental
over \(k\). We'll do this by arguing \(t\not\in k\) and then that \(t\)
is not algebraic over \(k\) either.
Since we're assuming \(v\) is \(k{\hbox{-}}\)regular,
\(t\in k \implies 1/t\in k\) and so \(v(1/t) \geq 0\), since every
element in \(k\) should have nonnegative valuation. But we're supposed
to have \(v(t) = 1\) by definition of being a uniformizer, so \(t\) can
not be in \(k\).
Suppose that \(t\) is algebraic over \(k\), then \(k(t)/k\) is an
integral extension, since we're adjoining one algebraic element. By the
previous proposition we have that \(v\) is \(k(t){\hbox{-}}\)regular,
since being regular is preserved by integral extensions. But now
rerunning the argument in the previous paragraph shows that this is a
contradiction: being \(k(t){\hbox{-}}\)regular would force
\(v(1/t) \geq 0\), but we'd still need \(v(1/t) = -1\).
So \(t\) is transcendental over \(k\), and \(k[t]\) is a polynomial
ring.
\end{proof}
\begin{proof}[Second step]
Let
\begin{itemize}
\tightlist
\item
\(A\) be the integral closure of \(k[t]\) in \(K\), and
\item
\(B\) be the integral closure of \(k[t]\) in \(L\).
\end{itemize}
Instead of a NTI square, we'll have the following 3-step diagram:
\begin{center}
\begin{tikzcd}
k[t] \ar[d, hook, "\subset"] \ar[r, hook, "\subset"] & A\ar[d, hook, "\subset"] \ar[r, hook, "\subset"] & B\ar[d, hook, "\subset"] \\
k(t) \ar[r, hook, "\subset"] & K\ar[r, hook, "\subset"] & L \\
\end{tikzcd}
\end{center}
So \(A\) is a Dedekind domain with \(\operatorname{ff}(A) = K\), as is
\(B\) with \(\operatorname{ff}(B) = L\), making both \(A\) and \(B\)
finitely generated \(k[t]{\hbox{-}}\)modules. Why? This comes from the
theorem of finiteness of integral closure when the downstairs domain is
grounded to a field. Since \(k[t]\) is finitely generated as a
\(k{\hbox{-}}\)algebra, this finiteness applies, which tells us that
\(A\) finitely generated as a \(k[t]{\hbox{-}}\)module, as is \(B\). But
if \(B\) is finitely generated as a \(k[t]{\hbox{-}}\)module and
\(A\supseteq k[t]\) is an even larger ring, then \(B\) is finitely
generated as an \(A{\hbox{-}}\)module (potentially with fewer
generators).
Thus \(B\) is a finitely generated \(A{\hbox{-}}\)module, and \(v\) is
\(k[t]{\hbox{-}}\)regular since \(t\) was a uniformizing element, making
\(v\) regular on both \(k\) and \(t\) and thus \(k[t]\). Then \(v\) is
also \(A{\hbox{-}}\)regular by the proposition, and thus
\(v = v_{\mathfrak{p}}\) for some
\(\mathfrak{p}\in \operatorname{mSpec}(A)\) coming from our
classification of \(A{\hbox{-}}\)regular valuations on a Dedekind
domain.
So the valuation on \(K\) is just the \(\mathfrak{p}{\hbox{-}}\)adic
valuation on this Dedekind domain. This means there is an equality of
valuation rings \(R = A_{\mathfrak{p}}\)\footnote{This is the
localization at \(\mathfrak{p}\).}, the valuation ring of the Dedekind
domain. So we now consider \(S\), the integral closure of \(R\) in
\(L\). This is a NTI situation, but the downstairs Dedekind domain is a
DVR, so it's local downstairs. We thus have compatibility between
integral closure and localization in the form of
\(S = B_{\mathfrak{p}} = B \otimes_A A_{\mathfrak{p}}\). This comes from
taking the whole integral closure \(B\), and only looking at the primes
lying over \(\mathfrak{p}\). Base change preserves finite generation,
and we know that \(B\) was finitely generated as an
\(A{\hbox{-}}\)module, so \(S\) is finitely generated as an
\(A_{\mathfrak{p}}{\hbox{-}}\)module and equality holds.
\end{proof}
\begin{remark}
If \(A_{\mathfrak{p}}\) was a \emph{complete} DVR, as opposed to just
some localization of an affine domain, \(B\) will be a semilocal
Dedekind domain and thus a PID, and again the number of primes it has
will be the number of primes in the original Dedekind domain lying over
the fixed prime \(\mathfrak{p}\).
\end{remark}
\begin{remark}
We're not really using valuation theory here, and this could have been
phrased purely in NTI language. But even then, the degree inequality for
extensions of Dedekind domains needs finite generated of the Dedekind
domain as a module over the bottom Dedekind domain to ensure equality.
You'd need a suitably algebraic text that considers not necessarily
separable \(L/K\), and you really do want finite generation of \(B\)
over \(A\) to make this work. See Dino Lorenzini's textbook!
\end{remark}
\begin{exercise}[?]
Let \(K/k\) be a one variable function field, and show that the
cardinality of the set of points is given by
\begin{align*}
{\left\lvert {\Sigma (K/k)} \right\rvert} = {\left\lvert {\left\{{ \text{monic irreducible polynomials } p \in k[t] }\right\}} \right\rvert} = \max( {\left\lvert {k} \right\rvert}, \aleph_0 )
.\end{align*}
\end{exercise}
\begin{remark}
If you know that \(r\) is surjective with finite fibers, where the image
is infinite (which it is here), the domain should be infinite of the
same cardinality by an easy set-theoretic exercise. Note that using
Möbius inversion, over a finite field there is at least one irreducible
polynomial of every degree, and finitely many of a fixed degree. So the
cardinality is \(\aleph_0\) when \(k\) is a finite field. If we took a
one variable function field over \({\mathbb{C}}\), we would get the
cardinality of the continuum. In this case, \(\Sigma(K/k)\) really is
the set of points on some compact Riemann surface, although the Zariski
topology will be too coarse to coincide with the induced Euclidean
topology.
\end{remark}
\begin{remark}
Note that affine Dedekind domains are important for us because every
finitely generated field extension of \(k\) are precisely the fraction
fields of affine domains over \(k\), where the transcendence degree of
the function field equals the Krull dimension of the affine domain.
We're especially interested in affine domains of dimension 1 over \(k\).
We established something particularly important in this proof:
\end{remark}
\hypertarget{affine-grounding-and-residue-fields}{%
\subsection{Affine Grounding and Residue
Fields}\label{affine-grounding-and-residue-fields}}
\begin{lemma}[Affine Grounding]
Let \(K/k\) be a one variable function field and \(v\in \Sigma(K/k)\) be
a place on that function field. Then there exists an affine Dedekind
domain \(A\) with \(\operatorname{ff}(A) = K\) and a maximal ideal
\(\mathfrak{p}\in \operatorname{mSpec}(A)\) such that
\(R_v = A_{\mathfrak{p}}\).
\end{lemma}
Thus we should think of the set of places as the
\(\operatorname{mSpec}\) of finitely many affine Dedekind domains glued
together. For each point (place), the basic open set around that point
is the affine Dedekind domain.
\begin{corollary}[?]
For \(v \in \Sigma(K/k)\), define the \textbf{residue field} of the
local ring \(R_v\) as \(k(v) \coloneqq R_v / \mathfrak{m}_v\). Then
\(k(v)/ k\) is a finite degree extension.
\end{corollary}
\begin{proof}[of corollary]
If \(R\) is a domain with maximal ideal \(\mathfrak{p}\), then the
quotient map factors through the localization, giving
\(R/\mathfrak{p} = R_{\mathfrak{p}} / \mathfrak{p} R_{\mathfrak{p}}\):\footnote{This
is a truly standard fact from commutative algebra.}
\begin{center}
\begin{tikzcd}
R \ar[r]\ar[dr] & R_{\mathfrak{p}}\ar[d] \\
& R/\mathfrak{p}
\end{tikzcd}
\end{center}
So by affine grounding, \(k(v)\) is also \(A/\mathfrak{p}\) where \(A\)
is an affine Dedekind domain and
\(\mathfrak{p}\in \operatorname{mSpec}(A)\). This is Zariski's
lemma\footnote{A field extension that is finitely generated as an
algebra is necessarily a finite degree extension.} : we showed that
\(k(v) \cong A/\mathfrak{p}\), where \(A\) is a finitely generated
algebra and thus so are its quotients. Thus \(k(v)\) is not just
finitely generated as a field extension, but also as a
\(k{\hbox{-}}\)algebra, making \(k(v)/k\) a finite extension.
\end{proof}
\begin{definition}[Degree of a Place]
The \textbf{degree} of \(v\in \Sigma(K/k)\) is
\([k(v) : k] \in {\mathbb{Z}}^{\geq 0}\).
\end{definition}
We are especially interested in degree 1 places, i.e.~those for which
the residue field is equal to \(k\) itself, so we denote these by
\(\Sigma_1(K/k)\). In any other course, we'd call this \(C(k)\), the
rational points on the associated curve.
\begin{exercise}[Some motivation]
Let \(f\in k[x, y]\) be irreducible, so that
\(A \coloneqq k[x, y] / \left\langle{f}\right\rangle\) is a
1-dimensional affine domain\footnote{This may not necessarily be a
Dedekind domain, since it need not be integrally closed.}. As above,
the resude fields of maximal ideals are finite extensions of \(k\). Show
that there is a correspondence
\begin{align*}
\left\{{\substack{\text{ Maximal ideals } \\ \mathfrak{p} \in \operatorname{mSpec}(A)}}\right\}
&\iff
\left\{{\substack{(x, y) \in k\times k {~\mathrel{\Big|}~}f(x, y) = 0}}\right\}
.\end{align*}
\end{exercise}
\begin{remark}
Note that the polynomial above may not define a smooth geometry, there
may instead be singular points:
\begin{figure}
\centering
\includegraphics{figures/image_2020-11-27-21-49-40.png}
\caption{Image}
\end{figure}
These singular points are what stops \(A\) from being integrally closed,
which is literally true when \(k\) is a perfect field.
\end{remark}
Whereas \(\Sigma(K/k)\) is always infinite, \(\Sigma_1(K/k)\) may be
finite or even empty. When \(k = {\mathbb{Q}}\), it may in fact be empty
``most of the time'' When \(k = {\mathbb{Q}}\), it may in fact be empty
``most of the time''.
\begin{exercise}[?]
For all \(v\in \Sigma(K/k)\), the degree of the point \(\deg(v)\) will
be divisible by \([\kappa(K) : k]\). Thus if \(\kappa(L) \supsetneq k\),
then \(\Sigma_1(K/k) = \emptyset\).\footnote{Use the fact that the
degree will be bigger than 1 when the constant field is bigger than
\(k\).}
\end{exercise}
Note that before we were writing the residue field as an extension of
\(k\), and it's worth checking that the constant subfield embeds as a
subfield of the residue field as well.
\begin{remark}
There is a tie to CM points on modular curves: if you have a function
field over \({\mathbb{Q}}\) which is not regular due to some proper
algebraic subextension, the residue fields of all of the points on the
curve will contain the algebraic closure of the field of definition.
Pete had some \({\mathbb{Q}}(X_n)\) function field, whose constant
subfield was \({\mathbb{Q}}(\zeta_n)\) (adjoining the \(n\)th roots of
unity), and none of these modular curves over \({\mathbb{Q}}\) have
closed points except when the residue fields contain
\({\mathbb{Q}}(\zeta_n)\).
\end{remark}
\begin{remark}
This is a way for there to not be points on the curve, so
\(\Sigma_1(K/k)\) is empty, but it's not the deepest reason -- this is a
cheap trick to produce ``pointless'' function fields. It can fail to
have degree 1 places in many different ways!
\end{remark}
\begin{exercise}[?]
Show that for for a one variable function field \(K/k\) TFAE:
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
Every \(v\in \Sigma(K/k)\) has degree 1,
\item
\(k\) is algebraically closed.
\end{enumerate}
\end{exercise}
\begin{remark}
One half is easy, since by definition the degree of the residue field is
the degree of some finite extension of the base field, but if \(k\) is
algebraically closed, the degree of any finite extension is one.
\end{remark}
\begin{exercise}[?]
For a field \(k\), set \({\mathbb{P}}^1(k) \coloneqq{\mathbb{P}}(k^2)\),
the projectivization of \(k\times k\), i.e.~the lines through the origin
in \({\mathbb{A}}^2/k\). By taking slopes of lines,
\({\mathbb{P}}^1(k) = k {\coprod}\left\{{\infty}\right\}\).
\begin{figure}
\centering
\includegraphics[width=3.64583in,height=\textheight]{figures/image_2020-11-27-23-57-12.png}
\caption{Image}
\end{figure}
Show that \(\Sigma_1(k(t)/k) = {\mathbb{P}}^1(k)\), and deduce that
\begin{align*}
\Sigma(k(t)/k) = {\mathbb{P}}^1(k) \iff k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu
.\end{align*}
\end{exercise}
Next up we'll talk about how the set of places is built from affine
Dedekind domains. After this, we'll be ready for chapter 2: divisors and
Riemann-Roch.
\hypertarget{lecture-6-affine-domains-and-places-sigmakk}{%
\section{\texorpdfstring{Lecture 6: Affine Domains and Places
\(\Sigma(K/k)\)}{Lecture 6: Affine Domains and Places \textbackslash Sigma(K/k)}}\label{lecture-6-affine-domains-and-places-sigmakk}}
The aim of this lecture is to explain the difference (including some
technicalities) between \(\Sigma(K/k)\) and affine Dedekind domains
\(R\) such that \(K = \operatorname{ff}(R)\).
Recall that
\begin{itemize}
\item
An \textbf{affine domain} over a field \(k\) is a domain that is
finitely generated as a \(k{\hbox{-}}\)algebra,\footnote{These are
very rich but easier to understand: take a polynomial ring in
finitely many variables and mod out by a prime ideal.}
\item
An \textbf{affine Dedekind domain} is an affine domain that is also a
Dedekind domain, so it is integrally closed and of Krull dimension 1,
\item
An \textbf{affine \(k{\hbox{-}}\)order} is a one-dimensional affine
domain. \footnote{These will be Noetherian by the Hilbert basis
theorem, but may not be integrally closed.}
\end{itemize}
\begin{example}[?]
If \(f \in k[x, y]\) is irreducible, then
\(k[x, y]/\left\langle{f}\right\rangle\) is an affine
\(k{\hbox{-}}\)order. It is an affine Dedekind domain if \(f\) is
\emph{nonsingular} over \(k\), i.e.~for all
\(a, b\in \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
such that \(f(a, b) = 0\), the usual partial derivatives in the sense of
Calculus \({\frac{\partial f}{\partial x}\,}\) and
\({\frac{\partial f}{\partial y}\,}\) do not simultaneously vanish at
\((a, b)\). This is a sufficient condition, although it's not far from
being necessary as well.
\end{example}
\begin{remark}
Let \(A/k\) be an affine Dedekind domain such that
\(\operatorname{ff}(A) = K\). Then
\(\operatorname{mSpec}(A) = \Sigma(K/A) \hookrightarrow\Sigma(K/k)\).
This follows because \(\Sigma(K/A)\) are the valuations that are not
just regular on \(k\), but also on \(A\), (i.e.~\(A{\hbox{-}}\)regular
valuations) so the valuation ring contains the entirety of \(A\). It's
thus natural to ask what its complement is, i.e.~those valuations which
are not regular on \(A\) and give its elements negative valuation. So
define
\begin{align*}
\Sigma(A, \infty) \coloneqq\Sigma(K/k) \setminus\Sigma(K/A)
,\end{align*}
the set of places at infinity with respect to \(A\).
\end{remark}
\begin{example}[?]
\(\Sigma(k[t], \infty) = \left\{{v_ \infty}\right\}\), which is the
infinite place, so the terminology at least matches up!
\end{example}
\begin{proposition}[Key]
For any affine Dedekind domain \(A\), \(\Sigma(A, \infty)\) is finite
and nonempty.
\end{proposition}
\begin{remark}
This is striking! This says that one affine Dedekind domain is giving
almost all of this infinite set of places, but never all of it.
\end{remark}
\begin{proof}[?]
By Noether Normalization \footnote{This says that if you have an affine
domain \(R\) of a certain Krull dimension, then it is finitely
generated as a module over a subring which is a polynomial ring in
\(\operatorname{trdeg}(R)\) variables. This is like a stronger
integral version of taking a finitely generated field extension and
writing it as a finite degree field extension of a purely
transcendental extension.}
there exists a \(t\in A\) that that \(A\) is a finitely generated (and
thus integral) \(k[t]{\hbox{-}}\)module, and \(A\) is the integral
closure of \(k[t]\) in \(K\). Why must this be the integral closure? Any
ring finitely generated over a subring will be an integral extension,
and \(A\) is a Dedekind domain and thus integrally closed. So let
\begin{align*}
r: \Sigma(K/k) \to \Sigma(k(t)/k)
\end{align*}
denote the restriction map; then by the regularity property we
established in \cref{lemma:regularity}, we have
\begin{align*}
\Sigma(K/A) = r^{-1}\qty{\Sigma(k(t)/k[t])}
.\end{align*}
Why? A valuation upstairs in the NTI square is regular with respect to
the integral extension upstairs iff it's regular with respect to the
ring it is the integral extension \emph{of}. So regularity is preserved
both ways by integral extensions. This means you can check regularity
either upstairs or downstairs, allowing us to identify the above
preimage.
This means that the places where are \emph{not} \(A{\hbox{-}}\)regular
upstairs are precisely those which are not \(k[t]{\hbox{-}}\)regular
downstairs, and so we have
\begin{align*}
\Sigma(A, \infty) = r^{-1}\qty{\Sigma(k[t], \infty)} = r^{-1}(v_ \infty)
,\end{align*}
since we now there is exactly one such non-regular valuation. But we
showed that \(r\) was surjective with finite nonempty fibers, so we're
done since our set is one of the fibers.
\end{proof}
\begin{remark}
Thus is \(K/k\) is a one variable function field and \(A\) is an affine
Dedekind domain with fraction field \(K\), then
\(\Sigma(K/k) = \operatorname{mSpec}(A) {\coprod}S\) where \(S\) is
finite and nonempty. Earlier we saw by affine grounding that for each
\(v\in \Sigma(K/k)\) there exists an affine Dedekind domain \(A\) with
\(v\in \Sigma(K/A)\), and thus \(\Sigma(K/k)\) admits a \emph{finite}
covering by \(\operatorname{mSpec}\) of affine Dedekind domains. The
picture of what's happening is that we have \(\Sigma(K/k)\) which is
quasicompact with respect to the Zariski topology, which contains many
\(\operatorname{mSpec}\), at least one of which contains \(v\). Note
that these \(\operatorname{mSpec}(A_j)\) for affine Dedekind domains
\(A_j\) is literally an open cover in this topology. But the open sets
are so large that they all have finite complement. However, this means
that instead of just an arbitrary open covering, one can choose a finite
open covering: one \(\operatorname{mSpec}(A_j)\) will cover all but
finitely many, and we can always find at least one
\(\operatorname{mSpec}(A_{j'})\) covering all of the remaining points.
\begin{figure}
\centering
\includegraphics{figures/image_2020-12-01-01-43-43.png}
\caption{Image}
\end{figure}
It will in fact turn out that we only need \textbf{two} domains to cover
everything.
\end{remark}
\begin{definition}[Holomorphy Rings]
For a set \(S \subseteq \Sigma(K/k)\), define the \textbf{holomorphy}
ring as
\begin{align*}
R^s \coloneqq\bigcap_{v\in \Sigma(K/k)\setminus S} R_v
.\end{align*}
\end{definition}
\begin{remark}
This is the intersection of a bunch of valuation rings, so this contains
elements that are simultaneously regular for this subset of valuations.
If \(S \subseteq S'\), then \(R^S \subseteq R^{S'}\), due to the fact
that we're taking complements and
\(\Sigma(K/k) \setminus S \supset \Sigma(K/k)\setminus S'\), so we're
removing bigger sets and thus intersecting over fewer things. This can
be thought of as relaxing some regularity conditions.
\end{remark}
\begin{remark}
How to think about holomorphy rings: if you take \(S = \emptyset\), you
intersect over all \(R_v\) and obtain \(R^\emptyset = \kappa(k)\). You
get a field that is algebraic over \(k\), so it's very small compared to
the other types of field extensions that arise. We'll see that this is
``unrepresentably'' small.
\end{remark}
\begin{exercise}[Every affine Dedekind domain is a unique holomorphy ring]
If \(A\) is an affine Dedekind domain with fraction field \(k\), then
\begin{align*}
A = R^S
= \bigcap_{v\in \operatorname{mSpec}(A)} R_v
&& S = \Sigma(K/k) \setminus\operatorname{mSpec}(A)
.\end{align*}
\end{exercise}
\begin{remark}
This is a fact for any Dedekind domain, which is the intersection over
all of its DVRs. You obtain the integral closure for a Dedekind domain
by intersecting all of the valuation rings, but here it is already
integrally closed. Its tautological that \(A \subset R^S\) here, so
\(R^S\) is an \emph{overring} of a Dedekind domain: for \(R\) a domain,
an \textbf{overring} is any ring \(T\) such that
\(R \subseteq T \subseteq \operatorname{ff}(R)\). When \(R\) is a PID,
the overrings are in bijective correspondence with subsets of prime
ideals (prime elements mod associates), so you get all overrings by
inverting such subsets. For Dedekind domains it's more complicated. Can
we classify all overrings of \(R\) when it is a Dedekind domain? The
answer will eventually be yes. Under what condition is every overring a
localization? When the class group is torsion. What are the
relationships between the class groups of the ring \(R\) and its
overrings \(\widehat{R}\)? It turns out that
\(\operatorname{cl}(\widehat{R})\) is a quotient of
\(\operatorname{cl}(R)\). We will show that all such overrings are of
the form \(R^W\) for some \(W\), i.e.~they're obtained by intersecting
some subset of the localizations of \(R\) at its maximal ideals.
Note that the holomorphy ring in the exercise is obtained from a finite
set of places. Conversely, given any finite nonempty set of places, then
the holomorphy of ring of all of the elements of \(K\) that are regular
with respect to all but this finite number of valuations will always be
an affine Dedekind domain with fraction field \(K\).
\end{remark}
\hypertarget{holomorphy-rings-are-affine-dedekind-domains-with-fraction-field-k}{%
\subsection{\texorpdfstring{Holomorphy Rings are Affine Dedekind Domains
with Fraction Field
\(K\)}{Holomorphy Rings are Affine Dedekind Domains with Fraction Field K}}\label{holomorphy-rings-are-affine-dedekind-domains-with-fraction-field-k}}
Next up is the main theorem of this lecture.
\begin{theorem}[Holomorphy rings on subsets are synonymous with affine Dedekind domains with fraction field $K$]\label{thm:main_lec_6}
Let \(K/k\) be a one variable function field and
\(S \subset \Sigma(K/k)\) finite and nonempty. Then \(R^S\) is an affine
Dedekind domain with \(\operatorname{ff}(R^S) = K\) and
\(\operatorname{mSpec}(R^S) = \Sigma(K/k)\setminus S\).
\end{theorem}
\begin{exercise}[?]
If \(S \subset \Sigma(K/k)\) is infinite, then \(R^S\) is Dedekind with
fraction field \(K\) but is \emph{not} finitely generated as a
\(k{\hbox{-}}\)algebra.
\end{exercise}
\begin{remark}
So what happens when you allow elements to fail regularity at an
infinite set of places instead of just a finite set? From the theory of
Dedekind domains, this will again be a Dedekind domain, but will be more
exotic than an affine Dedekind domain. What if it were finitely
generated as a \(k{\hbox{-}}\)algebra? Then it would be an affine
Dedekind domain, and we have a good understanding of
\(\operatorname{mSpec}\) of these types of rings, and it would have to
be a holomorphy ring with respect to some finite set. Note that
holomorphy rings for different subsets are distinct.
\end{remark}
\begin{remark}
We have an interesting class of rings: Dedekind domain which are
holomorphy rings with respect to an infinite set of places. What are
they good for? They're used in Pete's paper ``Elliptic Dedekind domains
revisited'' to give a new proof of a theorem of Clayborne (60s, at least
the third proof) that every commutative group is the ideal class group
of some \(R^S\). \todo[inline]{Get citation.} The Dedekind domain used
was a holomorphy ring \(R^S\) with respect to some infinite set \(S\).
He starts out with an elliptic function field \(K\) (so of genus 1 with
a degree 1 place), and taking the standard affine coordinate ring of the
curve is \(R^S\) for \(S\) the single degree 1 place. This is
particularly nice, since its class group is canonically isomorphic to
\(C(k)\), the \(k{\hbox{-}}\)rational points of the elliptic curve. When
you pass from a Dedekind domain to an overring you get some quotient of
the class group. Note that there are three degrees of freedom here: you
get to pick \(k\) to be any field, then \(K/k\) some function field, and
then \(S\). For this paper, \(k\) was already some weird transfinitely
iterated field. The upshot here is that not only is every commutative
group isomorphic to \(\operatorname{cl}(T)\) for \(T\) \emph{some}
Dedekind domain, \(T\) is in particular a holomorphy ring of the form
\(R^S\). This is pretty useful, but not nearly as much as \(R^S\) for
\(S\) a finite set of places.
\end{remark}
\begin{definition}[Poles and Zeros]
Let \(f\in K^{\times}\), then a place \(v\in \Sigma(K/k)\) is a
\textbf{pole} of \(f\) iff \(f\not\in R_v\), and \(v\) is a
\textbf{zero} of \(f\) iff \(f\in \mathfrak{m}_v\).
\end{definition}
\begin{lemma}[The divisor of a rational function is well-defined.]\label{lem:poles_and_zeros}
Let \(f\in K^{\times}\) be nonzero, then
\begin{align*}
{\left\lvert {\left\{{v\in \Sigma(K/k) {~\mathrel{\Big|}~}f\not\in R_vS}\right\}} \right\rvert} &< \infty && \text{(finite poles)}\\
{\left\lvert {\left\{{v\in \Sigma(K/k) {~\mathrel{\Big|}~}f\in \mathfrak{m}_v S}\right\}} \right\rvert} &< \infty && \text{(finite zeros)}
.\end{align*}
So \(f\) is not regular at only a finite set of places, as as the set of
points such that ``\(f(p) = 0\)'', i.e.~\(f\) is in the maximal ideal
which makes it zero in the residue field.
\end{lemma}
\begin{remark}
Thinking of \(f\) as a rational function, this says that the sets of
points which are poles or zeros are both finite.
\end{remark}
\begin{proof}[of first statement]
If \(f \in \kappa(K)\), then both sets are empty, so assume otherwise
that \(f\) is transcendental. This is because if \(f\) is a nonzero
constant function, i.e.~it is algebraic over \(k\), and both
\(f, f^{-1}\) lie in all of the valuation rings and none of the maximal
ideals. Then the integral closure \(A\) of \(k[f]\) in \(K\) is an
affine Dedekind domain containing \(f\).\footnote{Pete's Commutative
Algebra, Theorem 18.4 (a normalization theorem).} But we're done: for
all \(v\in \Sigma(K/A)\), we have \(f\in R_v\) and thus
\begin{align*}
\Sigma(A, \infty) = \Sigma(K/k) \setminus\Sigma(K/A)
,\end{align*}
which is finite by affine grounding. This is because \(\Sigma(K/A)\)
already has finite complement, so all but finitely many valuations are
\(A{\hbox{-}}\)regular, and \(f\in A\). Conversely, if \(f\) is
nonconstant it can not be regular at all places since it would then lie
\(\kappa(K)\).
\end{proof}
\begin{proof}[of second statement]
Note that
\(f\in \mathfrak{m}_v \iff v_p(f) > 0 \iff v_p(1/f) < 0 \iff 1/f\not\in R_v\),
so we can just apply the first statement to \(1/f\).
\end{proof}
\begin{exercise}[Function fields are always covered by $\mspec$ of two affine Dedekind domains (too easy!)]
Show that there exist \(A_1, A_2\) affine Dedekind domain such that
\begin{align*}
\Sigma(K/k) = \Sigma(K/A_1) \cup\Sigma(K/A_2)
.\end{align*}
\end{exercise}
\begin{remark}
This will follow from a theorem we haven't proved yet. If we think of
\(\Sigma(K/k)\) as a compact Riemann surface, the theorem is saying that
pulling out a single point (or any finite number) then what's left is
\(\operatorname{mSpec}(A)\) for \(A\) an affine Dedekind domain. So just
pull out two different points.
\end{remark}
\begin{remark}
The lemma is allowing us to define the divisor of a rational function.
We'll define \(\operatorname{Div}K\) as the free
\({\mathbb{Z}}{\hbox{-}}\)module with bases \(\Sigma(K/k)\). Any divisor
will be of the form
\begin{align*}
D = \sum_{p\in \Sigma(K/k)} n_p [p]
,\end{align*}
where all but finitely many of the \(n_p\) are zero. If we have a
rational function \(f\in K^{\times}\), we'll define
\begin{align*}
\operatorname{Div}f = \sum_{p\in \Sigma(K/k)} v_p(f) [p]
.\end{align*}
How do we know this is well-defined? We need \(v_p(f) = 0\) for all but
finitely many places \(p\). But
\(v_p(f) > 0 \implies f \not \in \mathfrak{m}_p\), and one part of the
lemma said \(f\) can only lie in finitely many \(\mathfrak{m}_p\). On
the other hand, \(v_p(f)\) can't be negative, since this would imply
\(f\not\in R_v\).
This is extremely important: the map that sends a rational function to
its divisor is multiplicative and additive, so this yields a subgroup of
\(\operatorname{Div}K\) called the \textbf{principal divisors}. The
quotient is the \textbf{class group} of \(K\), and now we are cooking
with gas (as Pete's undergraduate instructor used to say).
\end{remark}
\begin{theorem}[Strong Approximation]
Let \(X \subsetneq \Sigma(K/k)\) be proper and let
\(p_1, \cdots, p_r\in X\). Let
\(\left\{{x_j}\right\}_{j=1}^r \subset K\) and
\(\left\{{n_j}\right\}_{j=1}^r \subset {\mathbb{Z}}\). Then there exists
a single \(x\in K\) such that
\begin{align*}
\forall 1\leq j \leq r,\, v_{p_j} (x-x_j) &= n_j \\
\forall p\in X\setminus\left\{{p_j}\right\}_{j=1}^r,\, v_p(x) &\geq 0
.\end{align*}
\end{theorem}
\begin{remark}
Note that \(X\) is allowed to be infinite, so the statement only gets
stronger if we allow a maximal proper subset where its complement is
just a point. If we only had the first statement, this would be
\emph{weak approximation}. The conclusion is weaker, but it applies much
more generally. One first learns this in NTII, and it applies to any
finite set of inequivalent norms on a field. The second statement is a
requirement that \(x\) is regular. If \(X\) were \emph{not} all but one
place, we should replace it by that since it'd still satisfy the
hypotheses. Enlarging \(X\) only makes the conclusion of the second
statement stronger, since this is enforcing more integrality conditions.
\end{remark}
\begin{proof}[?]
Without loss of generality, assume that the complement
\(\Sigma(K/k) \setminus X = \left\{{p_0}\right\} \coloneqq S\) is a
single place. We know that \(R^S\) is an affine Dedekind domain (by a
theorem stated but not proved yet), so apply the \emph{Dedekind
Approximation Theorem}\footnote{Pete's NTII, Proposition 1.17}.
\end{proof}
\begin{remark}
Note that Stichtenoth uses Weil's proof of Riemann-Roch to prove this.
Too bad he doesn't have several hundred pages of lecture notes to draw
on! The difference between weak and strong approximation: weak applies
to a finite set of places, and strong applies to all but one place.
Later in NTII there's an adelic statement of strong approximation, which
works in the more general setting of a linear algebraic group over a
global field. You can take the adelic points of that group, remove one
place, and ask if strong approximation holds. It turns out to depend on
what kind of algebraic group you have.
\end{remark}
\hypertarget{proof-of-main-theorem}{%
\subsection{Proof of Main Theorem}\label{proof-of-main-theorem}}
We return now to the proof of \cref{thm:main_lec_6}.
We're trying to show that \(R^S\) for \(S\) a finite and nonempty set of
places is an affine Dedekind domain. So we need to show that it's
Dedekind, and that it's finitely generated over a field.
If \(\emptyset \subsetneq S_1 \subsetneq S_2\) are finite subsets of
\(\Sigma(K/k)\), then \(R^{S_1} \subseteq R^{S_2} \subset K\). By the
structure theory of Dedekind domains, \footnote{Pete's CA, Section 23.2.}
every overring of a Dedekind domain is again a Dedekind domain. This
allows us to restrict to the case where
\({\left\lvert {S} \right\rvert} = 1\).
\hypertarget{case-1}{%
\subsubsection{Case 1}\label{case-1}}
We start with the case where \(K\) is a one variable function field,
since it should certainly be true there. So assume \(K = k(t)\). If
\(S = \left\{{v_\infty}\right\}\) is just the infinite place, then
\(R^S = k[t]\) from a previous discussion. This is definitely a Dedekind
domain, since it's an affine PID.
\hypertarget{case-2}{%
\subsubsection{Case 2}\label{case-2}}
The next case is one place of degree 1, so
\(S= \left\{{v_{t-a}}\right\}\) corresponds to a monic irreducible
polynomial, where we use the fact that the degree of the residue field
is the degree of the polynomial. Then
\(R^S = k\left[{1\over t-a}\right]\). This is holomorphic at \(\infty\),
since the degree in the denominator is bigger than that of the
numerator. So it lies in \(R_{v_\infty}\) as well as \(R_{v_q}\) for
every monic irreducible polynomial \(q\) except for \(t-a\). This is a
PID since it's isomorphic to a polynomial ring, and has fraction field
\(K\). (?) We certainly have a containment \(\supseteq\), but the RHS is
already an affine Dedekind domain whose \(\operatorname{mSpec}\) is
everything but this single place. By the theory of overrings, the only
other possibility is that the RHS is bigger, but going from a Dedekind
domain to a larger Dedekind domain removes elements from
\(\operatorname{mSpec}\).
\hypertarget{case-3}{%
\subsubsection{Case 3}\label{case-3}}
Now consider the case \(S = \left\{{v_p}\right\}\) with
\(\deg p \coloneqq d > 1\). This corresponds to a monic irreducible
polynomial of degree bigger than 1. Note that \(k[t] \cong k[\alpha]\)
for any transcendental \(\alpha\), so we can take
\(k[1/p] \subset R^S\). This is an affine PID, and the containment
follows from the fact that \(1/p\) is holomorphic at \(\infty\) (for the
same reason as above). The only way it could \emph{not} be regular with
respect to some polynomial \(q\) would be that after cancelling the
numerator and denominator, \(q\) appears in the denominator, and that
happens precisely at \(p\). Now taking fraction fields, we have
\(\operatorname{ff}{k[t]} = k(t)\) and \([k(t) : k(1/p)] = d\), the
degree of the denominator, which follows from this exercise:
\begin{exercise}[Basic but important]
If \(p(t)/q(t) \in k(t)\) is a nonconstant rational function, then what
is the degree \(d \coloneqq[k(t) : k(p/q)]\)? Show that
\(d = \max\left\{{\deg p, \deg q}\right\}\).
\end{exercise}
So \(k[1/p] \subset R^S\) must be proper, since
\(\operatorname{ff}(R^S) = K\) but \(\operatorname{ff}(k[1/p])\) is a
proper extension. We can't have equality, so instead let \(A\) be the
integral closure of \(k[1/p]\) in \(k(t)\). Then \(A\) is a Dedekind
domain with \(\operatorname{ff}(A) = k(t)\) and
\(\operatorname{mSpec}(A) = \Sigma(k(t)/k) \setminus\left\{{v_p}\right\}\)
from the following NTI square:
\begin{center}
\begin{tikzcd}
{A} && {k(t)} \\
\\
{k[1/p]} && {k(1/p)} \\
\\
{k}
\arrow[from=5-1, to=3-1, no head]
\arrow[from=3-1, to=1-1, hook]
\arrow[from=3-3, to=1-3, hook]
\arrow["{\subseteq}" description, from=1-1, to=1-3, hook]
\arrow["{\subseteq}" description, from=3-1, to=3-3, hook]
\end{tikzcd}
\end{center}
\href{https://q.uiver.app/?q=WzAsNSxbMCwyLCJrWzEvcF0iXSxbMiwyLCJrKDEvcCkiXSxbMCwwLCJBIl0sWzIsMCwiayh0KSJdLFswLDQsImsiXSxbNCwwLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzAsMiwiIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMSwzLCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFsyLDMsIlxcc3Vic2V0ZXEiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDEsIlxcc3Vic2V0ZXEiLDEseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dXQ==}{Link
to diagram}
By affine grounding, we know \(k[1/p]\) is an affine Dedekind domain,
and by the second normalization theorem we know that \(A\) is finitely
generated as a module over \(k[1/p]\), which is in turn finitely
generated as an algebra over \(k\), making \(A\) a finitely generated
\(k{\hbox{-}}\)algebra. The key ingredient in identifying
\(\operatorname{mSpec}(A)\) is that \(1/p\in A\). By a previous
exercise, we can conclude that \(A\) is a holomorphy ring, and since we
know the exact excluded set is \(S\), we can conclude \(A = R^S\). This
makes \(A\) an affine Dedekind domain.\\
For the final case, suppose \(S\) is finite and nonempty. Choose
\(v\in S\) and define \(S_1 \coloneqq\left\{{v}\right\}\). Then
\(R^{S_1} \subseteq R^S \subset K\), so \(R^S\) is a Dedekind domain
since it's an overring of a Dedekind domain. A surprising fact is that
\(A \coloneqq R^S\) is not a PID when the degree is greater than 1, and
instead \(\operatorname{cl}(A) \cong {\mathbb{Z}}/d{\mathbb{Z}}\) and is
thus torsion. It'll be enough to show that \(R^S\) is finitely generated
as an algebra (but not a module?) over \(R^{S_1}\), which will make it a
finitely generated \(k{\hbox{-}}\)algebra, and we'd really like it to be
a localization. We examined this before: is every overring of a Dedekind
domain a localization? A theorem of Clayborne shows that this is true
when the class group is torsion.
Let \(v_2 \in S\setminus S_1\), so that every such \(v_2\) yields an
ideal \(\mathfrak{p}_{v_2} \in \operatorname{mSpec}R^{S_1}\). Since
\(\operatorname{cl}(R^S) = {\mathbb{Z}}/d{\mathbb{Z}}\), we don't know
that \(\mathfrak{p}_{v_2}\) is principal, but we do know that
\(\mathfrak{p}_{v_2}^\alpha\) is for some power \(\alpha\) Note that
localization is forgiving in the sense that inverting an element \(x\)
is equivalent to inverting any power \(x^k\) (e.g.~using that
\(1/x = x^{k}/x^{k+1})\). So we can write
\(\mathfrak{p}_{v_2}^\alpha = \left\langle{f}\right\rangle\), and it
follows that \(R^{\left\{{v, v_2}\right\}} = R_v[1/f]\) is an affine
domain which is obtained by localizing \(f\). Note that we can think of
this overring as puncturing or removing one place (a certain maximal
ideal) at a time, i.e.~intersecting over all of the maximal ideals
except one in order to go from \(R^{S_1} = R^{\left\{{v_1}\right\}}\) to
\(R^{\left\{{v_1, v_2}\right\}}\). You can continue this inductively
using the fact that \(R_v[1/f]\) is a different Dedekind domain -- since
it's an overring, the corresponding class group is a quotient and thus
still torsion. You could also continue this inductively by just
puncturing one point at a time. You can also do it all at once: for each
element in \(S\) not equal to \(v_1\), obtain an \(f_j\), and invert the
product \(\prod_j f_j\).\footnote{Alternatively, see Pete's Commutative
Algebra, Corollary 23.6.}
\textbf{Key fact}: We're in a lucky situation where we don't have a PID,
but we have a torsion class group. Anytime you pass to an overring by
puncturing finitely many maximal ideals, it will always be a
localization and thus monogenic as an algebra over the smaller Dedekind
domain.
\hypertarget{case-3-fixed-proof}{%
\subsection{Case 3: Fixed Proof}\label{case-3-fixed-proof}}
The remainder of the proof will go toward reducing to the first step of
a function field and exactly one place. We'll apply the Riemann-Roch
theorem, however this does not rely on results on holomorphy rings, so
there's no logical circularity. As usual, we lose no generality by
replacing \(k\) with \(\kappa(K)\) and just assuming that
\(\kappa(k) = k\).\\
Let \(S \subset \Sigma(K/k)\) be finite and nonempty. Then by
Riemann-Roch there exists an \(f\in K^{\times}\) having poles precisely
at the elements of \(S\), i.e.~\(f\) is regular away from \(S\).
\begin{quote}
Recall that poles were defined as elements not in \(R_v\). This is
motivated by considering meromorphic functions \(f\) on
\({\mathbb{C}}\), then the order of vanishing of \(f\) at \(p\) is a
discrete valuation, and if that valuation is negative then \(p\) is a
pole.
\end{quote}
Note that we're specifying the poles but not their orders, and allowing
poles of arbitrary orders would still allow us such a rational function
by a result like the Riemann inequalities, which is easier to prove than
the Riemann-Roch theorem. You can also obtain such a function from the
Strong Approximation theorem.\\
Since \(f\) has poles, it's nonconstant, so we have a nontrivial map
\(r: \Sigma(K/k) \to \Sigma(k(f)/k)\) to a rational function field and
thus \(r^{-1}(\infty) = S\) since the poles all like above the place at
\(\infty\). The analogy here is a holomorphic function \(f\) from a
compact Riemann surface to \({\mathbb{P}}^1_{/{\mathbb{C}}}\), in which
case \(f^{-1}(\infty)\) is the set of poles. Since \(k[f]\) is a
polynomial ring, we can take the integral closure of \(k[f]\) in \(K\),
say \(B\), in which case \(B\) is an affine Dedekind domain and
\(\operatorname{mSpec}(B) = \Sigma(K/k)\setminus S\).
The picture is as follows: think of \(k(f)\) as the Riemann sphere with
the point \(\infty\) and \(\Sigma(K/k)\) as a Riemann surface above it,
then \(S\) is the preimage of \(\infty\).
\begin{figure}
\centering
\includegraphics{figures/image_2020-12-01-22-09-07.png}
\caption{Image}
\end{figure}
If you have an upstairs valuation \(v \in \Sigma(K/k)\) in an integral
extension, then it is regular upstairs iff its restriction downstairs is
regular. Completing the NTI square yields
\begin{center}
\begin{tikzcd}
{B} && {K} \\
\\
{k[f]} && {k(f)}
\arrow["{\subseteq}" description, from=3-1, to=1-1, hook]
\arrow["{\subseteq}" description, from=3-3, to=1-3, hook]
\arrow["{\subseteq}" description, from=1-1, to=1-3, hook]
\arrow["{\subseteq}" description, from=3-1, to=3-3, hook]
\end{tikzcd}
\end{center}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJrW2ZdIl0sWzAsMCwiQiJdLFsyLDAsIksiXSxbMiwyLCJrKGYpIl0sWzAsMSwiXFxzdWJzZXRlcSIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzMsMiwiXFxzdWJzZXRlcSIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzEsMiwiXFxzdWJzZXRlcSIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzAsMywiXFxzdWJzZXRlcSIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV1d}{Link
to diagram}
Here \(B\) is the integral closure. So if we take a valuation in \(K\),
it is \(B{\hbox{-}}\)regular iff its restriction to \(k(f)\) is
\(A{\hbox{-}}\)regular. But \(R_v\) contains \(k[f]\) iff it contains
\(f\), since it's already a \(k{\hbox{-}}\)valuation, so the non-regular
valuations are those that restrict to \(\infty\).
Now if you have an extension of Dedekind domains, then the maximal
ideals upstairs are everything which restricts to a finite place
downstairs. So those that don't restrict to a finite place restrict to
\(\infty\), which is precisely the preimage of \(\infty\). With the
identification of
\(\operatorname{mSpec}(B) = \Sigma(K/k) \setminus S\),we have
\(B \subset R^S\) since every valuation in the complement of \(S\) is
regular at \(B\) by this argument. Since \(B \subset R^S \subset K\), we
can use the classification of overrings of Dedekind domains, and the
\(\operatorname{mSpec}\) corresponds to precisely the maximal ideals
that are being intersected. The only way this could be a \emph{proper}
extension would be if the \(\operatorname{mSpec}\) shrank, but then
\(R^S\) would be the holomorphy ring attached to a larger set than
\(S\). So we obtain an equality.
\(\hfill\blacksquare\)
\hypertarget{lecture-7-riemann-roch}{%
\section{Lecture 7: Riemann-Roch}\label{lecture-7-riemann-roch}}
\hypertarget{divisors-1}{%
\subsection{Divisors}\label{divisors-1}}
\begin{definition}[Divisor group]
The \textbf{divisor group} \(\operatorname{Div}K\) is the free
\({\mathbb{Z}}{\hbox{-}}\)module with basis \(\Sigma(K/k)\), so
\begin{align*}
\operatorname{Div}K \coloneqq\bigoplus_{p\in \Sigma(K/k)} {\mathbb{Z}}
.\end{align*}
Thus every \(D\in \operatorname{Div}K\) is of the form
\(D = \sum_{p\in \Sigma(K/k)} n_p p\) where \(n_p\in {\mathbb{Z}}\) and
are almost all zero, recalling that a point \(p\in \Sigma(K/k)\) is an
equivalence class of valuations.
\end{definition}
\begin{definition}[Effective Divisor]
A divisor \(D = \sum n_p p\) is \textbf{effective} iff \(n_p \geq 0\)
for all \(p\) and write \(D\geq 0\).
\end{definition}
\begin{definition}[Support of a divisor]
The \textbf{support} of a divisor \(D\) is the set of places
\(p\in \Sigma(K/k)\) such that \(n_p(D) \neq 0\). Note that this is
always a finite set, and the zero divisor is the unique divisor
supported on \(\emptyset\).
\end{definition}
\begin{definition}[Partial order on divisor]
We write \(D_1 \leq D_2\) iff \(D_2 - D_1 \geq 0\) is effective. Note
that this holds iff for all places \(p\in \Sigma(K/k)\), if
\(D_1 = \sum_p m_p p\) and \(D_2 = \sum_p n_p p\), then \(m_p\leq n_p\)
for all \(p\).
\end{definition}
This is a partially ordered commutative group, which came up when we
were talking about groups of divisibility. It's a reasonable group when
studying domains with nice factorization properties: if \(R\) is a UFD
with a set of principal prime ideals\footnote{Note that primes in a UFD
are principal.} denoted \(\Sigma(R)\), then the group of divisibility
\(G(R)\) is isomorphic to \(\bigoplus_{(p) \in \Sigma(R)} {\mathbb{Z}}\)
as a partially ordered commutative group.
There is an analogy: comparing UFDs to Dedekind domains, we trade unique
factorization of elements for factorization of ideals, and the group of
all fractional ideals in a Dedekind domain is a free commutative group
on its set of prime ideals. So \(\operatorname{Div}K\) is analogous to
the group of divisibility of a UFD and to the group of fractional ideals
of a Dedekind domain, the latter of which is the closer analogy. So
\(\operatorname{Div}K\) is a geometric or projective analog of the group
of fractional ideals, and is more than an analogy as we'll see later.
\begin{definition}[Degree of a Divisor]
There is a group morphism
\begin{align*}
\deg: \operatorname{Div}K &\to {\mathbb{Z}}\\
D = \sum_p n_p p &\mapsto \sum_p n_p \deg p
.\end{align*}
Its kernel is denoted \(\operatorname{Div}^0 K\), the \textbf{degree
zero divisors}. Note that if
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\),
then \(\deg p = 1\) for all \(p\).
\end{definition}
\begin{remark}
Note that this is similar to the augmentation in a group ring. This
construction can be done with any free \({\mathbb{Z}}{\hbox{-}}\)module,
and makes sense because only finitely many terms are nonzero. Recall
that to define the degree of a place \(v\in \Sigma(K/k)\), we consider
\(R_v \coloneqq\left\{{x\in K {~\mathrel{\Big|}~}v(x) \geq 0}\right\}\)
and
\(\mathfrak{m}_v \coloneqq\left\{{x\in K {~\mathrel{\Big|}~}v(x) >0}\right\}\),
and \(k(v) \coloneqq R_v / \mathfrak{m}_v\) is the residue field. Note
that \(k(v)\) is a field extension of \(k\) by composing
\(k\hookrightarrow R_v \twoheadrightarrow k(v)\), and we proved used
affine grounding and Zariski's lemma that this was a finite degree
extension. We can then define \(\deg v \coloneqq[k(v) : k]\). Note that
it's more natural to think of valuations \(v\) as points \(p\).
\end{remark}
\begin{definition}[Index of a divisor]
The \textbf{index} of \(K\) is defined as
\begin{align*}
I(K) \coloneqq{\left\lvert {\operatorname{coker}\deg} \right\rvert}
.\end{align*}
\footnote{This quantity made an appearance near the end of Pete's
advanced course on elliptic curves.}
\end{definition}
\begin{remark}
Note that \(I(K)\) is nonzero, since we can think of
\(p\in \operatorname{Div}K\) as the divisor with \(n_q = \indic{q=p}\),
so the image contains a subset consisting of all degrees of all places,
so the image is of the form \(d{\mathbb{Z}}\) for some \(d\). Some other
characterizations:
\begin{itemize}
\tightlist
\item
\(\deg\qty{\operatorname{Div}K} = I(K){\mathbb{Z}}\), so \(I(K)\) is
the generator of the degree ideal.
\item
\(I(K)\) is the least positive degree of a divisor on \(K\).
\item
\(I(K) = \gcd\qty{\left\{{\deg p {~\mathrel{\Big|}~}p\in \Sigma(K/k)}\right\}}\),
i.e.~the \(\gcd\) of the closed points.
\end{itemize}
The last characterization follows because we have generators of
\(\operatorname{Div}K\) given by ``skyscraper'' divisors \(p\) where
\(n_q = 1 \iff p=q\), so the image is the subgroup of \({\mathbb{Z}}\)
generated by the degrees of the points, i.e.~the \(\gcd\) of the
degrees.
\end{remark}
\begin{exercise}[?]
Let \(K/k\) be a one variable function field.
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that if \(\Sigma_1(K/k) \neq \emptyset\) then \(I(K) = 1\).
\item
Later we will show that if \({\left\lvert {k} \right\rvert} < \infty\)
then \(I(K) = 1\) but \(\Sigma_1(K/k)\) may be empty. Try to show
this.
\item
Show that if
\(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
then \(I(K) = 1\).
\end{enumerate}
\end{exercise}
\begin{remark}
\begin{enumerate}
\def\labelenumi{(\alph{enumi})}
\tightlist
\item
follows from the Riemann hypothesis for curves over a finite field,
although this is not how you should prove it. It was proved by F.K.
Schmidt much earlier in the 20th century, and this is the basic way of
understanding the zeta function of a curve.
\item
says that over a finite ground field, you may not have any degree 1
places. You can try constructing a hyperelliptic curve over a finite
field \({\mathbb{F}_{q}}\) with no rational points, which is always
possible if the genus is large compared to the size of
\({\mathbb{F}_{q}}\).
\end{enumerate}
\end{remark}
\begin{lemma}[?]
For a nonzero rational function \(f\in K^{\times}\) we have
\(v_p(f) = 0\) for almost every place \(p\in \Sigma(K/k)\).
\end{lemma}
\begin{proof}[?]
See previous lecture, in particular \cref{lem:poles_and_zeros}.
\end{proof}
This says that the set of places for which the valuation is nonzero is
finite, so except for finitely many places the valuation is zero. This
allows us to define the divisor of a rational function:
\begin{align*}
({\,\cdot\,}): K^{\times}&\to \operatorname{Div}K \\
f &\mapsto (f) \coloneqq\sum_p v_p(f) p
,\end{align*}
which is a group morphism.
\begin{exercise}[?]
Show that \((f) = 0 \iff f \in \kappa(K)\), which we're assuming is
equal to \(k\). This happens when it has neither zeros nor poles, so
it's an intersection of all of the \(R_v\), which is the integral
closure of \(k\) in \(K\). In general, this would mean that \(f\) is
algebraic over \(k\). So \(\ker ({\,\cdot\,}) = k^{\times}\).
\end{exercise}
\begin{definition}[Poles and Zeros of Elements of $K$]
For any \(D\in \operatorname{Div}K\) one may uniquely write it as
\(D = D_+ - D_-\), which are both effective divisors and so
\(D_+, D_- \geq 0\), and the uniqueness follows from requiring
\({\operatorname{supp}}(D_+) \cap{\operatorname{supp}}(D_-) = \emptyset\).
Note that this is just collecting positive and negative \(n_p\) into
each term, and leaving out all divisors for which \(n_p = 0\).
For \(f\in K^{\times}\), we define
\begin{align*}
(f)_+ &\coloneqq\text{the divisor of zeros of } f \\
(f)_- &\coloneqq\text{the divisor of poles of } f
,\end{align*}
where we can note that \((f) = (f)_+ - (f)-\).
\end{definition}
The next proposition shows that these geometric divisors can be
interpreted in terms of \({\mathbb{F}_{q}}\) points.
\begin{proposition}[?]
Let \(f\in K\setminus k\) be transcendental.
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
Let \(B_0\) be the integral closure of \(k[f]\) in \(K\), which is an
affine Dedekind domain of \(K\), i.e.~its fraction field is
\(K\).\footnote{As usual for an extension of Dedekind domains, we push
forward an ideal (maybe principal) into its integral closure and see
how it factors.}
\end{enumerate}
Then
\begin{align*}
fB_0 = \prod_{j=1}^r p_j^{a_j} \implies (f)_+ = \sum_{j=1}^r a_j p_j
.\end{align*}
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
Let \(B_ \infty\) be the integral closure of \(k[1/f]\) in \(K\),
which is an affine Dedekind domain of \(K\). Then
\begin{align*}
\qty{1\over f} B_ \infty = \prod_{j=1}^s q_j^{b_j} \implies (f)_+ = \sum_{j=1}^s b_j q_j
.\end{align*}
\end{enumerate}
\end{proposition}
\begin{exercise}[?]
Prove this proposition.
\end{exercise}
\begin{remark}
This says that pushing forward an ideal and looking at the factorization
is precisely what's needed to determine the divisor of zeros. There
aren't many new ideas for this proof, the point is that the set of
places upstairs is being controlled by \(\operatorname{mSpec}\) of
Dedekind domains.
\end{remark}
\begin{slogan}
In any affine coordinate chart, the divisor of a function is a principal
fractional ideal.
\end{slogan}
\hypertarget{the-degree-of-the-divisor-of-a-rational-function-is-zero}{%
\subsection{The Degree of the Divisor of a Rational Function is
Zero}\label{the-degree-of-the-divisor-of-a-rational-function-is-zero}}
\begin{corollary}[Excruciatingly Important: the degree of the divisor of any rational function is zero.]
Let \(f\in K\setminus k\) be transcendental, then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
\(\deg(f)_+ = [K : k(f) ] = \deg (f)_-\)
\item
\(\deg (f) = 0\).
\end{enumerate}
\end{corollary}
\begin{remark}
Here think of \(f\) as a holomorphic map from a curve to
\({\mathbb{P}}^1_{/{\mathbb{C}}}\), and the degree of this extension is
the degree of the corresponding branched cover. For \({\mathbb{C}}\),
this is literally the cardinality of any finite fibers. Note that (a)
follows by symmetry sense \(k(f) \cong k(1/f)\).
\end{remark}
\begin{proof}[?]
This comes down to NTI. We know
\(\deg(f)_+ = \sum{j=1}^r a_j \deg p_j\). In \(K/k(f)\), the places
\(p_1, \cdots, p_r\) all lie over the degree 1 place \(v_f\) of
\(k(f)\). The places where upstairs you have a zero are the places where
to coordinate downstairs is equal to zero, which corresponds to the
irreducible polynomial in \(f\) given by \(f\) itself. Since the residue
field at \(v_f\) downstairs is \(k\) itself, since it is
\(k[f]/\left\langle{f}\right\rangle\). So the downstairs places has
degree 1, and so the degree of the upstairs places, whatever the residue
field is, its degree over \(k\) is equal to its degree over the
downstairs residue field. Thus the geometric \(\deg p_j\) coincides with
the residual degree \(f_i\), and \(a_i\) is the ramification index in
the extension of Dedekind domains \(B_0 / k[f]\).\\
So we have a degree equality,
\begin{align*}
\sum_{j=1}^r a_j \deg p_j = \sum e_j f_j = [K: k(f)]
,\end{align*}
where the second equality follows from having an extension of Dedekind
domains with this nice finite generation hypothesis. We similarly get
\([k: k(f)] = \deg(f)_-\).\\
Note that part (b) follows immediately, since \((f) = (f)_+ - (f)_-\)
implies that\\
\begin{align*}
\deg(f) = \deg(f)_+ - \deg(f)_- = [k: k(f)] - [k: k(f)] = 0
.\end{align*}
\end{proof}
\begin{remark}
We have two different things that sound like the degree of a rational
function. We define the degree of a rational function
\(f\in K\setminus k\) as \([K: k(f)]\), otherwise it is the degree
(number of sheets) of the corresponding branched covering of
\({\mathbb{P}}^1\). But note that we also attached a divisor to \(f\),
which may be confusing, be hard to confuse in practice because we found
that \(\deg(f) = 0\) always.
\end{remark}
\begin{definition}[Principal Divisors]
The divisor of a rational function is called \textbf{principal}, we
define \(\operatorname{Prin}K\) to be the group of principal divisors.
\end{definition}
\begin{exercise}[$\Prin K$ is a group]
For \(f, g\in K^{\times}\), show that
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
\((1/f) = -(f)\),
\item
\((fg) = (f) + (g)\),
\item
\(\operatorname{Prin}K \leq \operatorname{Div}^0 K\) is a subgroup
(since we know they're degree zero).
\end{enumerate}
\end{exercise}
\begin{definition}[Linear Equivalence]
For \(D_i \in \operatorname{Div}K\), we set
\(D_1 \sim D_2 \iff D_1 - D_2 \in \operatorname{Prin}K\), in which case
we say these divisors are \textbf{linearly equivalent}.
\end{definition}
\begin{remark}
Near the end of the course we'll see why this is good terminology: it's
related to morphisms of projective space attached to linear systems.
\end{remark}
\begin{definition}[Divisor Class Group]
We define the \textbf{divisor class group} as
\begin{align*}
\operatorname{cl}K \coloneqq\operatorname{Div}K/\sim = \operatorname{Div}K / \operatorname{Prin}K
.\end{align*}
\end{definition}
But note that there's something between \(\operatorname{Prin}K\) and
\(\operatorname{Div}K\), namely \(\operatorname{Div}^0 K\):
\begin{definition}[Degree 0 Divisor Class Group (Important! Fundamental!)]
We define the \textbf{degree 0 divisor class group} as
\begin{align*}
\operatorname{Cl}^0 K \coloneqq\operatorname{Div}^0 K / \sim = \operatorname{Div}^0 K / \operatorname{Prin}K
.\end{align*}
\end{definition}
\begin{remark}
This is extremely important! Attached to a curve is a Jacobian abelian
variety, a nice group variety whose dimension is equal to the genus of
the curve, and the \(k{\hbox{-}}\)rational point of the Jacobian will
become a commutative group that is isomorphic to
\(\operatorname{Div}^0 K\).
\end{remark}
\begin{exercise}[?]
Show that we have the following exact sequences:
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
\begin{align*}
1 \to k^{\times}\to K^{\times}\xrightarrow{({\,\cdot\,})} \operatorname{Prin}K \to 0
.\end{align*}
\item
\begin{align*}
0 \to \operatorname{Cl}^0 K \to \operatorname{Cl}K \xrightarrow{\deg} I(K) {\mathbb{Z}}\to 0
.\end{align*}
\end{enumerate}
Deduce that
\(\operatorname{Cl}K \cong \operatorname{Cl}^0 K \oplus {\mathbb{Z}}\).
\end{exercise}
\begin{remark}
For (a), we saw that rational functions that have zero divisors are
constants, assuming that \(\kappa(K) = k\). For (b), because principal
divisors have degree zero, the degree map factors through the quotient.
The deduction comes from that fact that we have a free and hence project
\({\mathbb{Z}}{\hbox{-}}\)module, yielding a splitting.
\end{remark}
\begin{exercise}[Very important, Pete insists that someone solves it!]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that \(\operatorname{Div}^0 k(t) = \operatorname{Prin}k(t)\).
\item
Deduce that
\(\deg: \operatorname{Cl}k(t) \xrightarrow{\sim} {\mathbb{Z}}\) and
\(\operatorname{cl}^0 k(t) = 0\).
\end{enumerate}
\end{exercise}
\begin{remark}
Note that \(I(K) = 1\) in this case since both the \(t{\hbox{-}}\)adic
or \(\infty{\hbox{-}}\)adic valuation have degree one. Moral: the class
groups are not interesting on rational function fields. You have to take
a degree zero divisor on a rational function field and build a rational
function whose divisor is any given degree. This is extremely useful!
\end{remark}
\begin{remark}
More general if \(K/k\) has genus zero (e.g.~a rational function field),
then working over \({\mathbb{C}}\) we would have
\(\operatorname{Cl}^0 K\) equal to the points of some compact complex
Lie group of \({\mathbb{C}}{\hbox{-}}\)dimension \(g\), so a large
complex torus, unless \(g=0\). So if
\(k= \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\),
\(\operatorname{Cl}^0 K\) will be uncountably infinite when \(g>0\). If
not, it might trivial, or it might be anything in between.
\end{remark}
The following result appears in a 1973 paper of Rosen, where he
attributes it to F. K. Schmidt. It gives a close relationship between
\(\operatorname{Cl}^0 K\) and the class groups \(\operatorname{Cl}R^S\)
of the affine Dedekind domains of \(K\). This shows that instead of
\(\operatorname{Cl}^0 K\) just being an analogue of the class group of a
Dedekind domain, there's almost the same. If you fix \(K\),
\(\operatorname{Cl}^0 K\) is just one group attached to it, but there
are infinitely many \(R^S\) since there are infinitely many places. So
these groups can not be equal, since we could change the size of \(S\)
to obtain overrings of Dedekind domains, where the resulting class
groups are quotients. So you could kill finitely many elements in the
class group of the Dedekind domain by just passing to an overring by
adding finitely more places.
\begin{theorem}[Rosen]
Let \(S \subset \Sigma(K/k)\) be nonempty and finite, and recall that
the holomorphy ring was defined as
\begin{align*}
R^S = \cap_{v\in \Sigma(K/k)} R_v
.\end{align*}
Define the following:
\begin{itemize}
\tightlist
\item
\(D^0(S)\): the degree 0 divisors with support in \(S\).
\item
\(P(S) \coloneqq\operatorname{Prin}K \cap D^0(S)\), the principal
divisors supported in \(S\).
\begin{itemize}
\tightlist
\item
Divisors of rational functions all of whose zeros and poles lie in
\(S\).
\end{itemize}
\item
\(d_S\): The least positive degree of a divisor supported on \(S\).
\begin{itemize}
\tightlist
\item
Note that this is different to the index in that we restrict to
\(S\), and is thus a multiple of \(I(K)\).
\end{itemize}
\end{itemize}
Then there is an exact sequence
\begin{align*}
0 \to D^0(S) / P(S) \xrightarrow{\iota} \operatorname{Cl}^0 K \xrightarrow{\alpha} \operatorname{Cl}R^S \xrightarrow{\beta} C(d/ I(K)) \to 0
.\end{align*}
\end{theorem}
\begin{proof}[?]
See NTII, Theorem 3.27.
\end{proof}
\begin{remark}
Note that the kernel \(D(S)/P(S)\) could be infinite but is always
finitely generated. The map \(\alpha\) is induced by
\begin{align*}
\alpha': \operatorname{Div}K &\to \operatorname{Frac}R^S \\
\sum n_p p &\mapsto \prod_{p\in \operatorname{mSpec}R^S} p^{n_p}
,\end{align*}
where we note that \(\operatorname{mSpec}R^S \subset \Sigma(K/k)\), and
in fact \(\Sigma(K/k) = \operatorname{mSpec}R^S {\coprod}S\). We can do
this because if \(p\) is already in \({\operatorname{maxSpec}}R^S\), we
raise it to an appropriate power, and otherwise, for the finitely many
\(p\in S\) we just get rid of them. But this kills of some elements,
namely those things supported in \(S\), hence the kernel in the exact
sequence.\\
Note that the last group appearing is finite cyclic of order \(d/I(K)\).
If you just looked at \(D^0(S)\) before modding out by principal
divisors, if you didn't impose degree zero, the subgroup would be
isomorphic to \({\mathbb{Z}}^{{\left\lvert {S} \right\rvert}}\). But
there's a linear condition that the degree is equal to zero, which cuts
down the dimension by 1, yielding
\({\mathbb{Z}}^{{\left\lvert {S} \right\rvert} - 1}\). It's hard to say
how much \(P(S)\) is cutting down the size.
\end{remark}
\begin{remark}
The moral is that there is a map, but the kernel and cokernel both
depend on \(S\). If you understand \(\operatorname{Cl}^0 K\), however,
you have a good handle on all \(\operatorname{Cl}R^S\).
\end{remark}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that
\(D^0(S) \cong {\mathbb{Z}}^{{\left\lvert {S} \right\rvert} - 1}\).
\item
Suppose \(S\) consists of a single place whose degree is the quantity
\(d_S\) appearing in the previous theorem, the least positive degree
of a divisor supported on \(S\). Show that there is an exact sequence
\begin{align*}
0 \to \operatorname{Cl}^0 K \xrightarrow{\alpha} \operatorname{Cl}R^S \xrightarrow{\beta} C(d_S/I(K)) \to 0
.\end{align*}
\item
Deduce that \(\alpha\) is an isomorphism iff \(I(K) = d\).
\item
Deduce that if \(p\in \Sigma(k(t)/k)\) has degree \(d\), then
\(\operatorname{Cl}R^{\left\{{p}\right\}} \cong {\mathbb{Z}}/d{\mathbb{Z}}\).
\item
Deduce that if \(S = \left\{{p}\right\}\) and \(\deg p = 1\), then
\(\alpha: \operatorname{Cl}^0 K \to \operatorname{Cl}R^S\) is an
isomorphism.
\end{enumerate}
\end{exercise}
\begin{remark}
Note that if you're given a finite set of places and ask for all of the
rational functions that have zeros and poles only at those places, it is
difficult to determine how close that is to filling out the entire
degree zero divisor class group? If you have two degree 1 points
\(p_i\), so \({\left\lvert {S} \right\rvert} = 2\), do you have a
rational function whose divisor is \(p_1 - p_2\)? Probably not, because
then the divisor of such a function would have degree 1. You can
continue this line of thought, but already using elliptic function
fields you can see that all of these algebraic possibilities can occur.
\end{remark}
\begin{remark}
Note that in the case where \(S\) is a single point of degree \(d\),
then \(d_S\) is equal to the degree of the point \(d\). On the other
extreme, consider what happens when \(I(K) = 1\). Then \(C(d_S)\) is
cyclic of order \(d\), so in (c) if we have a rational function field,
we know it has degree 1 places (like \(0, \infty\)), and the class group
is zero. So if you take one place on \({\mathbb{P}}^1\) of degree \(d\)
and look at the correspond affine Dedekind domain of functions that are
regular away from that one place \(R^{\left\{{p}\right\}}\), then the
class group is nontrivial and it's thus not a PID. Note that
\(\operatorname{Cl}^0 {\mathbb{P}}^1\) is trivial, and puncturing it has
an effect on the divisor class.
\end{remark}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Suppose \(\operatorname{Cl}^0 K\) is finite, and show that every
\(\operatorname{Cl}R^S\) is finite. \footnote{Later we will show that
\(\operatorname{Cl}^0 K\) is finite when \(k\) is finite.}
\item
Suppose \(\operatorname{Cl}^0 K\) is finitely generated, and show that
for all finite nonempty \(S \subset \Sigma(K/k)\), there exists a
finite \(S' \supset S\) such that \(\operatorname{Cl}R^{S'}\) is
trivial.
\end{enumerate}
\end{exercise}
\begin{remark}
This is the positive characteristic version of one of the basic
finiteness theorems from NTI: the ring of integers of any number field
has finite class group. But the \(S{\hbox{-}}\)class group is always
finite, since it's a quotient of the class group, and that's what's
happening here. It's enough to show that the \(\operatorname{Cl}^0 K\)
and \(C(d_S/ I(K))\) appearing in the SES in the previous theorem are
finite, since the first term can only cut down the size. The groups
\(\operatorname{Cl}R^S\) when \(k\) is finite are analogues of the
\(S{\hbox{-}}\)class groups of number fields. In the function field
case, you can't get away from the \(S{\hbox{-}}\)class group, since if
\(S= \emptyset\) then \(R^S\) is not an interesting Dedekind domain:
it's just \(\kappa(K)\). So you have to put something at \(\infty\) to
even get a 1-dimensional domain, whereas in the number field case, you
always have a finite nonempty set of nonarchimedean places.
This allows us to deduce from the finiteness of this one geometric group
the finiteness of \(S{\hbox{-}}\)class groups in the characteristic
\(p\) case. If done correctly, this can be used to prove the finiteness
of class groups of all number fields, e.g.~if you do things in an adelic
way in NTII.
\end{remark}
\begin{theorem}[Trotter, 1988]
The ring \(R[\cos \theta, \sin \theta]\) of real trigonometric
polynomials is not a UFD, while
\({\mathbb{C}}[\cos \theta, \sin \theta]\) is a PID.
\end{theorem}
\begin{remark}
Trotter shows that using
\(\sin(\theta)\sin(\theta) = (1+\cos(\theta))(1-\cos(\theta))\) exhibits
non-unique factorization, since the terms appearing are non-associate
irreducible elements in an integral domain. See Pete's list of
exercises. Note that given an affine Dedekind domain how one figures out
what the infinite places are concretely, but this will come up when
discussing hyperelliptic curves.
\end{remark}
\begin{remark}
One exercise applies Rosen's theorem to show that
\(\operatorname{Cl}{\mathbb{R}}[\cos(\theta), \sin(\theta)] = {\mathbb{Z}}/2{\mathbb{Z}}\)
while \(\operatorname{Cl}{\mathbb{C}}[\cos(\theta), \sin(\theta)] = 1\).
What's happening is that over \({\mathbb{R}}\), there is perhaps one
degree 2 place at \(\infty\), but after extending scalars to
\({\mathbb{C}}\) it breaks up into two degree 1 places.
\end{remark}
\hypertarget{lecture-8-riemann-roch-spaces-part-1}{%
\section{Lecture 8: Riemann-Roch Spaces (Part
1)}\label{lecture-8-riemann-roch-spaces-part-1}}
\hypertarget{setup-for-the-riemann-roch-theorem}{%
\subsection{Setup for the Riemann-Roch
Theorem}\label{setup-for-the-riemann-roch-theorem}}
Setting up for the single most important theorem in the course: the
Riemann-Roch theorem. We start by motivating this by considering the
following property of \(K\coloneqq k(t)\): for any degree 1\footnote{So
the residue field of the corresponding DVR is \(k\) itself rather than
some proper finite degree extension.} place \(p \in \Sigma(K/k)\),
there exists an \(f\in K^{\times}\) such that \((f)_- = p\). In other
words, \(f\) is a rational function with a simple pole at the given
place, and no other poles. Why? We just know precisely what all of the
places are for this function field.
If \(p= \infty\), we can just take \(f(t) = t\), since any polynomial is
regular away from \(\infty\) and the valuation is \(-\deg(f) = -1\) The
other places \(p\) correspond to \(t-\alpha\) (the uniformizing element)
for \(\alpha\in k\), since they correspond to other points on
\({\mathbb{A}}^1_{/k}\), and so we can take \(f(t) = 1/(t-\alpha)\).
This \(f\) is regular at infinity since the degree of the numerator is
larger than the degree of the denominator, and the denominator doesn't
vanish at any other place.
\begin{remark}
With some thought, it can be found that this is a \emph{characteristic}
property of rational function fields: if \(f\in K\), a one variable
function field, and \(\deg(d)_- = 1\)\footnote{Recall that this is the
divisor pole.} then the degree of the function is equal to the degree
of the divisor of the zeros and the divisor of the poles, and thus the
degree of the extension \([K: k(t)] = 1\) and thus \(K = k(t)\) is
rational. So having a rational with a simple pole at only one point
\emph{only} happens in you're in a rational function field.
On the other hand, we both wanted and used in our discussion of
holomorphy rings the fact that given a nonempty finite subset
\(S \subset \Sigma(K/k)\), we want to find a rational function
\(f\in K^{\times}\) has poles at all of the points in \(S\), so
\({\operatorname{supp}}(f)_- = S\). Better yet, we'd like a bound on the
degree of any such \(f\), i.e.~the orders of all of these poles. If
\(S\) is a single place, unless the function field is rational, we can't
require the function to have a pole of degree 1 at that point. But can
it admit a pole of degree at most 10, for example? This is what
motivates the Riemann-Roch spaces and the Riemann-Roch theorem. If
you're trying to give a quantitative bound on how high of an order of a
pole you have to allow in order to have a rational function, this comes
from a key invariant called the \emph{genus} of the function field. The
theorem that will tell us about the existence of rational functions with
poles of prescribed degrees in terms of the genus is precisely the
Riemann-Roch theorem, so that's where we are headed.
\end{remark}
\hypertarget{the-riemann-roch-space}{%
\subsection{The Riemann-Roch Space}\label{the-riemann-roch-space}}
\begin{definition}[Riemann-Roch Space of $D$ (Key Definition)]
For \(D\in \operatorname{Div}K\), the \textbf{Riemann-Roch space} of
\(D\) is defined as
\begin{align*}
\mathcal{L}(D) \coloneqq\left\{{ f\in K^{\times}{~\mathrel{\Big|}~}(t) \geq - D}\right\} \cup\left\{{0}\right\}
.\end{align*}
\end{definition}
\begin{remark}
This will turn out to be a \(k{\hbox{-}}\)vector space, and is a sub
\(k{\hbox{-}}\)vector space of \(K\). One of the first things we'll
prove is that it's always finite dimensional. This is only interesting
when \(D\) is linearly equivalent to an effective divisor, so we should
think of \(D\) as having a nonnegative degree, and in fact itself being
an effective divisor. So this is the space of rational functions that
have prescribes poles of a prescribed order.
\end{remark}
\begin{question}
Does \(\mathcal{L}(D)\) contain any rational functions other than zero?
\end{question}
\begin{answer}
For any nonzero \(f\in \mathcal{L}(D)^{\bullet}\), the divisor
\(D + (f)\) is effective, since \((f) \geq -D\), and also linearly
equivalent to \(D\). If \(D\) is not linearly equivalent to an effective
divisor, this is just the zero vector space.
\end{answer}
\begin{exercise}[?]
Let \(K = k(t)\) and \(n\in {\mathbb{Z}}^{\geq 0}\). Show that
\begin{align*}
L(n\infty) = \left\{{f\in k[t] {~\mathrel{\Big|}~}\deg f \leq n}\right\}
\end{align*}
and in particular is a \(k{\hbox{-}}\)vector space of dimension
\(n+1\).\footnote{Recall that \(\infty\) is the \(1/t{\hbox{-}}\)adic
place.}
\end{exercise}
\begin{remark}
Note that \(\infty\) is a degree 1 place, and multiplying it by \(n\)
yields an effective divisor. The Riemann-Roch space here is comprised of
rational functions that regular away from \(\infty\), which are
polynomials, whose pole at \(\infty\) has order at worst \(n\). But the
order of a pole at infinity is its degree as a polynomial, since the
\(\infty{\hbox{-}}\)adic valuation is the negative degree, so this
yields polynomials of degree at most \(n\).
\end{remark}
\hypertarget{working-with-divisors}{%
\subsection{Working with Divisors}\label{working-with-divisors}}
\begin{lemma}[?]
For \(D\in \operatorname{Div}K\),
\begin{align*}
\mathcal{L}(D) \neq \left\{{0}\right\} \iff 0 \text{ is equivalent to an effective divisor}
.\end{align*}
\end{lemma}
\begin{proof}[?]
\(\implies\): If \(f\in \mathcal{L}(D)^{\bullet}\), then \(D + (f)\) is
effective and linearly equivalent to zero.
\(\impliedby\): If \(D' \geq 0\) and \(D' \sim D\), then
\(D' = D + (f) \geq 0\). So \((f) \geq -D\) and thus
\(f\in \mathcal{L}(D)\).
\end{proof}
\begin{example}[?]
\(\mathcal{L}(0) = \left\{{f {~\mathrel{\Big|}~}(f) \geq 0}\right\} \cup\left\{{0}\right\}\),
which consists of rational functions with no poles (so their divisor is
the zero divisor), and thus \(\mathcal{L}(0) = \kappa(K)\). I.e., these
are the constants: they are regular everywhere and have no zeros or
poles. We would like this space to have \(k{\hbox{-}}\)dimension 1, so
we impose \(\kappa(K) = k\).
\end{example}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
Show that for all \(D\),
\(\mathcal{L}(D) \in {\operatorname{Vect}}_k\).
\item
\begin{align*}
D\sim D' \implies \mathcal{L}(D) \cong_{{\operatorname{Vect}}_k} \mathcal{L}(D')
.\end{align*}
\end{enumerate}
\end{exercise}
\begin{remark}
You can frame the above as taking rational functions with poles of
certain orders, and analyzing the orders of poles of their sums. If you
take \(D'\) and write it as \(D + (f)\) for \(f\) a rational function,
then \(f\) should produce this isomorphism. The moral:
\(\mathcal{L}(D)\) only depends on the linear equivalence class of
\(D\).
\end{remark}
\begin{exercise}[?]
Let \(D\in \operatorname{Div}^0 K\) be a degree zero divisor, then TFAE:
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
\(\dim \mathcal{L}(D) \geq 1\)
\item
\(\dim \mathcal{L}(D) = 1\),
\item
\(D\) is principal, i.e.~the divisor of a rational function or
linearly equivalent to zero.
\end{enumerate}
\end{exercise}
\begin{slogan}
The only way a degree zero divisor can have a nontrivial Riemann-Roch
space is if it's linearly equivalent to zero.
\end{slogan}
\hypertarget{subspaces-and-dimension-of-riemann-roch-spaces}{%
\subsection{Subspaces and Dimension of Riemann-Roch
Spaces}\label{subspaces-and-dimension-of-riemann-roch-spaces}}
\begin{lemma}[?]\label{lemma:divisor_order_to_subspaces}
Let \(A \leq B\)\footnote{These are formal linear combinations of
places, so the coefficients in front of each place in \(A\) should be
less than the corresponding coefficient for \(B\), or equivalently
\(B-A\) is effective.} in \(\operatorname{Div}K\), then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
\(\mathcal{L}(A) \leq_{{\operatorname{Vect}}_k} \mathcal{L}(B)\) is a
subspace,
\item
\(\dim \mathcal{L}(B) / \mathcal{L}(A) \leq \deg B - \deg A = \deg(B - A)\).
\end{enumerate}
\end{lemma}
\begin{remark}
Since \(B \geq A\), you can think of this as starting with \(A\) and
adding an effective divisor to get \(B\), namely \(A + (B-A) = B\). How
much does that decrease the dimension of the Riemann-Roch space? At
most, by the degree of \(B-A\) as a divisor.
\end{remark}
\begin{corollary}[?]
For \(D\in \operatorname{Div}K\),
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
If \(\deg D < 0\) then \(\mathcal{L}(D) = 0\).
\item
If \(\deg (D) \geq 0\) then
\(\dim_k \mathcal{L}(D) \leq \deg(D) + 1 < \infty\).
\end{enumerate}
\end{corollary}
\begin{remark}
This shows that Riemann-Roch spaces are always finite dimensional, and
also gives a simple upper bound on that dimension.
\end{remark}
\begin{proof}[of corollary]
For (a), a divisor of negative degree is not linearly equivalent to an
effective divisor, so we might as well assume it's effective.
For (b), the dimension of \(\mathcal{L}(D)\) doesn't change if \(D\) is
replaced by a linearly equivalent divisor, so wlog assume \(D\) is
effective. Now write \(D = \sum_{i=1}^r p_i\) as a sum of not
necessarily distinct places, and use the lemma: each time you add an
effective divisor, the dimension either stays the same or increases by
at most the degree of the added divisor. So start with the zero divisor,
use the fact that \(\dim_k \mathcal{L}(0) = 1\), and apply the lemma
\(r\) times. This yields a space of dimension at most
\(1 + \sum \deg p_i = \deg D\).
\end{proof}
\begin{proof}[of lemma, part (a)]
If \(A\leq B\) and \(f\in \mathcal{L}(A)\), then \((f) \geq - A\). Since
\(-A \geq -B\), we have \((f) \geq -A \geq -B\), so
\(f\in \mathcal{L}(B)\).
\end{proof}
For the next part, it's perhaps easiest to consider the case
\(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\) so
everything has degree 1. If you go from a divisor to adding a single
degree 1 place, this lemma says that if you increase your Riemann-Roch
space by either allowing a pole at a point you didn't allow before or
allowing a pole of order 1 greater, then the dimension increases by at
most 1.
\begin{proof}[of lemma, part (b)]
From the previous argument, we see that it's enough to do this one place
at a time. So we can easily reduce to the case \(B = A + P\) for \(P\)
some place of degree not necessarily equal to 1 (since we're not
assuming
\(k=\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)),
using that fact that \(B \geq A\). So choose an element \(t\in K\) such
that
\begin{align*}
v_p(t) = v_p(B) = v_p(A) + 1
,\end{align*}
since \(B\) is built from \(A\) by adding a single copy of \(P\). For
\(f\in \mathcal{L}(B)\), we have by definition\footnote{Note that
\(v_p\) is the \(p{\hbox{-}}\)adic valuation, i.e.~the coefficient of
\(P\) in the divisor as a formal linear combination of points.}
\begin{align*}
v_p(f) \geq -v_p(B) = -v_p(t)
,\end{align*}
and so by bringing \(t\) to the other side we get \(v_p(ft) \geq 0\) and
thus \(ft\in R_p\) (the corresponding local ring). This allows us to
define a \(k{\hbox{-}}\)linear map
\begin{align*}
\psi: \mathcal{L}(B) &\to k(P) = R_p/\mathfrak{m}_p \\
f & \mapsto ft \pmod\mathfrak{m}_p
.\end{align*}
In words, we multiply \(f\) by \(t\) to make it \(p{\hbox{-}}\)adically
regular, then look at its image in the residue field. The kernel is
precisely those elements \(x\) such that multiplying by \(t\) lands in
the maximal ideal \(\mathfrak{m}_p\), which means that \(v(x)\) as 1
more than it could have been. So the kernel is all elements such that
multiplying by \(t\) and taking the valuation gives at least one, thus
\begin{align*}
\ker \psi = \left\{{f\in \mathcal{L}(B) {~\mathrel{\Big|}~}v_p(f) \geq -v_p(t) + 1 = -v_p(A)}\right\} = \mathcal{L}(A)
,\end{align*}
which follows since \(B\) and \(A\) only differ at \(P\), since
\(B = A+P\), so the divisors \(A, B\) have the same coefficient at every
other place. We thus have the following diagram:
\begin{center}
\begin{tikzcd}
{0} & {\mathcal{L}(A)} & {\mathcal{L}(B)} & {\mathcal{L}(B)/\mathcal{L}(A)} & {0} \\
\\
{} & {} & {} & k(P) = {R_p / \mathfrak{m}_p} & {\cdots}
\arrow[from=1-1, to=1-2, hook]
\arrow[from=1-2, to=1-3, hook]
\arrow[from=1-3, to=1-4, two heads]
\arrow[from=1-4, to=1-5, two heads]
\arrow[from=1-4, to=3-4, dotted, hook, "\exists \iota"]
\arrow[from=1-2, to=3-4, "\psi"]
\arrow[from=3-4, to=3-5]
\end{tikzcd}
\end{center}
\href{https://q.uiver.app/?q=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}{Link
to diagram}
where we can conclude that the indicated injection exists, and thus
\begin{align*}
\dim \mathcal{L}(B) / \mathcal{L}(A) \leq [k(p) : k] = \deg P
.\end{align*}
\end{proof}
\begin{fact}
For \(p\in \Sigma(K/k)\) with residue field \(k_p\) and
\([k_p: k] = d\), defining \(K_p\) as the completion of \(K\) with
respect to \({\left\lvert {{\,\cdot\,}} \right\rvert}_p\), there is an
isomorphism \(K_p \cong k_p((t))\), a formal Laurent series field. One
issue is that if \(d =1\) then \(k \subset k_p\), but not for general
\(d\geq 2\). However, taking the completion results in \(k \subset K_p\)
again. This shouldn't be too surprising from the perspective of local
fields in NTII. There is a structure theory of complete discretely
valued fields. This is an \emph{equicharacteristic} such field, i.e.~the
characteristic of the field agrees with that of the residue field, and
all equicharacteristic discretely valued fields will be isomorphic to a
ring of formal Laurent series. This isn't a fact of the geometry of
curves.
\end{fact}
\hypertarget{bounds-on-dimensions}{%
\subsection{Bounds on Dimensions}\label{bounds-on-dimensions}}
\begin{definition}[$\ell(D)$: The dimension of a Riemann-Roch space]
For \(D\in \operatorname{Div}K\), define
\begin{align*}
\ell(D) \coloneqq\dim_k \mathcal{L}(D)
.\end{align*}
\end{definition}
\begin{exercise}[?]
If \(D\in \operatorname{Div}k(t)\), show that
\begin{align*}
\ell(D) =
\begin{cases}
\deg(D) + 1 & \deg D \geq 0 \\
0 & \text{else}.
\end{cases}
\end{align*}
\end{exercise}
\begin{remark}
Recall that in a rational function field, every degree zero divisor is
principal, and if you adjust by a principal divisor, you don't change
\(\ell(D)\). This means that in any rational function field, any two
divisors of the same degree are going to be linearly equivalent, and
thus \(\ell(D)\) will only depend on \(\deg D\). So rational function
fields are much simpler than the fully general case.
\end{remark}
\begin{problem}[The Riemann-Roch Problem]
Give good upper and lower bounds on \(\ell(D)\) and especially
\(\ell(nD)\) as a function of \(n\).
\end{problem}
\begin{remark}
The stronger version of knowing \(\ell(D)\) in all cases is unsolvable.
If we knew the dimension of every Riemann-Roch space, then we would know
too much! E.g. about Weierstrass points on elliptic curves. (?) Looking
at positive multiples \(nD\) of a single divisor is common. If \(D\) is
a single point, then the support of the divisor is the collection of
places that appear with nonzero coefficients, \(nD\) has the same
support. This is analogous to not allowing poles at new points, but
rather allowing poles at the same points of higher order. So it's
reasonable to ask about asymptotic behavior of \(\ell(nD)\) in \(n\).
Secretly this is a kind of Hilbert function computation: if you have a
graded algebra and you look at dimensions of its graded pieces, then
there is a theorem that the Hilbert function is a polynomial for
\(n\gg 1\). Here, \(\ell(nD)\) will be a linear polynomial for
\(n\gg 1\) by the Riemann-Roch theorem, so there are some stabilization
phenomena, but given a random divisor of low degree it is difficult to
determine \(\ell(D)\).
\end{remark}
\begin{remark}
The last corollary gave us a lower bound:
\begin{align*}
\deg(D) \geq 0 \implies \deg(D) - \ell(D) \geq -1
.\end{align*}
This can also be thought of as an lower bound on \(\ell(D)\) in terms of
\(\deg(D)\), and next up we'll try to find an upper bound:
\end{remark}
\begin{proposition}[?]
There exists a \(\delta = \delta(K/k) \in {\mathbb{Z}}\) such that for
all \(A\in \operatorname{Div}K\), we have
\begin{align*}
\deg A - \ell(A) \leq \delta
.\end{align*}
\end{proposition}
\hypertarget{lecture-8-riemann-roch-spaces-part-2}{%
\section{Lecture 8: Riemann-Roch Spaces (Part
2)}\label{lecture-8-riemann-roch-spaces-part-2}}
Recall the proposition we ended with last time:
\begin{proposition}[?]\label{prop:deg_bounded_above}
There exists a \(\delta = \delta(K/k) \in {\mathbb{Z}}\) such that for
all \(A\in \operatorname{Div}K\), we have
\begin{align*}
\deg A - \ell(A) \leq \delta
.\end{align*}
\end{proposition}
\begin{exercise}[?]
This proposition is enough to show the existence of rational functions
whose polar divisor has as its support any finite subset
\(S \subset \Sigma(K/k)\).
\end{exercise}
Most of the lecture will be the proof of this statement.
\hypertarget{proof-of-upper-bound}{%
\subsection{Proof of Upper Bound}\label{proof-of-upper-bound}}
Rewriting \cref{lemma:divisor_order_to_subspaces} yields
\begin{align*}
A_2, A_2 \in \operatorname{Div}K,\, A_1 \leq A_2 \implies \deg A_1 - \ell(A_1) \leq \deg A_2 - \ell(A_2)
.\end{align*}
We now proceed to prove \cref{prop:deg_bounded_above} in several steps.
\hypertarget{step-1}{%
\subsubsection{Step 1}\label{step-1}}
Choose an \(x\in K\setminus k\) and set \(B\coloneqq(x)_-\).
\begin{claim}
There exists a \(C\geq 0\) such that for all \(n\geq 0\),
\begin{align*}
\ell(nB + C) \geq (n+1) \deg B
.\end{align*}
\end{claim}
So we give ourselves a certain effective divisor: the divisor of poles
of an arbitrary nonconstant element. We can then get a preliminary
asymptotic lower bound, not on the same Riemann-Roch space, but on a new
one after augmenting the space by some fixed effective divisor \(C\).
\begin{proof}[?]
Since \(K/k(x)\) has finite degree, let \(u_1,\cdots, u_d\) be a basis
for \(K\) consisting of finitely many rational functions. Note that
\(d = [K: k(x)]\), and is also equal to \(\deg B\) since \(B\) was a
divisor of poles. Noting that the divisor groups are free commutative
groups, so taking any finite number of elements in
\(\bigoplus {\mathbb{Z}}\), we can find an element that is less than or
equal to all of them. Thus we can choose a \(C\geq 0\) such that
\begin{align*}
(u_i) \geq -C && \forall 1\leq i \leq d
.\end{align*}
Since the \(u_i\) are \(k(x){\hbox{-}}\)linearly independent in \(K\),
the functions
\(\left\{{x^i u_j {~\mathrel{\Big|}~}\, 0\leq i \leq n,\,\, 1\leq j \leq d}\right\}\)
are \(k{\hbox{-}}\)linearly independent, since any \(k{\hbox{-}}\)linear
relation would immediately yield a \(k(x){\hbox{-}}\)linear relation
among the \(u_i\).
\begin{exercise}[?]
If \(f_i\in \mathcal{L}(D_i)\), so the poles of \(f\) are no worse than
\(D_i\), then the poles of \(f_1 f_2\) are bounded by \(D_1 + D_2\) and
thus \(f_1 f_2 \in \mathcal{L}(D_1 + D_2)\).
\end{exercise}
Now we can note that there are \((n+1)d = \deg B\) many elements here,
and moreover, these all lie in \(\mathcal{L}(nB + C)\) since each
\((u_j) \geq -C\) and \((x) \geq - B\) and \(i\leq n\). From this we can
conclude
\begin{align*}
\ell(nB+c) \geq (n+1) d = (n+1) \deg B
.\end{align*}
\end{proof}
\hypertarget{step-2}{%
\subsubsection{Step 2}\label{step-2}}
We'll now show that throwing in the fixed divisor \(C\) can't increase
the Riemann-Roch space that much, and in fact
\begin{align*}
\ell(nB + C) \leq \ell(nB) + \deg C
,\end{align*}
and so we get a bound
\begin{align*}
\ell(nB)
&\geq \ell(nB + C) - \deg C \\
&\geq (n+1) \deg B - \deg C \\
&= \deg(nB) + \qty{ [K:k(x)] - \deg C} \\
&\coloneqq\deg(nB) \pm \gamma
,\end{align*}
which shows that
\begin{equation}
\forall n\geq 0,\, \, \deg(nB) - \ell(nB) \leq \gamma
\label{eq:step_2_bound}
.\end{equation}
A problem here is that \(\gamma\) depends upon everything that we've
done so far, and this inequality only holds for multiples of a fixed
divisor (an infinite ray emanating from \(B\)).
\hypertarget{step-3}{%
\subsubsection{Step 3}\label{step-3}}
\begin{claim}
For all \(A\in \operatorname{Div}K\), there exist
\(A_1, D \in \operatorname{Div}K\) and \(n\geq 0\) such that
\(A \leq A_1\), \(A_1 \sim D\), and \(D\leq nB\). I.e. although it can't
literally be true that \(A \leq nB\), it will be up to linear
equivalence.
\end{claim}
To see this, set \(A_1 \coloneqq\max(A, 0)\). Using the bound from
\cref{eq:step_2_bound}, for \(n\gg 0\) we have
\begin{align*}
\ell(nB - A_1)
&\geq \ell(nB) - \deg A_1 \\
&\geq \deg(n B) - \gamma - \deg A_1 \\
&> 0
,\end{align*}
and so there exists a \(z\in \mathcal{L}(nB - A_1)^{\bullet}\), a
nontrivial element in the linear system.
\begin{remark}
The first inequality is an application of our lemma because \(A_1\) is
effective, which was the point of this maneuver. I.e., in order to get
from \(nB - A_1\) to \(nB\), we added \(A_1\), which can only increase
the dimension of the space by at most \(\deg A_1\). Finally, in the last
inequality, we use the fact that \(B\) has positive degree since it's a
divisor of poles of a nonconstant rational function, and the remaining
terms don't depend on \(n\), so we can make \(\deg(nB)\) arbitrarily
large.
\end{remark}
So now set \(D \coloneqq A_1 - (z)\), then \(A_1 \sim D\) and since it's
in the linear system,
\begin{align*}
(z) \geq - (nB - A_1) = A_1 - nB
\end{align*}
so \(-(z) \leq nB - A_1\) and by adding \(A_1\) to both sides, we obtain
\begin{align*}
0 = A_1 - (z) \leq nB
.\end{align*}
What have we shown? For any divisor \(D\), we can make it less than
\(nB\) for some \(n\), up to linear equivalence.
\hypertarget{step-4}{%
\subsubsection{Step 4}\label{step-4}}
Finally, for \(A\in \operatorname{Div}K\), choose \(A_1, D\) as in the
previous step, so \(A\leq A_1 \sim D \leq nB\). Then
\begin{align*}
\deg A - \ell(A)
&\leq \deg(A_1) - \ell(A) && \text{using } A\leq A_1\\
&= \deg(D) - \ell(D) && \text{changing within linear equivalence class}\\\
&\leq \deg(nB) - \ell(nB) \\
&\leq \gamma
.\end{align*}
\(\hfill\blacksquare\)
\hypertarget{genus}{%
\subsection{Genus}\label{genus}}
\begin{definition}[Genus (Important!)]
The \textbf{genus} of \(K/k\) is defined as
\begin{align*}
g \coloneqq\max_{A\in \operatorname{Div}K}(\deg(A) - \ell(A) + 1)
.\end{align*}
This exists by the \cref{prop:deg_bounded_above}, since this set is
bounded above.
\end{definition}
\begin{exercise}[?]
Show that \(g\geq 0\) always and
\begin{align*}
g(k(t)/k) = 0
.\end{align*}
\end{exercise}
\begin{remark}
Note that if the \(+1\) is mostly a correction factor to match up with
the topological genus of \({\mathbb{P}}^1_{/{\mathbb{C}}}\). That the
genus is non-negative should come from the lower bound we had from
before. It turns out that over \(k={\mathbb{C}}\), this genus will agree
on the nose with the topological genus of the corresponding compact
Riemann surface.
\end{remark}
\begin{theorem}[Riemann's Inequality]
If \(K/k\) is a function field of genus \(g\),
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
For all \(A\in \operatorname{Div}K\),
\begin{align*}
\ell(A) \geq \deg(A) + 1 - g
.\end{align*}
\item
There exists a \(c = c(K) \in {\mathbb{Z}}\) such that for all
\(A \in \operatorname{Div}K\),
\begin{align*}
\deg(A) \geq c \implies \ell(A) = \deg(A) - g + 1
.\end{align*}
\end{enumerate}
\end{theorem}
\begin{remark}
This says that the dimension of the linear system is very close to the
degree of the corresponding divisor, and is only off by a constant
factor \(g\). Part (a) is literally just a rearrangement of the
definition of the genus. Part (b) says that if you assume \(A\) has
sufficiently large degree, this upper bound becomes an equality.
\end{remark}
\begin{proof}[of b]
By the definition of \(g\), since it is a maximum there exists an
\(A_0\) such that
\begin{align*}
g = \deg(A_0) - \ell(A_0) + 1
.\end{align*}
Set \(c\coloneqq\deg(A_0) + g\). Then if \(\deg(A) \geq c\), we have
\begin{align*}
\ell(A - A_0)
&\geq \deg(A-A_0) - g + 1 \\
&\geq c - \deg(A_0) - g + 1 \\
&= 1
,\end{align*}
so there exists a \(z\in \mathcal{L}(A- A_0)^{\bullet}\) since the
dimension is at least 1.
Now set \(A' \coloneqq A + (z)\), and note that \(A' \geq A_0\). Thus
\begin{align*}
\deg(A) - \ell(A)
&= \deg(A') - \ell(A') \\
&\geq \deg(A_0) - \ell(A_0) && \text{by the lemma}\\
&=g-1
.\end{align*}
By maximality of the genus, we have \(\deg(A) - \ell(A) \leq g-1\),
which forces equality
\end{proof}
Next up: how to we make this inequality into an equality? It turns out
that there is some different divisor \(D'\) and we can subtract off
\(\ell(D')\), and that will be the Riemann-Roch theorem.
\hypertarget{lecture-9}{%
\section{Lecture 9}\label{lecture-9}}
Last time: we proved the Riemann Inequality
(\cref{thm:riemann_inequality}), the content of which is that there is a
non-negative quantity called the \emph{genus} for which the stated
inequality holds. The next step will be to try to improve this
inequality to an equality, for which we introduce the following
definition:
\begin{definition}[Index of Speciality]
\begin{align*}
\iota(A) \coloneqq\ell(A) - \deg A = g - 1 \geq 0
.\end{align*}
\end{definition}
Tautologically, this yields
\begin{align*}
\ell(A) - \iota(A) = \deg A - g + 1
.\end{align*}
\hypertarget{riemann-roch-theorem-and-applications}{%
\subsection{Riemann-Roch Theorem and
Applications}\label{riemann-roch-theorem-and-applications}}
\begin{theorem}[Riemann-Roch Theorem]
If \(K/k\) is a function field of genus \(g\), there exists a divisor
\(\mathfrak{k} \in \operatorname{Div}K\) such that for all
\(D\in \operatorname{Div}K\), \(\iota(D) = \ell(\mathfrak{k} - D)\), and
thus
\begin{align*}
\ell(D) - \ell(\mathfrak{k} - D) = \deg D - g + 1
.\end{align*}
\end{theorem}
\begin{remark}
As a result, \(\deg(D) > \deg(\mathfrak{k})\), so \(\mathfrak{k} - D\)
has negative degree. We know that \(\ell(D) > 0\) iff there is a
\(D' \sim D\) linearly equivalent to \(D\), and there's no effective
divisor equivalent to a divisor of negative degree. Thus
\(\iota(A) = 0\) as soon as \(\deg(D) > \deg(\mathfrak{k})\).
\end{remark}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
\(\ell(\mathfrak{k}) = g\) and \(\deg \mathfrak{k} = 2g-2\).
\item
\(g\geq 0\)
\item
If \(g\geq 1\), then the least \(\alpha\in {\mathbb{Z}}\) such that
whenever \(\deg(D) > \alpha\) we have \(\iota(D) = 0\) is given by
\(\alpha \coloneqq 2g-2\).
\end{enumerate}
\end{exercise}
\begin{remark}
Try taking \(D=0\) and \(\mathfrak{k} = D\) respectively, and compute
\(\iota(\mathfrak{k}) = 1\).
\end{remark}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that the Riemann-Roch theorem uniquely characterizes \(g\),
i.e.~there is at most one quantity \(g\) for which it holds.
\item
Show that if Riemann-Roch holds for \(\mathfrak{k}\), then it holds
for any \(\mathfrak{k}' \sim \mathfrak{k}\).
\item
Show that if Riemann-Roch holds for \(\mathfrak{k}_1, \mathfrak{k}_2\)
then \(\mathfrak{k}_1 \sim \mathfrak{k}_2\).
\end{enumerate}
\end{exercise}
\begin{definition}[Canonical Class]
Thus the Riemann-Roch theorem singles out a distinguished \emph{class}
of divisors \([\mathfrak{k}]\) of degree \(2g-2\), which is called the
\textbf{canonical class}. Any divisor \(D\in [\mathfrak{k}]\) is called
a \textbf{canonical divisor}.
\end{definition}
\begin{exercise}[?]
Let \(K/k\) be a genus zero function field.
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that \(\operatorname{Cl}^0(K) = 0\), so degree zero divisors are
principal.
\item
Show that \(D\in \operatorname{Div}K\) is canonical iff
\(\deg D = -2\).
\item
Show that the index \(I(K)\) can only be 1 or 2.
\item
Show that \(K \cong k(t)\) iff \(\Sigma _1 (K/k) \neq \emptyset\) iff
\(I(K) = 1\).
\end{enumerate}
\end{exercise}
\begin{exercise}[?]
If \(K/k\) is genus 1, show that a divisor \(D \in \operatorname{Div}K\)
is canonical iff \(D\) is principal.
\end{exercise}
\hypertarget{applications-of-riemann-roch}{%
\subsection{Applications of
Riemann-Roch}\label{applications-of-riemann-roch}}
\hypertarget{genus-zero-function-fields}{%
\subsubsection{Genus Zero Function
Fields}\label{genus-zero-function-fields}}
Let \(K/k\) be a genus zero function field with \(\mathfrak{k}\) a
canonical divisor, so \(\deg \mathfrak{k} = -2\). Then
\(\deg(-\mathfrak{k}) = 2\) and by Riemann-Roch every effective divisor
is not principal and
\begin{align*}
\ell(-\mathfrak{k}) = \deg(-\mathfrak{k}) -g + 1 = 3
.\end{align*}
One can choose \(-\mathfrak{k}\) to be effective, so
\(1 \in \mathcal{L}(-\mathfrak{k})\). So \(\mathcal{L}(\mathfrak{k})\)
has a basis of the form \(\left\{{1,x,y}\right\}\) for some elements
\(x, y\).
\begin{claim}
\begin{align*}
K = k(x, y)
.\end{align*}
\end{claim}
\begin{proof}[?]
\textbf{Case 1}: If either \(x\) or \(y\) has degree 1, this is also the
degree of the function field \(K / k(x)\), and then
\(\mathfrak{k} = k(x)\) or \(k(y)\) and the function field is rational.
\textbf{Case 2}: Since \((x) \geq -\qty{ -\mathfrak{k}}\), we have
\((x)_- \leq - \mathfrak{k} + (x)_+\), so
\(\deg(x) \leq \deg(-\mathfrak{k}) = 2\). So \(\deg(x) = 2\). Similarly,
\(deg(y) = 2\) and so \([K: k(x)] = 2\). Thus it's enough to show that
\(y\not\in k(x)\). Toward a contradiction, if \(y\in k(x)\), since
\([K: k(y)] = 2\), we get \(k(x) = k(y)\). In this case \(y\) is a
degree 1 rational function in \(x\), and so is of the form
\(y = {ax+ b \over cx + d}\) with \(a,b,c,d \in k\). This forces
\((x)_- = (y)_- = -\mathfrak{k}\), since the only possibilities for
these divisors are having degrees zero or one, and we've ruled out the
degree zero possibility.
So \(y\) only has poles where \(x\) has poles. This follows because for
a map \(f: k \to k(t)\), the places over the point at infinity are
places over infinity:
\begin{figure}
\centering
\includegraphics{figures/image_2020-12-23-00-32-06.png}
\caption{Image}
\end{figure}
Since \((x), (y)\) has the same divisor of poles in \(K\), all of their
poles like over \(\infty\in k(t)\). Moreover, since \(y\) is regular
away from \(\infty \in k(x)\), this forces \(c= 0\) and \(y=ax+b\). But
this exhibits a \(k{\hbox{-}}\)linear dependence between \(x, y\).
\end{proof}
Now consider \(1,x,y,x^2, xy, y^2 \in \mathcal{L}(-2 \mathfrak{k})\).
Since \(\ell(-2 \mathfrak{k}) = 5\), these is necessarily a linear
dependence in this set, so there is a tupoe
\((a,b,c,d,e,f) \subset k^6\) such that
\begin{align*}
f(x, y) \coloneqq ax^2 + bxy + xy^2 + dx + ey + f = 0
.\end{align*}
Not all of \(a,b,c\) can be zero, since this yields the linear
dependence \(dx + ey + 1 = 0\). Moreover \(f\) must be irreducible: if
not, it would have a linear factor, which would again yield a linear
dependence between \(x\) and \(y\).
\begin{theorem}[Genus Zero Function Fields are Quadratic Extensions]
For \(K/k\) a genus zero function field, there exists
\(a,b,c,d,e,f\in k\) with \(a,b,c\) not all zero such that
\begin{align*}
f(x, y) \coloneqq ax^2 + bxy + xy^2 + dx + ey + f = 0 {\quad \operatorname{and} \quad} K = K_f
.\end{align*}
\end{theorem}
\begin{exercise}[?]
Suppose \(\operatorname{ch}(k) \neq 2\), and show that every genus zero
function field is regular and there exist \(a,b,c\in k^{\times}\) such
that \(K = K_f\) where \(f(x, y) = ax^2 + by^2 + c\).
\end{exercise}
\hypertarget{lecture-10a-todo}{%
\section{Lecture 10A (Todo)}\label{lecture-10a-todo}}
\hypertarget{lecture-10b-todo}{%
\section{Lecture 10B (Todo)}\label{lecture-10b-todo}}
\hypertarget{lecture-10c-todo}{%
\section{Lecture 10C (Todo)}\label{lecture-10c-todo}}
\hypertarget{lecture-11a-weils-proof-of-riemann-roch}{%
\section{Lecture 11A: Weil's Proof of
Riemann-Roch}\label{lecture-11a-weils-proof-of-riemann-roch}}
Let \(K_{/k}\) be a one variable function field, finitely generated of
transcendence degree one, with \(\kappa(K) = k\), so \(k\) is
algebraically closed in \(K\). Define the \emph{small Adele ring}
associated to \(K\), as the restricted direction product with respect to
\(\left\{{R_v {~\mathrel{\Big|}~}v\in \Sigma(K/k)}\right\}\):
\begin{align*}
A_k \coloneqq\prod^{\text{res}}_{v\in \Sigma(K/k)} K
= \left\{{(x_v) \in K^{\Sigma(K/k)} {~\mathrel{\Big|}~}x_v \in R_v \text{ for a.e. } v }\right\}
,\end{align*}
where each factor is a copy of \(K\). Note that a \emph{restricted}
direct product is when you have a family of sets, and for each set you
also attach a subset. Then if you have a tuple in the entire direct
product, it's in the restricted direct product iff for all but finitely
many coordinates lie in a given subset. Here the subset is the valuation
ring \(R_v\). So these are tuples of elements of \(K\), indexed by
places, where each element has a \(p{\hbox{-}}\)adic valuation and the
only restriction is that (except for finitely many cases) we want this
valuation to be nonnegative.
\begin{remark}
To get the \emph{big} Adele ring, you'd replace \(K\) with its
completion with respect to the \(p{\hbox{-}}\)adic valuation. If \(k\)
is finite, then this is equal to the positive characteristic Adele ring
from NTII. If you complete, then you get a complete discretely valued
field whose residue field equals the residue field at the place \(v\).
So for finite extensions of \(k\), the residue will be finite iff \(k\)
is finite, and from the structure theory of discretely valued fields,
this field has a natural topology: the adic topology, coming from the
inverse limit. This will be locally compact iff the residue field is
finite. Here, since the ground field is infinite, even passing to the
completion wouldn't yield anything locally compact. So there's no
advantage to passing to the completion, although there's no harm either.
\end{remark}
Note that \(A_k\) is a ring, and in fact a \(K{\hbox{-}}\)algebra, but
we will only need its structure as a \(K{\hbox{-}}\)vector space. This
structure comes from embedding \(K\hookrightarrow A_k\) diagonally, so
\(x \mapsto [x, x, \cdots]\), and pull back \(v\in \Sigma(K/k)\),
remembering that every element of \(K\) (a rational function) is regular
except for finitely many \(v\).
If we have a valuation on \(K\), we can consider a place \(p\) and
projecting onto the \(k\)th factor:
\begin{align*}
A_j
\xrightarrow[]{\pi_p}\mathrel{\mkern-14mu}\rightarrow
K \xrightarrow{v_p} {\mathbb{Z}}\cup\left\{{\infty}\right\}
.\end{align*}
So we now attach an adelic version of the Riemann-Roch space: for
\(D\in \operatorname{Div}K\), we set
\begin{align*}
\mathcal{A}_k(D) \coloneqq\left\{{ \alpha \in \mathcal{A}_K {~\mathrel{\Big|}~}v_p(\alpha) \geq -v_p(D) \, \forall p \in \Sigma(K/k)}\right\}
.\end{align*}
The only difference here is that the usual space is over \(K\), and here
we're over \(\mathcal{A}_K\), which is a much larger space. This makes
things easier, however, in the same sense that studying a large
collection of local fields is easier than studying the corresponding
global field. Note that the \(p{\hbox{-}}\)adic valuation \(v_p\) is
just the coefficient of \(p\) in the divisor, and
\(\mathcal{A}_K \cap K\) yields the usual Riemann-Roch space.
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that \(\mathcal{A}_K(D)\) is an \(k{\hbox{-}}\)subspace of
\(\mathcal{A}_K\).\footnote{Consider scaling by nonzero constants,
where the valuation of constants are zero.}
\item
Show that (just as for the Riemann-Roch space)
\(D_1 \leq D_2 \implies \mathcal{A}_K(D_1) \subseteq \mathcal{A}_K(D_2)\).
\end{enumerate}
\end{exercise}
\begin{lemma}[?]
\begin{align*}
D_1 \leq D_2 \implies \dim_k \mathcal{A}_K(D_2) / \mathcal{A}_K(D_1) = \deg D_2 - \deg D_1
.\end{align*}
\end{lemma}
Note that this is the adelic analogue of our first lemma on Riemann-Roch
spaces, now with an equality instead of being bounded above.
\begin{proof}[?]
As we did before, by induction we can assume \(D_2 = D_1 + p\) for some
\(p\in \Sigma(K/k)\), i.e.~we can go from the smaller divisor to the
bigger one by repeatedly adding closed points. Then choose an element
\(t\in k^{\times}\) such that \(v_p(t) = v_p(D_2)\), and define a
similar map
\begin{align*}
\phi \mathcal{A}_K(D_2) &\to k_p \\
\alpha &\mapsto (t\alpha p) \pmod{\mathfrak{m}}_p
.\end{align*}
Why? Once you multiply by \(t\), note that we're looking in the \(p\)th
component. The condition before was that the valuation at the \(p\)th
component was at least \(-v_p(D_2)\), but now we're adding \(v_p(D_2)\).
This yields a nonnegative valuation, making the image lie inside the
corresponding local ring, so it makes sense to consider it modulo the
maximal ideal to get an element of the residue field. As before, it
should be clear that this is \(k{\hbox{-}}\)linear,
\(\ker \phi = \mathcal{A}_K(D_1)\), and is surjective. The kernel are
exactly those elements such that multiplying by \(t\) makes the
\(p{\hbox{-}}\)adic valuation at least 1, since that's what the maximal
ideal is. This is indeed \(\mathcal{A}_K(D_1)\), since \(D_1\) and
\(D_2\) are the same except for the added condition \(D_2 = D_1 + p\) at
\(p\).\\
So the main difference is that the map is now \emph{surjective}, which
was not true for the original Riemann-Roch space. Why? This is a purely
local situation. Take an element which is zero away from the \(p\)
component, which is easy to do since zero is in \(R_v\) for any \(v\).
So can you find an element of \(k\) such that multiplying by \(t\) and
reducing modulo the maximal ideal yields every element of the residue
field?
\end{proof}
\begin{theorem}[2.13]
For all \(D\),
\begin{align*}
\dim_k \mathcal{A}_K / \qty{\mathcal{A}_K(D) + K } = \iota(D) \coloneqq\ell(D) - \deg(D) + g - 1
,\end{align*}
where \(\iota(D)\) is the index of speciality of the divisor, which
measures the discrepancy between the degree and the dimension.
\end{theorem}
\begin{remark}
This says that adding \(K\) into the adelic Riemann-Roch space results
in a big \(k{\hbox{-}}\)vector space, having high dimension in the
infinite dimensional \(k{\hbox{-}}\)vector space \(\mathcal{A}_K\).
\end{remark}
\begin{proof}[Step 1]
For divisors \(A_1 \leq A_2\), we have a short exact sequence of
\(k{\hbox{-}}\)vector spaces
\begin{align*}
0 \to \mathcal{L}(A_2) / \mathcal{L}(A_1) \xrightarrow{\sigma_1} \mathcal{A}_K(A_2) / \mathcal{A}_K(A_1) \xrightarrow{\sigma_2} \qty{\mathcal{A}_K(A_2) + K} / \qty{\mathcal{A}_K(A_1) + K } \to 0
.\end{align*}
The first thing we did was compute the dimension of the middle quotient
space, which was \(\deg D_2 - \deg D_1\). Note that \(\sigma_2\) is a
quotient map, but \(\sigma_1\) just comes from embedding
\(K\hookrightarrow\mathcal{A}_K\). To show exactness, the only
nontrivial part is that
\(\ker(\sigma_2) \subset \operatorname{im}(\sigma_1)\). So take an
element \(\alpha\in \mathcal{A}_K(A_1) \pmod\mathcal{A}_K(A_1)\) such
that \(\sigma_2(\alpha) = 0\), so there exists an \(x\in K\) such that
\(\alpha-x \in \mathcal{A}_K(A_1)\) by definition of being zero in the
last quotient. Since
\(\mathcal{A}_K(A_1) \subseteq \mathcal{A}_K(A_2)\), we have that
\(x\in \mathcal{A}_K(A_2) \cap K \coloneqq\mathcal{L}(A_2)\). This
follows because \(\alpha, \alpha-x\) are both in \(\mathcal{A}_K(A_2)\).
Thus we have
\begin{align*}
\alpha + \mathcal{A}_K(A_1) = x + \mathcal{A}_K(A_1) = \sigma\qty{x + \alpha(A_1)}
.\end{align*}
\end{proof}
\begin{proof}[Step 2]
We can now compute the dimension of this quotient. Using step 1 and
Lemma 2.12, we get
\begin{align*}
\dim_k\qty{ \mathcal{A}_K(A_2) + K } / \qty{ \mathcal{A}_K(A_1) + K }
&= \dim_k \mathcal{A}_K(A_2) / \mathcal{A}_K(A_1) - \dim_k \mathcal{L}(A_2) / \mathcal{L}(A_1) \\
&= \qty{\deg A_1 - \ell(A_2) } - \qty{\deg A_1 - \ell(A_1) } \\
&= \iota(A_1) - \iota(A_2)
,\end{align*}
where the last step follows from adding and subtracting \(g-1\).
\end{proof}
\begin{proof}[Step 3]
By step 2, it is enough to show that for all
\(A_1 \in \operatorname{Div}K\), there exists a bigger divisor
\(A_2 \geq A_1\) such that \(\iota(A_2) = 0\) (by just adding closed
points) and \(\mathcal{A}_K(A_2) + K = \mathcal{A}_K\). By Riemann's
inequality, we have \(\iota(A_2) =0\) if \(\deg A_2 \gg 0\), so choose
such an \(A_2 \geq A_1\). Thus we're reduced to showing that if
\(\iota(B) = 0\) for all \(B\in \operatorname{Div}K\), then
\(\mathcal{A}_K = \mathcal{A}_K(B) + K\). We'll do this by choosing
another large effective divisor.\footnote{This ``cone structure'' on
divisors is very useful!}
Let \(B_1 \geq B\), then we have
\begin{align*}
\ell(B_1)
&\leq \deg(B_1) + \ell(B) - \deg(B) \\
&= \deg(B_1) - g + 1
.\end{align*}
Also, Riemann's inequality gives \(\ell(B_1) \geq \deg(B_1) - g + 1\),
so we have equality. Thus any divisor greater than or equal to a
non-special divisor is again non-special.\\
We want to take an arbitrary element of the Adele ring and show that it
differs from an element of the adelic Riemann-Roch space associated to
\(B\) by an element of \(K\), so we'll cleverly choose a divisor in
order to do this. So take an arbitrary element
\(\alpha\in \mathcal{A}_K\) of the Adele ring, then we may choose
\(B_1 \geq B\) such that \(\alpha\in \mathcal{A}_K(B_1)\). I.e.,
choosing \(B_1\) large enough is allowing the poles to be however bad
you want them to be, and \(\alpha\) is a fixed element, all but finitely
many elements have valuation \(\geq 0\).
We understand the relative situation well, based on what we proved. By
step 2, since \(B, B_1\) are non-special, the dimension of the quotient
is zero:
\begin{align*}
\dim_k(\mathcal{A}_K(B_1) + K) / (\mathcal{A}_K(B) + K)
&= \deg \qty{B_1 - \ell(B_1)} - \qty{\deg B - \ell(B) } \\
&= (g-1) - (g-1) \\
&= 0
.\end{align*}
But then these spaces are equal to each other, so
\(\mathcal{A}_K(B_1) + K = \mathcal{A}_K + K\). But we chose \(B_1\)
arbitrarily large so it contained \(\alpha\), and we found that the
resulting space is no bigger than the original. Note that \(B_1\) was
chosen so that \(\alpha\in \mathcal{A}_K(B_1)\) before adding \(K\),
which remains true when adding \(K\). But this says \(\alpha\) is in the
LHS, which equals the RHS. Then \(\alpha \in \mathcal{A}_K(B)\), where
\(\alpha\) was arbitrary, so \(\alpha\in \mathcal{A}_K(B) + K\).
\end{proof}
\begin{corollary}[2.14]
This can be applied to the zero divisor:
\begin{align*}
\dim_k \mathcal{A}_K( \mathcal{A}_K(0) + K ) \iota(0) = g
.\end{align*}
\end{corollary}
\begin{exercise}[?]
Corollary 2.14 shows that if \(K = k(t)\) is the rational function
field, then we have \(\mathcal{A}_K(0) + K = \mathcal{A}_K\).\footnote{So
every Adele differs from a rational function by an effective Adele.}
Show this directly.
\end{exercise}
\begin{remark}
Note that analogy to consider \(\mathcal{A}({\mathbb{Q}})\), where you
get \(\mathcal{A}_{\mathbb{Q}}= \widehat{{\mathbb{Z}}}+ {\mathbb{Q}}\),
where \(\widehat{{\mathbb{Z}}}\) denotes the profinite completion.
Recall that
\({\mathbb{A}}_{\mathbb{Q}}= \prod_p' {\mathbb{Q}}_p \times{\mathbb{R}}\),
and inside of this we have
\({\mathbb{A}}(0) \coloneqq\prod_p {\mathbb{Z}}_p \times{\mathbb{R}}\).
Not too crazy of a fact: given an Adele, it has finitely many places
where its \(p{\hbox{-}}\)adic valuation is negative, so it shouldn't be
hard to find a rational number as a correction term which doesn't change
the valuation. The fact that this works for \({\mathbb{Q}}\) is related
to \({\mathbb{Z}}\) being a PID.
\end{remark}
\hypertarget{lecture-11b-weils-proof-of-riemann-roch-todo}{%
\section{Lecture 11B: Weil's Proof of Riemann-Roch
(TODO)}\label{lecture-11b-weils-proof-of-riemann-roch-todo}}
\hypertarget{lecture-11c-weils-proof-of-riemann-roch-todo}{%
\section{Lecture 11C: Weil's Proof of Riemann-Roch
(TODO)}\label{lecture-11c-weils-proof-of-riemann-roch-todo}}
\hypertarget{lecture-12-chapter-3-curves-over-a-finite-field}{%
\section{Lecture 12: Chapter 3, Curves Over a Finite
Field}\label{lecture-12-chapter-3-curves-over-a-finite-field}}
\hypertarget{finiteness-of-class-groups}{%
\subsection{Finiteness of Class
Groups}\label{finiteness-of-class-groups}}
We consider \(k= {\mathbb{F}}_q\) a finite field, which by definition is
a one variable global function field. Idea: we've defined some affine
dedekind domains (the holomorphy rings) had a finite nonempty set of
places of the function field. These are analogous to the ring of
integers of a number field, or more generally \(S{\hbox{-}}\)integer
rings. Recall some basic results from NT1: the finiteness of the class
group, and the finite generation of the unit group. Here we have a class
groups of affine Dedekind domain, and by Rosen's theorem, there are
infinitely many as you vary over nonempty subsets of places of the
function field, and they're all closely connected to a geometric class
group: the degree zero divisor class group. Thus by this analogy, when
the field is finite, we'd expect that \(\operatorname{Cl}^0(K)\) is
finite as well, which is the main result we'll prove today.
\hypertarget{base-extension-1}{%
\subsection{Base Extension}\label{base-extension-1}}
Let \(K_{/{\mathbb{F}}_q}\) be a one variable function field with
constant field \({\mathbb{F}}_q\), so that the only elements of \(K\)
that are algebraic over \({\mathbb{F}}_q\) are already in
\({\mathbb{F}}_q\). Since \({\mathbb{F}}_q\) is a perfect field
(\(x\mapsto x^p\) is a surjection), every such function field is
regular.
Let
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_q\)
be an algebraic closure, then for all \(r\in {\mathbb{Z}}^+\) there
exists a unique degree \(r\) extension, which we'll denote
\({\mathbb{F}}_{q^r}\). The extension
\({\mathbb{F}}_{q^r}/{\mathbb{F}}_q\) is a cyclic galois extension
(i.e.~it's galois group is cyclic) with a canonical generator: the
Frobenius map.
The galois theory of the constant field comes in when trying to study
constant extensions of the function field. There is a general theory of
constant extensions, but in our case, every such extension will be
cyclic or procyclic, so we don't need the entire theory.
For any positive integer \(r\), define the extension
\(K_r \coloneqq K {\mathbb{F}}_{q^r}\) given by extending scalars, which
is a regular function field over \({\mathbb{F}_{q^r}}\). There are two
ways to obtain this: either take an algebraic closure of \(K\) and take
the compositum, or take
\(K\otimes_{{\mathbb{F}_{q}}}{\mathbb{F}_{q^r}}\), which we proved was
again a field. This \(K_r\) is what we get by extending constants, and
the way regular function fields work is that if you make an arbitrary
extension of the ground field, then you retain a regular function field
over this new extension. On the other hand, note that \(K_R/K\) is a
degree \(r\) arithmetic extension of function field, whose galois group
is also generated by Frobenius. If we take any regular function field
over \(k\) and then take a finite galois extension \(l/k\), then
extending scalars in this way would give an extension of the upstairs
fields which is galois and has the same galois group as the constant
extension. This is \emph{arithmetic} because the only thing that changes
going from \(K\) to \(K_r\) is the field of constants.
In the analogy of function fields as the meromorphic functions on a
Riemann surface, this type of extension has no analog: since
\({\mathbb{C}}\) is algebraically closed, there are no constant
extensions. So arithmetic extensions are just extending scalars, and
\emph{geometric} extensions don't change the constant field at all and
instead have the property that if you extended scalars to the algebraic
closure, you'd have an extension of the same degree. Note that the étale
fundamental group also has a similar decomposition into an arithmetic
part and a geometric part (see Daniel Litt's course).
\hypertarget{splitting-of-places}{%
\subsubsection{Splitting of Places}\label{splitting-of-places}}
\begin{question}
Given a place in \(K\), how does it split (or not) in \(K_r\)?
\end{question}
\begin{remark}
We can ask this question in whenever we have an extension of function
fields. This reduces to the usual ATI type of question: for
\(v\in \Sigma(K/{\mathbb{F}_{q}})\), choose an affine Dedekind domain
\(R\) such that \(v\in \Sigma(K/R)\), i.e.~the place is regular. Let
\(S\) be the integral closure of \(K\) in \(K_r\); this place
corresponds to a maximal ideal \(\mathfrak{p}\), we then want to factor
its pushforward \(\mathfrak{p}_v S\). So this question is a special case
of how a prime ideal factors in an extension of Dedekind domains.
\end{remark}
We'll temporarily black-box the following lemma:
\begin{lemma}[?]
Suppose \(v\) is the downstairs place, \(r\) is the degree of the
extension, and \(d\coloneqq\deg(v)\). Then
\begin{itemize}
\item
\(K_r/K\) is galois and we have \(efg = r\).\footnote{\(e\) is the
prime ramification index, \(f\) is the prime residual degree, and
\(g\) is the number of distinct primes. This result essentially
comes from ANTI, replacing \(\sum e_i f_i = r\).}
\item
This extension will be unramified: we in fact have \(e=1\), so
\(g = \gcd(d, r)\) and \(f = r/\gcd(d, r)\), and
\item
Each place \(w\in \Sigma(K_r/{\mathbb{F}_{q^r}})\) lying over \(v\)
has degree \(d/\gcd(d, r)\).
\end{itemize}
\end{lemma}
\begin{remark}
Note that having an extension of Dedekind domains coming from a galois
extension of fields simplifies things: this makes the inertial degree
and ramification indices coincide.
\end{remark}
\begin{example}[?]
\envlist
\begin{itemize}
\item
The extension is inert \(\iff\) \(\gcd(d, r) = 1\)
\begin{itemize}
\tightlist
\item
I.e. \(d, r\) are coprime and \(g=e=1, f=r\).
\end{itemize}
\item
The extension splits completely \(\iff\) \(r \divides d\).
\begin{itemize}
\tightlist
\item
If \(r=d\), i.e.~we take a degree \(d\) place and extend scalars to
\(K_d\), it splits completely into \(d\) degree 1 places.
\end{itemize}
\item
All \(w\divides v\) have degree 1 \(\iff\) \(d\divides r\).
\end{itemize}
\end{example}
\begin{remark}
Suppose we have \(w\) over \(v\) with
\(w\in \Sigma({\mathbb{F}_{q^r}})\) and
\(v\in \Sigma(K/{\mathbb{F}_{q}})\). If \(v\) has degree \(d\), this
means that the residue field satisfies
\(k(v) \cong {\mathbb{F}}_{q^{d}}\), since we have unique extensions in
each degree. If \(f\) is the \(f\) from ANTI, it is also the degree of
the residual extension, so we know \([k(w): k(v)] = f\) and thus
\(k(w) \cong {\mathbb{F}}_q^{f d}\).
On the other hand, \(k(w)\) is an extension of \({\mathbb{F}_{q^r}}\) of
degree \(\deg(w)\), so
\(k(w) \cong {\mathbb{F}}_{ \qty{q^r}^{\deg(w)} } = {\mathbb{F}}_{q^{r\deg(w)}}\).
Thus \(r=fg\) and
\begin{align*}
q^{f\deg(v)} = q^{r\deg(w)} \implies \deg(w) = \qty{f\over r}\deg(v) = { \deg(v) \over g}
.\end{align*}
The residue field, if it changes at all, can only increase in size,
since any extension of Dedekind domains induces an extension of residue
fields. So the size of the residue field of \(w\) is at least as big as
the size of the residue field of \(v\). But the degree of \(w\) is
measured relative to the extended field \({\mathbb{F}_{q^r}}\), since
it's the degree of the residue field as an extension of
\({\mathbb{F}_{q^r}}\). So consider \(\deg(w) = \deg(v)/g\), we see that
even as the residue field is increasing by a factor of \(f\), the degree
of the point is decreasing by a factor of \(g\).
\textbf{Upshot}: The residue field grows, but its degree can only
shrink. Thus making an extension forces the degrees of the upstairs
places to \emph{decrease}.
\end{remark}
We're trying to find out in how many ways a discrete valuation extends
to a finite degree field extension. From ANTII, we have a result that
describes this: if \(v\) is a rank 1 valuation on \(k\) and \(L/K\) is a
finite degree extension, then the extensions of \(v\) to \(L\)
correspond with \(\operatorname{mSpec}(\widehat{K}_v \otimes_K L)\),
where the hat denotes completing \(K\) with respect to the valuation.
The \(e,f,g\) can all be computed as well.\footnote{See Pete's NTII
notes, Theorem 1.64.}
This is some finite degree \(\widehat{K}_v\) algebra, and if \(L/K\) is
separable then this decomposes as a finite product of finite degree
field extensions of \(K\) and \(\widehat{K}_v\), the number of which
will be \(g\). The \(e\) and \(f\) can be read off because each
extension will have a ramified and unramified part.
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that
\({\mathbb{F}}_{q^d} \otimes_{{\mathbb{F}}_q} {\mathbb{F}}_{q^r} \cong {\mathbb{F}}_{q^l}^{d'}\)
where \(l = \operatorname{lcm}(d, r)\) and \(d' = \gcd(d, r)\).
\item
Generalize this to the case when \(k_p / k\) and \(\ell / k\) are both
cyclic galois extensions.
\end{enumerate}
\end{exercise}
\hypertarget{degree-1-places-and-rational-points-on-a-curve}{%
\subsection{Degree 1 Places and Rational Points on a
Curve}\label{degree-1-places-and-rational-points-on-a-curve}}
Taking the lemma as a black box, for \(r\in {\mathbb{Z}}^+\) let
\(N_r \coloneqq{\left\lvert {\Sigma_1(K_r/ {\mathbb{F}_{q^r}}) } \right\rvert}\),
i.e.~the number of degree 1 places of the function field after making a
degree \(r\) extension. Equivalently,
\(N_r = {\left\lvert {C({\mathbb{F}_{q^r}})} \right\rvert}\) where \(C\)
is a unique complete nonsingular curve over \({\mathbb{F}_{q}}\)
corresponding to \(K\), and this denotes the number of
\({\mathbb{F}_{q^r}}\) rational points. We'll eventually see these are
finite.
\begin{remark}
Important way of thinking about these: degree one places of a function
field over \(k\) correspond to \(k{\hbox{-}}\)rational points of a
curve.
\end{remark}
\begin{corollary}[Equivalence of data: places and rational points]
\begin{align*}
N_r = \sum_{d\divides r} d\cdot {\left\lvert {\Sigma_d(K/ {\mathbb{F}_{q}})} \right\rvert}
,\end{align*}
so knowing the number of closed points of each degree is equivalent to
knowing the \({\mathbb{F}_{q^r}}{\hbox{-}}\)points for all \(r\).
\end{corollary}
\begin{proof}[?]
Let \(w\in \Sigma_1(K_r/{\mathbb{F}_{q^r}})\) be a degree 1 point and
set \(v \coloneqq w\cap K\) so \(w\) lies over \(v\). What is the degree
of \(v\)? Setting \(d \coloneqq\deg(v)\), the lemma gives
\begin{align*}
1 = \deg(w) = {d\over \gcd(d, r)}
,\end{align*}
which implies that \(\gcd(d, r) = d\) and thus \(d\divides r\). So for
each \(d\) dividing \(r\), every degree of
\(v\in \Sigma(K/{\mathbb{F}_{q}})\) contributes \(\gcd(d, r) = d\)
degree 1 points on \(K_r\), i.e.~every downstairs degree \(d\) place
splits into \(d\) degree one places. So for every such \(d\), every
degree \(d\) closed point contributes \(d\) degree 1 closed points lying
above it, and conversely if \(d\) does not divide \(r\) then the
upstairs point would not have degree 1, so this accounts for all of the
degree 1 points.
\end{proof}
\begin{remark}
We saw that the degree 1 places and the rational points are the same
information, and there is a third equivalently quantity: \(A_n\),
defined to be the number of effective divisors of degree \(n\).
\end{remark}
\hypertarget{finiteness-of-places-and-rational-points}{%
\subsection{Finiteness of Places and Rational
Points}\label{finiteness-of-places-and-rational-points}}
\begin{lemma}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
For all \(d\), the number of degree \(d\) closed points
\(\Sigma_d(K/{\mathbb{F}_{q}})\) is finite (and therefore \(N_r\) is
finite), and
\item
For all \(n\), \(A_n\) is finite.
\end{enumerate}
\end{lemma}
\begin{proof}[of a]
Let \(L/K\) be a degree \(n\) extension of regular function fields over
\({\mathbb{F}_{q}}\). We then have a restriction map
\begin{align*}
r: \Sigma(L/{\mathbb{F}_{q}}) \twoheadrightarrow\Sigma(K/{\mathbb{F}_{q}})
\end{align*}
which we showed is surjective with finite fibers. We can say a little
bit more: for all places \(w \in \Sigma(L/{\mathbb{F}_{q}})\), we have
an inequality
\begin{align*}
\qty{1\over n}\deg(w) \leq \deg(r(w)) \leq \deg(w)
,\end{align*}
noting that we're now measuring all degrees over a common ground field
\({\mathbb{F}_{q}}\). So things are now what you'd expect: the degree of
the upstairs point is a multiple of the degree of the downstairs point.
The upper bound comes from the fact that the residue of the upstairs
point is a finite extension of the residue field of the downstairs
points. The opposite inequality comes from ANTI: the degree of the
residual extension is at most the degree of the entire extension.
So \(r\) doesn't preserve degrees exactly, but preserves them up to a
bounded factor, and thus \(\Sigma_{\leq d}(L/{\mathbb{F}_{q}})\) is
finite for all \(d\) \(\iff\) \(\Sigma_{\leq d}(K/{\mathbb{F}_{q}})\) is
finite for all \(d\). Because of this, we can reduce the situation by
exchanging the function field \(L/{\mathbb{F}_{q}}\) with any other
function field for which \(L\) is a finite extension, and in particular
we can take the rational function field \(K = {\mathbb{F}}_q(t)\). What
are the degree \(d\) places of a rational function field? There is
exactly one place at infinity, and the remaining ones correspond to
monic irreducible polynomials. Since \({\mathbb{F}_{q}}\) is finite,
there are only finitely many such polynomials of any fixed
degree.\footnote{There is an exact formula for this quantity.}
\end{proof}
\begin{proof}[of b]
Left as an exercise.
Some remarks: how do you build an effective divisor of degree \(n\)?
Take closed points (places) and start adding them up with positive
coefficients, then the degree of the divisor is the sum of the degrees
of the places. But if you only have finitely many places, each of which
can only be used a bounded number of times (certainly no more than \(n\)
times!), thus one can only build finitely many effective divisors of
each degree.
\end{proof}
\hypertarget{finiteness-of-class-group}{%
\subsection{Finiteness of Class Group}\label{finiteness-of-class-group}}
\begin{proposition}[Finiteness of class group]
The degree 0 divisor class group \(\operatorname{Cl}^0(K)\) is finite.
\end{proposition}
This is a geometric analog of the finiteness of the class group of the
ring of integers of a number field. By Rosen's theorem, as an immediate
corollary, the class group of any affine dedekind domain over a finite
ground field is finite. This follows from looking at the exact sequence:
a finite index subgroup of the class group of any dedekind domain is a
quotient of \(\operatorname{Cl}^0(K)\), and a finite index subgroup of a
finite group is finite.
\begin{proof}[?]
Set \(\delta \coloneqq I(K)\) to be the index, i.e.~the least possible
degree of a divisor.\footnote{By a theorem of Schmidt, we'll later prove
that \(\delta = 1\).}
In any case, for all \(n\in {\mathbb{Z}}\), we have
\begin{align*}
\operatorname{Cl}^n K =
\begin{cases}
0 & \delta\nmid n \\
{\left\lvert {\operatorname{Cl}^0 K} \right\rvert} & \delta\divides n
\end{cases}
.\end{align*}
If you have any degree \(n\) divisors, then \(\operatorname{Cl}^n K\)
will be a coset of \(\operatorname{Cl}^0 K\). Here we just look at the
degree map, which is a group morphism onto its image, of which all
nonempty fibers have the same size. Thus we may work with
\(\operatorname{Cl}^n K\) for \(n\gg 0\).\\
In particular, choose \(n\geq g\) the genus such that
\(\delta \divides n\), and let \(D \in \operatorname{Div}^n K\). A
Riemann-Roch computation shows that \(\ell(D)\), the dimension of the
linear system, is at least \(n-g+1\), and so we have \(\ell(D) \geq 1\)
and \(D\) is linearly equivalent to an effective divisor. This shows
that the map taking effective degree \(n\) divisors to
\(\operatorname{Cl}^n K\) taking a divisor to its divisor class
(restricted to effective divisors) is surjective. But we just saw that
the set of effective degree \(n\) divisors is finite -- it was built out
of finitely many closed points of bounded degrees -- forcing
\(\operatorname{Cl}^n K\) to be finite. The result follows because
\(\operatorname{Cl}^n K\) is a coset of \(\operatorname{Cl}^0 K\), all
of which have the same size, and the index is finite.
\end{proof}
\begin{definition}[Class Number of $K$]
The \textbf{class number} of \(K\) is defined as
\begin{align*}
h \coloneqq{\left\lvert {\operatorname{Cl}^0 K} \right\rvert}
.\end{align*}
\end{definition}
\begin{remark}
There is a much fancier proof: there exists a \(g{\hbox{-}}\)dimensional
abelian variety \(A / {\mathbb{F}}_q\), the \emph{Jacobian variety} of
\(C/{\mathbb{F}}_q\), such that
\(\operatorname{Cl}^0 K + A({\mathbb{F}}_q)\). It is built out of the
degree 0 divisor class group in some functorial way. In particular,
\(A\) is a projective variety, and thus embeds into some
\({\mathbb{P}}^N_{/{\mathbb{F}}_q}\), and so
\({\left\lvert {A({\mathbb{F}}_q)} \right\rvert} \leq {\left\lvert {{\mathbb{P}}^N_{/{\mathbb{F}}_q}} \right\rvert} < \aleph_0\).\\
As one varies over all function fields over all finite fields, there
will only be finitely many whose class number is bounded by some fixed
\(h_0\). E.g. there are only finitely many function fields of class
number 1, and these can be explicitly listed. So \(h\to \infty\) in some
sense, which is not proved by showing that
\({\left\lvert {A({\mathbb{F}_{q}})} \right\rvert} \to \infty\), and
we'll instead prove it using methods closer to what we're seeing in this
course.
\end{remark}
Up next: setting up the zeta function.
\hypertarget{lecture-13-splitting-places}{%
\section{Lecture 13: Splitting
Places}\label{lecture-13-splitting-places}}
Recall that we previously looked at the regular function fields: we took
a function field in one variable and considered the class of function
fields for which we could take any extension of the constant field that
we wanted. As long as the ground field is perfect, being regular is
equivalent to the constant subfield being \(k\) itself. However, we
haven't done anything with them yet!
If you take an algebraic closure of the finite ground field
\({\mathbb{F}}_q\), there is a unique subextension of degree \(r\) for
every \(r\), so we call that \({\mathbb{F}}_{q^r}\). The extension
\({\mathbb{F}}_{q^r}/{\mathbb{F}}_q\) is cyclic galois, with a geometric
Frobenius \(x\to x^q\). Note that \({\mathbb{F}}_{q^r}\) is the fixed
field of \(F^r\), the \(r\)th power of the Frobenius map. We set
\(K_r \coloneqq K {\mathbb{F}}_{q^r}\), which is a regular function
field over \({\mathbb{F}}_{q^r}\). Note that we could view this as a
function field just over \({\mathbb{F}}_q\), but it would not be
regular. Then \(K_r/K\) is a degree \(r\) arithmetic extension of
function fields.
\begin{question}
What happens to places when making this scalar extension? I.e., how to
places in \(K\) decompose in \(K_r\)?
\end{question}
\begin{remark}
This is related to an Algebraic Number Theory I problem: for
\(v\in \Sigma(K_{/\mathbb{F}_q})\) above an affine Dedekind domain \(R\)
such that \(v\in \Sigma(K/R)\), let \(S\) be the integral closure of
\(K\) in \(K_r\). Then we want to factor \(p_v S\)?
\todo[inline]{Not quite sure.}
\end{remark}
\hypertarget{how-places-split}{%
\subsection{How Places Split}\label{how-places-split}}
\begin{lemma}[Key lemma about how places split.]
Suppose \(d\coloneqq\deg(v)\). Then \(K_r/K\) is galois, so we have
\(efg=r\). In fact, \(c=1\), so \(f = {r\over \gcd(d, r)}\) and
\(g = \gcd(d, r)\) and each place
\(w\in \Sigma(K_r / {\mathbb{F}}_{q^r})\) has degree
\({d\over \gcd(d, r)}\).
\end{lemma}
\begin{remark}
We have the following cases:
\begin{itemize}
\item
The extension is \emph{inert} iff \(\gcd(d, r) = 1\),
\item
The extension \emph{splits completely} iff \(r\divides d\),
\item
All \(w\) dividing \(v\) have degree 1 iff \(d\divides r\).
\end{itemize}
\end{remark}
The last thing we proved was that the degree zero divisor class group is
finite when we're over a finite ground field. Why is this true? Whenever
there is a divisor of degree \(n\), then the set of degree \(n\)
divisors is a coset of the degree zero divisors, all of which have the
same cardinality. We proved finiteness using the Riemann-Roch theorem,
using the fact that the set of \emph{effective} degree \(n\) divisors is
finite for all \(n\).
The next main topic will be the \textbf{zeta function}, which keeps
track of three equivalent packets of information: \(A_n\), the number of
effective divisors of degree \(n\), the number of places of degree \(d\)
(since an effective divisor is a linear combination of these), and
\(N_r\) the number of degree 1 points in the degree \(r\) extension,
i.e.~the number of \({\mathbb{F}}_{q^r}\) rational points.
\hypertarget{counting-effective-divisors}{%
\subsection{Counting Effective
Divisors}\label{counting-effective-divisors}}
\begin{lemma}[?]
Suppose \(C\in \operatorname{Cl}(K)\), then
\begin{itemize}
\item
The number of effective divisors \(D \in [C]\) is given by
\begin{align*}
{q^{\ell(C)} -1 \over q-1}
,\end{align*}
where \(\ell(C)\) is the dimension of the linear system associated to
the divisor class \(C\), and this is the dimension of a projective
space over \({\mathbb{F}}_q\).
\item
For all \(n>2g-2\) with \(\delta \divides n\), we have
\begin{align*}
A_n = h \qty{ q^{n+1-g} - 1\over q-1}
.\end{align*}
\end{itemize}
\end{lemma}
\begin{proof}[?]
\envlist
\textbf{Proof of (a)}: The set of effective divisors linearly equivalent
to \(D\) is naturally viewed as the projectivization
\({\mathbb{P}}\mathcal{L}(D)\) of the one-dimensional subspaces of the
linear system of that divisor class. It is then a fact that the number
of elements in a \(d{\hbox{-}}\)dimensional vector space over
\({\mathbb{F}}_q\) has dimension precisely \(\frac{q^d-1}{q-1}\)
elements. The projectivization comes in because two different functions
have the same divisor if one of them is a constant multiple of the
other. Note that the number of elements is computed as the number of
nonzero elements divided by the number of nonzero scalars.\\
\textbf{Proof of (b)}: This will come out of the Riemann-Roch theorem.
In order to compute the number of divisors in a divisor class, you need
to know the dimension of the linear system, which is not easy in
general. However, if the divisor class has sufficiently large degree,
the Riemann-Roch theorem tells you exactly what it is. As long as
\(n > 2g-2\), there is no correction term, and the dimension of the
linear system is equal to its degree minus the genus plus one. So by
Riemann-Roch, since \(\deg(D) > 2g-2\), \(D\) is non-special and
\(\ell([D]) = n-g+1\), which yields the desired formula for \(A_n\).
\end{proof}
\begin{remark}
This is the sharpest result possible: the canonical divisor has degree
\(2g-2\) and is special, so this fails for the canonical class.
The upshot: there are three piece of information:
\begin{itemize}
\item
\(N_r\), the number of \({\mathbb{F}}_{q^r}\) rational points,
\item
\({\left\lvert {\Sigma_d(K_{/\mathbb{F}_q})} \right\rvert}\) the
number of closed points / places of degree \(d\),
\item
\(A_n\) the number of effective divisors of degree \(n\),
\end{itemize}
and there are simple formulas relating these. Moreover, it is enough to
know only finitely many of these quantities, where the number depends on
\(g\).
\end{remark}
\hypertarget{hasse-weil-zeta-functions}{%
\subsection{Hasse-Weil Zeta Functions}\label{hasse-weil-zeta-functions}}
There is a general theory that will unify
\begin{itemize}
\item
The Riemann zeta function, thought of as the zeta function of
\({\mathbb{Z}}\),
\item
The Dedekind zeta function, attached to the ring of integers over a
number field,
\item
The Hasse-Weil zeta function of a one variable function field over a
finite field,
\end{itemize}
all of which will be special cases of a \emph{Serre zeta function} which
can be attached to a finite type scheme over \({\mathbb{Z}}\).
Note that we aren't specifically discussing schemes in this course, but
you don't need to know much about what a scheme is to define the
Hasse-Weil zeta function. Just note that an affine finite-type
\({\mathbb{Z}}{\hbox{-}}\)scheme corresponds to a finitely generated
\({\mathbb{Z}}{\hbox{-}}\)algebra, and a general finite-type
\({\mathbb{Z}}{\hbox{-}}\)scheme will be covered by finitely many affine
ones, the zeta function will be determined by these finitely many
\({\mathbb{Z}}{\hbox{-}}\)algebras and some kind of inclusion-exclusion
principle (since the scheme is a not necessarily disjoint union of
affine schemes).
Recall that \(A_n = A_n(K)\) is the number of effective divisors of
degree \(n\), which we've proved is finite. We have a formula when
\(n> 2g-2\), namely
\begin{align*}
Z(t) = \sum_{n=0}^\infty A_n t^n = \sum_{D\in \operatorname{Div}^+(K)} t^{\deg(D)} \in {\mathbb{Z}}[[t]]
,\end{align*}
where \(\operatorname{Div}^+\) are the effective divisors and we've
collected terms based on their degree. This is analogous to the Dedekind
zeta function of a number field \(K\), a formal Dirichlet series which
is given by
\begin{align*}
\zeta_K(s) = \sum_{I \in \mathcal{I}\qty{{\mathbb{Z}}_K^\bullet}} {\left\lvert { {\mathbb{Z}}_K / I} \right\rvert}^{-s}
.\end{align*}
where the sum is now over all of the nonzero ideals of the ring of
integers, where we measure the size using the \emph{norm}, i.e.~the size
of the residue field. There's an analogy between integral ideals (vs
fractional ideals) and effective divisors. We could get an Euler product
decomposition for the Dedekind zeta function by only considering prime
ideals, since in a Dedekind domain all ideals factor uniquely into prime
ideals. In fact, any nonzero ideal is a linear combination of prime
ideals. Similarly, the effective divisors are linear combinations of
effective divisors, so an Euler product expansion is possible here too.
If we take a prime ideal, since we're in a discrete valuation ring, we
can consider the local ring at that point. We can take the residue
field, which in general won't be finite, but will be a finite extension.
So a reasonable measure of the size of a prime divisor would be the
dimension of its residue field as a vector space over \(K\).
Note that if we wanted to make these look even more similar to each
other, we could define \(a_n\) (depending on \({\mathbb{Z}}_K\)) as
\begin{align*}
a_n = {\left\lvert {\left\{{I {~\trianglelefteq~}{\mathbb{Z}}_K {~\mathrel{\Big|}~}{\left\lvert {{\mathbb{Z}}_K/I} \right\rvert} = n}\right\}} \right\rvert}
,\end{align*}
which allows us to write
\begin{align*}
\zeta_K(s) = \sum_{n=1}^\infty {a_n \over n^s}
.\end{align*}
\begin{question}
Where we're going: how does \(Z(t)\) depend on \(K\)?
\end{question}
\begin{answer}
It turns out that it only depends on \(A_0, A_1, \cdots, A_{2g-2}\), and
thus \(Z(t)\) depends on only finitely much information. We will
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
Show that \(Z(t) \in {\mathbb{Z}}(t)\), i.e.~it is a rational function
and can be written \(Z(t) = P(t)/ Q(t)\).
\end{enumerate}
\begin{quote}
Note: the denominator will always be the same, \((1-t)(1-qt)\), and
we'll always have \(\deg P = 2g\). This is essentially coming from
\(\ell{\hbox{-}}\)adic cohomology. We'll also determine the leading
coefficient.
\end{quote}
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\item
Understand \(\deg P\) and \(\deg Q\) in terms of the genus \(g\).
\item
Ask about the roots of \(P(t)\), and establish a Riemann hypothesis
for Dedekind zeta functions (and in particular, the Riemann zeta
function).
\end{enumerate}
\begin{quote}
In particular, what are their magnitudes? This is what Weil did, this is
the big theorem in this area. Note that we'll need to consider
reciprocal roots, which will end up having magnitude \(\sqrt{q}\). We'll
see why this happens, and it turns out to be analogous to fact that the
nontrivial zeros of the Riemann zeta function have real part \(1/2\).
\end{quote}
\end{answer}
These are approximately in order of difficulty. The first two will
follow from Riemann-Roch, but the third will be much deeper. This is
essentially a positive characteristic analogue of the usual Riemann
hypothesis. Note that we're in a global field, the positive
characteristic analog of a number field, and for number fields the
Riemann hypothesis is the single outstanding problem. In the function
field case, it is a theorem!
\begin{proposition}[Formula for the zeta function exhibiting rationality]
Let \(K_{/\mathbb{F}_q}\) have genus \(g\) and \(\delta = I(K)\) the
index, the least positive degree of a divisor.\footnote{It will turn out
(by a theorem of Schmidt) that \(\delta = 1\) in the case of a finite
ground field.}
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
If \(g=0\), then
\begin{align*}
Z(t) = {1\over q-1} \qty{{q \over 1-q^\delta t^\delta} - {1 \over 1-t^\delta}}
.\end{align*}
\item
If \(g\geq 1\), then \(Z(t) + F(t) + G(t)\) where
\begin{align*}
F(t)
&= {1\over q-1} \sum_{0\leq \deg C \leq 2g-2} q^{\ell(C)} t^{\deg(C)} \\
G(t)
&= {h \over g-1} \qty{
{ q^{1-g} (qt)^{2g-2+\delta} \over 1 - (qt)^\delta } - {1 \over 1 - t^\delta}
}
,\end{align*}
so \(F\) involves summing over all divisor classes of degree at most
\(2g-2\), and \(G\) is a term coming from Riemann-Roch involving the
class number \(h\).
\end{enumerate}
\end{proposition}
\begin{remark}
Note that as a consequence, it will definitely be rational in \(q\), and
will have a simple pole at \(t=1\). There's no major idea for the proof:
when the degree of the divisor class is sufficiently large, we just have
an exact formula. If it is smaller, than the formula involves the
dimension of the linear system.
\end{remark}
\hypertarget{proof-of-rationality}{%
\subsection{Proof of Rationality}\label{proof-of-rationality}}
\begin{proof}[of rationality of $Z(t)$]
Recall that \(\ell(C)\) is the dimension of the associated Riemann-Roch
space.
When \(g=0\), by Riemann-Roch we have \(\operatorname{Cl}^0(K) = 0\)
over any ground field \({\mathbb{k}}\) (see exercises), and so \(h=1\).
Since every \(n\geq 0\) satisfies \(n\geq 2g-2\) when \(g=0\), if
\(\delta\divides n\) we have
\begin{align*}
A_n = h \qty{ q^{n+1 - g} - 1 \over q-1} = {q^{n+1} - 1 \over q-1}
,\end{align*}
and since \(A_n=0\) unless \(n\) is divisible by \(\delta\), we have
\begin{align*}
Z(t) = \sum_{n=0}^\infty A_n t^n = \sum_{n=0}^{\infty} A_{\delta n} t^{\delta n} = \sum_{n=0}^\infty {q^{\delta n + 1} -1 \over q-1} t^{\delta n}
.\end{align*}
This can now be split into two terms, each of which will have a
geometric series to sum.
Now let \(g\geq 1\), and write
\begin{align*}
\sum_{{n=0}}^\infty A_n t^n = \sum_{\deg(C) \geq 0} {\left\lvert {\left\{{ A\in C {~\mathrel{\Big|}~}A\geq 0}\right\}} \right\rvert}t^{\deg(C)}
,\end{align*}
where we instead count the number of divisors in each divisor class (a
consequence of the previous lemma). Continuing this computation, we
separate out the part where \(\deg(C) \leq 2g-2\) and pull out the
\(-1\) in the numerator:
\begin{align*}
\cdots
&= \sum_{\deg(C) \geq 0} {q^{\ell(C)} - 1 \over q - 1}t^{\deg(C)} \\
&= \qty{1\over q-1} \qty{ \sum_{0\leq \deg(C) \leq 2g-2} q^{\ell(C)} t^{\deg(C)}
+ \sum_{\deg(C) > 2g-2} q^{\deg(C) - g + 1} t^{\deg(C)} - \sum_{\deg(C) \geq 0} t^{\deg(C)}
} \\
&\coloneqq F(t) + G(t)
,\end{align*}
so we can write
\begin{align*}
F(t) &= {1\over q-1} \sum_{0\leq \deg(C) \leq 2g-2} q^{\ell(C)} t^{\deg(C)}
\\
(q-1)G(t) &= \sum_{n = {2g-2 \over \delta} + 1}^\infty h q^{n\delta + 1 - g} t^{n\delta} - \sum_{{n=0}}^\infty ht^{n\delta}
.\end{align*}
Note that \(\delta\divides 2g-2\) since the canonical divisor has degree
\(2g-2\) and \(\delta\) is a gcd. Note that for \(g=1\), the index
divides zero, which tells you nothing about it! This now reduces to some
geometric series that can be summed, which shows these are rational
functions in \(t\).
\end{proof}
\begin{exercise}[?]
Let \(K = {\mathbb{F}}_q(t)\), then \(g=0, \delta = 1\), and
\begin{align*}
Z(t) = {1\over (1-qt)(1-t)}
.\end{align*}
We will see in general that the numerator is of the form \(L(t)\) where
\(L\in {\mathbb{Z}}[t]\) has degree \(2g\).
\end{exercise}
Note that this all generalized to higher dimensional projective
varieties \(X_{/{\mathbb{F}}_q}\), for which these properties were
proved by the work of Deligne. In general, \(Z(t)\) will be of the form
\begin{align*}
Z_X(t) = {L_1(t) \cdots L_{2d-1}(t) \over L_0(t) \cdots L_{2d}(t)}
,\end{align*}
where \(d = \dim(X)\) and \(\deg L_i\) will be the dimension of the
\(i\)th \(\ell{\hbox{-}}\)adic cohomology. Moreover, if
\(X_{/{\mathbb{F}}_q}\) is a reduction mod \(q\) of a variety in
characteristic zero, these will be the Betti numbers of
\(X_{/{\mathbb{C}}}\). If we take a compact Riemann surface, which has a
honest topological genus of \(g\), the Betti numbers are \(1, 2g, 1\),
and this recovers the formula above for \(L(t)\) and its degree.
The next result will be the following theorem:
\begin{theorem}[Schmidt, 1910ish]
For all \(K_{/\mathbb{F}_q}\),
\begin{align*}
\delta = I(K) = 1
.\end{align*}
\end{theorem}
This will greatly simplify the previous formulas. A useful application
is if you have a genus zero curve of index 1, applying Riemann-Roch to a
divisor of degree 1 shows that the function field is rational. Thus the
only genus zero function field over \({\mathbb{F}}_q\) is the rational
function field. Useful aside: the Riemann hypothesis here gives an
estimate of the number of \({\mathbb{F}}_{q^r}\) rational points.
\hypertarget{lecture-14-the-hasse-weil-zeta-function}{%
\section{Lecture 14: The Hasse-Weil Zeta
Function}\label{lecture-14-the-hasse-weil-zeta-function}}
Recall the that \emph{Hasse-Weil zeta function} of a one-variable
function field \(K/{\mathbb{F}}_q\) over a finite ground field is
defined in the following way: let \(A_n = A_n(K)\) be the number of
effective divisors of degree \(n\). We have proved that \(A_n\) is
finite, and for \(n>2g-2\) we have a formula
\begin{align*}
Z(t) = \sum_{n=0}^\infty A_n t^n
= \sum_{D\in \operatorname{Div}^+(K)} t^{\deg(D)} \in {\mathbb{Z}}[[t]]
,\end{align*}
which is a formal power series with integer coefficients.
\begin{remark}
Recall that we have proved that it is a rational function of \(t\), and
in particular when \(g=0, \delta = 1\) \footnote{The \emph{index} of the
function field, least positive degree of a divisor.} we get
\begin{align*}
Z(t) = {1 \over (1-qt)(1-t)}
.\end{align*}
We got another expression which isn't fantastic: it involves this
\(\delta\), which we'll work toward proving is equal to 1. When \(g>1\),
we broke the zeta function into two pieces \(Z(t) = F(t) + G(t)\). For
divisors of sufficiently high degree, Riemann-Roch tells you what the
dimension of the Riemann-Roch space is, and \(G(t)\) explains the part
coming from divisors of large degree. We obtained a formula previously
for \(F(t)\) and \(G(t)\), and once we show \(\delta=1\) the formula for
\(G\) will simplify. For \(F(t)\), we specifically had
\begin{align*}
F(t) = {1\over q-1} \sum_{0\leq \deg(c) \leq 2g-2} q^{\ell(c) t^{\deg(c)}}
,\end{align*}
where the sum is over divisor classes and \(\ell\) is the dimension of
linear system corresponding to a divisor. But this isn't a great
formula: what are these classes, dhow many are in each degree, and what
is the dimension of the Riemann-Roch space?
\end{remark}
\begin{remark}
This is analogous to the Dedekind zeta function of a number field \(K\),
in which case
\begin{align*}
\zeta_K(s) = \sum_{T\in \ell({\mathbb{Z}}_k)}^\bullet {\left\lvert {{\mathbb{Z}}_k/I} \right\rvert}^{-s}
,\end{align*}
which will be covered in a separate lecture on \emph{Serre zeta
functions}.
\end{remark}
\begin{theorem}[F.K. Schmidt]
For all \(K/{\mathbb{F}}_q\), we have \(\delta = I(K) = 1\) where \(I\)
is the index.
\end{theorem}
This will follow from the associated, but it much weaker. However, this
is one of the facts we'd like to establish to use to \emph{prove} the
Riemann hypothesis.
\begin{remark}
Pete studied this in 2004 and found that every \(I\in {\mathbb{Z}}^+\)
arises as the index of a genus one function field \(K/{\mathbb{Q}}\).
\end{remark}
Notation: for \(n\in{\mathbb{Z}}^+\), let \(\mu_n\) denote the \(n\)th
roots of unity in \({\mathbb{C}}\).
\begin{lemma}[?]
For \(m, r\in {\mathbb{Z}}^+\), set \(d \coloneqq\gcd(m, r)\). Then
\begin{align*}
\qty{1-t^{mr/d}}^d = \prod_{\xi\in \mu_r} 1 - (\xi t)^m
.\end{align*}
\end{lemma}
\begin{proof}[?]
In \({\mathbb{C}}[x]\), we have
\begin{align*}
(X^{r/1} - 1)^d = \prod_{\xi\in \mu_r}(X - \xi^m)
,\end{align*}
where both sides are monic polynomials whose roots include the
\((r/d)\)th roots of unity, each with multiplicity \(d\). On the LHS,
the distinct roots are the \(r/d\)th roots of unity, then raising to the
\(d\)th power gives them multiplicity \(d\). On the RHS, this is an
exercise in cyclic groups: consider the \(n\)th power map on
\({\mathbb{Z}}/r{\mathbb{Z}}\) and compute its image and kernel. As
\(\xi\) ranges over \(r\)th roots of unity, \(\xi^m\) ranges over all
\(r/d\)th roots of unity, each occurring with multiplicity \(d\).
Substituting \(X= t^{-m}\) and multiplying both sides by \(t^r\) yields
the original result.\footnote{Special case: set \(m=r\), so \(d=r\),
then the RHS is \(r\) copies of 1.}
\end{proof}
\hypertarget{comparing-zeta-functions-after-extending-scalars}{%
\subsection{Comparing Zeta Functions After Extending
Scalars}\label{comparing-zeta-functions-after-extending-scalars}}
Next up, we want to compare the zeta function \(Z(t)\) for a function
field over \({\mathbb{F}}_q\) to the zeta function obtained when
extending scalars to \({\mathbb{Q}}^r\).
\begin{proposition}[Factorization identity for the zeta function]
Let \(K/{\mathbb{F}}_q\) be a function field, \(r\in {\mathbb{Z}}^+\),
and take the compositum \(K_r\) of \(K\) and \({\mathbb{F}}_q^r\) viewed
as a function field over \({\mathbb{F}}_q^r\). Let \(Z(t)\) be the zeta
function of \(K/{\mathbb{F}}_q\) and \(Z_r(t)\) the zeta function of
\(K_r/{\mathbb{F}}_q^r\). Then
\begin{align*}
Z_r(t^r) = \prod_{\xi \in \mu_r} Z(\xi t)
.\end{align*}
\end{proposition}
\begin{proof}[?]
We have an Euler product formula
\begin{align*}
Z(t) = \prod_{p\in \Sigma(K/{\mathbb{F}}_q)} (1 - t^{\deg(p)})^{-1}
.\end{align*}
where the sum is over places of the function field.\footnote{Proving
this Euler product formula might show up in a separate lecture, but it
is not any more difficult than proving it for the Riemann zeta
function.}
\begin{exercise}
Why is this product expansion true? Write as a geometric series with
ratio \(t^{\deg(p)}\). Here just expand each summand to get
\begin{align*}
Z(t) = \prod_p \sum_{j=1}^\infty t^{j\deg(p)}
.\end{align*}
Multiplying this out and collecting terms is in effect multiplying out
the prime divisors to get effective divisors.
\end{exercise}
We now use the result about splitting that was stated (but not proved):
\begin{claim}
If \(p\in \Sigma_m(K/{\mathbb{F}}_q)\) is a degree \(n\) place and
\(r\in {\mathbb{Z}}^+\), then there exist precisely
\begin{align*}
d\coloneqq\gcd(m, r)
\end{align*}
places \(p^r\) of \(K_r\) lying over \(p\), where each place \(p^r\) has
degree \(m/d\).
\end{claim}
In order to compare \(Z_r(t)\) to \(Z(t)\), we collect the \(p'\) into
ones that have the same fiber. We then can range over all \(p\) first,
then over all \(p'\) in the fiber above \(p\), yielding
\begin{align*}
Z_r(t^r) = \prod_{p\in \Sigma(K_{/{\mathbb{F}}_q})} \prod_{p'/p} {1 \over 1 - t^{r\deg(p')}}
.\end{align*}
Using the Euler product identity, we have for
\(p\in \Sigma_m(K_{/{\mathbb{F}}_q})\) and \(d\coloneqq\gcd(m, r)\) we
can express the innermost product as
\begin{align*}
\prod_{p'/p} {1 \over 1 - t^{r\deg(p')}} = (1 - t^{rm/d})^{-d} = \prod_{\xi\in \mu_r} (1- (\xi t)^m)^{-1}
,\end{align*}
where we've used the fact that we know there are exactly \(d\) places
and each contributes the same degree in the first expression. By using
\(-d\) in the previous lemma, we get the last term. Combining all of
this yields
\begin{align*}
Z_r(t^r)
= \prod_{\xi \in \mu_r} \prod_{p\in \Sigma(K_{/{\mathbb{F}}_q})} (1- (\xi t)^{\deg p})^{-1}
= \prod_{\xi \in \mu_r} Z(\xi t)
.\end{align*}
\end{proof}
\begin{remark}
Similar to taking an abelian extension of number fields and noting that
the Dedekind zeta function factors into a finite product: the original
zeta function, and in general, Hecke \(L\) functions. If you do this for
an abelian number field over \({\mathbb{Q}}\), then the Dedekind zeta
function of the upstairs number field will be a finite product where one
of the terms in the Riemann zeta function and the others are Dirichlet
\(L\) functions associated to certain Dirichlet characters. So this is
some (perhaps simpler) version of that.
\end{remark}
\hypertarget{proof-that-delta-1}{%
\subsection{\texorpdfstring{Proof That
\(\delta = 1\)}{Proof That \textbackslash delta = 1}}\label{proof-that-delta-1}}
We can finally prove Schmidt's theorem that \(\delta = 1\):
\begin{proof}[$\delta = 1$]
Take a \(\delta\)th root of unity \(\xi \in \mu_\delta\). Then for all
places \(p \in \Sigma(K_{/{\mathbb{F}}_q})\), \(\delta\) divides
\(\deg p\) by definition since it is a gcd, and so we have
\begin{align*}
Z(\xi t)
= \prod_{p\in \Sigma(K_{/{\mathbb{F}}_q})} (q - (\xi t)^{\deg p} )^{-1}
= \prod_{p\in \Sigma_{K_{{\mathbb{F}}_q}}} {1 \over 1 - t^{\deg p}} = Z(t)
,\end{align*}
using the fact that \(\xi^{\deg p} = 1\).
We're now in a situation where we can apply the previous proposition,
which gives the following identity for the zeta function over the degree
\(\delta\) extension:
\begin{align*}
Z_{\delta}(t^\delta) = \prod_{\xi \in \mu_\delta} Z(\xi t) = Z(t)^\delta
.\end{align*}
Our previous formulas show that any zeta function for a 1-variable
function field over a finite field has a simple pole at \(t=1\), and
since \({\operatorname{Ord}}_{t-1}(t^\delta) = 0\), we get
\begin{align*}
-1 = {\operatorname{Ord}}_{t-1} Z_\delta(t^\delta) = {\operatorname{Ord}}_{t-1} Z(t)^\delta) = -\delta
,\end{align*}
where for the first equality we're using the fact that the
\((t-1){\hbox{-}}\)adic valuation of \(Z_\delta(t^\delta)\) is one, and
for the RHS, the ordinary zeta function has a simple pole at \(t=1\) and
since we have a valuation, raising something to the \(\delta\)th power
is just \(\delta\) times the original valuation.
\end{proof}
There is some modest representation theory (character theory) that shows
up when looking at zeta functions of abelian extensions.
\begin{remark}
We can also conclude that every genus zero function field
\(K_{/\mathbb{F}_q}\) is isomorphic to \({\mathbb{F}}_q(t)\) and thus
rational, since such a function field rational iff it has index one.
Why? By Riemann-Roch, index one implies existence of a divisor of degree
one, and taking a genus zero curve says that every divisor of
nonnegative degree is linearly equivalent to an effective divisor. Thus
if you have a divisor of degree one, you have an effective divisor of
degree one, which makes the function field a degree one extension of a
rational function field.
\end{remark}
\begin{exercise}[?]
Let \(K = {\mathbb{F}}_q(t)\), then show that \(g=0, \delta = 1\), and
\begin{align*}
Z(t) = {1 \over (1-qt)(1-t)}
.\end{align*}
\begin{quote}
Hint: go back to complicated formulas and substitute \(\delta=1\) to
simplify things.
\end{quote}
\end{exercise}
Thus for rationality of the zeta function, we can get rid of the
\(\delta\) cluttering up formulas.
\hypertarget{the-functional-equation}{%
\subsection{The Functional Equation}\label{the-functional-equation}}
Going back to the plan, we wanted to show
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
Rationality: \(Z(t) \in {\mathbb{Q}}(t)\) and thus
\(Z(t) = P(t) / Q(t)\),
\item
Understand the degrees of \(P\) and \(Q\) in terms of the genus, and
\item
Ask about the roots of \(P(t)\) to understand the analog of the
Riemann Hypothesis for Dedekind zeta functions
\end{enumerate}
We'll want to establish a functional equation, as is the usual yoga for
zeta functions, since it helps establish a meromorphic continuation to
\({\mathbb{C}}\). The algebraic significance of the functional equation
is that it aids in understand several equivalent packets of data:
\begin{itemize}
\item
The number of effective divisors of a given degree,
\item
The number of places of a given degree,
\item
The number of rational points over each finite degree extension of the
base field.
\end{itemize}
\begin{theorem}[Functional Equation]
Let \(K_{/\mathbb{F}_q}\) be a function field of genus \(g\), then
\begin{align*}
Z(t) = q^{g-1} t^{2g-2} Z\qty{1\over qt}
.\end{align*}
\end{theorem}
\begin{proof}[?]
For \(g=0\), we know that
\begin{align*}
Z(t) = {1 \over (1-t)(1-qt)}
,\end{align*}
and plugging in \({1\over qt}\) is a straightforward calculation. So
assume \(g\geq 1\).
The idea was that we wrote \(Z(t) = F(t) + G(t)\). The \(F(t)\) piece
came from summing over divisor classes of degree between \(0\) and
\(2g-2\) and recording the dimension of the associated linear system.
The tricky piece \(G(t)\) came from summing an infinite geometric series
to get a more innocuous closed-form expression of a rational function.
So the strategy here is to separately establish the functional equation
for each of \(F\) and \(G\) separately. How to do this: for \(g=0\),
there was no \(F(t)\) piece. If we have a closed form it's just a
computational check. For \(F(t)\), we'll use our greatest weapon and
dearest ally, the Riemann-Roch theorem. This will provide the extra
symmetry we need.
We essentially already applied Riemann-Roch to \(G(t)\) to get the
closed-form expression, but we haven't applied it to the small degree
divisors. This doesn't tell you what the dimension is, but rather gives
you a duality result: ti gives the dimension in terms of the dimension
of a complementary divisor.
Take a canonical divisor \(\mathcal{K} \in \operatorname{Div}(K)\), so
\(\deg \mathcal{K} = 2g-2\). As \(C\) runs through all divisor classes
of \(\mathcal{K}\) of degree \(d\) with \(0\leq d \leq 2g-2\), so does
the complementary divisor \(\mathcal{K}-C\).
We can thus write
\begin{align*}
(q-1) F(t)
&= \sum_{0 \leq \deg C \leq 2g-2 } q^{\ell(C)} t^{\deg(C)} \\
(q-1)G(t)
&= h \qty{ {q^g t^{2g-1} \over 1-qt} - {1 \over 1-t} }
.\end{align*}
We can thus compute
\begin{align*}
(q-1) F\qty{1\over qt}
&= \sum_{0\leq \deg C \leq 2g-2} q^{\ell(C)} \qty{1\over {qt} }^{\deg C} \\
&= \sum_{0\leq \deg C \leq 2g-2} q^{\ell(\mathcal{K} - C)} \qty{1\over {qt} }^{2g-2-\deg C}
,\end{align*}
where in the second step we've exchanged \(C\) for \(\mathcal{K}- C\)
and noted that \(\deg(\mathcal{K}-C) = 2g-2-\deg(C)\). We now do the
calculation another way,
\begin{align*}
(q-1) F(t)
&= \sum_{0\leq \deg C \leq 2g-2} q^{\ell(C)} t^{\deg C} \\
&=
q^{g-1} t^{2g-1} \sum_{0\leq \deg C \leq 2g-2} q^{\deg(C) - (2g-2) + \ell(\mathcal{K}-C)} t^{\deg(C) - (2g-2)} && \text{by Riemann-Roch} \\
&= q^{g-1} t^{2g-2} \sum_{0 \leq \deg C \leq 2g-2} q^{\ell(\mathcal{K} - C)} \qty{1\over qt}^{\deg(\mathcal{K} - C)} \\
&= q^{g-1} t^{2g-2} (q-1) F\qty{1\over qt}
.\end{align*}
where we've used Riemann-Roch to find that
\(\ell(C) = \ell(\mathcal{K}-C) + \deg(C) - g + 1\). Cancelling the
common factor of \((q-1)\) establishes the functional equation for
\(F(T)\).
Now using the fact that \(\delta = 1\), we have
\begin{align*}
(q-1)G(t) = h \qty{ {q^g t^{2g-1} \over 1-qt} - {1\over 1-t} }
,\end{align*}
and thus
\begin{align*}
(q-1) q^{g-1} t^{2g-2} G\qty{1\over qt}
&= hq^{g-1} t^{2g-2} \qty{q^g \qty{1\over qt}^{2g-1} - {1\over 1 - q \qty{1\over qt}} - {1\over 1 - {1\over qt}} } \\
&=
h\qty{ {-1\over 1-t} + {q^g t^{2g-1} \over 1-qt}} \\
&= (q-1) G(t)
,\end{align*}
which establishes the functional equation for \(G(t)\).
\end{proof}
\hypertarget{the-l-polynomial}{%
\subsection{\texorpdfstring{The \(L\)
Polynomial}{The L Polynomial}}\label{the-l-polynomial}}
\begin{definition}[The $L$ Polynomial]
\begin{align*}
L(t) \coloneqq(1-t) (1-qt) Z(t) \in {\mathbb{Z}}[t]
.\end{align*}
\end{definition}
This clears the denominators in \(Z(t)\), so this is now a polynomial of
degree at most \(2g\). We can thus rewrite
\begin{align*}
Z(t) = {L(t) \over (1-t)(1-qt)} = {a_{2g} t^{2g} + \cdots + a_1 t + a_0 \over (1-t)(1-qt)}
.\end{align*}
Note that if we know \(L(t)\), then we know \(Z(t)\), and in particular
we would like to know what the coefficients \(a_j\) are. We'll be able
to determine \(a_0 = 1\) in all cases, as well as \(a_{2g}\) in all
cases pretty easily. So it looks like it only remains to compute
\(a_1, \cdots, a_{2g-1}\), but the functional equation will give a
``mirror'' relation between pairs of coefficients. The upshot is that
the functional equation shows that we only need to know
\(a_1, \cdots, a_g\) to completely determine \(Z(t)\). If \(g=1\), just
one coefficient suffices. It turns out that \(a_1\) will be \(q+1\)
minus the number of degree one places.
\begin{question}
\envlist
\begin{itemize}
\item
What are the constraints on these quantities?
\item
Can we write the zeta function in a nice way?
\item
Exactly what do we need to compute to determine it?
\end{itemize}
\end{question}
It will turn out that computing the number of rational points over
\({\mathbb{F}}_{q}, {\mathbb{F}}_{q^2}, \cdots, {\mathbb{F}}_{q^g}\)
will be possible. For example, for a hyperelliptic curve, we'll have an
explicit defining equation and can make an explicit point count, and you
only need \(g\) of them.
\hypertarget{lecture-15-the-lhbox-polynomial}{%
\section{\texorpdfstring{Lecture 15: The
\(L{\hbox{-}}\)Polynomial}{Lecture 15: The L\{\textbackslash hbox\{-\}\}Polynomial}}\label{lecture-15-the-lhbox-polynomial}}
\hypertarget{big-list-of-important-facts}{%
\subsection{Big List of Important
Facts}\label{big-list-of-important-facts}}
Recall that we had \(Z(t) + F(t) + G(t)\):
\begin{align*}
(q-1) F(t)
&= \sum_{0 \leq \deg C \leq 2g-2 } q^{\ell(C)} t^{\deg(C)} \\
(q-1)G(t)
&= h \qty{ {q^g t^{2g-1} \over 1-qt} - {1 \over 1-t} }
.\end{align*}
Note that \(F(t)\) is a polynomial of degree at most \(2g-2\), and
clearing denominators in \(G(t)\) yields a polynomial of degree at most
\(2g\)
\begin{definition}[The $L\dash$polynomial]
The \(L{\hbox{-}}\)polynomial is defined as
\begin{align*}
L(t) \coloneqq(1-t)(1-qt) Z(t) = (1-t)(1-qt) \sum_{n=0}^\infty A_n t^n \in {\mathbb{Z}}[t]
.\end{align*}
\end{definition}
It turns out that the degree bound of \(2g\) is sharp, and the
coefficients closer to the middle are most interesting:
\begin{theorem}[?]
Let \(K/{\mathbb{F}}_q\) be a function field of genus \(g\geq 1\), then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
\(\deg L = 2g\).
\item
\(L(1) = h\)
\item
\(L(t) = q^g t^{2g} L\qty{1\over qt}\).
\item
Writing \(L(t) = \sum_{j=1}^{2g} a_j t^{j}\),
\end{enumerate}
\begin{itemize}
\item
\(a_0 = 1\) and \(a_{2g} = q^g\).
\item
For all \(0\leq j \leq g\), we have \(a_{2g-j} = q^{g-j}a_j\).
\item
\(a_1 = {\left\lvert {\Sigma_1(K/{\mathbb{F}_{q}})} \right\rvert} - (q+1)\),
which notably does not depend on \(g\).
\item
Write \(L(t) = \prod_{j=1}^{2g} (1 - \alpha_j t) \in {\mathbb{C}}[t]\)
\footnote{The polynomial isn't monic, but rather has a constant
coefficient, so this expansion is somewhat more natural than (say)
\(\prod (t-\alpha)\).}
\end{itemize}
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{4}
\item
The
\(\alpha_j \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
\footnote{\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
denotes the algebraic integers.} (which were \emph{a priori} in
\({\mathbb{C}}\)) and can be ordered such that for all
\(1\leq j \leq g\), we have \(a_j a_{g+j} = q\). \footnote{This is the
first hint at the Riemann hypothesis: if for example they all had
the same complex modulus, this would force
\({\left\lvert {a_j} \right\rvert} = \sqrt q\). Thus proving that
they all have the same absolute value is 99\% of the content!}
\item
If \(L_r(t) = (1-t)(1-q^rt) Z_r(t)\) then
\(L_r(t) = \prod_{j=1}^{2g}(1-\alpha_j^r t)\), where \(K_r\) is the
constant extension \(K {\mathbb{F}_{q^r}}/{\mathbb{F}_{q^r}}\)
\end{enumerate}
\end{theorem}
Note that the \(\alpha_j\) are reciprocal roots.
\hypertarget{proofs}{%
\subsection{Proofs}\label{proofs}}
\hypertarget{the-degree-of-l-and-l1}{%
\subsubsection{\texorpdfstring{The degree of \(L\) and
\(L(1)\)}{The degree of L and L(1)}}\label{the-degree-of-l-and-l1}}
\begin{proof}[of a]
We saw from \(Z(t) = F(t) + G(t)\) that \(\deg L \leq 2g\). Equality
will follow from the proof of (d) part 1, since this would imply that
\(a_{2g} = q^g \neq 0\).
\end{proof}
\begin{proof}[of b]
Our formula \(Z(t) = F(t) + G(t)\) and Schmidt's theorem (showing
\(\delta = 1\)) gives
\begin{align*}
L(t) = (1-t) (1-qt) F(t) + {h \over q-1} \qty{ q^g t^{2g-2} (1-t) - (1-qt)}
,\end{align*}
where we've expanded \(G\) but not \(F\) because it involves various
\(\ell(D)\) which are difficult to compute. It is some polynomial
though, and we can evaluate \(L\) at 1 to get \(L(1) = h\). Thus the
class number is the sum of the coefficients!
\end{proof}
\hypertarget{functional-equation}{%
\subsubsection{Functional Equation}\label{functional-equation}}
\begin{proof}[of c]
This follows easily from the functional equation for \(Z(t)\), which we
already established using the Riemann-Roch theorem:
\begin{align*}
Z(t) = q^{g-1} t^{2g-2} Z\qty{1\over qt}
.\end{align*}
We can compute
\begin{align*}
q^g t^{2g} L\qty{1\over qt}
&= q^g t^{2g} \qty{1 - {1\over qt}} \qty{1 - {1\over t}} Z\qty{1\over qt} \\
&= q^{g-1} t^{2g-2} (1-t) (1-qt) Z\qty{1\over qt} \\
&= (1-t) (1-qt) Z(t) \\
&\coloneqq L(t)
,\end{align*}
where we've distributed one \(q\) and two \(t\)s in the first steps.
\end{proof}
\hypertarget{coefficients-a_j-for-j0-1-2g-and-duality}{%
\subsubsection{\texorpdfstring{Coefficients \(a_j\) for \(j=0, 1, 2g\)
and
Duality}{Coefficients a\_j for j=0, 1, 2g and Duality}}\label{coefficients-a_j-for-j0-1-2g-and-duality}}
\begin{proof}[of d]
Using the functional equation from (c), we can write
\begin{align*}
L(t) = q^g t^{2g} L\qty{1\over qt} = \qty{a_{2g} \over q^g} + \qty{a_{2g-1} \over q^{g-1}}t + \cdots + \qty{a_0 q^g} t^{2g}
,\end{align*}
where we're correcting by enough in \(t\) but not enough in \(q\) and
seeing what we get. Equating coefficients, for \(0\leq j \leq g\) we
have
\begin{equation}
a_{2g-j} = q^{g-j} a_j
\label{eq:sym_formula_proofc}
.\end{equation}
Using the fact that \(A_0\) is the number of effective degree zero
divisors, which is only zero, we have \(A_0 = 1\) and we can multiply
formal power series to obtain
\begin{align*}
L(t) = a_0 + a_1 t + \cdots + a_{2g} t^{2g}
&= (1-t)(1-qt) \sum_{n=0}^\infty A_n t^n \\
&= \qty{ 1 - (q+1)t + qt^2 }(1 + A_1 t + A_2 t^2 + \cdots)\\
&= 1 + \qty{A_1 - (q+1) }t + \cdots
.\end{align*}
From this, we can read off
\begin{itemize}
\tightlist
\item
\(L(0) = a_0 = 1\)
\item
\(a_1 = A_1 - (q+1) = \Sigma_1(K/k) - (q+1)\)
\item
\(a_{2g} = a_{2g-0} = q^{g-0}a_0 = a^g\) by taking \(j=0\) in
\cref{eq:sym_formula_proofc}, and thus \(\deg L = 2g\).
\end{itemize}
\end{proof}
\hypertarget{absolute-values-of-roots-rh}{%
\subsubsection{Absolute Values of Roots /
RH}\label{absolute-values-of-roots-rh}}
\begin{proof}[of e (the most interesting!)]
Consider the \textbf{reciprocal polynomial}
\begin{align*}
L^{\perp}(t) \coloneqq t^{2g} L\qty{1\over t}
= t^{2g} + a_1 t^{2g-1} + \cdots + q^g
.\end{align*}
The original polynomial had \({\mathbb{Z}}\) coefficients and constant
term 1, so this polynomial is monic and has a nonzero constant term.
Thus its roots are patently nonzero algebraic integers in
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu^{\bullet}\).
If \(L^{\perp}(t) = \prod_{j=1}^{2g} (t-\alpha_j)\), then
\begin{align*}
L(t) = t^{2g} L^{\perp}\qty{1\over t} = \prod_{j=1}^{2g} (1 - \alpha_j t)
\end{align*}
and if the roots of \(L(t)\) are \(r_j\), then the roots of
\(L^{\perp}(t)\) are the reciprocal roots \(1/r_j\) and vice-versa. This
shows the first assertion that
\(r_j \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Z}}\mkern-1.5mu}\mkern 1.5mu\)
as well.
The most interesting part is what follows. Making the substitution
\(t=qu\) and using (c) we get
\begin{align*}
L^{\perp}(t)
&= \prod_{j=1}^{2g} (t- \alpha_j) \\
&\coloneqq t^{2g} L\qty{1\over t} \\
&= q^{2g} u^{2g} L\qty{1\over qu} && \text{by (c)}
.\end{align*}
Using \(u = t/q\), we can write
\begin{align*}
q^g L(u)
&= q^g \prod_{j=1}^{2g} (1 - \alpha_j u) \\
&= q^g \prod_{j=1}^{2g} \qty{ 1 - {\alpha_j \over q}t} \\
&= q^g \prod_{j=1}^{2g} {\alpha_j \over q} \prod_{j=1}^{2g}\qty{ t - {1\over \alpha_j} } \\
&= \prod_{j=1}^{2g} \qty{t - {q\over \alpha_j}}
,\end{align*}
where we've pulled out a factor of \(-\alpha_j/q\) and in the last step
we've used that \(\prod_{j=1}^{2g} \alpha_j = q^g\). This follows
because the \(\alpha_j\) are the roots of \(L^{\perp}\), which has even
degree, so the product of all of the roots is equal to the constant term
of \(L^{\perp}\), which is the leading term of \(L\), which we showed
was \(q^g\).
This says that if we take these roots \(\alpha_j\) as a multiset and
replace each \(\alpha_j\) with \(q/\alpha_j\), we get the same multiset
back. I.e., this multiset is stable under the involution
\begin{align*}
{\mathbb{C}}^{\times}&\to {\mathbb{C}}^{\times}\\
z &\mapsto {q\over z}
.\end{align*}
This almost pairs up the elements of this finite set of roots, except it
may have fixed points. The complex numbers \(\alpha\) such that
\(\alpha = q/\alpha\) are precisely \(\pm \sqrt q\). So group the
\(\alpha_i^{-1}\) into
\begin{itemize}
\tightlist
\item
\(k\) \textbf{pairs} of nonfixed points, where
\(\alpha_i \neq q/\alpha_i\),
\item
\(m\) points such that \(\alpha_i = \sqrt q\),
\item
\(n\) points such that \(\alpha_i = -\sqrt q\).
\end{itemize}
So we'd like to show that \(m\) and \(n\) are both even, so when we're
pairing roots with reciprocals these get paired with themselves. We know
\(2k + m + n = 2g\), so \(m+n\) is even. We also know that
\begin{align*}
q^g
&= \prod_{j=1}^{2g} \alpha_j \\
&= q^k \qty{\sqrt{q}}^m \qty{-\sqrt q}^n \\
&= (-1)^n q^{k + {m \over 2} + {n\over 2}} \\
&= (-1)^n q^g
.\end{align*}
This forces \(n\) to be even, and since \(m = 2g-2k-n\), \(m\) must be
even as well.
\end{proof}
\begin{proof}[of f]
We used Dirichlet's character-style decomposition of \(Z(t)\) in
Schmidt's theorem, and we'll use it again here. Write
\begin{align*}
L_r(t^r)
&= (1-t^r) (1-q^r t^r) Z_r(t^r) \\
&= (1-t^r) (1-q^r t^r) \prod_{\xi \in \mu_r} Z(\xi t) \\
&= (1-t^r) (1-q^r t^r) \prod_{\xi \in \mu_r} {L(\xi t) \over (1-\xi t)(1-q\xi t) } \\
&= \prod_{\xi \in \mu_r} L(\xi t)
,\end{align*}
where we've used that
\begin{align*}
\prod_{\xi \in \mu_r} {1\over 1 - \xi t} &= 1-t^r \\
\prod_{\xi \in \mu_r} {1\over 1 - q\xi t} &= 1-q^rt^r \\
\end{align*}
which leads to all of the denominators canceling. We can then expand
\(L_r(t^r)\) as a product to compute
\begin{align*}
L_r(t^r)
&= \prod_{\xi \in \mu_r} L(\xi t) \\
&= \prod_{\xi\in \mu_r} \prod_{j=1}^{2g} (1- \alpha_j qt) \\
&= \prod_{j=1}^{2g} \prod_{\xi\in \mu_r} (1- \alpha_j qt) && \text{since these are finite products}\\
&= \prod_{j=1}^{2g} (1 - \alpha_j^r t^r)
.\end{align*}
From this we can conclude that
\(L_r(t) = \prod_{j=1}^{2g} (1- \alpha_j^r t)\), since \(t^r\) is just
an indeterminate and these are all identities of polynomials.
\end{proof}
\hypertarget{applications-and-corollaries}{%
\subsection{Applications and
Corollaries}\label{applications-and-corollaries}}
\hypertarget{counting-rational-points}{%
\subsubsection{Counting Rational
Points}\label{counting-rational-points}}
\begin{corollary}[?]
Suppose \(K/{\mathbb{F}_{q}}\) is genus \(g\geq 1\) and
\(L(t) = \prod_{j=1}^{2g}(1- \alpha_j t)\). Then for all
\(r\in {\mathbb{Z}}^{\geq 0}\), we have a nice expression for \(N_r\):
\begin{align*}
N_r \coloneqq{\left\lvert {\Sigma_1(K_r/{\mathbb{F}_{q^r}})} \right\rvert} = q^r + 1 - \sum_{j=1}^{2g} \alpha_j^r
.\end{align*}
\end{corollary}
\begin{proof}[?]
Let
\(L_r(t) = \sum_{j=1}^{2g} a_{j, r} = \prod_{j=1}^{2g} (1 - \alpha_j^r t)\),
so \(a_{1, r} = -\sum_{j=1}^{2g} \alpha_j^r\). Then using (d) part 3, we
can write
\begin{align*}
{\left\lvert {\Sigma_1(K_r/{\mathbb{F}_{q^r}})} \right\rvert} = q^r + 1 + a_{1, r} = q^r + 1 - \sum_{j=1}^{2g} \alpha_j^r
.\end{align*}
This follows from consider \(\prod (1-\alpha_j^r t)\), where extracting
the \(t^1\) coefficient involves choosing \(-\alpha_j^r\) once and 1
from all of the remaining terms, and then you sum over the disjoint
possibilities.
\end{proof}
\begin{remark}
We'd really like to compute the coefficients of the \(L\) polynomials,
since we can solve a polynomial equation to get the roots. But the
Galois groups of these polynomials may not be solvable, so the term
\(\sum \alpha_j^r\) will in general be some symmetric function in the
complex roots. Note that any symmetric polynomial in the roots is also a
symmetric polynomial in the coefficients.
\end{remark}
\hypertarget{relating-rational-points-to-coefficients}{%
\subsubsection{Relating Rational Points to
Coefficients}\label{relating-rational-points-to-coefficients}}
\begin{corollary}[?]
For \(K/{\mathbb{F}_{q}}\) a function field, define
\begin{align*}
S_r \coloneqq N_r - (q^r + 1) = - \sum_{j=1}^{2g} \alpha_j^r
.\end{align*}
Note that
\(N_r = {\left\lvert {\Sigma(K_r/{\mathbb{F}_{q^r}})} \right\rvert}\) is
the number of \({\mathbb{F}_{q^r}}{\hbox{-}}\)rational point. Then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
\(L'(t)/L(t) = \sum_{r=1}^\infty S_r t^{r-1}\).
\item
\(a_0 = 1\), and for all \(1\leq i \leq g\),
\begin{align*}
ia_i = S_i a_0 + S_{i-1} a_1 + \cdots + S_1 a_{i-1}
.\end{align*}
\end{enumerate}
\end{corollary}
\begin{remark}
What's the usefulness here? If you only have the coefficients of the
\(L\) polynomials, taking the logarithmic derivative gives access to
these quantities \(S_r\). The second formula is a recursive expression
for the \(a_i\) in terms of the \(S_i\). So you can compute the
coefficients of the \(L\) polynomial by counting
\({\mathbb{F}_{q^r}}{\hbox{-}}\)rational points on your curve (or places
on your function field) for \(r=1,2,\cdots, g\). Similarly, if you have
all of the coefficients for a \(Z\) polynomial, you can solve for the
\(S_i\).
\end{remark}
\begin{proof}[of a]
Essentially just a computation. Logarithmically differentiating both
sides of \(L(t) = \prod_{j=1}^{2g} (1-\alpha_j t)\) and expanding in a
geometric series yields
\begin{align*}
{L'(t) \over L(t) }
&= \sum_{j=1}^{2g} {-\alpha_j \over a -\alpha_j t} \\
&= \sum_{j=1}^{2g} (-\alpha_j) \sum_{r=0}^\infty \qty{\alpha_j t}^r \\
&= \sum_{r=1}^{\infty} \qty{\sum_{j=1}^{2g} (-\alpha_j^r) }t^{r-1} \\
&= \sum_{r=1}^{\infty} S_r t^{r-1}
.\end{align*}
\end{proof}
\begin{proof}[of b]
Clearing denominators and equating coefficients in
\(L'(t) = L(t) \sum_{r=1}^{\infty} S_r t^{r-1}\) yields the result
immediately, since the \(ia_i\) are what appear as coefficients in the
derivative of a formal power series, whereas the RHS is a Cauchy
product.
\end{proof}
\begin{remark}
The moral: to compute zeta functions, you don't have to enumerate
divisors and compute dimensions of Riemann-Roch spaces. Note that the
Riemann-Roch theorem tells us something interesting about these
dimensions, but doesn't compute the dimension outright! Instead, it
suffices to compute \({\mathbb{F}_{q^r}}{\hbox{-}}\)rational points for
\(r\leq g\).
A few lectures ago we discussed the places on a hyperelliptic function
field, including a place at infinity. Computing the zeta function of a
hyperelliptic curve involves plugging in \(x{\hbox{-}}\)values and
determining if it is
\begin{itemize}
\tightlist
\item
A nonzero non-square: no \(y{\hbox{-}}\)values,
\item
Zero: exactly one \(y{\hbox{-}}\)value,
\item
A nonzero square: two \(y{\hbox{-}}\)values.
\end{itemize}
This is what happens at the finite places. To handle the place at
\(\infty\), there is a recipe for the degree of the polynomial in terms
of the coefficients. So for any hyperelliptic function field (and in
particular, for any elliptic function field) we have a concrete
algorithm for computing their zeta functions. Note that this is not
necessarily a \emph{good} algorithm: it still involves plugging in many
values and checking if things are squares in finite values. It seems
that most people who compute a lot of zeta functions mostly focus on
hyperelliptic function fields.
How are you going to compute zeta functions or even places for more
complicated function fields? The Riemann-Hurwitz formula says that since
any function field is a finite degree extension of a rational function
field, the curve is given as a degree 2 branched cover of
\({\mathbb{P}}^1\), it suffices to compute the fibers of this cover in
order to get point counts.
\end{remark}
\hypertarget{lecture-16}{%
\section{Lecture 16}\label{lecture-16}}
\hypertarget{weil-bounds}{%
\subsection{Weil Bounds}\label{weil-bounds}}
Last time: we finished a discussion of the Hasse-Weil zeta function over
a finite ground field \({\mathbb{F}_{q}}\).
\begin{exercise}[?]
Let \(K/{\mathbb{F}_{q}}\) be a function field of genus 1.
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show
\begin{align*}
Z(t) = { 1-at + qt^2 \over (1-t)(1-qt)}
,\end{align*}
where
\(a = q+1 - {\left\lvert {\Sigma_1(K/{\mathbb{F}}_q)} \right\rvert}\).
\item
Let \(L(t) \coloneqq(1-\alpha_1 t)(1-\alpha_2 t)\). Show that
\(a = \alpha_1 + \alpha_2\), and that for all
\(r\in {\mathbb{Z}}^{+}\),
\begin{align*}
{\left\lvert {\Sigma_1 (K_r / {\mathbb{F}}_{q^r}} \right\rvert} = q^r +1 - \alpha_1^r - q^r/\alpha_1^r
.\end{align*}
Thus for elliptic curves \(E/{\mathbb{F}}_q\), knowing
\({\left\lvert {E({\mathbb{F}}_q)} \right\rvert}\) determines
\({\left\lvert {E({\mathbb{F}}_{q^r})} \right\rvert}\) for all
\(r\in {\mathbb{Z}}^+\).
\item
Suppose \(a^r = 0\) and show
\end{enumerate}
\begin{align*}
r \text{ odd} &\implies {\left\lvert {\Sigma_1 (K_r/{\mathbb{F}}_{q^r}) } \right\rvert} = q^r + 1 \\
r \equiv 2\pmod 4 &\implies {\left\lvert {\Sigma_1 (K_r/{\mathbb{F}}_{q^r}) } \right\rvert} = (q^{r/2} + 1)^2 \\
r \equiv 0\pmod 4 &\implies {\left\lvert {\Sigma_1 (K_r/{\mathbb{F}}_{q^r}) } \right\rvert} = (q^{r/2} - 1)^2 \\
.\end{align*}
\end{exercise}
\begin{theorem}[?]
Let \(K/{\mathbb{F}}_q\) be a function field of genus \(g\) with
\(L{\hbox{-}}\)polynomial \(L(t) = \prod_{i=1}^g (1 - \alpha_i t)\).
Then \({\left\lvert {\alpha} \right\rvert}_i = \sqrt{q}\) for all \(i\).
\end{theorem}
\begin{remark}
In order to prove this, Weil had to develop foundations for algebraic
geometry in positive characteristic. His original proof used
intersection theory on algebraic surfaces.
\end{remark}
\begin{corollary}[Weil Bounds]
If \(K/{\mathbb{F}}_q\) is a function field of genus \(g\), then
\begin{align*}
{\left\lvert { {\left\lvert {\Sigma_1 (K/{\mathbb{F}}_q)} \right\rvert} - (q+1) } \right\rvert} \leq 2g \sqrt{ q}
.\end{align*}
\end{corollary}
\begin{remark}
This says that the number of \({\mathbb{F}}_q\) points is approximately
\(q+1\)\textless{} where the error is controlled by the genus.
\end{remark}
\begin{proof}[?]
We know
\begin{align*}
{\left\lvert {\Sigma_1(K/{\mathbb{F}}_q)} \right\rvert} = q + 1 - \sum \alpha_i
.\end{align*}
Thus
\begin{align*}
{\left\lvert {
\Sigma_1(K/{\mathbb{F}}_q) - (q+1)
} \right\rvert}
=
{\left\lvert {
\sum \alpha_i
} \right\rvert}
\leq
\sum_{i=1}^{2g} {\left\lvert {\alpha_i} \right\rvert}
=2g\sqrt q
.\end{align*}
\end{proof}
\begin{corollary}[?]
For \(r\gg 0\),
\(N_r \coloneqq{\left\lvert {\Sigma_1(K_r/ {\mathbb{F}}_{q^r})} \right\rvert} \geq 1\).
\end{corollary}
\begin{proof}[?]
The Weil bounds yield
\begin{align*}
N_r \geq q^r + 1 - 2g q^{r/2}
\geq
q^{r/2} \qty{q^{r/2} - 2g} \overset{r\to \infty} \infty
.\end{align*}
This can be alternatively phrased as
\begin{align*}
N_r = q^r + O_g(q^{r/2})
.\end{align*}
Note that we've used the fact that making any separable extension of a
function field will preserve the genus, and so \(g\) is fixed.
\end{proof}
\begin{remark}
So for \(r\) large enough, there is an
\({\mathbb{F}}_{q^r}{\hbox{-}}\)rational point, and \(N_r \to \infty\)
exponentially fast in \(r\).
\end{remark}
\begin{remark}
As a consequence, if \(\omega\in \Sigma_1(K_r/ {\mathbb{F}}_{q^r})\),
let \(v\coloneqq\omega\cap K\). We saw that \(\deg(v) = d\) which
divides \(r\), so we can form the divisor
\(r/d[p] \in \operatorname{Div}^r K\) for any place \(p\) below \(v\).
So any degree 1 place yields a degree \(r\) divisor, which shows that
\(\operatorname{Div}^r K \neq \emptyset\) for all \(r\) large enough and
thus \(\delta=1\) (which is Schmidt's theorem).
\end{remark}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that there is a constant \(C\) depending only on \(g\) such that
if \(r\geq C\) then for all \(n\geq 2\), \(N_{nr} > N_r\).
\item
Use the following fact to show that there exists a constant \(D\)
depending on \(g\) such that for all \(d\geq D\),
\(\Sigma_d(K/{\mathbb{F}}_q) \neq \emptyset\):
\begin{align*}
N_r = \sum_{d\divides r} d {\left\lvert { \Sigma_d(K/{\mathbb{F}}_q) } \right\rvert}
.\end{align*}
\end{enumerate}
\end{exercise}
\begin{remark}
Note that this is stronger than Schmidt's theorem: it implies that not
only do you have a divisor of degree \(d\), but also a place of degree
\(d\).
\end{remark}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Use the Weil bounds to show that when \(g=0\),
\({\left\lvert {\Sigma_1(K/{\mathbb{F}}_q)} \right\rvert} = q+1\).
Deduce that every genus zero function field is rational.
\item
Use the Weil bounds to show that if \(g=1\), then
\({\left\lvert {\Sigma_1(K/{\mathbb{F}}_q)} \right\rvert} \geq \qty{\sqrt q - 1}^2 > 0\),
and thus every genus 1 function field over \({\mathbb{F}}_q\) is
elliptic.\footnote{By definition, a genus 1 function field with a
degree 1 place, which can be used as the origin for the group
structure.}
\end{enumerate}
\end{exercise}
\begin{corollary}[Serre Bounds]
Let \(K/{\mathbb{F}}_q\) be a function field of genus \(f\), then
\begin{align*}
{\left\lvert {
\Sigma_1(K/{\mathbb{F}}_q) - (q + 1)
} \right\rvert}
\leq g {\left\lfloor 2 \sqrt q \right\rfloor}
.\end{align*}
\end{corollary}
\begin{remark}
We write
\begin{align*}
M_q(g) &\coloneqq\text{The maximal } {\left\lvert {\Sigma(K/{\mathbb{F}}_q)} \right\rvert} \text{ as $K$ ranges over genus $g$ function fields} \\
m_q(g) &\coloneqq\text{The minimal } {\left\lvert {\Sigma(K/{\mathbb{F}}_q)} \right\rvert} \text{ as $K$ ranges over genus $g$ function fields} \\
A(q) &\coloneqq\limsup_{g\to \infty} {M_q(g) \over g}
.\end{align*}
This is essentially the best constant that can be put in front of \(g\)
in the bounds. With these definitions, we have
\begin{align*}
A(q) \leq 2\sqrt q && \text{by Weil bounds }\\
A(q) \leq {\left\lfloor 2\sqrt q \right\rfloor} && \text{by Serre bounds }\\
.\end{align*}
Note that we can do much better, e.g.~\(A(2) \leq \sqrt{2} - 1\), but it
is not known if this is sharp.
\end{remark}
\begin{proof}[of Serre's bound]
We may assume \(g \geq 1\), and write
\(L(t) = \prod_{i=1}^{2g} (1-\alpha_i t)\), where the \(\alpha_i\) may
be ordered such that \(\alpha_i \alpha_{g+-} = g\) for all \(i\). By the
Riemann hypothesis, we have
\({\left\lvert {\alpha_i} \right\rvert} = \sqrt q\), and so for each
\(i\) we have
\(\mkern 1.5mu\overline{\mkern-1.5mu\alpha_i\mkern-1.5mu}\mkern 1.5mu = q/\alpha_i = \alpha_{g+i}\).
We now pair in the following way: set
\begin{align*}
\gamma_i &\coloneqq\qty{ \alpha_i + \mkern 1.5mu\overline{\mkern-1.5mu\alpha_i\mkern-1.5mu}\mkern 1.5mu } + {\left\lfloor 2\sqrt q \right\rfloor} + 1 \\
\delta &\coloneqq-\qty{ \alpha_i + \mkern 1.5mu\overline{\mkern-1.5mu\alpha_i\mkern-1.5mu}\mkern 1.5mu } + {\left\lfloor 2\sqrt q \right\rfloor} + 1 \\
.\end{align*}
These are real algebraic integers and by the Riemann hypothesis they are
positive. Since
\(L^{\perp}(t) \coloneqq\prod t-\alpha_i \in {\mathbb{Z}}[t]\), take any
complex embedding
\begin{align*}
\sigma: {\mathbb{Q}}\qty{\left\{{\alpha_i}\right\}} \hookrightarrow{\mathbb{C}}
,\end{align*}
which preserves the \(\alpha_i\). If \(\sigma(\alpha_i) = \alpha_j\),
then we have
\begin{align*}
\sigma(\mkern 1.5mu\overline{\mkern-1.5mu\alpha_i\mkern-1.5mu}\mkern 1.5mu) = \sigma\qty{q/\alpha_i} = q/\sigma(\alpha_i) = \mkern 1.5mu\overline{\mkern-1.5mu\sigma(\alpha_i)\mkern-1.5mu}\mkern 1.5mu = \alpha_j
,\end{align*}
and thus \(\sigma\) preserves the multisets of the \(\gamma_i\) and
\(\delta_i\).
Now set \(\gamma\coloneqq\prod \gamma_i\) and
\(\delta \coloneqq\prod \delta_i\), making these both positive real
integers that are fixed by every embedding \(\sigma\), and thus
\(\gamma, \delta\in {\mathbb{Z}}^+\). We can now apply the AM-GM
inequality:
\begin{align*}
{1\over g} \sum_{i=1}^g \gamma_i
&\geq \qty{ \prod_{i=1}^g \gamma_i}^{1\over g}
= \gamma^{1/g} \geq 1
,\end{align*}
and thus
\begin{align*}
g \leq \sum \gamma_i
= \sum_{\alpha_i + \mkern 1.5mu\overline{\mkern-1.5mu\alpha_i\mkern-1.5mu}\mkern 1.5mu} + g{\left\lfloor 2\sqrt q \right\rfloor} + g
= \sum \alpha_i + g{\left\lfloor 2\sqrt q \right\rfloor} + g
,\end{align*}
and we can conclude that
\(-\sum \alpha_i \leq g{\left\lfloor 2\sqrt g \right\rfloor}\).
Repeating the argument with the \(\delta_i\) yields
\(\sum \alpha_i \leq g {\left\lfloor 2\sqrt q \right\rfloor}\), meaning
that
\({\left\lvert {\sum \alpha_i } \right\rvert} \leq g {\left\lfloor 2\sqrt q \right\rfloor}\).
Thus
\begin{align*}
{\left\lvert { {\left\lvert {\Sigma_1(K/{\mathbb{F}}_q)} \right\rvert} - (q+1) } \right\rvert} = {\left\lvert {\sum \alpha_i} \right\rvert} \leq g {\left\lfloor 2\sqrt q \right\rfloor}
.\end{align*}
\end{proof}
\begin{remark}
An application to class numbers: since \(g = L(1)\), by the Riemann
hypothesis we have
\begin{align*}
\sqrt{q} - 1 \leq {\left\lvert {\alpha_i - 1} \right\rvert} \leq \sqrt{q} + 1
\end{align*}
and thus
\begin{align*}
\qty{\sqrt q - 1}^{2g} \leq h \leq \qty{\sqrt q + 1}^{2g}
.\end{align*}
There is a slight improvement:
\begin{align*}
\qty{\sqrt q - 1}^{2g} \leq \qty{q+1 - {\left\lfloor 2\sqrt q \right\rfloor} }^g \leq h \leq {\left\lfloor \qty{\sqrt q + 1}^2 \right\rfloor}^g = \qty{q+1 + {\left\lfloor 2\sqrt q \right\rfloor}}^g
.\end{align*}
\end{remark}
\begin{corollary}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
For all \(q\geq 5\), we have
\begin{align*}
h\ geq \qty{\sqrt 5 - 1}^{2g} \geq 1.236^{2g}
.\end{align*}
\item
The class number grows exponentially in the genus not just for each
fixed \(q\) but uniformly over all \(q\geq 5\). In particular, if
\(g\geq 1\), this forces \(h> 1\).
\end{enumerate}
\end{corollary}
What exactly is happening at small \(q\), such as
\(q\in \left\{{2,3,4}\right\}\)?
\begin{theorem}[?]
For \(K/{\mathbb{F}}_q\) of genus \(g\geq 1\),
\begin{align*}
h \geq \qty{q-1 \over 2} \qty{q^{2g} + 1 - 2gq^g \over g\qty{q^{g+1} - 1}}
.\end{align*}
\end{theorem}
\begin{exercise}[?]
\envlist
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
Show that
\begin{align*}
h\geq \qty{q-1 \over 2}\qty{{ q^{g-1} \over g} - {2\over q}}
,\end{align*}
which still grows exponentially in \(g\).
\item
Show that for any \(1 2g-2\). Let
\(Q\in \Sigma_1(K_{2g} / {\mathbb{F}}_{q^{2g}})\) be a degree one place
and restrict to \(K\) to obtain \(P \coloneqq Q\cap K\) which has degree
\(\ell\) dividing \(2g\). Then \(\qty{2g \over \deg(P)} P\) has degree
\(2g\), and this yields a map
\begin{align*}
\Sigma_1(K_{2g} / {\mathbb{F}}_{q^{2g}}) &\to \text{Effective degree $2g$ divisors} \\
Q &\mapsto {2g \over \deg P} P
.\end{align*}
This is not necessarily surjective, but how far is it from being
injective? The fibers have size at most \(2g\) since we have a degree
\(2g\) extension of Dedekind domains. Thus
\begin{align*}
{\left\lvert {\Sigma_1 (K_{2g} / {\mathbb{F}}_{q^{2g}}) } \right\rvert} \leq A_{2g}
,\end{align*}
and substituting the known value of \(A_{2g}\) and rearranging yields
\begin{align*}
h
&\geq {1\over 2g} \qty{q-1 \over q^{g+1} - 1} N_{2g} \\
&\geq \qty{1\over 2g} \qty{q-1 \over q^{g+1} - 1} \qty{ q^{2g} + 1 - 2g q^g}
&& \text{by the Weil bounds} \\
&= \qty{q-1 \over 2} \qty {q^{2g} + 1 - 2gq^g \over g\qty{q^{g+1} - q}}
.\end{align*}
\end{proof}
\hypertarget{lecture-17-todo}{%
\section{Lecture 17 (Todo)}\label{lecture-17-todo}}
\hypertarget{lecture-18-todo}{%
\section{Lecture 18 (Todo)}\label{lecture-18-todo}}
\hypertarget{lecture-19-todo}{%
\section{Lecture 19 (Todo)}\label{lecture-19-todo}}
\hypertarget{lecture-20-todo}{%
\section{Lecture 20 (Todo)}\label{lecture-20-todo}}
\hypertarget{lecture-21-todo}{%
\section{Lecture 21 (Todo)}\label{lecture-21-todo}}
\hypertarget{lecture-22-todo}{%
\section{Lecture 22 (Todo)}\label{lecture-22-todo}}
\hypertarget{lecture-23-sketch}{%
\section{Lecture 23 (Sketch)}\label{lecture-23-sketch}}
\todo[inline]{What is an isogeny?}
\todo[inline]{What is an Artin-Schreier extension?}
\todo[inline]{What is Kummer theory?}
\todo[inline]{What are Weil differentials?}
\todo[inline]{What are Kahler differentials?}
\todo[inline]{What is the Riemann Hurwitz formula?}
\begin{corollary}[?]
Let \(k\) be a perfect field of characteristic \(p>0\),
\(d\in {\mathbb{Z}}^{\geq 0}\) with \(\gcd(d, p) = 1\), and let
\(f\in k[x]\) and \(L\coloneqq K(p^{-1} (f))\). Then \([L:K] = p\) and
\(L/k\) is a regular function field of genus
\(g = {1\over 2}(p-1)(d-1)\) that is unramified away from \(\infty\).
\end{corollary}
\hypertarget{artin-schreier-extensions-of-function-fields}{%
\subsection{Artin-Schreier Extensions of Function
Fields}\label{artin-schreier-extensions-of-function-fields}}
\begin{fact}
For \(k\) a field, \(\operatorname{ch}(k) = p > 0\), and \(a, b\in k\),
TFAE:
\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
\(k(p^{-1}(a)) = k(p^{-1}(b))\)
\item
\(a\) and \(b\) generate the same cyclic subgroup of \(k/p(k)\).
\end{enumerate}
In particular, if \(K(p^{-1}(u))/k\) is an Artin-Schreier extension,
then for all \(x\in k\), \(k(p^{-1}(u - (x^p - x))) = k(p^{-1}(u))\).
\end{fact}
\begin{lemma}[?]
Let \(k\) a \emph{perfect} field of characteristic \(p>0\), \(K/k\) a
function field, \(u\in K\), and \(p\in \Sigma(K/k)\).
\begin{itemize}
\item
There exists a \(z\in K\) such that
\(z_v \coloneqq v_p(u - (z^p - z))\) satisfies either
\begin{itemize}
\tightlist
\item
\(z_v \geq 0\), or
\item
\(z_v\leq 0\) and \(z_p\) is prime to \(p\).
\end{itemize}
\item
There exists a most one \(m\in {\mathbb{Z}}\) that is negative and
prime to \(p\) such that for some \(z\in K\) we have
\(v_p(u - (z^p - z)) = m\). If such an \(m\) exists, it is given by
\(m = \max\left\{{v_p(u - (z^p - z)) {~\mathrel{\Big|}~}z\in K }\right\}\).
\item
It follows that precisely \emph{one} of the two alternatives in the
first statement holds.
\end{itemize}
\end{lemma}
\begin{theorem}[Genus Formula for Artin-Schreier Extensions]
Let \(k\) a \emph{perfect} field of characteristic \(p>0\), \(K/k\) a
function field, \(u\in K\), \(L\coloneqq K(p^{-1}(u))\),
\(p\in \Sigma(K/k)\), and set
\begin{align*}
M_p \coloneqq
\begin{cases}
{\left\lvert {m} \right\rvert} & \text{if there exists a $z\in K$ such that } v_p(u - (z^p - z)) = m \\
-1 & \text{if there exists a $z\in K$ such that } v_p(u - (z^p - z)) \geq 0 \\
\end{cases}
.\end{align*}
Then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
If \(M_p = -1\), then \(p\) is unramified in \(L\).
\item
If \(M_p \geq 1\), then \(p\) is totally ramified in \(L\). Letting
\(\tilde p\) be the unique place lying over \(p\), then
\begin{align*}
d(\tilde p / p) = (p-1)(M_p + 1) && \text{(wild ramification)}
.\end{align*}
\item
Suppose there exists a \(p\) such that \(M_p \geq 1\). Then
\([L: K] = p\), \(L/k\) is regular, and we have a genus formula
\begin{align*}
g_L = p g_K + \qty{p-1 \over 2} \qty{ -2 + \sum_{p\in \Sigma(K/k)}(M_p+1)\deg p }
.\end{align*}
\end{enumerate}
\end{theorem}
\hypertarget{lecture-24-hermitian-function-fields-sketch}{%
\section{Lecture 24: Hermitian Function Fields
(Sketch)}\label{lecture-24-hermitian-function-fields-sketch}}
\todo[inline]{What is an elementary $p{\hbox{-}}$group?}
\todo[inline]{What is geometrically irreducible?}
\begin{theorem}[Stichtenoth Prop 6.4.1]
Let \(k\) be a perfect field of characteristic \(p>0\),
\(q \coloneqq p^s\) some power of \(p\), \(K \coloneqq k(x)\). Let
\(u \in k^{\times}\) and suppose \(T^q + \mu T\) splits in
\(k\).\footnote{When \(u=-1\), this recovers
\(q{\hbox{-}}\)Artin-Schreier extensions.} Let \(f\in k[x]\) with
\(\deg(f) = M\) where \(p\not\divides M\). Then
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
The polynomial
\begin{align*}
P(x, y) \coloneqq y^q + uy -f(x) \in k[x]
\end{align*}
is geometrically irreducible, and so
\(L\coloneqq\operatorname{ff}(k[x,y]/\left\langle{p}\right\rangle)\)
is a regular function field over \(k\).
\item
We have \([L: K] = q\).
\item
\(A\coloneqq\left\{{\gamma\in k {~\mathrel{\Big|}~}\gamma^q + u\gamma = 0}\right\}\)
is an order \(q\) subgroup of \({\mathbb{G}}_a/k \coloneqq(k, +)\).
Moreover, for all \(\sigma\in {\operatorname{Aut}}(L/K)\), there
exists a unique \(\gamma(\sigma)\in A\) such that
\(\sigma(y) = y + \gamma(\sigma)\) and
\(\sigma \mapsto \gamma(\sigma)\) yields an isomorphism
\({\operatorname{Aut}}(L/K) \xrightarrow{\sim} A\).
\item
No finite place of \(K\) ramifies in \(L\), while \(p_\infty\) is
totally ramified. If \(\tilde p_ \infty / p_ \infty\), then
\(d(\tilde p_ \infty / p_ \infty) = (q-1)(M+1)\).
\item
We have
\begin{align*}
g_L = \qty{1\over 2}(q-1)(m-1)
.\end{align*}
\end{enumerate}
\end{theorem}
Next up: one of the most important function fields of all time!
\begin{definition}[Hermitian Function Field]
Set \(A_q \coloneqq{\mathbb{F}}_{q^2}(x, y)\) and consider the
polynomial
\begin{align*}
y^q + y = x^{q+1}
.\end{align*}
Then \(u=1, M = q+1\), and
\(g = \qty{1\over 2}(q)(q-1) = {q \choose 2}\).
\end{definition}
\begin{theorem}[?]
\begin{align*}
{\left\lvert {\Sigma_1(A_q / {\mathbb{F}}_{q^2})} \right\rvert} = q^3 + 1
.\end{align*}
\end{theorem}
\todo[inline]{What are the Weil bounds?}
\begin{corollary}[Ihara]
If \(K/{\mathbb{F}}_q\) is a maximal function field of genus \(g\), then
\begin{align*}
g\leq \qty{1\over 2}\qty{q - \sqrt q}
.\end{align*}
\end{corollary}
\begin{fact}
If \(K/{\mathbb{F}}_{q^2}\) is maximal, then
\(N_1 = q^2 + 1 + 2gq = q^2 + q - \sum_{j=1}^{2g} \alpha_j\). Applying
the RH, \({\left\lvert {\alpha_j} \right\rvert} = q\), and it follows
that \(\alpha_j = -q\) for all \(j\) and thus
\begin{align*}
L(t) = \qty{1+qt}^{2g}
.\end{align*}
\end{fact}
\begin{theorem}[Kleiman, Serre]
If \(K/{\mathbb{F}}_q \subset L/{\mathbb{F}}_q\) is a finite extension
of function fields, then \(L_K(t)\) divides \(L_L(t)\).
\end{theorem}
\begin{corollary}[?]
If \(L/{\mathbb{F}}_{q^2}\) is maximal, so is \(K/{\mathbb{F}}_{q^2}\).
\end{corollary}
\begin{theorem}[Stichtenoth?]
\begin{align*}
{\operatorname{Aut}}(A_1 / {\mathbb{F}}_{q^2}) \cong \operatorname{PGU}_3({\mathbb{F}}_{q^2})
,\end{align*}
the \emph{projective unitary group}, which is of order
\(q^3(q^2-1)(q^3+1)\).
\end{theorem}
\begin{remark}
The size of this group \(G_q\) is asymptotically \(G_q \sim q^8\), while
\(g(A_q) \sim {q^2\over 2}\), so this is a lot of automorphisms compared
to the sizes of automorphism groups of Riemann surfaces. More precisely,
\(G_q > 16 g(q)^4\).
\end{remark}
\begin{theorem}[Stichtenoth]
For any other function field \(K/k\) for any field \(k\),
\({\left\lvert {{\operatorname{Aut}}(K/k)} \right\rvert} < 16g^4\).
\end{theorem}
\begin{remark}
This only happens in positive characteristic, when
\(\operatorname{ch}(k), g, q\) match up in a very specific way. So
Hermitian function fields are the algebraic curves with the most
symmetries.
\end{remark}
\begin{theorem}[Hurwtiz]
In characteristic zero, if \(g\geq 2\) then
\({\left\lvert {{\operatorname{Aut}}(K/k)} \right\rvert} \leq 84(g-1)\).
\end{theorem}
\hypertarget{lecture-25-differential-pullback-theorem-sketch}{%
\section{Lecture 25: Differential Pullback Theorem
(Sketch)}\label{lecture-25-differential-pullback-theorem-sketch}}
This will recover the Riemann-Hurwitz formula by taking degrees.
\begin{lemma}[?]
Let \(K/k \subset L/\ell\) be a finite degree extension of function
fields, and suppose \(K/k\) is regular and \(L/K\) is separable. Then
\(\ell/k\) and \(L/\ell\) are separable and \(L\ell\) is regular.
\begin{center}
\begin{tikzcd}
& {L} \\
{K} && {\ell} \\
& {k}
\arrow["{\scriptscriptstyle\text{separable}}", from=2-1, to=1-2, no head]
\arrow[from=1-2, to=2-3, no head]
\arrow[from=2-3, to=3-2, no head]
\arrow["{\scriptscriptstyle\text{regular}}"', from=2-1, to=3-2, no head]
\end{tikzcd}
\end{center}
\href{https://q.uiver.app/?q=WzAsNCxbMSwyLCJrIl0sWzIsMSwiXFxlbGwiXSxbMSwwLCJMIl0sWzAsMSwiSyJdLFszLDIsIlxcc2NyaXB0c2l6ZVxcdGV4dHtzZXBhcmFibGV9IiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dLFsyLDEsIiIsMCx7InN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMSwwLCIiLDAseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJub25lIn19fV0sWzMsMCwiXFxzY3JpcHRzaXplXFx0ZXh0e3JlZ3VsYXJ9IiwyLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==}{Link
to diagram}
\end{lemma}
Recall some facts/definitions:
\begin{itemize}
\item
The \textbf{adele ring of \(K\)} is defined as
\begin{align*}
\mathcal{A}_K \coloneqq\prod_{v\in \Sigma(K/k)}' K
\end{align*}
which is a \emph{restricted direct product}, i.e.~each element
\(\alpha\in \mathcal{A}_K\) has the property that for almost every
\(p\), the \(p{\hbox{-}}\)adic valuation of the \(p\)th coordinate
\(v_p(\alpha_p) \geq 0\). There is a diagonal embedding
\begin{align*}
K &\hookrightarrow\mathcal{A}_K \\
f &\mapsto (f, f, \cdots)
.\end{align*}
\item
For any divisor \(D\in \operatorname{Div}K\), define
\begin{align*}
\mathcal{A}_K(D) \coloneqq\left\{{\alpha\in \mathcal{A}_K {~\mathrel{\Big|}~}v_p(\alpha_p) \geq - v_p(D)\,\, \forall p}\right\}
,\end{align*}
the adelic analog of the Riemann-Roch space.
\item
A space of linear forms
\begin{align*}
\Omega(D) \coloneqq\left\{{\omega: \mathcal{A}_K \to A {~\mathrel{\Big|}~}\ker \omega \supseteq K + \mathcal{A}_K(D)}\right\}
\end{align*}
where \(D_1 \leq D_2 \implies \Omega_K(D_2) \leq \Omega_K(D_1)\).
\item
\(\Omega_K \coloneqq\varinjlim_{D} \Omega_K(D)\).
\item
For any \(\omega \in \Omega_K^{\bullet}\),
\((\omega) \coloneqq\max\left\{{D {~\mathrel{\Big|}~}\omega=0 \text{ on } \mathcal{A}_K(D) + K}\right\}\).
\item
\(\mathcal{A}_{L/K} = \left\{{\alpha\in \mathcal{A}_L {~\mathrel{\Big|}~}\alpha q_1 = \alpha q_2 \text{ if } Q_1, Q_2 / p}\right\} \leq_{{\operatorname{Vect}}_\ell} A_L\)
\item
The \textbf{adelic trace map}
\begin{align*}
\operatorname{Tr}_{L/K} : \mathcal{A}_{L/K} & \to \mathcal{A}_K \\
\alpha &\mapsto \operatorname{Tr}_{L/K}(\alpha) / p = \operatorname{Tr}_{L/K}(\alpha_Q) && \text{for any }Q/p
.\end{align*}
\end{itemize}
\begin{theorem}[?]
Let \(\omega\in \Omega_K\), then there exists a unique
\(\omega^* \in \mathcal{A}_L\) such that
\begin{itemize}
\tightlist
\item
For all \(\alpha\in \mathcal{A}_{L/K}\), we have
\(\operatorname{Tr}_{\ell/k} \omega^*(\alpha) = \omega(\operatorname{Tr}_{L/K}(\alpha))\).
\end{itemize}
\(\omega^*\) is formally denoted \(\mathrm{Cotr}_{L/K}(\omega)\) and
called the \textbf{cotrace} of \(\omega\).
\end{theorem}
\begin{theorem}[?]
If \(K/k \subset L/\ell\) is a finite extension of function fields with
\(K/k\) regular, then for all \(\omega\in \mathcal{A}_K^{\bullet}\), we
have \(\omega^* \in \mathcal{A}_L^{\bullet}\). Moreover,
\begin{align*}
(\omega^*) = \iota_{L/K}((\omega)) + D(L/K)
.\end{align*}
Taking degrees yields the Riemann-Hurwitz formula.
\end{theorem}
\addsec{ToDos}
\listoftodos[List of Todos]
\cleardoublepage
% Hook into amsthm environments to list them.
\addsec{Definitions}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]
\cleardoublepage
\addsec{Theorems}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]
\cleardoublepage
\addsec{Exercises}
\renewcommand{\listtheoremname}{}
\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]
\cleardoublepage
\addsec{Figures}
\listoffigures
\cleardoublepage
\printbibliography[title=Bibliography]
\end{document}