\input{"preamble.tex"}

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







\begin{document}

\date{}
\maketitle
\begin{flushleft}
\textit{D. Zack Garza} \\
\textit{University of Georgia} \\
  \textit{\href{mailto: dzackgarza@gmail.com}{dzackgarza@gmail.com}} \\
{\tiny \textit{Last updated:} 2021-12-02 }
\end{flushleft}


\newpage

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

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

Possible topics announcement from Daniel

\begin{quote}
The course will loosely follow Poonen's book on rational points,
available here: \url{https://math.mit.edu/~poonen/papers/Qpoints.pdf}
Planned topics include: the Hasse principle for quadratic forms,
obstructions to the Hasse principle (i.e.~the Brauer-Manin obstruction
and beyond), finding rational points and some effective methods
(e.g.~Chabauty), as well as some conjectural aspects of rational points.
I plan to cover topics in the second half of the semester which depend
on student interest; i.e.~if there's interest I can say some things
about Faltings's proof of the Mordell conjecture.
\end{quote}

\hypertarget{thursday-august-19}{%
\section{Thursday, August 19}\label{thursday-august-19}}

\begin{remark}

Some useful prerequisites:

\begin{itemize}
\tightlist
\item
  Number theory (e.g.~places)
\item
  Class field theory

  \begin{itemize}
  \tightlist
  \item
    See Cassels-Frolich (up through ch.~5 and 6)
  \end{itemize}
\item
  AG (although we'll avoid the language of schemes)
\item
  Galois and group cohomology
\item
  Bjorn Poonen's book
\end{itemize}

\end{remark}

\begin{remark}

On notation:

\begin{itemize}
\tightlist
\item
  \(k{ {}^{\cdot n} }\) will denote \(n\)th powers in \(k\), and
  similarly for \(k^{\times}\).
\item
  \(k^{\scriptscriptstyle\mathrm{un}}\) denotes an unramified extension.
\end{itemize}

\end{remark}

\begin{remark}

Setup: let \(k = {\mathbb{Q}}\) or more generally a number field or a
function field over \({\mathbb{F}}_q\). Consider a system of polynomial
equations over \(k[x_1, \cdots, x_{m}]\):
\begin{align*}
\begin{cases}
f_1(x_1, \cdots, x_m) &  = 0
\\
\quad \vdots & \vdots \\
f_n(x_1, \cdots, x_m)& =0 .
\end{cases}
\end{align*}
Some natural questions:

\end{remark}

\begin{remark}[Topic 1: Are there any common solutions?]

More generally, does \(X \coloneqq V(f_1, \cdots, f_n)\) have any
rational points? How many rational points are there? Finitely many, or
infinitely many?

\end{remark}

\begin{remark}[Topic 2: what is the distribution of points?]

\begin{itemize}
\tightlist
\item
  How many points are there of height at most \(N\), where
  \(\operatorname{ht}(a/b) = \max({\left\lvert {a} \right\rvert}, {\left\lvert {b} \right\rvert})\)?
\item
  Are they Zariski dense? I.e. are there solutions outside of the ideal
  \(\left\langle{f_i}\right\rangle\)?
\item
  Are they \emph{potentially} dense, i.e.~dense after some finite
  extension \(k\hookrightarrow k'\)?
\item
  Choosing \(k\hookrightarrow{\mathbb{C}}\) or \({ {\mathbb{Q}}_p }\),
  are the solutions dense in the analytic topology on
  \(X({\mathbb{C}}), X({ {\mathbb{Q}}_p })\)? If not, what is the
  closure?
\end{itemize}

There are many conjectures around these questions, but few general
results!

\end{remark}

\begin{remark}[Topic 3: Local to Global Principles]

Topic 3: local to global principles. Given \(X_{/ {{\mathbb{Q}}}}\), if
\(X({ {\mathbb{Q}}_p })\neq \emptyset\) for all \(p\) and
\(X({\mathbb{R}}) \neq \emptyset\), does this imply that
\(X({\mathbb{Q}}) \neq \emptyset\)? More generally, for \(X_{/ {k}}\)
with \(X(K_v) \neq \emptyset\) for all places \(v\) of \(K\), is this
enough to imply \(X(k)\neq \emptyset\) If so, we say \(X\) satisfies the
\textbf{Hasse principle}. If not, are there obstructions?

\end{remark}

\begin{remark}[Topic 3': Weak and Strong Approximation]

As an example,
\begin{align*}
X(k) \hookrightarrow\prod_{v\in P(k)} X(k_v)
\end{align*}
where \(p(k)\) are the places of \(k\). Is this map dense? Note the
topology is the product topology, so a basis for opens are sets with
finitely factors with opens, and the remaining are the entire space.
Strong approximation is an adelic version of this.

Obstructions to this principle: if this is not dense, what is the
closure \(X(k)\) in \(\prod X(k_v)\) or \(X({\mathbb{A}})\) for
\({\mathbb{A}}\) the adeles? One example we'll consider is the
Brauer-Manin obstruction.

\end{remark}

\begin{remark}[Topic 4: effectiveness and decidability questions.]

Given a variety \(X_{/ {{\mathbb{Q}}}}\), is there an actual algorithm
that decides if \(X({\mathbb{Q}})= \emptyset\)? This is known over
\({\mathbb{Z}}\), but open for \({\mathbb{Q}}\) and most (not all)
number fields. Are there special classes of varieties where the answer
of yes? For curves, this is only known contingent on open problems (the
abc conjecture, the section conjecture, Birch-Swinnerton-Dyer, etc).

Given a special \(X_{/ {k}}\) can you find \(X(k)\)?

\end{remark}

\begin{remark}

Other possible topics:

\begin{itemize}
\tightlist
\item
  The Mordell-Weil theorem for \(X\) an abelian variety, and a
  generalization, the Néron-Lang theorem which works over other fields.
\item
  Falting's theorem, that curves of genus 2 have finitely many rational
  points.
\end{itemize}

\end{remark}

\hypertarget{examples-of-hasse-principles}{%
\subsection{Examples of Hasse
Principles}\label{examples-of-hasse-principles}}

\begin{example}[?]

Let \(a\in {\mathbb{Q}}\), does \(x^2 = a\) satisfy a local to global
principle? This is related to Chebotarev density.

Claim: any positive number \(a\) such that \(v_p(a)\) is even for all
\(p\) is necessarily a square. This follows from writing
\(a = \pm \prod p_i^{n_i}\) where \(n_i\in {\mathbb{Z}}\) and is equal
to zero for all but finitely many \(i\), then its square root is
obtained by halving all of the \(n_i\). Note that
\(a \in ({\mathbb{R}}^{\times})^2\) implies \(a\) is positive, and
\(a\in ({ {\mathbb{Q}}_p }^{\times})^2\) implies that \(n_p\) is even.

\end{example}

\begin{example}[?]

Let \(a\in {\mathbb{Q}}\) and take \(x^n=a\), or more generally
\(f(x) = a\) for \(f\in {\mathbb{Q}}[x]\), where \(f(x) - a\) is
irreducible. Corollary of Chebotarev density: the set of primes where
\(f-a\operatorname{mod}p\) has no linear factors has positive density.
This means that an even stronger theorem is true: there exists a \(c<1\)
such that if \(f-a\) has no roots \(\operatorname{mod}p\) for a set of
primes of density \(d>c\), then \(f-a\) has no roots. So this satisfies
the Hasse principle.

\end{example}

\begin{example}[Conics]

Take \(X \coloneqq V(ax^2+by^2 +cz^2) \subseteq {\mathbb{P}}^2\) for
\(a,b,c\in {\mathbb{Q}}\). This also satisfies the Hasse principle, but
the proof is harder. Note that \(x^2+y^2+z^2 = 0\) has no rational
points (excluding zero since we're in \({\mathbb{P}}^2\)) since it has
no solutions over \({\mathbb{R}}\). It is potentially dense, noting that
one can take \({\mathbb{Q}}[i]\) over \({\mathbb{Q}}\) and get rational
points \(0, 1, \infty\). Given one point, one can stereographically
project to yield infinite many points by just taking lines through the
fixed point and letting slopes vary.

\todo[inline]{Something about using ${\mathcal{O}}(1)$ to give an embedding into ${\mathbb{P}}^1$. Start with ${\mathcal{O}}(-1)$, dualize, project?}

\end{example}

\begin{example}[Severi-Brauer varieties]

Taking \(X_{/ {k}}\) such that
\(X_{/ {\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu}} \cong {\mathbb{P}}^n_{/ {\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu}}\)
satisfy the Hasse principle.

\end{example}

\begin{example}[Quadrics]

A theorem by Hasse-Minkowski shows that these also satisfy the Hasse
principle.

\end{example}

\begin{example}[Genus 1 curves]

The \textbf{Selmer curve} \(3x^3 + 4y^3 + 5z^3=0\) does \emph{not}
satisfy the Hasse principle, which can be understood in terms of the
Tate-Shafarevich group or Brauer-Manin obstructions.

\end{example}

\begin{remark}

Note that it doesn't make sense to say a single variety satisfies the
Hasse principle, but rather a class. But it makes sense to say a single
variety \emph{doesn't}.

\end{remark}

\begin{remark}

A common generalization is that these are all torsors for an algebraic
group, i.e.~a homogeneous space, for which there are cohomological
methods to understand the Hasse principle.

\end{remark}

\begin{remark}

A variety \(X_{/ {k}}\) is \emph{geometrically integral} in the affine
case if when \(X = V(f_1,\cdots, f_n)\), the ring
\(\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu[x_1,\cdots, x_n]\)
is an integral domain.

\end{remark}

\begin{theorem}[?]

Suppose \(K\) is a number field and \(X_{/ {K}}\) is geometrically
integral. Then \(X(K_v) \neq \emptyset\) for all but finitely many
\(v\).

\end{theorem}

\begin{proof}[Sketch/idea]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Write \(X = V(f_1, \cdots, f_n)\) with a nonempty smooth locus
  \(X^{{\mathsf{sm}}} \subseteq X\) which is a variety (just adjoin
  inverses of partial derivatives appearing in minors of Jacobian
  matrices). So
  \(X^{{\mathsf{sm}}}/{\mathcal{O}}_{K, S} = {\mathcal{O}}_K\left[ {1\over N} \right]\)
  which is smooth over \({\mathcal{O}}_{K, S}\)
\item
  Use Lang-Weil to show that
  \(X^{\mathsf{sm}}({\mathcal{O}}_{K, S} / {\mathfrak{p}}) \neq \emptyset\)
  for almost all \({\mathfrak{p}}\).
\item
  Use smoothness and Hensel's lemma to get
  \(X^{{\mathsf{sm}}}({\mathcal{O}}_{K, S}{ {}^{ \widehat{{\mathfrak{p}}} } })\).
\end{enumerate}

\end{proof}

\hypertarget{tuesday-august-24}{%
\section{Tuesday, August 24}\label{tuesday-august-24}}

\begin{remark}

Last time: if \(K\) is a number field and \(X_{/ {K}}\) is geometrically
irreducible, then \(X(K_v) \neq \emptyset\) for almost all \(v\).

\end{remark}

\begin{proof}[?]

Choose
\(X_{/ {{\mathcal{O}}_K{ \left[ { \scriptstyle \frac{1}{N} } \right] }}}\)
such that \(X\) has geometrically integral fibers. It's enough to show
that \(X(K(v)) \neq \emptyset\) for almost all \(v\), where \(K(v)\) is
the residue field at finite places \(v\).

Now use the following theorem:

\end{proof}

\begin{theorem}[Lang-Weil Estimates]

If \(X\) over
\({\mathcal{O}}_K{ \left[ { \scriptstyle \frac{1}{N} } \right] }\) is
geometrically integral, then
\begin{align*}
\# X({\mathbb{F}}_{q^k}) = ( 1 + O(q^{1\over 2}) ) q^{k\dim X}
.\end{align*}

\end{theorem}

\begin{claim}

If \(X_{/ {{\mathcal{O}}_{K_v}}}\) is smooth then
\begin{align*}
X(K(v)) \neq \emptyset
\implies
X(K_v) \neq \emptyset
.\end{align*}

\end{claim}

\begin{proof}[?]

Use

\begin{itemize}
\tightlist
\item
  Slice and Hensel, or the formal smoothness criterion, i.e.~
\end{itemize}

\begin{center}
\begin{tikzcd}
    {\operatorname{Spec}R} && X \\
    \\
    {\operatorname{Spec}R'} && Y
    \arrow["{\mathsf{sm}}", from=1-3, to=3-3]
    \arrow["{ \operatorname{cl}} ", from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\exists"{description}, dashed, from=3-1, to=1-3]
    \arrow[from=3-1, to=3-3]
\end{tikzcd}
\end{center}

Taking \(R \coloneqq R'/I\) with \(I\) nilpotent.

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXHNwZWMgUiJdLFswLDIsIlxcc3BlYyBSJyJdLFsyLDIsIlkiXSxbMiwwLCJYIl0sWzMsMiwiXFxzbW9vdGgiXSxbMCwxLCJcXGNsIl0sWzAsM10sWzEsMywiXFxleGlzdHMiLDEseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMSwyXV0=}{Link
to Diagram}
\end{quote}

See Hartshorne chapter 3, in the exercises!

\end{proof}

\begin{remark}

As a black box, we'll use that this is true for
\(\dim_{{{\mathcal{O}}_{K_v}}} X = 1\), i.e.~for curves. This follows
from the Weil conjectures for curves, see Severi/Bombieri. If \(X\) is
genus \(g\), then in fact we have a finer estimate:
\begin{align*}
{\left\lvert { \# X({\mathbb{F}}_{q^k}) - q^n } \right\rvert} \leq q^{1\over 2} + 1
.\end{align*}

\end{remark}

\begin{proof}[?]

We'll show this for
\(\dim_{{\mathcal{O}}_K{ \left[ { \scriptstyle \frac{1}{n} } \right] }} = 2\).
Idea: try to fiber with curves.

\begin{itemize}
\item
  Suppose \(\operatorname{reldim}X = 1\) for \(X \to S\) over
  \({\mathcal{O}}_K{ \left[ { \scriptstyle \frac{1}{n} } \right] }\)
  where \(S\) is a curve with geometrically integral fibers.
\item
  Without loss of generality, \(X\to S\) where

  \begin{itemize}
  \item
    \(S\) is smooth of genus \(g'\),
  \item
    \(X/S\) is smooth with fibers of genus \(g\).
  \item
    Now take the count
  \end{itemize}

  \begin{align*}
  X({\mathbb{F}}_{q^k}) 
  &= (1 + O_{g'}(q^{- {k \over 2} }) )q\cdot (1 + O_g(q^{- {k\over 2} }))q \\
  &= (1 + O_{g, g'}(q^{- {k\over 2} }))q^2
  .\end{align*}
\item
  Such an \(X\to S\) after replacing \(X\) by an open subvariety. The
  proof of this follows from
  \href{https://www.wikiwand.com/en/Theorem_of_Bertini}{Bertini}: for
  \(X \subseteq {\mathbb{P}}^n\), take geometric projections and delete
  the singular locus. The fibers are slices by hyperplanes, and thus the
  fibers are geometrically integral.
\end{itemize}

\end{proof}

\hypertarget{brauer-groups}{%
\subsection{Brauer Groups}\label{brauer-groups}}

\begin{remark}

Some upcoming topics:

\begin{itemize}
\item
  Severi-Brauer varieties (so \(X_{/ {K}}\) where
  \(X_{/ {\mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu}} \cong {\mathbb{P}}^n\))
  satisfy the Hasse principle. Implies Hasse-Minkowski!
\item
  The Brauer-Manin obstruction to the Hasse principle.
\end{itemize}

\end{remark}

\hypertarget{the-brauer-manin-obstruction}{%
\subsubsection{The Brauer-Manin
Obstruction}\label{the-brauer-manin-obstruction}}

\begin{remark}

Setup:

\begin{itemize}
\tightlist
\item
  \(X\) is a variety,
\item
  \(\mathop{\mathrm{Br}}(X)\) is an abelian group
\item
  Given \(X \xrightarrow{f} Y\), there is an induced map
  \(f^*: \mathop{\mathrm{Br}}(Y)\to \mathop{\mathrm{Br}}(X)\).
\end{itemize}

For \(K\) a number field (which we can view as a variety with a single
point), we have
\begin{align*}
\mathop{\mathrm{Br}}(K_v)
= 
\begin{cases}
{\mathbb{Q}}/{\mathbb{Z}}&  v \text{ finite}
\\
{\mathbb{Z}}/2 & v \text{ real}
\\
0 & v \text{ complex},
\end{cases}
\end{align*}
which fits into a SES
\begin{align*}
0 \to \mathop{\mathrm{Br}}(K) \to \bigoplus_v \mathop{\mathrm{Br}}(K_v) \xrightarrow{\Sigma} {\mathbb{Q}}/{\mathbb{Z}}\to 0
.\end{align*}
Note that most of the terms in the middle sum are
\({\mathbb{Q}}/{\mathbb{Z}}\), making \(\mathop{\mathrm{Br}}(K)\) a
large group.

\end{remark}

\begin{remark}

The yoga of the Hasse principle says we should try to solve things in
adelic points first. Write
\begin{align*}
{\mathbb{A}}_K = \prod_v' (K_v, {\mathcal{O}}_v) \subseteq \prod_v K_v
\end{align*}
where we take the restricted product. There is a map
\(X(K)\to X({\mathbb{A}}_K)\), and taking
\(\alpha\in \mathop{\mathrm{Br}}(X)\) one gets a map
\(\alpha^*: X(K) \to \mathop{\mathrm{Br}}(K)\). This yields a diagram

\begin{center}
\begin{tikzcd}
    {X(K)} && {X({\mathbb{A}}_K)} \\
    \\
    {\mathop{\mathrm{Br}}(X)} && {\mathop{\mathrm{Br}}({\mathbb{A}}_k) \cong \bigoplus_v \mathop{\mathrm{Br}}(K_v)}
    \arrow["{\alpha^*}", from=1-1, to=3-1]
    \arrow["{\tilde\alpha^*}", from=1-3, to=3-3]
    \arrow[from=1-1, to=1-3]
    \arrow[from=3-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJYKEspIl0sWzAsMiwiXFxCcihYKSJdLFsyLDAsIlgoXFxBQV9LKSJdLFsyLDIsIlxcQnIoXFxBQV9rKSBcXGNvbmcgXFxiaWdvcGx1c192IFxcQnIoS192KSJdLFswLDEsIlxcYWxwaGFeKiJdLFsyLDMsIlxcdGlsZGVcXGFscGhhXioiXSxbMCwyXSxbMSwzXV0=}{Link
to Diagram}
\end{quote}

Using that
\(\Sigma: \mathop{\mathrm{Br}}({\mathbb{A}}_K) \to {\mathbb{Q}}/{\mathbb{Z}}\),
for a fixed \(\alpha\in \mathop{\mathrm{Br}}(K)\),
\begin{align*}
X(K) \subseteq (\Sigma \circ \tilde \alpha)^{-1}(0) \subseteq X({\mathbb{A}}_K)
,\end{align*}
and \((\Sigma \circ \tilde\alpha)^{-1}(0) = X({\mathbb{A}}_K)^\alpha\).
Thus the Hasse principle is violated if \(X({\mathbb{A}}_K)\) is
nonempty but \(X({\mathbb{A}}_K)^\alpha\) is empty. More generally, it's
violated if
\begin{align*}
X({\mathbb{A}}_K)^{\mathop{\mathrm{Br}}}
\coloneqq
\displaystyle\bigcap_{\alpha\in \mathop{\mathrm{Br}}(X)} X({\mathbb{A}}_K)^{\alpha} = \emptyset
.\end{align*}

\end{remark}

\hypertarget{the-hasse-principle-for-severi-brauers}{%
\subsubsection{The Hasse Principle for
Severi-Brauers}\label{the-hasse-principle-for-severi-brauers}}

\begin{remark}

Let \(X_{/ {K}}\) be a Severi-Brauer, then
\([X] \in \mathop{\mathrm{Br}}(K)\) and
\(X\cong {\mathbb{P}}^n_{/ {K}} \iff [X] = 0\). Using that
\begin{align*}
\oplus \iota_v: \mathop{\mathrm{Br}}(K) \hookrightarrow\bigoplus_v \mathop{\mathrm{Br}}(K_v)
,\end{align*}
we have
\begin{align*}
[X] = 0 \iff \iota_v(X) = 0\,\, \forall v
&& \text{since } \iota_v(X) = [X_{K_v}] \in \mathop{\mathrm{Br}}(K_v)
.\end{align*}

\end{remark}

\begin{fact}

It turns out that \(X\cong {\mathbb{P}}^n \iff X(K) \neq \emptyset\).

\end{fact}

\hypertarget{brauer-groups-and-galois-cohomology}{%
\subsection{Brauer Groups and Galois
Cohomology}\label{brauer-groups-and-galois-cohomology}}

\begin{definition}[Brauer Groups]

Let \(K \in \mathsf{Field}\), then
\begin{align*}
\mathop{\mathrm{Br}}(K) \coloneqq H^2_{ \mathsf{Gal}} (K, \mkern 1.5mu\overline{\mkern-1.5muK\mkern-1.5mu}\mkern 1.5mu^{\times}) = H^2_{{\mathsf{Grp}}}({ \mathsf{Gal}} (K^s/K), (K^s)^{\times})
.\end{align*}

\end{definition}

\begin{remark}

Let \(G \in {\mathsf{Grp}}\) be discrete, so we're not considering any
topology on it. Let \(M \in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\), or
equivalently \(M\in \mathsf{{\mathbb{Z}}[G]}{\hbox{-}}\mathsf{Mod}\).

We can take invariants and coinvariants:
\begin{align*}
M^G &\coloneqq\left\{{m\in M {~\mathrel{\Big\vert}~}gm=m \,\, \forall g\in G }\right\} = \mathop{\mathrm{Hom}}_{{\mathbb{Z}}[G]}({\mathbb{Z}}, M) \\
M_G &\coloneqq M /\left\langle{\left\{{gm-m {~\mathrel{\Big\vert}~}g\in G}\right\}}\right\rangle = {\mathbb{Z}}\otimes_{{\mathbb{Z}}[G]} M
.\end{align*}
These are the largest submodules/quotient modules respectively on which
\(G\) acts trivially.

\end{remark}

\begin{exercise}[?]

Why are these equal to homs and tensors respectively?

\end{exercise}

\begin{definition}[Group cohomology]

\begin{align*}
H^i(G; M) &\coloneqq\operatorname{Ext} ^i_{{\mathbb{Z}}[G]}({\mathbb{Z}}; M)\\
H_i(G; M) &\coloneqq\operatorname{Tor}_i^{{\mathbb{Z}}[G]}({\mathbb{Z}}; M)
.\end{align*}

\end{definition}

\begin{example}[Cyclic groups]

For \(G\coloneqq{\mathbb{Z}}\), we have
\({\mathbb{Z}}[G] = {\mathbb{Z}} { \left[ {x, x^{-1}} \right] }\). Take
a projective resolution
\begin{align*}
0 \to {\mathbb{Z}}[G] \xrightarrow{\cdot (x-1)} {\mathbb{Z}}[G] \xrightarrow{x\mapsto 1} {\mathbb{Z}}\to 0
.\end{align*}

Deleting the augmentation and applying
\(\mathop{\mathrm{Hom}}_{{\mathbb{Z}}[G]}({-}, {\mathbb{Z}})\) yields
\(0\to {\mathbb{Z}}\xrightarrow{f: \cdot (x-1)} {\mathbb{Z}}\to 0\), and
noting that \(x\) acts by 1, \(f\) is the zero map. This yields
\begin{align*}
H^*(G; {\mathbb{Z}}) &=
\begin{cases}
{\mathbb{Z}}&  * = 0, 1
\\
0 & \text{else}.
\end{cases}
\\
H_*(G; {\mathbb{Z}}) &=
\begin{cases}
{\mathbb{Z}}&  * = 0, 1
\\
0 & \text{else}.
\end{cases}
\end{align*}

\end{example}

\hypertarget{group-cohomology-thursday-august-26}{%
\section{Group Cohomology (Thursday, August
26)}\label{group-cohomology-thursday-august-26}}

\begin{quote}
See Cassels-Frohlich, Stein, etc for group cohomology.
\end{quote}

\hypertarget{computing-examples}{%
\subsection{Computing Examples}\label{computing-examples}}

\begin{example}

For \(G={\mathbb{Z}}\), take the resolution
\begin{align*}
0 \to {\mathbb{Z}}[x, x^{-1}] \xrightarrow{x-1} {\mathbb{Z}}[x, x^{-1}] \to 0
.\end{align*}
Then \(H_*(G; {\mathbb{Z}})= H^*(G;{\mathbb{Z}})\) is \({\mathbb{Z}}\)
in degrees 0 and 1, and 0 otherwise. For
\(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\), we have
\begin{align*}
H^*(G; M) = H^*(M \xrightarrow{x-1} M)
=
\begin{cases}
M^G &  *=0
\\
M_G & * = 1
\\
0 & \text{else},
\end{cases}
\\
H_*(G; M) = H_*(M \xrightarrow{x-1} M)
=
\begin{cases}
M_G &  *=0
\\
M^G & * = 1
\\
0 & \text{else}.
\end{cases}
\end{align*}

\end{example}

\begin{example}[?]

For \(G = {\mathbb{Z}}/n\), write \(\sigma\) as the generator so that
\({\mathbb{Z}}[G] = {\mathbb{Z}}[\sigma] / \left\langle{\sigma^n - 1}\right\rangle\)
We can take a resolution
\begin{align*}
\cdots \to
{\mathbb{Z}}[\sigma]/ \left\langle{\sigma - 1}\right\rangle
\xrightarrow{\sigma - 1} 
{\mathbb{Z}}[\sigma]/ \left\langle{\sigma - 1}\right\rangle
\xrightarrow{1 + \sigma + \cdots + \sigma^{n-1}} 
{\mathbb{Z}}[\sigma]/ \left\langle{\sigma - 1}\right\rangle
\xrightarrow{\sigma - 1} 
{\mathbb{Z}}[\sigma]/ \left\langle{\sigma - 1}\right\rangle
\to 0
.\end{align*}

Now apply
\(\mathop{\mathrm{Hom}}_{{\mathbb{Z}}[G]}({-}, {\mathbb{Z}})\), use that
\(\mathop{\mathrm{Hom}}_{{\mathbb{Z}}[G]}({\mathbb{Z}}[G], {\mathbb{Z}}) = {\mathbb{Z}}\),
and take homology of the complex
\begin{align*}
{\mathbb{Z}}\xrightarrow{\sigma - 1} {\mathbb{Z}}\xrightarrow{\sum \sigma^i} \to {\mathbb{Z}}\xrightarrow{\sigma - 1} \cdots
{\mathbb{Z}}\xrightarrow{0} {\mathbb{Z}}\xrightarrow{n} \to {\mathbb{Z}}\xrightarrow{0} \cdots
.\end{align*}
This yields
\begin{align*}
H^*(G; {\mathbb{Z}}) = 
\begin{cases}
{\mathbb{Z}}& *=0 
\\
0 & * \text{ odd}
\\
{\mathbb{Z}}/n & * \text{ even}.
\end{cases}
\end{align*}

\end{example}

\begin{remark}

For the free abelian group \({\mathbb{Z}}^n\), we get
\(H^*({\mathbb{Z}}^n; {\mathbb{Z}}) = \bigwedge\nolimits^* ({\mathbb{Z}}^n)\).
For the free group \(F_n\), we get \(H^*(F_n; {\mathbb{Z}})\) is
\({\mathbb{Z}}\) in degree zero (always true for the trivial module,
since the invariants are everything) and \({\mathbb{Z}}^n\) in degree 1.

\end{remark}

\begin{fact}

If \(X\) is a CW complex with
\(\pi_0(X) = 0, \pi_1(X) = G, \pi_{>2}(X) = 0\), then
\(H^*_{{\mathsf{Grp}}}(G; {\mathbb{Z}}) = H^*_{{\operatorname{Sing}}}(X; {\mathbb{Z}})\).
Note that \(X \xrightarrow{\sim} {\mathbf{B}}G\) in this case, and the
proof is easy: take the universal cover, then the simplicial/cellular
cohomology resolves \({\mathbb{Z}}\) as a
\({\mathbb{Z}}[G]{\hbox{-}}\)module.

\end{fact}

\begin{proposition}[?]

Suppose \(G\) is finite and \(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\),
then \(H^{>n}(G; M)\) is torsion. 1. It suffices to show this for
\(*=1\) by using dimension shifting. Choose \(M \hookrightarrow I\) into
an injective object to get a SES
\begin{align*}
  0 \to M\to I \to M/I \to 0
 \end{align*}
to get a LES in cohomology, and use that \(\operatorname{Ext}\) into
injectives vanishes to get \(H^*(G; M) \cong H^{*}(G; M/I)[-1]\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  We want to show
  \(H^1(G; M) = \operatorname{Ext} _{{\mathbb{Z}}[G]}^1({\mathbb{Z}}; M)\)
  is torsion, and it suffices to show
  \(\operatorname{Ext} ^1_{{\mathbb{Z}}[G]}({\mathbb{Z}}; M)\otimes{\mathbb{Q}}= 0\),
  which we can replace with
  \(\operatorname{Ext} _{{\mathbb{Z}}[G]}^1({\mathbb{Q}}, M \otimes{\mathbb{Q}})\).
  So we consider SESs of the form
  \begin{align*}
    0 \to M\otimes Q \to W \to {\mathbb{Q}}
    ,\end{align*}
  which we'd like to split as a SES of \(G{\hbox{-}}\)representations
  over \({\mathbb{Q}}\).
\end{enumerate}

See uniquely divisible groups?

This splits by Maschke's theorem: all SESs of irreducible
representations of \(G\) for \(G\) finite over
\(\operatorname{ch}k = 0\) split. The usual proof over \({\mathbb{C}}\)
doesn't work for \({\mathbb{Q}}\), but one uses a splitting instead of
an inner product.

\end{proposition}

\hypertarget{functoriality}{%
\subsection{Functoriality}\label{functoriality}}

\begin{remark}

Given \(M\to N \in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\) there are maps
\begin{align*}
H^*(G; M) &\to H^*(G; N) \\
H_*(G; M) &\to H_*(G; N)
.\end{align*}
Suppose \(\iota: G\to T\) with
\(M\in {\mathsf{T}{\hbox{-}}\mathsf{Mod}}\), then there are induced maps
\begin{align*}
\iota^*: H^*(T; M) &\to H^*(G; M) \\
\iota_*: H_*(T; M) &\to H_*(G; M)
\end{align*}
coming from the functoriality of Ext and Tor under change of rings.

We'll use the following as a black box: for \(G\leq T\) finite index,
there is a \emph{trace} map (or \emph{corestriction})
\begin{align*}
\operatorname{tr}_{G/T}: H^*(G; M) \to H^*(T; M)
.\end{align*}
It's functorial in \(M\), and \(\operatorname{tr}_{G/T} \circ \iota^*\)
is multiplication by \(m \coloneqq[G: T]\). This yields another proof of
the previous element: take \(G = 1\) to get \(H^*(G; M) = 0\) and check
\(\operatorname{tr}_{G/T} \circ \iota_*\) is multiplication by
\({\left\lvert {T} \right\rvert}\) and zero, making the group torsion.

\end{remark}

\begin{remark}

Some interpretations:

\begin{itemize}
\tightlist
\item
  \(H_1(G; {\mathbb{Z}}) = G^{\operatorname{ab}}= G/[G, G]\) is the
  abelianization (which can still be torsion).
\item
  \(H^1(G; {\mathbb{Z}}) = \mathop{\mathrm{Hom}}_{{\mathsf{Grp}}}(G;{\mathbb{Z}})\),
  which is always torsionfree.
\item
  \(H^2(G; M)\) classifies extensions of \(G\) by \(M\) in the following
  sense: \(G'\) occurring in a ``SES'' \(\xi: 0\to M\to G' \to G\to 1\)
  such that the action of \(G\) on \(M\) by conjugation is the given
  \(G{\hbox{-}}\)module structure on \(M\). Moreover \(\xi = 0\) in
  \(H^2(G; M)\) iff \(\xi\) splits, then \(G' \cong G \rtimes M\). For
  \(M\) a trivial \(G{\hbox{-}}\)module, these are \emph{central
  extensions}.
\end{itemize}

\end{remark}

\begin{warnings}

Note all SESs yield semidirect products: take
\(0\to {\mathbb{Z}}\xrightarrow{\cdot n} {\mathbb{Z}}\to {\mathbb{Z}}/n \to 0\),
which has no sections since \({\mathbb{Z}}\) has no
\(n{\hbox{-}}\)torsion. This in fact represents a generator
\(H^2({\mathbb{Z}}/n; {\mathbb{Z}})\).

\end{warnings}

\begin{definition}[Galois cohomology]

Let \(L_{/ {k}}\) be a finite Galois extension,
\(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\) for
\(G\coloneqq{ \mathsf{Gal}} (L_{/ {k}} )\). Then
\begin{align*}
H^*_{ \mathsf{Gal}} (L_{/ {k}} ; M) \coloneqq H^*_{\mathsf{Grp}}(G; M)
.\end{align*}
If \(M\) is a discrete continuous
\({ \mathsf{Gal}} (k^s/K){\hbox{-}}\)module, then
\begin{align*}
H^i(k; M) \coloneqq\colim_{U {~\trianglelefteq~}{ \mathsf{Gal}} (k^s/k)} H^*( { \mathsf{Gal}} (k^s/k) / U; M)
.\end{align*}

\end{definition}

\begin{quote}
The stabilizer of any point is open (and finite index).
\end{quote}

\begin{definition}[Brauer Groups]

\begin{align*}
\mathop{\mathrm{Br}}(k) = H^2(K; (k^s )^{\times})
.\end{align*}

\end{definition}

\begin{example}[?]

Consider \(\mathop{\mathrm{Br}}({\mathbb{F}}_q)\), then
\({ \mathsf{Gal}} ({\mathbb{F}}_q^s/{\mathbb{F}}_q) = \widehat{{\mathbb{Z}}} \left\langle{\mathop{\mathrm{Frob}}_q}\right\rangle\).
Then
\begin{align*}
\mathop{\mathrm{Br}}({\mathbb{F}}_q) 
&\coloneqq H^2\qty{\widehat{{\mathbb{Z}}}\left\langle{\mathop{\mathrm{Frob}}_q}\right\rangle; \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_q ^{\times}} \\
&=\colim_{U_n \subseteq \widehat{{\mathbb{Z}}}\to \widehat{{\mathbb{Z}}}\to {\mathbb{Z}}/n} H^2\qty{ {\mathbb{Z}}/n; (\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_q^{\times})^{U_n}}\\
&=\colim H^2\qty{ {\mathbb{Z}}/n \left\langle{\mathop{\mathrm{Frob}}_q}\right\rangle; \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_{q^n}^{\times}} \\
&=\colim H^2( { \mathsf{Gal}} ({\mathbb{F}}_{q^n} / {\mathbb{F}}_q); \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}\mkern-1.5mu}\mkern 1.5mu_{q^n}^{\times}) \\
&= \colim H^2 \qty{
  {\mathbb{F}}_{q^n}^{\times}\xrightarrow{\mathop{\mathrm{Frob}}- 1} {\mathbb{F}}_{q^n}^{\times}\xrightarrow{\operatorname{Nm}} {\mathbb{F}}_{q^n}^{\times}\to \cdots 
} \\
&= \colim {\mathbb{F}}_q^{\times}/ \operatorname{Nm}({\mathbb{F}}_{q^n}, {\mathbb{F}}_q) {\mathbb{F}}_{q^n}^{\times}\\
&= \colim 0 \\
&= 0
.\end{align*}

Note: we've used that
\begin{align*}
\ker (\mathop{\mathrm{Frob}}- 1: x\mapsto x^{q-1}) = {\mathbb{F}}_q^{\times}
.\end{align*}

\end{example}

\begin{exercise}[?]

Show that the norm is surjective.

\end{exercise}

\hypertarget{tuesday-august-31}{%
\section{Tuesday, August 31}\label{tuesday-august-31}}

\begin{remark}

Today: a systematic way to compute group cohomology by taking standard
resolution. For a fixed group \(G\), we want to resolve \({\mathbb{Z}}\)
by free \({\mathbb{Z}}[G]{\hbox{-}}\)modules, so take a simplicial
resolution

\begin{center}
\begin{tikzcd}
    \cdots
    \stackar{7}[r]
    &
    G{ {}^{ \scriptscriptstyle\times^{3} }  }
    \stackar{5}[r]
    &   
    G^{ {}^{ \scriptscriptstyle\times^{2} }  }
    \stackar{3}[r]
    &
    G
\end{tikzcd}
\end{center}

Taking free \({\mathbb{Z}}{\hbox{-}}\)modules yields

\begin{center}
\begin{tikzcd}
    \cdots
    \stackar{7}[r]
    &
    {\mathbb{Z}}[G{ {}^{ \scriptscriptstyle\times^{3} }  }]
    \stackar{5}[r]
    &   
    {\mathbb{Z}}[G { {}^{ \scriptscriptstyle\times^{2} }  }]
    \stackar{3}[r]
    &
    {\mathbb{Z}}[G]
\end{tikzcd}
\end{center}

Note that this is a simplicial set whose realization is \(EG\).

\end{remark}

\begin{proposition}[?]

\(C_\bullet(G)\) is exact, and
\({\mathbb{Z}}[G{ {}^{ \scriptscriptstyle\times^{n} } }]\) is free in
\({\mathsf{{\mathbb{Z}}[G]}{\hbox{-}}\mathsf{Mod}}\) where
\(G\curvearrowright G{ {}^{ \scriptscriptstyle\times^{n} } }\)
diagonally and this extends linearly.

\end{proposition}

\begin{proof}[?]

\({\mathbb{Z}}[G { {}^{ \scriptscriptstyle\times^{n} } }]\) is a free
\({\mathbb{Z}}[G]{\hbox{-}}\)module, using that
\(\left\{{(1, g_1, \cdots, g_{n-1}) {~\mathrel{\Big\vert}~}g_k \in G}\right\}\)
is a free basis, since these are representatives for
\(G{\hbox{-}}\)orbits on \(G{ {}^{ \scriptscriptstyle\times^{n} } }\).

That this is an exact complex will follow from a nullhomotopy
\(h: {\mathbb{Z}}[G{ {}^{ \scriptscriptstyle\times^{n-1} } }] \to {\mathbb{Z}}[G{ {}^{ \scriptscriptstyle\times^{n} } }]\)
so that \(hd + dh = \operatorname{id}\). Take the map
\(h(g_1, \cdots, g_n) = (e, g_1, \cdots, g_n)\), then
\begin{align*}
(hd)(g_1, \cdots, g_n) 
&= h \sum (-1)^i (g_1, \cdots, \widehat{g}_i, \cdots, g_n)\\
&= \sum (-1)^i (e, g_1, \cdots, \widehat{g}_i, \cdots, g_n)
.\end{align*}
and
\begin{align*}
(dh)(g_1, \cdots, g_n)
&= d (e, g_1, \cdots, g_n) \\
&= (g_1,\cdots, g_n) - \sum (-1)^i (e, g_1, \cdots, \widehat{g}_i, \cdots, g_n)
,\end{align*}
and adding these two cancels the two summed terms and yields the
identity.

Then just recall from homological algebra that \(x \in \ker d\) implies
\(x = hdx + dhx = dhx\), so \(x\in \operatorname{im}d\), so this makes
the complex exact.

\end{proof}

\begin{corollary}[?]

For \(G \in {\mathsf{Grp}}\) discrete and
\(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\),
\begin{align*}
H^*(G; M) &= H^*( \operatorname{Hom}^{\bullet}_{{\mathbb{Z}}[G]}(C_\bullet(G), M) ) \\
H_*(G; M) &= H^*( M \otimes_{{\mathbb{Z}}[G]} C_\bullet(G) )
.\end{align*}

\end{corollary}

\begin{remark}

Can we find a smaller way to represent this? Note that
\begin{align*}
{\mathbb{Z}}[G^{\times n}] = \bigoplus_{(g_1, \cdots, g_n) \in G^{n-1}} {\mathbb{Z}}[G](1, g_1, \cdots, g_{n-1})
,\end{align*}
and there is a free/forgetful adjunction between modules and sets that
yields
\begin{align*}
\mathop{\mathrm{Hom}}_{{\mathbb{Z}}[G]}( {\mathbb{Z}}[G { {}^{ \scriptscriptstyle\times^{n} }  } ], M ) \cong \mathop{\mathrm{Hom}}_{{\mathsf{Set}}}(G{ {}^{ \scriptscriptstyle\times^{n-1 } }  }, M)
.\end{align*}

\end{remark}

\begin{definition}[Reduced Complex]

For \(G\in {\mathsf{Grp}}\) discrete and
\(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\), set
\begin{align*}
\tilde C^r(G; M) \coloneqq\mathop{\mathrm{Hom}}_{{\mathsf{Set}}}(G { {}^{ \scriptscriptstyle\times^{r} }  }, M)
.\end{align*}
The boundary maps are given by
\begin{align*}
\delta: \tilde C^0(G, M) &\to \tilde C^1(G, M) \\
\delta f( \sigma) &= \sigma f({-}) - f({-})
\\ \\ 
\delta: \tilde C^1(G, M) &\to \tilde C^2(G, M) \\
\delta f( \sigma, \tau) &= \sigma f(\tau) - f(\sigma \tau) + f(\sigma)
\\ \\
\delta: \tilde C^2(G, M) &\to \tilde C^3(G, M) \\
\delta f( \sigma, \tau, \rho) &= \sigma f(\tau, \rho) - f(\sigma\tau, \rho) + f(\sigma, \tau\rho) - f(\sigma, \tau) 
.\end{align*}
The pattern is multiply by \(\sigma\) on the outside, cycle through
multiplying it to each argument, and for the last term leave \(\sigma\)
off.

\end{definition}

\begin{remark}

Punchline: in principle, group cohomology is computable -- however, the
complex is quite large and not practical for large groups.

\end{remark}

\hypertarget{some-formal-properties}{%
\subsection{Some Formal Properties}\label{some-formal-properties}}

\begin{proposition}[Spectral Sequences]

For \(H{~\trianglelefteq~}G\) and
\(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\), the
\textbf{Hochschild-Serre spectral sequence} reads
\begin{align*}
E_2^{p, q} = H^p( G/H; H^q(H; M)) \Rightarrow H^{p+q}(G; M)
.\end{align*}

\end{proposition}

\begin{remark}

This is useful for inducting on the lengths of composition series, since
e.g.~for solvable groups one can take \(G/H\) to be cyclic and \(H\) a
smaller solvable group.

\end{remark}

\begin{proposition}[Inflation/Restriction Exact Sequence]

This spectral sequence induces an \textbf{inflation/restriction exact
sequence}

\begin{center}
\begin{tikzcd}
    0 \\
    {H^1\qty{ {G \over H}; M^H}} && {H^1\qty{G; M}} && {H^1\qty{H; M}^{G\over H}} \\
    \\
    {H^2\qty{ {G \over H}; M^H}} && {H^2\qty{G; M}}
    \arrow[from=1-1, to=2-1]
    \arrow[from=2-1, to=2-3]
    \arrow[from=2-3, to=2-5]
    \arrow[from=2-5, to=4-1, out=0, in=180]
    \arrow[from=4-1, to=4-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwxLCJIXjFcXHF0eXsge0cgXFxvdmVyIEh9OyBNXkh9Il0sWzIsMSwiSF4xXFxxdHl7RzsgTX0iXSxbNCwxLCJIXjFcXHF0eXtIOyBNfV57R1xcb3ZlciBIfSJdLFswLDMsIkheMlxccXR5eyB7RyBcXG92ZXIgSH07IE1eSH0iXSxbMiwzLCJIXjJcXHF0eXtHOyBNfSJdLFswLDAsIjAiXSxbNSwwXSxbMCwxXSxbMSwyXSxbMiwzXSxbMyw0XV0=}{Link
to Diagram}
\end{quote}

\end{proposition}

\begin{remark}

This comes from the bottom-left corner of the HS spectral sequence,
which is a general principle for first quadrant spectral sequences. Note
that the \(G/H\) action comes from \(G\curvearrowright H\) by
conjugation, which yields a \(G{\hbox{-}}\)action on \(H^*\), and since
\(H\) acts trivially on \(H^*(H; M)\) (since e.g.~\(M^H\) has a trivial
action), this action factors through \(G/H\).

\end{remark}

\hypertarget{forms-torsors-and-h1}{%
\subsection{\texorpdfstring{Forms, Torsors, and
\(H^1\)}{Forms, Torsors, and H\^{}1}}\label{forms-torsors-and-h1}}

\begin{definition}[Forms/descent, a pseudo-definition]

Let \(X_{/ {k}}\) be an object (e.g.~a variety, a group scheme, a
variety with extra structure), then a \textbf{form} of \(X\) over \(k\)
is an object \(X'_{/ {k}}\) with an isomorphism
\(X'_{/ {k^s}} \xrightarrow{\sim} X\) (i.e.~a \textbf{descent} of
\(X\)).

\end{definition}

\begin{example}[?]

For \(X \coloneqq{\mathbb{P}}^n_{/ {k^s}}\) then a form of \(X_{/ {k}}\)
is a Severi-Brauer variety, for example a smooth conic.

\end{example}

\begin{example}[Severi Brauers]

Let \(E\) be a genus 1 curve, then \(E\) is a form for its Jacobian
\(\operatorname{Jac}(E)\), i.e.~it becomes isomorphic to its Jacobian if
it has a rational point. Not every curve has such a point, so they only
become isomorphic after base changing to a separable closure. Note that
\(\operatorname{Jac}(E) \curvearrowright E\) by addition of divisors
(since Jacobians have degree zero, curves have divisors of degree 1, and
adding them yields a degree 1 divisor). It is in fact a torsor.

\end{example}

\begin{example}[?]

If \(L_{/ {k}}\) is a finite separable extension then \(L\) is a form of
\((k^s){ {}^{ \scriptscriptstyle\times^{n} } }\).

\end{example}

\begin{example}[?]

The groups \({\operatorname{SO}}(p, q)_{/ {{\mathbb{R}}}}\), the
matrices preserving a quadratic form
\begin{align*}
h_{p, q}\coloneqq\operatorname{diag}(1, \cdots, 1, -1, \cdots, -1)
\end{align*}
with \(p\) copies of 1 and \(q\) copies of \(-1\), and these are all
forms of \({\operatorname{SO}}(p+q)_{/ {{\mathbb{C}}}}\).

\end{example}

\begin{proposition}[?]

Suppose \(X_{/ {k}}\) is some object (e.g.~a variety, then forms of
\(X_{k^s}\) over \(k\) are canonically in bijection with
\(H^1_{{ \mathsf{Gal}} }(k; \mathop{\mathrm{Aut}}(X_{k^s}))\) (recalling
that this was defined as a direct limit). Note that this automorphism
group may be nonabelian, which we still need to define.

\end{proposition}

\begin{proof}[?]

Suppose \(\mathop{\mathrm{Aut}}(X_{k^s})\) is abelian, then we'll show
the following stronger claim:

\begin{claim}

\(X'_L \xrightarrow{\sim} X_L\) since there is a bijection
\begin{align*}
\left\{{\substack{
  \text{Forms of } X_{k^s} \\
  \text{split by } L _{/ {k}} 
}}\right\}
&\rightleftharpoons
  H^1_{ \mathsf{Gal}} (L_{/ {k}} ; \mathop{\mathrm{Aut}}(X_L))
.\end{align*}

\end{claim}

\begin{proof}[?]

Recall that
\begin{align*}
H^1(L_{/ {k}} ; \mathop{\mathrm{Aut}}(X_L)) = H^1( \tilde C^\bullet({ \mathsf{Gal}} (L/k)); \mathop{\mathrm{Aut}}(X_L) )
.\end{align*}
Given \(X'_{/ {k}}\) split by \(L\), we want a map
\({ \mathsf{Gal}} (L/k) \to \mathop{\mathrm{Aut}}(X_L)\). Choose an
isomorphism \(X'_L \xrightarrow{\sim} X_L\), noting that Galois acts on
the LHS since it's defined over \(k\), which will be different from the
natural action on the right-hand side. So we can take a map
\begin{align*}
f: { \mathsf{Gal}} (L/k) \to \mathop{\mathrm{Aut}}(X'/L) \xrightarrow{\sim} \mathop{\mathrm{Aut}}(X_L)
,\end{align*}
although this is not generally a homomorphism.

Instead, \(f(\sigma \tau) = f(\sigma) f(\tau)^\sigma\), a
\textbf{crossed homomorphism} which involves acting on the coefficients
of defining equations (which come from \(L\)). This says that
\(f\in \ker \delta\), the differential for \(\tilde C^\bullet\). So we
now have a map from forms split by \(L\) to
\(H^1({ \mathsf{Gal}} (L/k), \mathop{\mathrm{Aut}}(X_L))\), and we'll
show it's injective and surjective.

~

\textbf{Injectivity:} Suppose \(X', X''\) are isomorphic forms of \(X\),
so we have an isomorphism defined over \(k\) of the form
\(X_L' \xrightarrow{\sim} X_L''\).

\begin{exercise}[?]

This changes \(f\) by an element of the form \(\delta(g)\) for
\(g\in \mathop{\mathrm{Aut}}(X_L)\).

\end{exercise}

\textbf{Surjectivity:} Given a crossed homomorphism
\(f: { \mathsf{Gal}} (L/k) \to \mathop{\mathrm{Aut}}(X_L)\), we want to
produce a form of \(X_{/ {k}}\) mapping to it. This is the hardest part
of the argument!

Suppose \(X_{/ {k}}\) is a variety. First suppose
\(X\in {\mathsf{Aff}}{\mathsf{Var}}\), so \(X = \operatorname{Spec}R\)
and
\({ \mathsf{Gal}} (L_{/ {k}} )\curvearrowright_f R_L = R \otimes_k L\),
which is only an \(L{\hbox{-}}\)semilinear action. Then
\(X' = \operatorname{Spec}(R_L)^{{ \mathsf{Gal}} (L/k)}\), and the claim
is that \(X_L' \cong X_L\). The proof of this is \textbf{Galois
descent}, i.e.~there is an equivalence of tensor categories
\begin{align*}
\adjunction{({-}) \otimes L}
{({-})^{{ \mathsf{Gal}} (L_{/ {k}} )} }
{{\mathsf{k}{\hbox{-}}\mathsf{Mod}}^{\otimes} }{{\mathsf{L}{\hbox{-}}\mathsf{Mod}}^{\otimes} + \text{ a semilinear action of } { \mathsf{Gal}} (L_{/ {k}} ) }
\end{align*}

Now for general \(X\), one reduces to the case of affines. One can
alternatively prove Galois descent without reference to affine
varieties.

\end{proof}

\end{proof}

\hypertarget{thursday-september-02}{%
\section{Thursday, September 02}\label{thursday-september-02}}

\hypertarget{correspondence-of-forms}{%
\subsection{Correspondence of Forms}\label{correspondence-of-forms}}

\begin{remark}

Last time: standard/reduced complexes, forms, and \(H^1\). A
meta-definition for today: let \(k, L \in \mathsf{Field}\) with
\(L_{/ {k}}\) finite and separable, and \(X_{/ {k}}\) an object over
\(k\) (e.g.~an algebraic variety, possibly with extra structure). A
\textbf{form} of \(X_{/ {k}}\) split by \(L\) is an object
\(X' _{/ {k}}\) of the same class as \(X\) such that
\(X_L \xrightarrow{\sim} X_L'\).

\end{remark}

\begin{theorem}[A meta-theorem]

The theorem was that there is a canonical bijection
\begin{align*}
\left\{{\substack{
  \text{Forms of } X \\ 
  \text{split by } L
}}\right\}
&\rightleftharpoons
H^1_{ \mathsf{Gal}} (L_{/ {k}} ; \mathop{\mathrm{Aut}}_k(X_L))
\end{align*}
Note that we didn't assume the coefficients formed an abelian group, so
we'll explain this today. It is true that
\(\mathop{\mathrm{Aut}}(X_L) \in {\mathsf{{ \mathsf{Gal}} (L_{/ {k}} )}{\hbox{-}}\mathsf{Mod}}\).
We'll say that \(X'\) is just \textbf{a form} of \(X\) if there exists
some \(L'\) finite separable that splits \(k\). In this case there is a
correspondence
\begin{align*}
\left\{{\substack{
  \text{Forms of }X
}}\right\}
&\rightleftharpoons
  H^1_{{ \mathsf{Gal}} }(L_{/ {k}} ; \mathop{\mathrm{Aut}}_k(X_{k^s}) )
\end{align*}

\end{theorem}

\begin{proof}[A meta-proof]

What is the map? Given a form \(X'\), we by definition have
\(F: X_L' \xrightarrow{\sim} X_L\), and we want a map
\({ \mathsf{Gal}} (L_{/ {k}} ) \to \mathop{\mathrm{Aut}}_k(X_L)\) such
that \(\delta f = 0\) for the differential in cohomology. Since \(X'\)
is defined over \(k\), we have an action
\({ \mathsf{Gal}} (L_{/ {k}} )\curvearrowright X_L'\), i.e.~a map
\({ \mathsf{Gal}} (L_{/ {k}} ) \to \mathop{\mathrm{Aut}}(X_L')\), which
we can compose with the given isomorphism to obtain
\begin{align*}
f: { \mathsf{Gal}} (L_{/ {k}} ) \to \mathop{\mathrm{Aut}}(X_L')\to \mathop{\mathrm{Aut}}(X_L)
.\end{align*}
We have \(f( \sigma \tau) = f(\sigma) f(\tau)^\sigma\). What happens if
we change the isomorphism \(F\) to some \(F'\), changing by some
\(g\in \mathop{\mathrm{Aut}}(X_L)\)

\begin{exercise}[?]

Here \(f\) changes by a map of the form \(\sigma \to g(g^{-1})^\sigma\).

\end{exercise}

We'll write an inverse map using Galois descent. Given
\(f: { \mathsf{Gal}} (L_{/ {k}} )\to \mathop{\mathrm{Aut}}_k(X_L)\) with
\(f(\sigma \tau) = f( \sigma) f( \tau)^{ \sigma}\), we want to construct
a form of \(X\). Assume \(X \in {\mathsf{Aff}}{\mathsf{Sch}}\), so
\(X = \operatorname{Spec}(A)\) for some \(A\in{\mathsf{Alg}_{/k} }\),
then define
\begin{align*}
X' \coloneqq\operatorname{Spec}(A\otimes_k L)^{{ \mathsf{Gal}} (L_{/ {k}} )}
\end{align*}
where the action is given by \(f\).

\end{proof}

\begin{remark}

What is \(\mathop{\mathrm{Aut}}_k(X_{/ {L}} )\) is nonabelian? Then we
just make this proof a definition, and set
\begin{align*}
H^1(L_{/ {k}}  ; G) \coloneqq\left\{{f: { \mathsf{Gal}} (L_{/ {k}} )\to G {~\mathrel{\Big\vert}~}f( \sigma \tau) = f( \sigma) f( \tau)^{ \sigma} }\right\}/ ( \sigma\to g(g^{-1}) ^{ \sigma})
.\end{align*}
Here the maps are of finite discrete groups. This is a pointed set,
using the constant map as a basepoint.

\end{remark}

\hypertarget{torsors}{%
\subsection{Torsors}\label{torsors}}

\begin{definition}[Torsor]

Recall that for \(G \in {\mathsf{Alg}}{\mathsf{Grp}}_{/ {k}}\), a
\textbf{torsor} for \(G\) (or a \emph{principal homogeneous space}) is

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  A form of \(G\) under the left action of \(G\) on itself, i.e.~a
  variety \(X\) with a left \(G{\hbox{-}}\)action \(G\times X\to X\)
  where \(X_L \xrightarrow{ \sim} G_L\) using the left-translation
  action.
\item
  A \(G{\hbox{-}}\)variety \(X\) such that
  \(G\times X \xrightarrow{\sigma, \pi_2} X\times X\) is an isomorphism.
\end{enumerate}

\end{definition}

\begin{claim}

Note that these are equivalent if \(G\) is smooth, which for us will
always happen in characteristic zero.

\end{claim}

\begin{theorem}[?]

If \(G\) is smooth, then \(G{\hbox{-}}\)torsors are canonically in
bijection with \(H^1(k; G(k^s))\), and \(G{\hbox{-}}\)torsors split by
\(L\) biject with \(H^1(L_{/ {k}} ; G(L))\).

\end{theorem}

\begin{exercise}[?]

Prove this! It suffices to show that
\(\mathop{\mathrm{Aut}}_{G_L}(G_L)\cong G_L\) as a
\({\mathsf{Grp}}{\mathsf{Sch}}_{/ {G_L}}\).

\end{exercise}

\hypertarget{example-kummer-theory}{%
\subsection{Example: Kummer Theory}\label{example-kummer-theory}}

\begin{example}[Kummer theory]

Suppose \(\mu_p \subseteq k\), so \(k\) contains all \(p\)th roots of
unity. Then a \(\mu_p{\hbox{-}}\)torsor is the same as a
\({\mathbb{Z}}/p\) Galois extension of \(k\), where we allow
\(k^p =\mu_p\) itself.

\end{example}

\begin{theorem}[?]

There is a bijection
\begin{align*}
\left\{{\substack{
  {\mathbb{Z}}/p {\hbox{-}}\text{extensions}
}}\right\}
&\rightleftharpoons
H^1(k; \mu_p)
\end{align*}

\end{theorem}

\begin{proof}[?]

Use the SES
\begin{align*}
1 \to \mu_p \to (k^s)^{\times}\xrightarrow{x\mapsto x^p} (k^s)^{\times}\to 1
,\end{align*}
which yields a LES
\begin{align*}
1 \to
H^0(k; \mu_p) \to
H^0(k; (k^s)^{\times}) \xrightarrow{x\mapsto x^p} 
H^0(k; (k^s)^{\times})\to
H^1(k; \mu_p) \to
H^1(k; (k^s)^{\times})
,\end{align*}
and identifying terms yields
\begin{align*}
0 \to k^{\times}/ (k^{\times})^p \to H^1(k; \mu_p) \to H^1(k; (k^s)^{\times})
.\end{align*}

\end{proof}

\begin{example}[?]

What is \(H^1(k; (k^s)^{\times})\)? Use that
\(L^{\times}= \mathop{\mathrm{Aut}}(V_{/ {L}} )\) where \(V\) is a
1-dimensional vector space over \(L\). The claim is that by Galois
descent, forms for a vector space split by \(L\) are precisely vector
spaces over \(k\), which makes them all trivial. This in fact implies
the more general fact that \(H^1(k; \operatorname{GL}_n(k^s)) = 1\).

\end{example}

\begin{remark}

Kummer theory gives us an explicit form of the map and identifying terms
yields
\begin{align*}
0 \to k^{\times}/ (k^{\times})^p \xrightarrow{x\mapsto k[x^{1\over p} ]} H^1(k; \mu_p) \to H^1(k; (k^s)^{\times})
.\end{align*}
This can be found by unwinding the definition of the map from the snake
lemma, or noting that the kernel of a map from the absolute Galois group
cuts out exactly this field.

\end{remark}

\hypertarget{geometry-of-brauer-groups}{%
\subsection{Geometry of Brauer Groups}\label{geometry-of-brauer-groups}}

\begin{example}[of $H^1$]

\(H^1(k; G)\) are forms of objects with automorphism groups \(G\).

\begin{itemize}
\tightlist
\item
  Vector spaces are obtained by taking \(G = \operatorname{GL}_n\).
\item
  Forms of \({\mathbb{P}}^n\), i.e.~Severi-Brauer varieties, come from
  taking \(G \coloneqq\operatorname{PGL}_{n+1}\).
\item
  For \(G\) finite, a form of \(G\) is an étale \(k{\hbox{-}}\)algebra
  (product of separable extensions of \(k\) with total Galois group
  \(G\)).

  \begin{itemize}
  \tightlist
  \item
    For \(G\) simple, these are Galois extensions with Galois group
    \(G\). For \(G\coloneqq{\mathbb{Z}}/p\), this is Kummer theory.
  \end{itemize}
\item
  For \(E\) an elliptic curve, all genus 1 curves are torsors for their
  Jacobian. So genus 1 curves \(C\) with
  \(\operatorname{Jac}(C) \cong E\) biject with \(H^1(k; E(k^s))\).
\end{itemize}

\end{example}

\begin{remark}

We'll now look at \(H^2\), and there is a correspondence

\includegraphics{figures/SectionsOne.pdf}

Given a set-theoretic section \(s: G\to G'\), we get a map
\begin{align*}
f_s: G{ {}^{ \scriptscriptstyle\times^{2} }  } &\to A \\
(g_1, g_2) &\mapsto s(g_1) s(g_2) s(g_1 g_2)^{-1}
.\end{align*}
Note that if \(s\) is a group morphism, this is just the constant map.

\begin{claim}

One needs to show the following:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\delta f_s = 0\), so one gets a cocycle.
\item
  Changing \(s\) changes \(f_s\) by a coboundary.
\item
  Make the inverse.
\end{enumerate}

\end{claim}

The group operation here is
\(G' \cdot G'' \coloneqq G' { \underset{\scriptscriptstyle {G} }{\times} } G'' / A\),
and the multiplication map is
\begin{align*}
(a_1, g_1) \cdot (a_2, g_2) \coloneqq(a_1 a_2 f_s(g_1, g_2), \, g_1 g_2)
.\end{align*}

\end{remark}

\begin{remark}

Suppose \(1\to Z\to H' \to H\to 1\) is a SES of groups with a
\(G{\hbox{-}}\)action such that \(Z\) is in the center of \(H'\). Then
there is a ``LES''

\begin{center}
\begin{tikzcd}
    1 \\
    \\
    {Z^G} && {(H')^G} && {H^G} \\
    \\
    {H^1(G; Z)} && {H^1(G; H')} && {H^1(G; H)} \\
    \\
    {H^2(G; Z)}
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow[from=5-1, to=5-3]
    \arrow[from=3-3, to=3-5]
    \arrow["{\delta_0}"', from=3-5, to=5-1, in=180,out=0]
    \arrow[from=5-3, to=5-5]
    \arrow["{\delta_1}"', from=5-5, to=7-1, in=180, out=0]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwwLCIxIl0sWzAsMiwiWl5HIl0sWzIsMiwiKEgnKV5HIl0sWzQsMiwiSF5HIl0sWzAsNCwiSF4xKEc7IFopIl0sWzIsNCwiSF4xKEc7IEgnKSJdLFs0LDQsIkheMShHOyBIKSJdLFswLDYsIkheMihHOyBaKSJdLFswLDFdLFsxLDJdLFs0LDVdLFsyLDNdLFszLDQsIlxcZGVsdGFfMCIsMl0sWzUsNl0sWzYsNywiXFxkZWx0YV8xIiwyXV0=}{Link
to Diagram}
\end{quote}

Note that some terms here are only sets, so exactness means that
differentials surject onto kernels, and
\(H^1(G; Z)\curvearrowright H^1(G;H')\) and \(H^1(G; H)\) is the
quotient by this action.

\end{remark}

\begin{remark}

\end{remark}

\begin{definition}[Brauer group]

Take
\(1\to {\mathbb{G}}_m \to \operatorname{GL}_n \to \operatorname{PGL}_n \to 1\),
then we get a map
\begin{align*}
H^1(k; \operatorname{PGL}_n(k^s)) \xrightarrow{\iota_n} H^2(k, (k^s)^{\times})
.\end{align*}
Then define the \textbf{Brauer group} of \(k\) to be
\begin{align*}
\mathop{\mathrm{Br}}(k) \coloneqq\displaystyle\bigcup_{n} \operatorname{im}(\iota_n)
.\end{align*}

\end{definition}

\begin{remark}

Studying \(H^2\) is hard in general, so this fact is the reason we can
actually study Brauer groups.

\todo[inline]{Something about Hilbert 90}

This surjection gives us geometric objects to work with. We'll show this
is a group next time, along with the following theorem:

\end{remark}

\begin{theorem}[?]

\begin{align*}
\displaystyle\bigcup_n \operatorname{im}(\iota_n) = H^2(k; (k^s)^{\times})
.\end{align*}

\end{theorem}

\hypertarget{tuesday-september-07}{%
\section{Tuesday, September 07}\label{tuesday-september-07}}

\hypertarget{intro-historical-pov-on-brauer-groups}{%
\subsection{Intro: Historical POV on Brauer
Groups}\label{intro-historical-pov-on-brauer-groups}}

\begin{remark}

Last time we defined
\(\mathop{\mathrm{Br}}(k) \coloneqq H^2(k; k^{\times})\) and had a SES
\begin{align*}
1 \to (k^s)^{\times}\to \operatorname{GL}_n(k^s)\to \operatorname{PGL}_n \to 1
.\end{align*}
We identified a subset of \(\operatorname{PGL}_n{\hbox{-}}\)torsors in
\(H^1(k; \operatorname{PGL}_n(k^s)) \xrightarrow{\iota_n} H^2(k; (k^s)^{\times})\),
and alternatively defined
\(\mathop{\mathrm{Br}}(k) = \cup_n \operatorname{im}(\iota_n)\). We'll
now look at geometric interpretations of elements of \(H^1\).

\end{remark}

\begin{example}[?]

\(\mathop{\mathrm{Aut}}(X) = \operatorname{PGL}_n\) for the following:

\begin{itemize}
\tightlist
\item
  \({\mathbb{P}}^{n-1}\)
\item
  \(\operatorname{GL}_n\)
\item
  \(\operatorname{Mat}(n\times n)\), by the Skolem-Noether theorem.
\end{itemize}

\end{example}

\begin{corollary}[?]

For any of the \(X\) above, there is an isomorphism:
\begin{align*}
H^1(k; \operatorname{PGL}_n(k^s)) 
\xrightarrow{\sim} \left\{{\text{Forms of $X$}}\right\}_{/ {\sim}} 
\xrightarrow{\sim} 
\left\{{\operatorname{PGL}_n{\hbox{-}}\text{torsors}}\right\}_{/ {\sim}} 
.\end{align*}

\end{corollary}

\begin{definition}[Severi-Brauers]

A \textbf{Severi-Brauer} variety over \(k\) is a form of
\({\mathbb{P}}^n_{/ {k}}\) for some \(n\).

\end{definition}

\begin{example}[?]

\envlist

\begin{itemize}
\tightlist
\item
  \(C\) a conic with no rational points, e.g.~\(x^2 + y^2 + z^2 = 0\)
  over \({\mathbb{R}}\).
\item
  \(\operatorname{Sym}^n C\) is a nontrivial Severi-Brauer if \(n\) is
  odd. It's difficult to write any down for even \(n\), e.g.~there are
  no Severi-Brauer surfaces over \({\mathbb{R}}\).
\end{itemize}

\end{example}

\begin{definition}[CSAs/Azumaya Algebras]

A finite dimensional \textbf{central simple algebra} or \textbf{Azumaya
algebra} over \(k\) is a associative algebra over \(k\) with no
nontrivial 2-sided ideals with center \(k\) which is finite-dimensional
as a \(k{\hbox{-}}\)vector space.

\end{definition}

\begin{theorem}[Classification of CSAs]

Let \(A\in {\mathsf{Alg}}_{/ {k}}\), then TFAE:

\begin{itemize}
\tightlist
\item
  \(\exists\) a finite separable extension \(L _{/ {k}}\) where after
  base-changing to \(L\) one obtains
  \(A\otimes_k L \cong \operatorname{Mat}(n\times n, L)\).
\item
  \(A\otimes_k k^s \cong \operatorname{Mat}(n\times n, k^s)\).
\item
  \(\exists\) a finite (not necessarily separable) extension
  \(L_{/ {k}}\) such that
  \(A\otimes_k L \cong \operatorname{Mat}(n\times n, L)\).
\item
  \(A\) is a finite dimensional central simple algebra / Azumaya
  algebra.
\item
  \(A\) is a matrix algebra over a finite-dimensional central
  \(k{\hbox{-}}\)division algebra.
\end{itemize}

This is essentially a classification theorem: they're all forms of
matrix algebras over division algebras. Moreover there is a bijection
\begin{align*}
\left\{{\text{Central simple $k{\hbox{-}}$algebras }}\right\} \to H^2(k; (k^s)^{\times})
.\end{align*}

\end{theorem}

\begin{definition}[Opposite algebra]

If \(A \in \mathsf{CSA}_{/ {k}}\), then
\(A^{\operatorname{op}}\in \mathsf{CSA}_{/ {k}}\) is an algebra with the
same underlying vector space as \(A\) with
\(a \cdot_{^{\operatorname{op}}} b \coloneqq ba\).

\end{definition}

\begin{definition}[Morita equivalence]

\(A, B\) are Morita equivalent if \(A\otimes_k B^{\operatorname{op}}\)
is isomorphic to a matrix algebra.

\end{definition}

\begin{theorem}[?]

Given \(A, B \in \mathsf{CSA}_{/ {k}}\) which correspond to elements
\([A], [B]\in H^2\), then

\begin{itemize}
\tightlist
\item
  \([A] = [B] \iff A, B\) are Morita equivalent.
\item
  \([A]^{-1}= [A^{\operatorname{op}}]\).
\item
  \([A] \cdot [B] = [A\otimes_k B]\).
\end{itemize}

\end{theorem}

\hypertarget{the-boundary-map-and-twisted-vector-space}{%
\subsection{The Boundary Map and Twisted Vector
Space}\label{the-boundary-map-and-twisted-vector-space}}

\begin{remark}

We'd now like to make the boundary map explicit:
\begin{align*}
H^1(k; \operatorname{PGL}_n(k^s)) \to H^2(k; (k^s)^{\times})
.\end{align*}
Given \([f] \in H^1\), choose a representable cocycle \(f\):

\begin{center}
\begin{tikzcd}
    {{ \mathsf{Gal}} (k^s/k)} && {\operatorname{PGL}(k^s)} \\
    \\
    {{ \mathsf{Gal}} (L/k)} && {\operatorname{PGL}_n(L)}
    \arrow["f", from=1-1, to=1-3]
    \arrow[from=1-1, to=3-1]
    \arrow[hook, from=3-3, to=1-3]
    \arrow["{\tilde f}"', from=3-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXEdhbChrXnMvaykiXSxbMiwwLCJcXFBHTChrXnMpIl0sWzAsMiwiXFxHYWwoTC9rKSJdLFsyLDIsIlxcUEdMX24oTCkiXSxbMCwxLCJmIl0sWzAsMl0sWzMsMSwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMiwzLCJcXHRpbGRlIGYiLDJdXQ==}{Link
to Diagram}
\end{quote}

To compute this boundary, we use the original SES:

\begin{center}
\begin{tikzcd}
    1 && {k^s} && {\operatorname{GL}_n(k^s)} && {\operatorname{PGL}_n(k^s)} && 1 \\
    \\
    &&&&&& {{ \mathsf{Gal}} (k^s/k)}
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-7, to=1-9]
    \arrow["f", from=3-7, to=1-7]
    \arrow["{\text{Choose a set-theoretic lift } \tilde f}"{pos=0.8}, dashed, from=3-7, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCIxIl0sWzIsMCwia15zIl0sWzQsMCwiXFxHTF9uKGtecykiXSxbNiwwLCJcXFBHTF9uKGtecykiXSxbOCwwLCIxIl0sWzYsMiwiXFxHYWwoa15zL2spIl0sWzAsMV0sWzEsMl0sWzIsM10sWzMsNF0sWzUsMywiZiJdLFs1LDIsIlxcdGV4dHtDaG9vc2UgYSBzZXQtdGhlb3JldGljIGxpZnQgfSBcXHRpbGRlIGYiLDAseyJsYWJlbF9wb3NpdGlvbiI6ODAsInN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

So \(\tilde f: { \mathsf{Gal}} (k^s/k) \to \operatorname{GL}_n(k^s)\) is
a lieft of \(f\), and \(\delta f\) measures the failure of \(\tilde f\)
to be a cocycle. We have
\begin{align*}
\delta \tilde f(\sigma, \tau) = \tilde f(\sigma \tau) \qty{ \tilde f( \sigma) \tilde f (\tau)^{ \sigma} }^{-1}\in (k^s)^{\times}
,\end{align*}
using exactness since for \(f\) it lands in \(\operatorname{PGL}_n\) and
is trivial.

\end{remark}

\begin{definition}[Twisted vector spaces]

For \(L_{/ {k}}\) a separable extension and
\(\alpha: G{ {}^{ \scriptscriptstyle\times^{2} } } \to L^{\times}\) a
2-cocycle, so \([\alpha] \in H^2(L_{/ {k}} ; L^{\times})\), a
\textbf{twisted vector space} is a twisted semilinear action of
\({ \mathsf{Gal}} (L_{/ {k}} )\) on \(L^n\). I.e. it is a map
\begin{align*}
\tilde f: { \mathsf{Gal}} (L_{/ {k}} ) &\to \mathop{\mathrm{Aut}}(L^n) = \operatorname{GL}_n(L) \\
\text{such that } \tilde f( \sigma \tau) &= \tilde f( \sigma) \tilde g( \tau)^{ \sigma} \alpha( \sigma, \tau)
.\end{align*}

\end{definition}

\begin{remark}

For each \(\sigma\in { \mathsf{Gal}} (L_{/ {k}} )\) we get a
\(\sigma{\hbox{-}}\)semilinear automorphism of \(L^n\), i.e.~a map
\begin{align*}
f_\sigma: L^n &\to L^n \\
\text{where } f_ \sigma(s \cdot v) &= \sigma(s) \cdot f_{ \sigma}(v)
,\end{align*}
which is just the definition of semilinearity, and moreover
\(f_{ \sigma \tau} = f_{ \sigma} f_{ \tau} \alpha( \sigma, \tau)\).

\end{remark}

\begin{remark}

If \(\alpha = \operatorname{id}\), an \(\alpha{\hbox{-}}\)twisted vector
space is the same as a \(k{\hbox{-}}\)vector space by Galois descent.

\end{remark}

\begin{proposition}[Properties of categories of twisted vector spaces]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(\alpha \in \operatorname{im}\qty{ H^1(k; \operatorname{PGL}_n(k^s)) \to H^2(k; (k^s)^{\times}) } \iff\)
  there exists an \(n{\hbox{-}}\)dimensional \(\alpha{\hbox{-}}\)twisted
  vector space.

  The proof of this is just unwinding definitions, it's literally the
  same data!
\item
  The category \(\mathsf{Tw}_\alpha\) of \(\alpha{\hbox{-}}\)twisted
  vector spaces is abelian -- the only nontrivial thing to check is that
  there are enough injectives.
\item
  There are natural functors
  \begin{align*}
  ({-}) \otimes({-}) &: 
  \mathsf{Tw}_\alpha \times \mathsf{Tw}_{\alpha'} \to \mathsf{Tw}_{\alpha \alpha'} \\
  \mathop{\mathrm{Hom}}({-}, {-})&:
  (\mathsf{Tw}_\alpha)^{^{\operatorname{op}}} \times \mathsf{Tw}_{\alpha'} \to \mathsf{Tw}_{\alpha' \alpha^{-1}} \\
  \operatorname{Sym}^n, \bigwedge\nolimits^n &:
  \mathsf{Tw}_\alpha \to \mathsf{Tw}_{\alpha^n}
  .\end{align*}
\item
  If \(F_{/ {k}}\) is a separable field extension, then
  \begin{align*}
  ({-})\otimes F: \mathsf{Tw}_{\alpha}{}_{/ {k}}  \to \mathsf{Tw}_{\alpha}{}_{/ {F}} 
  .\end{align*}
\item
  There is an equivalence of categories
  \begin{align*}
  \mathsf{Tw}_{\operatorname{id}}{}_{/ {k}}  \xrightarrow{\sim} {\mathsf{k}{\hbox{-}}\mathsf{Mod}}
  .\end{align*}
\end{enumerate}

\end{proposition}

\begin{proposition}[?]

There is a 1-dimensional \(\alpha{\hbox{-}}\)twisted vector space iff
\([\alpha] = 1\in H^1(k; (k^s)^{\times})\).

\end{proposition}

\begin{proof}[?]

\(\impliedby\): First suppose \(\alpha \equiv 1\), then
\(\mathsf{Tw}_\alpha \xrightarrow{\sim} { \mathsf{Vect} }_{/ {k}}\), so
just take the vector space \(k\). If \(\alpha = \delta g\) for some
\(g: { \mathsf{Gal}} (k^s/k) \to (k^s)^{\times}\). Then the action
\({ \mathsf{Gal}} (k^s/k)\curvearrowright k^s\) where
\(f_\sigma = g(\sigma)\) is a 1-dimensional \(\alpha{\hbox{-}}\)twisted
vector space by sending \(1\to g(\sigma)\) and extending semilinearly.

\(\implies\): Let \(V\) be a 1-dimensional \(\alpha{\hbox{-}}\)twisted
vector space. Choose an isomorphism \(V \xrightarrow{\sim} k^s\) For
each \(\sigma\in { \mathsf{Gal}} (k^s/k)\) set \(g(\sigma) = g(1)\) and
\(g( \sigma \tau) = g( \sigma) g( \tau)^{ \sigma} \alpha( \sigma, \tau)\),
then
\begin{align*} 
\alpha= \delta g = g( \sigma \tau )\qty{g( \sigma) g( \tau)^{ \sigma} }^{-1}
.\end{align*}

\end{proof}

\begin{theorem}[?]

Suppose \(\alpha\in H^2(k; (k^s)^{\times})\) is in
\(\operatorname{im}\qty{ H^1(k; \operatorname{PGL}_n) \to H^2(k; (k^s)^{\times})}\),
then \(\alpha^n = 1\).

\end{theorem}

\begin{proof}[?]

If \(\alpha\) is in the image, there exists an
\(n{\hbox{-}}\)dimensional \(\alpha{\hbox{-}}\)twisted vector space
\(V\in \mathsf{Tw}_\alpha\), and so
\(\bigwedge\nolimits^n V \in \mathsf{Tw}_{\alpha^n}\).

\end{proof}

\begin{definition}[Index and period]

Given \(H^2(k; (k^s)^{\times}) = \mathop{\mathrm{Br}}(k)\) (which we'll
prove soon), the \textbf{period} of \(\alpha\) is the order of
\(\alpha\), and the \textbf{index} is defined the minimal \(n\) such
that \(\alpha\) is in the above image. I.e.,
\begin{align*}
\operatorname{period}(\alpha) &\coloneqq{\operatorname{Ord}}(\alpha) \\
\mathrm{index}(\alpha) &\coloneqq\min\left\{{ n {~\mathrel{\Big\vert}~}\alpha \in \operatorname{im}( H^1 \to H^2)}\right\}
.\end{align*}

\end{definition}

\begin{corollary}[?]

Period divides index.

\end{corollary}

\begin{question}

An open question: how different are the period and index? See the
period-index problem.

\end{question}

\begin{remark}

There are some maps between the categories \(\mathsf{Tw}_\alpha\),
\(\mathsf{SB}\) (Severi-Brauers), and \(\mathsf{CSA}\):

\begin{center}
\begin{tikzcd}
    &&& V \\
    &&& {\mathsf{Tw}_\alpha} \\
    {{\mathbb{P}}(V)} &&&&& {\mathop{\mathrm{End}}_k(V)} \\
    {} & \mathsf{SB}&&& \mathsf{CSA}
    \arrow[from=2-4, to=4-5]
    \arrow[from=2-4, to=4-2]
    \arrow[dashed, maps to, from=1-4, to=3-1]
    \arrow[dashed, maps to, from=1-4, to=3-6]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNyxbMywxLCJcXFR3X1xcYWxwaGEiXSxbMSwzLCJcXFNCIl0sWzQsMywiXFxDU0EiXSxbMywwLCJWIl0sWzAsM10sWzAsMiwiXFxQUChWKSJdLFs1LDIsIlxcRW5kX2soVikiXSxbMCwyXSxbMCwxXSxbMyw1LCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dLFszLDYsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifSwiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d}{Link
to Diagram}
\end{quote}

An analogy is that in vector spaces, \({\mathbb{P}}^n\) is to
\(\mathop{\mathrm{End}}(V)\) as \(\mathsf{SB}\) is to \(\mathsf{CSA}\)
in twisted vector spaces. Note that
\({ \mathsf{Gal}} (L_{/ {k}} ) "\curvearrowright" V\), which isn't a
true action but only fails to be one up to a scalar. Thus projectivizing
yields a semilinear action
\({ \mathsf{Gal}} (L_{/ {k}} )\curvearrowright{\mathbb{P}}(V)\), and
Galois descent yields forms of \({\mathbb{P}}(V)_{/ {k}}\).

\end{remark}

\begin{remark}

Why is \(\mathop{\mathrm{End}}(V)\) a form of
\(\operatorname{Mat}(n\times n)\)? Since \(V\in \mathsf{Tw}_\alpha\),
split it: choose an \(L\) such that
\({ \left.{{ \alpha}} \right|_{{L}} }\) is trivial. Then
\(\mathsf{Tw}_{{ \left.{{\alpha }} \right|_{{L}} }}= { \mathsf{Vect} }_{/ {L}}\).

\end{remark}

\hypertarget{thursday-september-09}{%
\section{Thursday, September 09}\label{thursday-september-09}}

\begin{remark}

Last time: 3 geometric avatars of elements \(\alpha\) of a Brauer group:

\begin{itemize}
\tightlist
\item
  \(\alpha{\hbox{-}}\)twisted vector spaces \(\mathsf{Tw}_\alpha\)
\item
  After projectivizing: Severi-Brauer varieties
\item
  Taking endomorphisms: central simple algebras.
\end{itemize}

Here we set \(G \coloneqq{ \mathsf{Gal}} (L_{/ {k}} )\) and
\(\alpha: G{ {}^{ \scriptscriptstyle\times^{2} } } \to L^{\times}\)
representing \([\alpha]\in H^2(G; L^{\times})\), and defined an
\(\alpha{\hbox{-}}\)twisted vector space as a
\(V\in { \mathsf{Vect} }_{/ {L}}\) with a semilinear map
\(f_\sigma: V\to V\) for each \(\sigma\in G\) where
\(\sigma(\ell v) = \sigma(\ell) \sigma(v)\) such that
\(f_{ \sigma \tau} = f_{ \sigma} \circ f_{ \tau} \alpha(\sigma\tau)\).
Last time we used this to show that
\begin{align*}
\operatorname{im}\qty{H^1(k; \operatorname{PGL}_n) \to H^2(k; (k^s)^{\times}) }
\end{align*}
is \(n{\hbox{-}}\)torsion.

\end{remark}

\begin{theorem}[?]

The category \(\mathsf{Tw}_\alpha\) is \textbf{semisimple}, i.e.~every
SES splits, and every object is a direct sum of simple objects.

\end{theorem}

\begin{proof}[?]

Note that in vector spaces,
\(\mathop{\mathrm{Hom}}_k(A, B) \cong B\otimes_k A {}^{ \vee }\), so
\(\mathop{\mathrm{Hom}}_k({-}, B) = ({-})\otimes_k B {}^{ \vee }\) as
functors.

Take a SES
\begin{align*}
0 \to V_2 \to W \to V_1 \to 0 \in \mathsf{Tw}_\alpha
.\end{align*}
We want to split this, a good trick to try every time: apply
\(\mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}(V_1, \cdot)\):
\begin{align*}
0 \to 
\mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}(V_1, V_2)
\to
\mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}(V_1, W)
\to
\mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}(V_1, V_1)
\to 0
.\end{align*}

This sequence is exact since we can write
\begin{align*}
\mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}({-}, V_1) = ({-})\otimes_k V_1 {}^{ \vee }
.\end{align*}
It's enough to split this SES, since any splitting
\(s: \mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}(V_1, V_2) \to \mathop{\mathrm{Mor}}(V_1, W)\)
would allow taking \(s(\operatorname{id}_{V_1})\) to split the original.
But this sequence does split, since
\(\mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}(V_1, V_1)\) is free, thus
projective.

\end{proof}

\begin{theorem}[?]

Any two simple objects \(D_1, D_2\in \mathsf{Tw}_\alpha\) are
isomorphic.

\end{theorem}

\begin{remark}

This is an analog of showing that every vector space is a sum of
1-dimensional sub-vector spaces, i.e.~every vector space has a basis. In
this situation, it's essentially Schur's lemma.

\end{remark}

\begin{proof}[?]

\(\mathop{\mathrm{Mor}}_{\mathsf{Tw}_\alpha}(D_1, D_2)\in{ \mathsf{Vect} }_{/ {L}}\)
is of dimension \(d = \dim_L(d_1) \dim_L(d_2) > 0\), so there exists a
nonzero map \(f:D_1\to D_2\). The claim is that \(f\) is an isomorphism:
since both objects are simple, just use that \(\ker D_1 \leq D_1\) and
\(\operatorname{im}f \leq D_2\) are sub-objects.

\end{proof}

\begin{corollary}[?]

There exists a unique simple object \(D\) of \(\mathsf{Tw}_\alpha\), and
every other object is of the form
\(D{ {}^{ \scriptscriptstyle\oplus^{I} } }\).

\end{corollary}

\begin{corollary}[?]

Any CSA is a matrix algebra over a division algebra.

\end{corollary}

\begin{proof}[?]

\(\mathop{\mathrm{End}}(D{ {}^{ \scriptscriptstyle\oplus^{n} } }) = \operatorname{Mat}(n\times n, \mathop{\mathrm{End}}(D))\),
so it's enough to show \(\mathop{\mathrm{End}}(D)\) is a division
algebra. This follows by the previous argument, again using Schur's
lemma.

\end{proof}

\begin{corollary}[?]

For \(X_{/ {k}}\) a Severi Brauer,
\(X\cong {\mathbb{P}}^n_{/ {k}} \iff X(k) \neq \emptyset\).

\end{corollary}

\begin{proof}[?]

\(\implies\): Clear, since \({\mathbb{P}}^n\) has rational points!

\(\impliedby\): We'll do a variant of the proof that uses
\(\mathsf{Tw}_\alpha\). Let \(X = {\mathbb{P}}(V)\) for
\(V\in \mathsf{Tw}_{\alpha}\), then any point \(x\in X\) yields a
1-dimensional (twisted!) subspace \(R \subseteq V\). Then
\([\alpha] = 0 \in H^2(k; (k^s)^{\times})\), and by Hilbert 90 this
comes from a point in the following composition:

\begin{center}
\begin{tikzcd}
    {H^1(k; \operatorname{GL}_n)} && {H^1(k; \operatorname{PGL}_n)} && {0\in H^2(k; (k^s)^{\times})} \\
    {[\alpha]} && {[X]} && 0
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-1, to=1-3]
    \arrow[maps to, from=2-3, to=2-5]
    \arrow[maps to, from=2-1, to=2-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJIXjEoazsgXFxHTF9uKSJdLFsyLDAsIkheMShrOyBcXFBHTF9uKSJdLFs0LDAsIjBcXGluIEheMihrOyAoa15zKVxcdW5pdHMpIl0sWzAsMSwiW1xcYWxwaGFdIl0sWzIsMSwiW1hdIl0sWzQsMSwiMCJdLFsxLDJdLFswLDFdLFs0LDUsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XSxbMyw0LCIiLDAseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJtYXBzIHRvIn19fV1d}{Link
to Diagram}
\end{quote}

This forces \(X = {\mathbb{P}}^n\).

\end{proof}

\begin{proof}[$\impliedby$, classical proof]

Let \(X\in \mathsf{SB}\) with \(X(k) \neq \emptyset\), then Artin
defines \(X {}^{ \vee }\), a dual Severi Brauer variety. This is
constructed using that \(X_{k^s} = {\mathbb{P}}^n\) and sets
\(X_{k^s} {}^{ \vee }= ({\mathbb{P}}^n) {}^{ \vee }\), which comes with
descent data to \(k\). A rigorous construction is that if
\(X = {\mathbb{P}}(V)\), we set
\(X {}^{ \vee }= {\mathbb{P}}(V {}^{ \vee })\). If \(X\) has a
\(k{\hbox{-}}\)point, then \(X {}^{ \vee }\) has a rational hyperplane
\(H\). The claim is that \(X {}^{ \vee }= {\mathbb{P}}^n\): this follows
from the fact that \({\mathcal{O}}(H)\) is a line bundle on
\(X {}^{ \vee }\) which is isomorphic to \({\mathcal{O}}(1)\) on
\(({\mathbb{P}}^n) {}^{ \vee }\) after base changing to \(k^s\). This
follows from cohomology of base change, since
\begin{align*}
{{\Gamma}\qty{X {}^{ \vee }, {\mathcal{O}}(H)_{/ {k^s}} } } = {{\Gamma}\qty{X_{k^s} {}^{ \vee }, {\mathcal{O}}(H)_{/ {k^s}} } } = {{\Gamma}\qty{{\mathbb{P}}^n_{/ {Y}} , {\mathcal{O}}(1)} }
.\end{align*}
So \({\mathcal{O}}(H)\) yields a map \(X {}^{ \vee }\to {\mathbb{P}}^n\)
which is an isomorphism after passing to \(k^s\). Now we can write
\(X = (X {}^{ \vee }) {}^{ \vee }\) and
\(X {}^{ \vee }= {\mathbb{P}}^n\), so
\begin{align*}
X = (X {}^{ \vee }) {}^{ \vee }= ({\mathbb{P}}^n) {}^{ \vee }\cong {\mathbb{P}}^n
.\end{align*}

\end{proof}

\begin{definition}[Reduced norm and trace]

Let \(A\in \mathsf{CSA}_{/ {k}}\), then there are maps
\begin{align*}
\operatorname{Nm}_{A_{/ {k}} }: A &\to k && \text{multiplicative} \\
\operatorname{Tr}_{A_{/ {k}} }: A &\to k && \text{additive}
.\end{align*}

How they're constructed: let
\(A\in \mathop{\mathrm{End}}(V) = V\otimes V {}^{ \vee }\), then since
\(\bigwedge\nolimits^*({-})\) is a functor, there is a map
\begin{align*}
\operatorname{Nm}_{A_{/ {k}} }: \mathop{\mathrm{End}}(V) &\to \mathop{\mathrm{End}}\qty{ \bigwedge\nolimits^{\dim V} V} = k \\
\operatorname{Tr}_{A_{/ {k}} }: \mathop{\mathrm{End}}(V) &{ { \, \xrightarrow{\sim}\, }}V\otimes V {}^{ \vee }\xrightarrow{{\left\langle {{-}},~{{-}} \right\rangle}} k
.\end{align*}

\end{definition}

\begin{proposition}[?]

For \(A\in \mathsf{CSA}_{/ {k}}\), then if there exists a nonzero
\(f\in A\) with \(\operatorname{Nm}_{A_{/ {k}} }(f) = 0\), then \(A\) is
not a division algebra.

\begin{quote}
Algebra: nontrivial matrix algebra over a field implies existence of
matrices with determinant zero.
\end{quote}

\end{proposition}

\begin{proof}[?]

The norm is multiplicative, so if \(f\) is a unit then
\(\operatorname{Nm}(f f^{-1})= 1\neq 0\).

\end{proof}

\begin{theorem}[?]

There is a surjection
\begin{align*}
\displaystyle\bigcup_n H^1(k; \operatorname{PGL}_n) \twoheadrightarrow H^2(k; (k^s)^{\times})
.\end{align*}

\end{theorem}

\begin{proof}[sketch]

It's enough to show the following surjection:
\begin{align*}
\displaystyle\bigcup_n H^1(L_{/ {k}} ; \operatorname{PGL}_n) \to H^2(L_{/ {k}} ; L^{\times})
.\end{align*}
Given \(\alpha\) in the codomain, interpret it as a central extension:
\begin{align*}
1 \to L^{\times}\to M_\alpha \to { \mathsf{Gal}} (L_{/ {k}} ) \to 1
.\end{align*}

\begin{definition}[Semilinear group rings]

Define \(L[M_\alpha]\) to be the \textbf{semilinear group ring} of
\(M_\alpha\):
\begin{align*}
L[M_\alpha] \bigoplus_{\lambda \in M_\lambda} L[e_\lambda]
\end{align*}
where \(e_{\lambda_1} e_{ \lambda_2} = e_{ \lambda_1 \lambda_2}\) and
\(\ell e_{ \lambda} = e_{\lambda} \lambda(\ell)\).

\end{definition}

\begin{claim}

\(A_\alpha \coloneqq L[M_{ \alpha}] / \left\langle{ \lambda e_1 - 1 e_{ \lambda}}\right\rangle\)
is a CSA mapping to \([\alpha]\). See Serre's \emph{Local Fields}.

\end{claim}

\end{proof}

\begin{question}

Can this construction be done in \(\mathsf{SB}\) or
\(\mathsf{Tw}_\alpha\)?

\end{question}

\hypertarget{computing-brauer-groups}{%
\subsection{Computing Brauer Groups}\label{computing-brauer-groups}}

\begin{remark}

\end{remark}

\begin{claim}

\(\mathop{\mathrm{Br}}({\mathbb{F}}_q) = 0\).

\end{claim}

\begin{theorem}[?]

Let \(k\) be a \(C_1{\hbox{-}}\)field, so any homogeneous polynomial in
\(k\) with degree \(d < n\) has a nonzero solution. Then
\(\mathop{\mathrm{Br}}(k) = 0\).

\end{theorem}

\begin{remark}

Note that
\href{https://www.wikiwand.com/en/Chevalley\%E2\%80\%93Warning_theorem}{Chevalley-Warning}
exactly says that finite fields are \(C_1\).

\end{remark}

\begin{proof}[of theorem]

\envlist

\begin{claim}

Let \(A\in \mathsf{CSA}_{/ {k}}\), then
\(\operatorname{Nm}_{A_{/ {k}} }: A\to k\) is a polynomial function on
\(n^2\) variables of degree \(n\).

\end{claim}

\begin{proof}[?]

This is true for the actual determinant, and this is a claim that can be
checked after passing to \(k^s\) since the norm is a \emph{form} of the
determinant.

\end{proof}

\begin{corollary}[?]

If \(k\) is \(C_1\) and \(\operatorname{rank}A > 1\), there exists a
nonzero \(f\in A\) such that \(\operatorname{Nm}_{A_{/ {k}} }(f) = 0\).

\end{corollary}

But all \(k{\hbox{-}}\)division algebras are isomorphic to \(k\), here
all CSAs are of the form \(\operatorname{Mat}(n\times n, k)\), so the
Brauer group is trivial.

\end{proof}

\begin{theorem}[Tsem]

If \(k = \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu\)
and \(C_{/ {k}}\) is a smooth proper curve, then the function field
\(k(C)\) is \(C_1\).

\end{theorem}

\begin{proof}[?]

Let \(f\) be a homogeneous polynomial, \(\deg f = d\), in \(n\)
variables over \(k(C)\) with \(d < n\). Then regard
\(f: k(C)^n \to k(C)\), we want to show \(f^{-1}(0)\) is big. Let
\(p\in C\), and now \(f\) as a map
\begin{align*}
f: {{\Gamma}\qty{C; {\mathcal{O}}(r\cdot p)^n} } \to {{\Gamma}\qty{C; {\mathcal{O}}(r d \cdot p)} }
,\end{align*}
which is a polynomial map of finite dimensional vector spaces that are
subspaces of the previous domain/codomain. Using Riemann-Roch, the
dimension of the left-hand side grows like \(r\cdot n\) and the
right-hand side grows like \(r\cdot d\), and for \(r\) large enough,
\(rn > rd\). Since \(f\) is homogeneous, \(f^{-1}(0)\) contains 0, so
\(\dim f^{-1}(0)> 0\). But a positive-dimensional variety over an
algebraically closed field has lots of rational points!

\end{proof}

\hypertarget{tuesday-september-14}{%
\section{Tuesday, September 14}\label{tuesday-september-14}}

\begin{remark}

Goal: Severi-Brauer varieties satisfy the Hasse principle, and develop
the Brauer-Manin obstruction. We have the following theorem: if
\(X\in \mathsf{SB}_{/ {k}}\), then TFAE:

\begin{itemize}
\tightlist
\item
  \(X\) has a rational point,
\item
  \(X \cong {\mathbb{P}}^n\) for some \(n\),
\item
  \([X]\in \mathop{\mathrm{Br}}(k)\) is the trivial class.
\end{itemize}

We'll soon prove the following theorem:

\end{remark}

\begin{theorem}[Hasse principle for Severi Brauers]

For \(K\) a number field, there is an injective map
\begin{align*}
\mathop{\mathrm{Br}}(k) \hookrightarrow\bigoplus _{v\in  \mathrm{Pl}\qty{k}  } \mathop{\mathrm{Br}}(k_v)
,\end{align*}
which is a statement of the Hasse principle, since the previous theorem
shows that if \(\mathop{\mathrm{Br}}(k_v)\) is empty for all \(k_v\), it
will have to come from a zero class in \(\mathop{\mathrm{Br}}(k)\)

\end{theorem}

\begin{remark}

Note that the cokernel of this map is prominent in class field theory!
Today we'll compute \(\mathop{\mathrm{Br}}(k_v)\), or more generally
\(\mathop{\mathrm{Br}}(F)\) for \(F\) a local field.

\end{remark}

\hypertarget{cyclic-algebras}{%
\subsection{Cyclic Algebras}\label{cyclic-algebras}}

\begin{remark}

Setup: take \(k\in \mathsf{Field}\), \(L_{/ {k}}\) a
\(C_n{\hbox{-}}\)Galois extension, which is the data of
\begin{align*}
\chi_L: { \mathsf{Gal}} (k^s_{/ {k}} )\to C_n
.\end{align*}
For \(a\in K^s\), we'll consider pairs
\((\chi, a) = L[x]^\chi / \left\langle{x^n - a}\right\rangle\) where
commutation in \(L[x]^\chi\) is given by \(lx = x\sigma(\ell)\) for
\(l\in L\) where \(C_n = \left\langle{\sigma}\right\rangle\). This is a
\(k{\hbox{-}}\)vector space of dimension \(n^2\), and the claim is that
\((\chi, a) \in \mathsf{CSA}\).

\end{remark}

\begin{example}[?]

Take
\(\chi: { \mathsf{Gal}} ({\mathbb{C}}_{/ {{\mathbb{R}}}} ) \to C_2\)
with \(a=-1\), then
\((\chi, a) = {\mathbb{H}}= {\mathbb{R}}[i, j] / \left\langle{i^2,j^2,[ij]}\right\rangle\)
is the (Hamilton) quaternions.

\end{example}

\begin{fact}

One can view \(\chi \in H^1_{ \mathsf{Gal}} (k; C_n)\) and
\begin{align*}
a\in H^1_{ \mathsf{Gal}} (k; \mu_n) = k^{\times}/ (k^{\times}){ {}^{\cdot 2} }
.\end{align*}
In this case
\begin{align*}
(\chi, a) \coloneqq\chi \smile[a] \in H^2(k; \mu_n) \subseteq H^2(k; (k{ {}^{ \operatorname{sep} } })^{\times})
.\end{align*}
Note that this cup product can be computed explicitly from the product
on \(\operatorname{Ext}\) or using the standard resolution.

\end{fact}

\begin{remark}

Now to compute more Brauer groups! So far, we've only done relatively
trivial examples. We'll start with local fields: for algebraically
closed fields, Galois cohomology vanishes, so

\begin{itemize}
\tightlist
\item
  \(\mathop{\mathrm{Br}}({\mathbb{C}}) = 0\)
\item
  To compute
  \(\mathop{\mathrm{Br}}({\mathbb{R}}) = H^2({ \mathsf{Gal}} ({\mathbb{C}}_{/ {{\mathbb{R}}}} ); {\mathbb{C}}^{\times})\),
  take the resolution
\end{itemize}

\begin{center}
\begin{tikzcd}
    && \vdots \\
    {P^\bullet:} && {{\mathbb{Z}}[x]/\left\langle{x^2-1}\right\rangle} \\
    && {{\mathbb{Z}}[x]/\left\langle{x^2-1}\right\rangle} \\
    && {{\mathbb{Z}}[x]/\left\langle{x^2-1}\right\rangle} \\
    && {{\mathbb{Z}}[x]/\left\langle{x^2-1}\right\rangle} & 1 \\
    && {\mathbb{Z}}& 1
    \arrow[from=5-3, to=6-3]
    \arrow["{x-1}", from=4-3, to=5-3]
    \arrow[from=5-4, to=6-4]
    \arrow["{x+1}", from=3-3, to=4-3]
    \arrow["{x-1}", from=2-3, to=3-3]
    \arrow[from=2-3, to=3-3]
    \arrow[from=4-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOSxbMiwzLCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMiw0LCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMiw1LCJcXFpaIl0sWzMsNCwiMSJdLFszLDUsIjEiXSxbMiwyLCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMiwxLCJcXFpaW3hdL1xcZ2Vuc3t4XjItMX0iXSxbMCwxLCJQXlxcYnVsbGV0OiJdLFsyLDAsIlxcdmRvdHMiXSxbMSwyXSxbMCwxLCJ4LTEiXSxbMyw0XSxbNSwwLCJ4KzEiXSxbNiw1LCJ4LTEiXSxbNiw1XSxbMCw1XV0=}{Link
to Diagram}
\end{quote}

Then we can take
\(H^*(\mathop{\mathrm{Hom}}_{{\mathsf{{ \mathsf{Gal}} }{\hbox{-}}\mathsf{Mod}}}(P^\bullet, {\mathbb{C}}^{\times}))\):

\begin{center}
\begin{tikzcd}
    && 1 && {z\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu} \\
    {\mathbb{C}}^{\times}&& {\mathbb{C}}^{\times}&& {\mathbb{C}}^{\times}&& {\mathbb{C}}^{\times}\\
    z && {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5muz^{-1}} && z && {\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5muz^{-1}}
    \arrow[from=2-1, to=2-3]
    \arrow[from=2-3, to=2-5]
    \arrow[from=2-5, to=2-7]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-5, to=3-7]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMTAsWzAsMSwiXFxDQ1xcdW5pdHMiXSxbMiwxLCJcXENDXFx1bml0cyJdLFswLDIsInoiXSxbMiwyLCJcXGJhcnt6fXpcXGludiJdLFs0LDEsIlxcQ0NcXHVuaXRzIl0sWzQsMiwieiJdLFs2LDIsIlxcYmFye3p9elxcaW52Il0sWzYsMSwiXFxDQ1xcdW5pdHMiXSxbMiwwLCIxIl0sWzQsMCwielxcYmFye3p9Il0sWzAsMV0sWzEsNF0sWzQsN10sWzIsM10sWzUsNl0sWzgsOV1d}{Link
to Diagram}
\end{quote}

Check that
\(\mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu z^{-1}= 1\)
then
\(z = \mkern 1.5mu\overline{\mkern-1.5muz\mkern-1.5mu}\mkern 1.5mu\) so
\(z\in {\mathbb{R}}^{\times}\) and \(\ker d = {\mathbb{R}}^{\times}\).
Similarly, \(\operatorname{im}d = {\mathbb{R}}^{\times}_{>0}\), so
\begin{align*}
\mathop{\mathrm{Br}}({\mathbb{R}}) = {\mathbb{R}}^{\times}/ {\mathbb{R}}^{\times}_{>0} = \left\{{\pm 1}\right\}
.\end{align*}

\end{remark}

\begin{example}[?]

\({\mathbb{H}}\) represents \(-1\) in
\(\mathop{\mathrm{Br}}({\mathbb{R}})\), as does the corresponding Severi
Brauer
\begin{align*}
\left\{{x^2 + y^2 + z^2 = 0}\right\} \subseteq {\mathbb{P}}^2_{/ {{\mathbb{R}}}} 
.\end{align*}
Note that \(+1\) is represented by the field itself, regarded as a
\(1\times 1\) matrix algebra, or projective space.

\end{example}

\begin{remark}

Write \(k^{\scriptscriptstyle\mathrm{un}}\) for the maximal unramified
extensions, where an extension is \emph{ramified} if the degree of the
residue field changes (or the valuation remains an integer?) For
example, for \(k = { {\mathbb{Q}}_p }\), we have
\(k^{\scriptscriptstyle\mathrm{un}}= \operatorname{ff}(W(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}_p\mkern-1.5mu}\mkern 1.5mu))\)
(i.e.~the Witt vectors). In general,
\(k^{\scriptscriptstyle\mathrm{un}}= k(\mu_\infty')\) where
\(\mu_\infty'\) is the set of roots of unity of order prime to the
characteristic. As a corollary,
\({ \mathsf{Gal}} (k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ) = \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{F}}_q\mkern-1.5mu}\mkern 1.5mu/{\mathbb{F}}_1 = \widehat{{\mathbb{Z}}}\).

\end{remark}

\begin{theorem}[?]

For \(k\) a nonarchimedean local field (a finite extension of
\({ {\mathbb{Q}}_p }\)), then
\(\mathop{\mathrm{Br}}(k) = {\mathbb{Q}}/{\mathbb{Z}}\)

\begin{itemize}
\tightlist
\item
  \(H^2(k^{{\scriptscriptstyle\mathrm{un}}}_{/ {k}} ; (k^{\scriptscriptstyle\mathrm{un}})^{\times}) = {\mathbb{Q}}/{\mathbb{Z}}\)
\item
  \(H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} , (k^{\scriptscriptstyle\mathrm{un}})^{\times}) { { \, \xrightarrow{\sim}\, }}H^2(k; (k^s)^{\times}) = \mathop{\mathrm{Br}}(k)\)
  is an isomorphism.
\end{itemize}

\end{theorem}

\begin{remark}

Many proofs of this are delicate! We'll follow a mix of Cassels-Frolich
and Milne for this proof.

\end{remark}

\begin{proof}[of 1]

Take the SES coming from the valuation map:

\begin{center}
\begin{tikzcd}
    1 && {U_{k^{\scriptscriptstyle\mathrm{un}}}} && {(k^{\scriptscriptstyle\mathrm{un}})^{\times}} && {\mathbb{Z}}&& 0
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow["{\operatorname{val}}", from=1-5, to=1-7]
    \arrow[from=1-7, to=1-9]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCIxIl0sWzIsMCwiVV97a15cXHVucmFtfSJdLFs0LDAsIihrXlxcdW5yYW0pXFx1bml0cyJdLFs2LDAsIlxcWloiXSxbOCwwLCIwIl0sWzAsMV0sWzEsMl0sWzIsMywiXFx2YWwiXSxbMyw0XV0=}{Link
to Diagram}
\end{quote}

\begin{claim}

\envlist

\begin{itemize}
\tightlist
\item
  \(H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; {\mathbb{Z}}) = {\mathbb{Q}}/{\mathbb{Z}}\).
\item
  \(H^*(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}}) = 0\)
\end{itemize}

\end{claim}

\begin{remark}

Why this implies the theorem: take the LES in cohomology to get the
following:

\begin{center}
\begin{tikzcd}
    \textcolor{rgb,255:red,214;green,92;blue,92}{H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}})=0} && {H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; (k^{\scriptscriptstyle\mathrm{un}})^{\times})} && {H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; {\mathbb{Z}})} \\
    \\
    \textcolor{rgb,255:red,214;green,92;blue,92}{H^3(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}}) = 0}
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=1-5]
    \arrow[from=1-5, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJIXjIoa15cXHVucmFtIFxcc2xpY2UgazsgVV97a15cXHVucmFtfSk9MCIsWzAsNjAsNjAsMV1dLFswLDIsIkheMyhrXlxcdW5yYW0gXFxzbGljZSBrOyBVX3trXlxcdW5yYW19KSA9IDAiLFswLDYwLDYwLDFdXSxbMiwwLCJIXjIoa15cXHVucmFtIFxcc2xpY2UgazsgKGteXFx1bnJhbSlcXHVuaXRzICkiXSxbNCwwLCJIXjIoa15cXHVucmFtIFxcc2xpY2UgazsgXFxaWikiXSxbMCwyXSxbMiwzXSxbMywxXV0=}{Link
to Diagram}
\end{quote}

A claim is that
\(H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; {\mathbb{Z}}) = H^2(\widehat{{\mathbb{Z}}}; {\mathbb{Z}})\).
One can compute this colimit explicitly, but there is a SES
\begin{align*}
0\to {\mathbb{Z}}\to {\mathbb{Q}}\to {\mathbb{Q}}/{\mathbb{Z}}\to 0
.\end{align*}

Now note that \(H^{>0}(G; {\mathbb{Q}}) = 0\) for profinite groups,
since this is necessarily a torsion \({\mathbb{Q}}{\hbox{-}}\)vector
space. For a full proof, use that multiplication by \(n\) is an
isomorphism the annihilates it. As a corollary, taking the LES above
yields \(H^{i}(G; {\mathbb{Z}}) = H^{i-1}(G; {\mathbb{Q}}/Z)\) for
\(i\geq 2\). Thus
\begin{align*}
H^2(\widehat{{\mathbb{Z}}}; {\mathbb{Z}}) = H^1(\widehat{{\mathbb{Z}}}; {\mathbb{Q}}/{\mathbb{Z}}) = \mathop{\mathrm{Hom}}_{{\mathsf{Top}}}(\widehat{{\mathbb{Z}}}, {\mathbb{Q}}/{\mathbb{Z}}) = {\mathbb{Q}}/{\mathbb{Z}}
,\end{align*}
and in fact
\begin{align*}
H^1(\widehat{{\mathbb{Z}}}; {\mathbb{Q}}/{\mathbb{Z}}) = \colim_n \mathop{\mathrm{Hom}}(C_n; {\mathbb{Q}}/{\mathbb{Z}})
.\end{align*}

\end{remark}

\end{proof}

\begin{proof}[of b]

Here we'll have to use the structure of
\(U_{k^{\scriptscriptstyle\mathrm{un}}}\). It's enough to show
\begin{align*}
H^{>0}(k_n{} _{/ {k}} ; U_{k_n}) = 0
\end{align*}
for \(k_n{}_{/ {k}}\) unramified of finite degree \(n\), using that
these are unique. We'll use the following:

\begin{definition}[?]

There is a filtration
\({\operatorname{Fil}}_r U_{k_n} = \left\{{u\in U_{k_n} {~\mathrel{\Big\vert}~}u = 1 \operatorname{mod}\pi^r}\right\}\)
for \(\pi\) a uniformizer.

\end{definition}

\begin{fact}

We can identify
\begin{align*}
{\operatorname{Fil}}_r/{\operatorname{Fil}}_{r+1} 
=
\begin{cases}
\kappa_n^{\times}& r=0 
\\
\kappa_n^+ & r>0.
\end{cases}
,\end{align*}
where \(\kappa\) denotes residue fields, \(\kappa_n{}_{/ {\kappa}}\) is
the unique degree \(n\) extension, and \(\kappa^+\) is the additive
group. Why: use that these look like power series, and the associated
graded picks off the \(r\)th coefficient. Moreover, things like
\(1+\pi^2\) can be units by formally inverting using geometric series.

\end{fact}

Thus it's enough to show for residue fields that
\begin{align*}
H^{>0}(\kappa_n{}_{/ {\kappa}} ; \kappa_n^{\times}) &= 0 \\
H^{>0}(\kappa_n{}_{/ {\kappa}} ; \kappa_n^+) &= 0
,\end{align*}
since each graded piece of the associated grading having zero cohomology
implies the entire thing has zero cohomology.

For the first,

\begin{itemize}
\tightlist
\item
  \(i=1\) is Hilbert 90,
\item
  \(i=2\) follows from
  \(\mathop{\mathrm{Br}}(\kappa_n{} _{/ {\kappa}} ) = 0\),
\item
  \(i\geq 3\) uses that \(H^* = H^*[-2]\), since the resolution used for
  cohomology of a cyclic group was 2-periodic.
\end{itemize}

For the second, to compute the cohomology of a cyclic group we take the
2-periodic resolution:

\begin{center}
\begin{tikzcd}
    x && {x^{q}-x} \\
    {k_n} && {k_n} && {k_n} && \cdots \\
    && x && {\sum x^{q^i}}
    \arrow[from=2-1, to=2-3]
    \arrow[from=2-3, to=2-5]
    \arrow["{\mathop{\mathrm{Frob}}- 1}", from=1-1, to=1-3]
    \arrow["\operatorname{tr}_ { {\kappa_n} {}_{/ {\kappa}}  }", from=3-3, to=3-5]
    \arrow[from=2-5, to=2-7]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsOCxbMCwxLCJrX24iXSxbMCwwLCJ4Il0sWzIsMCwieF57cX0teCJdLFsyLDEsImtfbiJdLFs0LDEsImtfbiJdLFsyLDIsIngiXSxbNCwyLCJcXHN1bSB4XntxXml9Il0sWzYsMSwiXFxjZG90cyJdLFswLDNdLFszLDRdLFsxLDIsIlxcRnJvYiAtIDEiXSxbNSw2LCJcXHRleHR7XCJBZGRpdGl2ZSBub3JtXCJ9IiwyXSxbNCw3XV0=}{Link
to Diagram}
\end{quote}

Then

\begin{itemize}
\item
  \(H^2 = \ker/\operatorname{im}\), and \(\ker = k\) since Frobenius
  fixes everything, and use that
  \begin{align*}
  \sum x^{q^i} = x + x^q + x^{q^2} + \cdots = \operatorname{tr}_{\kappa_n {}_{/ {\kappa}} }(x)
  .\end{align*}
\item
  If \(n\) is invertible, so \(p\notdivides n\), writing
  \(\operatorname{Tr}(1) = n\) we can take
  \(\operatorname{Tr}(a/n) = a\).
\item
  It suffices to show this polynomial isn't identically zero, but it's a
  polynomial of degree \(q^{n-1}\) but \(\# \kappa_n = q^n\).
\item
  Now use that \(a = \operatorname{tr}(x)\) for some \(a^{\bullet}\),
  then take \(b = \operatorname{tr}(bx/a)\).
\end{itemize}

\end{proof}

\hypertarget{thursday-september-16}{%
\section{Thursday, September 16}\label{thursday-september-16}}

\hypertarget{computing-brauer-groups-1}{%
\subsection{Computing Brauer Groups}\label{computing-brauer-groups-1}}

\begin{remark}

Let \(k\) be a \(p{\hbox{-}}\)adic field, our goal is to show
\(\mathop{\mathrm{Br}}(k) = {\mathbb{Q}}/{\mathbb{Z}}\). We were trying
to show

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; (k^{\scriptscriptstyle\mathrm{un}})^{\times}) = {\mathbb{Q}}/{\mathbb{Z}}\),
\item
  Any Brauer class is split by an \emph{unramified} extension
\end{enumerate}

This says that we can split the computation of
\(\mathop{\mathrm{Br}}(k)\) into an interesting part (the ramified case)
and a trivial part (the unramified case).

\todo[inline]{Check 2!}

To prove 1, we used
\begin{align*}
1 \to U_{k^{\scriptscriptstyle\mathrm{un}}} \to (k^{\scriptscriptstyle\mathrm{un}})^{\times}\to {\mathbb{Z}}\to 0
,\end{align*}

and

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; {\mathbb{Z}}) = {\mathbb{Q}}/{\mathbb{Z}}\),
\item
  \(H^{>0}(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}}) = 0\),
\end{enumerate}

where we used a filtration
\begin{align*}
{\operatorname{Fil}}_r U_{k^{\scriptscriptstyle\mathrm{un}}} 
=
\begin{cases}
U_{k^{\scriptscriptstyle\mathrm{un}}} &  r=0
\\
 \left\{{x{~\mathrel{\Big\vert}~}x\equiv 1\operatorname{mod}\pi^r}\right\}& r\geq 1.
\end{cases}
\end{align*}
and
\begin{align*}
{\mathsf{gr}\,}^r({\operatorname{Fil}}_\bullet U_{k^{\scriptscriptstyle\mathrm{un}}}) 
=
\begin{cases}
\kappa^{\times}& r=0 
\\
\kappa & r\geq 1.
\end{cases}
.\end{align*}

We now want to show

\begin{itemize}
\tightlist
\item
  \(H^{>0}(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; \mkern 1.5mu\overline{\mkern-1.5mu\kappa\mkern-1.5mu}\mkern 1.5mu^{\times}) = H^{>0}(\kappa; \mkern 1.5mu\overline{\mkern-1.5mu\kappa\mkern-1.5mu}\mkern 1.5mu^{\times})\)
\item
  \(H^*( k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; \mkern 1.5mu\overline{\mkern-1.5mu\kappa\mkern-1.5mu}\mkern 1.5mu^{\times}) = H^*(\kappa; \mkern 1.5mu\overline{\mkern-1.5mu\kappa\mkern-1.5mu}\mkern 1.5mu^{\times}) = 0\),
  and we were working on \(*=2\).
\end{itemize}

\end{remark}

\begin{proposition}[?]

For \(k\) any field, \(H^1(k; (k{ {}^{ \operatorname{sep} } })^+) = 0\),
where \(k^+\) denotes taking the additive group.

\end{proposition}

\begin{proof}[?]

\(H^1\) here classifies forms of SESs
\begin{align*}
0\to k \to V\to k \to 0
,\end{align*}
since automorphisms of this SES correspond to matrices
\(\left\{{ { \begin{bmatrix}  {1} & {*} \\  {0} & {1} \end{bmatrix} } {~\mathrel{\Big\vert}~}*\in k^+}\right\} \cong k^+\).
But any form of this splits, since any SES of vector spaces splits.

\end{proof}

\begin{theorem}[?]

For \(k\) any field,
\(H^{>0}(k; (k{ {}^{ \operatorname{sep} } })^+) = 0\).

\end{theorem}

\begin{proof}[?]

It's enough to show this for finite extensions, so consider
\(H^{>0}(L_{/ {k}} ; L^+) = 0\). The normal basis theorem implies that
\(L^+ \cong k[G]\) as a \(G{\hbox{-}}\)module, since this is the regular
representation. We'll use the following common lemma:

\begin{lemma}[Shapiro's Lemma]

If \(H\leq G\) are finite groups and
\(M \in {\mathsf{H}{\hbox{-}}\mathsf{Mod}}\) then
\begin{align*}
H^*(G; \operatorname{Ind}_H^G M)\cong H^*(H; M), &&
\operatorname{Ind}_H^G M = M\otimes_{{\mathbb{Z}}[H]} {\mathbb{Z}}[G]
.\end{align*}

\end{lemma}

Now use that
\begin{align*}
H^*( { \mathsf{Gal}} (L_{/ {k}} ); k[{ \mathsf{Gal}} (L_{/ {k}} )] ) = H^*(1; k) = 0 && *>0
.\end{align*}

\end{proof}

\begin{proof}[of Shapiro's lemma]

Let
\(P^\bullet \rightrightarrows{\mathbb{Z}}\in {\mathsf{{\mathbb{Z}}[G]}{\hbox{-}}\mathsf{Mod}}\)
be a free resolution and use Frobenius reciprocity to write
\begin{align*}
H^*(G; \operatorname{Ind}_H^G M) 
&= H^*(\mathop{\mathrm{Hom}}(P^\bullet, \operatorname{Ind}_H^G M)) \\
&= H^*(\mathop{\mathrm{Hom}}( \mathop{\mathrm{Res}}^G_H P^\bullet, M)) \\
&= H^*(H; M)
,\end{align*}
where
\(\mathop{\mathrm{Res}}_H^G P^\bullet \rightrightarrows{\mathbb{Z}}\in {\mathsf{{\mathbb{Z}}[H]}{\hbox{-}}\mathsf{Mod}}\)
is a free resolution, since
\(P^\bullet = {\mathbb{Z}}[G]{ {}^{ \scriptscriptstyle\oplus^{I} } }\)
(using that it's free) and thus
\(\mathop{\mathrm{Res}}_H^G P^\bullet = {\mathbb{Z}}[H]{ {}^{ \scriptscriptstyle\oplus^{I'} } }\).

\end{proof}

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

We now want to prove (3),
\begin{align*}
H^*(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}} / {\operatorname{Fil}}^r) = 0 && *>0
.\end{align*}
By induction on \(r\), since we have a SES
\begin{align*}
0 \to{\operatorname{Fil}}^{r-1}/{\operatorname{Fil}}^r \to U_{k^{\scriptscriptstyle\mathrm{un}}}/{\operatorname{Fil}}^r \to U_{k^{\scriptscriptstyle\mathrm{un}}}/ {\operatorname{Fil}}_{r-1} \to 1
,\end{align*}
where \(H^*\) of the two outer terms vanishes and thus so does \(H^*\)
of the middle by the LES in cohomology.

For (4), we want to show
\(H^*(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}}) \to \lim_r H^*(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}} / {\operatorname{Fil}}_r)\).
We can move an inverse limit in:
\begin{align*}
\invlim_r \invlim_n H^*({k_n}_{/ {k}} ; U_{k^{\scriptscriptstyle\mathrm{un}}} / {\operatorname{Fil}}^r)
&= \invlim_r \invlim_n H^*(\mathop{\mathrm{Hom}}(P^\bullet, U_{k^{\scriptscriptstyle\mathrm{un}}} / {\operatorname{Fil}}^r)) \\
&= H^*(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; \invlim_r U_{k^{\scriptscriptstyle\mathrm{un}}} / {\operatorname{Fil}}^r)
.\end{align*}

This uses the Mittag-Leffler condition to show that \(\lim_1\) vanishes,
which applies because we actually have surjectivity.

\end{proof}

\begin{theorem}[Hasse]

If \(D_{/ {k}}\) is a division algebra over \(k\) a \(p{\hbox{-}}\)adic
field (or any local field) with \(\dim_k D = n^2\) (using that it's a
form of a matrix algebra), then \(D\) is split by the unique
\textbf{unramified} extension of \(k\) of degree \(n\).

\end{theorem}

\begin{remark}

That there is a unique such extensions follows from the fact that
\(\widehat{{\mathbb{Z}}}\) has a unique subgroup of every index.

\end{remark}

\hypertarget{proof-of-theorem}{%
\subsection{Proof of theorem}\label{proof-of-theorem}}

\begin{remark}

Write \(k_n\) for the unique unramified extension of degree \(n\). We'll
want to show

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Show that it's enough to show \(K_n \subseteq D\),
\item
  Actually show \(k_n \subseteq D\).
\end{enumerate}

\end{remark}

\begin{lemma}[?]

For \(k\) any field and \(D_{/ {k}}\) any division algebra of
\(\dim_k D = n^2\), then if \(L _{/ {k}} \subseteq D\) is a Galois
extension of degree \(n\), then \(D\) splits over \(L\).

\begin{quote}
This is true without the extension being Galois.
\end{quote}

\end{lemma}

\begin{proof}[of lemma, using $\Tw$]

Write \(D = \mathop{\mathrm{End}}(V) \in \mathsf{Tw}_{/ {k}}\) for some
\(V\in \mathsf{Tw}\) of dimension \(n\), then
\begin{align*}
D { \underset{\scriptscriptstyle {k} }{\times} {L} } = \mathop{\mathrm{End}}(V{ \underset{\scriptscriptstyle {k} }{\times} {L} }) \in \mathsf{Tw}_{/ {L}} 
.\end{align*}
Then since \(L \subseteq D\), we have \(L\curvearrowright V\) so
\(L\otimes_k L \curvearrowright V_L\), then use that
\(L\otimes_k L { { \, \xrightarrow{\sim}\, }}L^{n}\) for
\(n\coloneqq\# { \mathsf{Gal}} (L_{/ {k}} )\).

\begin{quote}
Why: write \(L\otimes_k L = k[x] / I\) and use the Chinese remainder
theorem!
\end{quote}

We can write \(L^n = \oplus Le_i\), so \(V_L = \oplus e_i V_L\) which is
dimension 1 and thus its Brauer class is trivial.

\end{proof}

\begin{remark}

Other proofs of this seem much more difficult!

So now let's show \(k_n\) splits \(D\). We'll need to develop some
valuation theory for division algebras.

\end{remark}

\begin{definition}[Valuations on division algebras]

Define a valuation \(v: D\to {\mathbb{Z}}\cup\left\{{\infty}\right\}\)
extending the valuation on \(K \subseteq D\) given by
\(1/n {\operatorname{val}}(\operatorname{Nm}_{D_{/ {k}} }(x))\).
Equivalently, for \(x\in D\), use that \(k(x) \subseteq D\) is a
finitely generated \(k{\hbox{-}}\)algebra in which every nonzero element
is a unit, so it's a field and carries a natural valuation.

\end{definition}

\begin{definition}[Valuation ring]

Define
\begin{align*}
{\mathcal{O}}_D &\coloneqq\left\{{x{~\mathrel{\Big\vert}~}v(x) \geq 0}\right\} \subseteq D\\
{\mathfrak{m}}_D &\coloneqq\left\{{x{~\mathrel{\Big\vert}~}v(x) > 0}\right\} \subseteq D \\
{\mathcal{I}}&\coloneqq{\mathcal{O}}_D/{\mathfrak{m}}_D
,\end{align*}
where \({\mathfrak{m}}_D \in \operatorname{mSpec}D\), and set
\begin{align*}
f &\coloneqq[{\mathcal{I}}: k] && \text{Degree of field extensions} \\
e &\coloneqq[{\operatorname{val}}(k) : {\operatorname{val}}(D)] && \text{Ramification index}
.\end{align*}

\end{definition}

\begin{remark}

Note that \({\mathcal{I}}\) is a field, since all division algebras over
finite fields are field extensions (using our computation of the Brauer
groups of fields).

\end{remark}

\begin{fact}

\(ef=n^2\), where the same proof for extensions of \(p{\hbox{-}}\)adic
fields goes through.

\end{fact}

\begin{claim}

\begin{align*}
e=f=n
.\end{align*}

\end{claim}

\begin{remark}

We'll show

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(e\leq n\),
\item
  \(f\leq n\),
\end{enumerate}

Then since \(ef=n^2\) this forces \(e=f=n\).

\end{remark}

\begin{lemma}[?]

Any commutative \(L \in {\mathsf{Alg}}_{/ {k}}\) with \(L \subseteq D\)
satisfies \(\dim_k L \leq n\).

\end{lemma}

\begin{proof}[?]

It's enough to prove this for \(\operatorname{Mat}(n\times n; k)\),
since the dimension won't change after passing to a finite extension,
and proving here is classical.

\begin{exercise}[?]

Prove this!

\end{exercise}

\end{proof}

\begin{proof}[of claim]

For (1): chose \(\pi\in {\mathcal{O}}_D\) with \(v(\pi) = 1/e\),
i.e.~something with minimal positive valuation. Then
\(k(\pi) \subseteq D\) is an extension over \(k\) of degree at most
\(n\), by the lemma.

For (2): Write \({\mathcal{I}}= \kappa( \alpha)\) for \(\alpha\) a
primitive element, and let \(\tilde \alpha \in D\) be a lift. Then
\(k( \tilde \alpha) \subseteq D\) is a field extension of degree
\(\leq n\) by the lemma, and its residue field is \({\mathcal{I}}\).

\end{proof}

\begin{corollary}[?]

We have an exact equality
\begin{align*}
[k( \tilde \alpha): k] = n
,\end{align*}
so \(k(\tilde \alpha)_{/ {k}}\) is unramified, and there's a unique such
extension, and since \(\kappa(\tilde \alpha) \subseteq D\).

\end{corollary}

\begin{remark}

A proof of this theorem using \(\mathsf{Tw}\) or \(\mathsf{SB}\) would
be clarifying.

\end{remark}

\begin{claim}

The following map is an isomorphism:
\begin{align*}
{\mathbb{Q}}/{\mathbb{Z}}\cong H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; (k^{\scriptscriptstyle\mathrm{un}})^{\times}) { { \, \xrightarrow{\sim}\, }}H^2(k; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu^{\times})
.\end{align*}

\end{claim}

\begin{proof}[of claim]

Use that the LHS is isomorphic to
\(H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^0(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times}))\),
and consider the Hochschild-Serre spectral sequence
\begin{align*}
H^p(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^q(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})) \Rightarrow H^{p+q}(k; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})
.\end{align*}

The spectral sequence reads:

\begin{center}
\begin{tikzcd}[column sep=0.05in]
    & {} \\
    \\
    \\
    \\
    && \textcolor{rgb,255:red,92;green,92;blue,214}{H^2(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^0(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})) = {\mathbb{Q}}/{\mathbb{Z}}} \\
    && \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^0(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})) =_{90} 0} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^1(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^1(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})) = 0} \\
    && \textcolor{rgb,255:red,92;green,92;blue,214}{H^0(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^0(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})) = k^{\times}} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^0(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^1(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})) =_{90}0} & \textcolor{rgb,255:red,214;green,92;blue,92}{H^0(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ; H^2(k^{\scriptscriptstyle\mathrm{un}}; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu ^{\times})) = 0} \\
    {} &&&&&& {} \\
    & {}
    \arrow[from=1-2, to=9-2]
    \arrow[from=8-1, to=8-7]
\end{tikzcd}
\end{center}

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

Then for degree reasons, there are no nontrivial differentials to kill
the two nonzero terms. One can alternatively use the SES
\begin{align*}
0 \to \mathop{\mathrm{Br}}(k^{\scriptscriptstyle\mathrm{un}}_{/ {k}} ) \to \mathop{\mathrm{Br}}(k) \to \mathop{\mathrm{Br}}(k^{\scriptscriptstyle\mathrm{un}})
.\end{align*}

\end{proof}

\hypertarget{tuesday-september-21}{%
\section{Tuesday, September 21}\label{tuesday-september-21}}

\begin{remark}

Last time: for \(k\) a \(p{\hbox{-}}\)adic field, we have
\(\mathop{\mathrm{Br}}(k) = {\mathbb{Q}}/{\mathbb{Z}}\). The plan for
today:

\begin{itemize}
\item
  Examples
\item
  A SES for \(L\) a number field:
  \begin{align*}
  0\to \mathop{\mathrm{Br}}(L) \to \bigoplus_{v\in  \mathrm{Pl}\qty{k}  } { L_{\widehat{v}} }\to {\mathbb{Q}}/{\mathbb{Z}}\to 0
  .\end{align*}
\item
  Possibly the Hasse-Minkowski theorem
\item
  The Brauer-Manin obstruction.
\end{itemize}

\end{remark}

\hypertarget{construction-of-brauer-classes-over-k}{%
\subsection{\texorpdfstring{Construction of Brauer classes over
\(K\)}{Construction of Brauer classes over K}}\label{construction-of-brauer-classes-over-k}}

\begin{remark}

Fix a character to a cyclic group
\begin{align*}
\chi: { \mathsf{Gal}} (k{ {}^{ \operatorname{sep} } }_{/ {k}} ) \to C_n = \left\langle{\sigma}\right\rangle
\end{align*}
and set \(k_\chi\) to be the fixed field.

\end{remark}

\begin{definition}[Cyclic Algebra]

For \(a\in k^{\times}/(k^{\times}){ {}^{\cdot n} }\), write
\begin{align*}
(\chi, a) = k_\chi \left\langle{\sigma}\right\rangle / \left\langle{ \sigma s = s^{\sigma} \sigma, \sigma^n - a}\right\rangle && s\in k_\chi
.\end{align*}

\end{definition}

\begin{remark}

We have
\begin{align*}
[(\chi, a)] \coloneqq[X] \smile[a] \in H^1(K; C_n)\cup H^1(K; \mu_n) = \mathop{\mathrm{Br}}(k)
.\end{align*}
There are cases where it's not known if these types of algebras are
generators of certain Brauer groups.

\end{remark}

\begin{remark}

For \(k\) a \(p{\hbox{-}}\)adic field and \(k_n\) the unique unramified
degree \(n\) extension, we can construct a character
\begin{align*}
\chi_n: { \mathsf{Gal}} (k{ {}^{ \operatorname{sep} } }_{/ {k}} ) \to { \mathsf{Gal}} (k_n{}_{/ {k}} ) { \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }C_n
,\end{align*}
where the isomorphism is canonical, sending the Galois group to the
Frobenius.

\end{remark}

\begin{theorem}[?]

Let \(\pi\) be a uniformizer of \({\mathcal{O}}_K\). Every CSA is
equivalent to one of the form
\begin{align*}
(\chi_n, \pi^m) \to {m\over n} \in {\mathbb{Q}}/{\mathbb{Z}}= \mathop{\mathrm{Br}}(k)
.\end{align*}

\end{theorem}

\begin{remark}

If \(m, n\) are coprime one gets a division algebra.

\end{remark}

\begin{proof}[Sketch]

This is mostly a computation that involves unwinding the isomorphism
\(\mathop{\mathrm{Br}}(k) \to {\mathbb{Q}}/{\mathbb{Z}}\). A sketch:

\begin{itemize}
\tightlist
\item
  The class \([(\chi_n, \pi)]\) has order \(n\),
\item
  The class \([(\chi_n, \pi)]^m = [(\chi_n, \pi^m)]\), which is given by
  a cup product.
\end{itemize}

\end{proof}

\begin{remark}[An algorithm to compute]

Let \(D_{/ {k}}\) be a division algebra.

\begin{itemize}
\tightlist
\item
  Find a copy of \(k_n\) in \(D\), which can be done since this is a
  division algebra of dimension \(n^2\).
\item
  There exists a \(\sigma \in D\) such that
  \(\sigma\curvearrowright K_n\) by conjugation is the canonical
  generator of
  \({ \mathsf{Gal}} (k_n {}_{/ {k}} ) { \underset{{\mathrm{can}}}{{ { \, \xrightarrow{\sim}\, }}} }C_n\)
  (where we take \(\mathop{\mathrm{Frob}}\) as the canonical generator).
\item
  Then
  \([D] \mapsto {v(\sigma) \over n} \in {\mathbb{Q}}/{\mathbb{Z}}= \mathop{\mathrm{Br}}(k)\),
  where \(v\) is the normalized valuation on \(D\) we constructed
  previously. Note that this is well defined since changing \(D\)
  changes the output by an integer.
\end{itemize}

\end{remark}

\begin{example}[The simplest case: $n=2$]

Using that there is in fact a canonical isomorphism \(\mu_2 \cong C_2\)
since there's only one nontrivial element in each group, we have
\begin{align*}
H^1(k; C_2) = H^1(k; \mu_2) = k^{\times}/(k^{\times}){ {}^{\cdot 2} }
.\end{align*}
Hence any character
\begin{align*}
\chi: { \mathsf{Gal}} (k{ {}^{ \operatorname{sep} } }_{/ {k}} ) \to C_2 = \mu_2
\end{align*}
is represented by some
\(b_\chi \in k^{\times}/(k^{\times}){ {}^{\cdot 2} }\). So we have an
identification
\begin{align*}
(\chi, a) \leadsto (b_\chi, a)_2 = (a, b_\chi)_2
=
\begin{cases}
0 & v(a) \equiv v(b) \operatorname{mod}2 
\\
{1\over 2} & \text{else}.
\end{cases}
\end{align*}
For the corresponding extension to be unramified, one needs the
valuation to be zero. So for example taking \(k(\pi)_{/ {k}}\) yields a
ramified extension since \(v(\pi ) = 1\).

\begin{quote}
Note that here \(({-}, {-})_n\) is generally a Hilbert or norm-residue
symbol.
\end{quote}

\end{example}

\begin{exercise}[?]

Prove that these cyclic algebras are CSAs.

\end{exercise}

\hypertarget{the-ses}{%
\subsection{The SES}\label{the-ses}}

\begin{remark}

Our goal for today: for \(k\) a number field, show the following
sequence is exact
\begin{align*}
0\to \mathop{\mathrm{Br}}(k) \to \bigoplus_{v\in  \mathrm{Pl}\qty{k}  } { k_{\widehat{v}} }\xrightarrow{\sum} {\mathbb{Q}}/{\mathbb{Z}}\to 0
.\end{align*}

\end{remark}

\begin{proposition}[?]

For \(\alpha \in \mathop{\mathrm{Br}}(k)\), using the pullback of
\(i_v\),
\begin{align*}
\mathop{\mathrm{Br}}(K) \xrightarrow{\prod_{i_v^*}} \prod_v \mathop{\mathrm{Br}}({ k_{\widehat{v}} })
\end{align*}
factors through
\(\bigoplus_v \mathop{\mathrm{Br}}({ k_{\widehat{v}} })\),
i.e.~\(i_v^*( \alpha) = 0\) for almost all \(v\).

\end{proposition}

\begin{proof}[of prop, proof 1]

First represent \(\alpha\) by \(X\in \mathsf{SB}\), so
\(X({ k_{\widehat{v}} }) \neq \emptyset\) for almost all \(v\). This
implies \(X_{{ k_{\widehat{v}} }} \cong {\mathbb{P}}^n_{/ {k}}\) for
almost all \(v\).

\end{proof}

\begin{definition}[Ideles]

\begin{align*}
{\mathbb{I}}_k \coloneqq\prod_v' ({ k_{\widehat{v}} }^{\times}, {\mathcal{O}}_{{ k_{\widehat{v}} }}^{\times}) = \left\{{ (x_v) \in \prod_v { k_{\widehat{v}} }^{\times}{~\mathrel{\Big\vert}~}x_v \in {\mathcal{O}}_{{ k_{\widehat{v}} }}^{\times}\, \text{ for almost all } v}\right\}
.\end{align*}
A basis of open sets is given by
\((x) \cdot \prod_v {\mathcal{O}}_{{ k_{\widehat{v}} }}^{\times}\).

\end{definition}

\begin{warnings}

There is a map
\begin{align*}
{\mathbb{I}}_k &\hookrightarrow{\mathbb{A}}_k^2 \\
x &\mapsto (x, x^{-1})
,\end{align*}
and there is a subspace topology -- but this is not equivalent to the
topology above, and is in fact a source of an infamous error!

\end{warnings}

\begin{definition}[S-ideles]

If \(S\) is a finite set of places of \(K\) containing all infinite
places, then define
\begin{align*}
{\mathbb{I}}_{k, S} = \prod_{v\in S} { k_{\widehat{v}} }^{\times}\times \prod_{v\not\in S} {\mathcal{O}}_{{ k_{\widehat{v}} }}^{\times}\subseteq {\mathbb{I}}_K
.\end{align*}

\end{definition}

\begin{fact}

\begin{align*}
{\mathbb{I}}_k = \colim_{S} {\mathbb{I}}_{k, s}
.\end{align*}

\end{fact}

\begin{remark}

The idea will be to study the following SES of Galois modules:
\begin{align*}
1 \to L^{\times}\to {\mathbb{I}}_L \to C_L \to 1
,\end{align*}
where \(C_L\) is the idele class group.

\end{remark}

\begin{proposition}[?]

\begin{align*}
H^2(L_{/ {k}} ; {\mathbb{I}}_L) &= \bigoplus_{v\in \mathrm{Pl}\qty{k} }\mathop{\mathrm{Br}}({ L_{\widehat{v}} }{}_{/ {{ k_{\widehat{v}} }}} ) \\
H^2(k; {\mathbb{I}}_{k{ {}^{ \operatorname{sep} } }}) &= \bigoplus_{v\in \mathrm{Pl}\qty{k} } \mathop{\mathrm{Br}}({ k_{\widehat{v}} })
,\end{align*}

\end{proposition}

\begin{theorem}[?]

\begin{align*}
H^1(L_{/ {k}} ; C_L) &= 0 \\
H^2(L_{/ {k}} ; C_L) &= [d] \in {\mathbb{Q}}/{\mathbb{Z}}, \quad d\coloneqq{1\over [L:k]}
.\end{align*}

This will imply
\begin{align*}
H^1(k; C_{k{ {}^{ \operatorname{sep} } }}) &= 0 \\
H^2(k; C_{k{ {}^{ \operatorname{sep} } }}) &= {\mathbb{Q}}/{\mathbb{Z}}
.\end{align*}

\end{theorem}

\begin{proof}[sketch]

We can write
\begin{align*}
H^2(L_{/ {k}} ; {\mathbb{I}}_L) 
&= H^2(L_{/ {k}} ; \directlim_T {\mathbb{I}}_{L, T}) \\
&= \directlim_T H^2(L_{/ {k}} ; {\mathbb{I}}_{L, T})
,\end{align*}
so it's enough to show that for \(S\) a finite set of places of \(K\)
and \(T\) a set of places over \(S\) that we have
\begin{align*}
H^2(L_{/ {k}} ; {\mathbb{I}}_{L, T}) = \bigoplus_{v\in S} \mathop{\mathrm{Br}}({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} )
.\end{align*}

\end{proof}

\begin{exercise}[?]

Try to prove this, it uses Shapiro's lemma and isn't too difficult.

\end{exercise}

\hypertarget{thursday-september-23}{%
\section{Thursday, September 23}\label{thursday-september-23}}

\begin{remark}

Let \(k\in \mathsf{Field}\), we have a SES
\(1\to k^{\times}\to {\mathbb{I}}_k \to C_k \to 1\). An exercise from
last time: for \(\mathrm{Pl}\qty{k}\) the places of \(k\), prove that
\begin{align*}
H^2(L_{/ {k}} ; {\mathbb{I}}_L) = \bigoplus_{v\in  \mathrm{Pl}\qty{k} } \mathop{\mathrm{Br}}({ L_{\widehat{v}} }/{ k_{\widehat{v}} })
,\end{align*}
where \({ L_{\widehat{v}} }\) was obtained by choosing any place above
\(v\) in \(L\) and completing.

\end{remark}

\hypertarget{proof-of-theorem-1}{%
\subsection{Proof of Theorem}\label{proof-of-theorem-1}}

\begin{remark}

For \(S \subseteq \mathrm{Pl}\qty{k}\) a finite set of places containing
all of the infinite places and \(T\) a set of places of \(L\) above
\(S\), we have
\begin{align*}
{\mathbb{I}}_{L, T} = \prod_{w\in T}L_w^{\times}\times \prod_{w\not\in T} {\mathcal{O}}_{{ L_{\widehat{v}} }}^{\times}
.\end{align*}
We can also write
\(H^2(L_{/ {k}} ; {\mathbb{I}}_L) = \directlim_T H^2(L_{/ {k}} ; {\mathbb{I}}_{L, T})\),
so it's enough to show the following:
\begin{align*}
H^2(L_{/ {k}} ; {\mathbb{I}}_{L, T}) &= \bigoplus_{v\in S} \mathop{\mathrm{Br}}({ L_{\widehat{v}} }/ { k_{\widehat{v}} }) \\
H^2(L_{/ {k}} ; {\mathbb{I}}_{\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu, T}) &= \bigoplus_{v\in S} \mathop{\mathrm{Br}}({ k_{\widehat{v}} })
.\end{align*}
We have
\begin{align*}
H^2(L_{/ {k}} ; {\mathbb{I}}_{L, T}) = \prod_{v\in S} H^2(L_{/ {k}} ; \prod_{w_{/ {v}} } L_w^{\times}) \times \prod_{v\in S} H^2(L_{/ {k}} ; \prod_{w_{/ {v}} } {\mathcal{O}}_{L_w} ^{\times}) 
,\end{align*}
noting that we need to take the entire product to actually get a Galois
module.

\end{remark}

\begin{claim}

\begin{align*}
H^2(L_{/ {k}} ; \prod_{w_{/ {v}} } { L_{\widehat{v}} }^{\times}) &= \mathop{\mathrm{Br}}({ L_{\widehat{v}} }{}_{/ { { k_{\widehat{v}} }}}  ) \\
H^2(L_{/ {k}} ; \prod_{w_{/ {v}} } { L_{\widehat{v}} }^{\times}) &= 0 
.\end{align*}

\end{claim}

\begin{proof}[of 1]

\begin{align*}
H^2(L_{/ {k}} ; \prod_{w_{/ {v}} } L_w^{\times}) 
&=
H^2(L_{/ {k}} ; \operatorname{Ind}_{{ \mathsf{Gal}} (L/k)}^{{ \mathsf{Gal}} ({ L_{\widehat{v}} }/{ k_{\widehat{v}} })} { L_{\widehat{v}} }^{\times}) \\
&= H^2({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} ; { L_{\widehat{v}} }^{\times}) \\
&\coloneqq\mathop{\mathrm{Br}}({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} )
.\end{align*}

\end{proof}

\begin{proof}[of 2]

Write
\begin{align*}
H^2(L_{/ {k}} ; \prod_{w_{/ {v}} } L_w^{\times}) 
&= 
H^2(L_{/ {k}} ; \operatorname{Ind}_{{ \mathsf{Gal}} (L/k)}^{{ \mathsf{Gal}} ({ L_{\widehat{v}} }/{ k_{\widehat{v}} })} {\mathcal{O}}_{{ L_{\widehat{v}} }}^{\times}) \\
&= 
H^2({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} ; {\mathcal{O}}_{{ L_{\widehat{v}} }}^{\times})\\
&= 0
.\end{align*}

\end{proof}

\begin{corollary}[?]

\begin{center}
\begin{tikzcd}
    {\mathop{\mathrm{Br}}(k)} && {\bigoplus_{v} \mathop{\mathrm{Br}}({ k_{\widehat{v}} })} \\
    \\
    {H^2(k; \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu^{\times})} && {H^2(k; \mathbb{I}_{\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu}) }
    \arrow[from=1-1, to=1-3]
    \arrow[from=3-1, to=3-3]
    \arrow[Rightarrow, no head, from=1-1, to=3-1]
    \arrow[Rightarrow, no head, from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJcXEJyKGspIl0sWzIsMCwiXFxiaWdvcGx1c197dn0gXFxCcihrX3YpIl0sWzAsMiwiSF4yKGs7IFxcYmFye2t9XFx1bml0cykiXSxbMiwyLCJIXjIoazsgXFxJSV97XFxiYXIga30pIl0sWzAsMV0sWzIsM10sWzAsMiwiIiwxLHsibGV2ZWwiOjIsInN0eWxlIjp7ImhlYWQiOnsibmFtZSI6Im5vbmUifX19XSxbMSwzLCIiLDEseyJsZXZlbCI6Miwic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoibm9uZSJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\end{corollary}

\hypertarget{injectivity}{%
\subsection{Injectivity}\label{injectivity}}

\begin{theorem}[Injectivity]

\(\mathop{\mathrm{Br}}(k) \hookrightarrow\bigoplus_v \mathop{\mathrm{Br}}({ k_{\widehat{v}} })\)
is injective, since \(H^1(L_{/ {k}} ; C_L) = 0\).

\end{theorem}

\begin{theorem}[Actual IRL application of Sylow theorems]

If \(G\) is a finite group and
\(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\) then \(H^i(G; M) = 0\) if
\(H^i(G_p; { \left.{{M}} \right|_{{G_p}} }) = 0\) for all \(p\) where
\(G_p\) is a \(p{\hbox{-}}\)Sylow subgroup of \(G\).

\end{theorem}

\begin{proof}[?]

There's a map
\begin{align*}
\qty{H^1(G; M) \xrightarrow{\operatorname{res}}H^1(G_p; { \left.{{M}} \right|_{{G_p}} }) \xrightarrow{\mathop{\mathrm{coRes}}} H^1(G; M) } = {\operatorname{mult}}_d, d \coloneqq[G: G_p]
.\end{align*}
Since \(d\) is prime to \(p\), \(\operatorname{res}\) is injective on
\(p{\hbox{-}}\)power torsion, making \(H^1(G; M)\) torsionfree. Then
since \(G\) is finite, \(H^i(G; M)\) is torsion, and the only torsion
torsionfree group is zero.

\end{proof}

\begin{remark}

There will be multiple steps:

\begin{itemize}
\item
  It's enough to prove this for \({ \mathsf{Gal}} (L_{/ {k}} )\) a
  \(p{\hbox{-}}\)group, using theorem on applications of Sylow. We know
  enough about the structure of \(p{\hbox{-}}\)groups to make induction
  arguments!
\item
  It's enough to show that \(H^i(L_{/ {k}} ; C_L) = 0\) for
  \(L_{/ {k}}\) cyclic. Letting \(L_{/ {k}}\) be Galois with
  \(G\coloneqq H^i(L_{/ {k}} )\) a \(p{\hbox{-}}\)group, then let
  \(H \leq G\) be a nontrivial normal cyclic subgroup. Then the
  inflation-restriction exact sequence yields
  \begin{align*}
  0 \to H^1(G/H; C_L^H) \to H^1(G; C_L) \to H^1(H; C_L) 
  ,\end{align*}
  using idele class groups and writing \(C_L^H\) for the class group of
  the fixed field by \(H\), and recalling that this comes from the
  Hochschild-Serre spectral sequence. By induction on the size of \(G\),
  we'll know the right-hand side is 0, and the left-hand side is 0 by
  induction on \(\# G\). However, note that we have to show this for
  \emph{all} cyclic extensions!
\item
  Prove the following theorem;
\end{itemize}

\end{remark}

\begin{remark}

Note that \(C_L\) will not even be finitely generated!

\end{remark}

\begin{theorem}[?]

If \(L_{/ {k}}\) is cyclic, then \(H^1(L_{/ {k}} ; C_L) = 0\), and
\(\# H^2(L_{/ {k}} ; C_L) = [L: k]\).

\end{theorem}

\begin{warnings}

Note that \(\# H^1 = 1\) in this case!

\end{warnings}

\begin{definition}[Herbrand Quotient]

If \(G\) is finite cyclic and
\(M\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\), define the \textbf{Herbrand
quotient} as
\begin{align*}
q(M) \coloneqq{ \# H^2(G; M) \over \# H^1(G; M) }
,\end{align*}
whenever this ratio is defined.

\end{definition}

\begin{remark}

Taking logs makes this look like an Euler characteristic.

\end{remark}

\begin{lemma}[Herbrand quotients are multiplicative]

Suppose \(0\to A\to B\to C\to 0\) is a SES of \(G{\hbox{-}}\)modules for
\(G\) cyclic. Then
\begin{align*}
q(A) q(C) = q(B)
.\end{align*}

\end{lemma}

\begin{exercise}[A fun one]

Prove this! It's the same proof that \(\chi(A) + \chi(C) = \chi(B)\).

\end{exercise}

\begin{lemma}[?]

If \(M\) is finite, then \(q(M) = 1\), so this invariant for infinite
modules.

\end{lemma}

\begin{proof}[?]

We first claim that \(\# M^G = \# M_G\), recalling that
\(M_G = M/\left\langle{g-1}\right\rangle = M/IM\) for \(I\) the
augmentation ideal. Note that in finite groups, for a SES
\(0\to A\to B\to C\to 0\) yields \(\#B = (\# A)\cdot (\# C)\), or
equivalently \((\#A)\cdot (\# B)^{-1}\cdot (\# C) = 1\) and this extends
to longer exact sequences.

Now use the exact sequence
\begin{align*}
0 \to M^G \to M \xrightarrow{g-1} M \to M_G \to 0
,\end{align*}
and so
\begin{align*}
(\# M^G) \cdot (\# M)^{-1}\cdot (\# M) \cdot (\# M_G)^{-1}= 1
.\end{align*}

Now to show that the sizes are equal, Recall that
\begin{align*}
H^*(G; M) = H^*\qty{ M \xrightarrow{g-1} M \xrightarrow{\sum g^i} M \to \cdots}
.\end{align*}
Thus we get
\begin{align*}
0 \to H^1(G; M) \to \operatorname{coker}( M \xrightarrow{g-1} M) \xrightarrow{\sum g^i} \ker(M \xrightarrow{g-1} M) \twoheadrightarrow H^2(G; M) \to 0 \\
\implies 0 \to H^1(G; M) \to M_G \xrightarrow{\sum g^i} M^G \twoheadrightarrow H^2(G; M) \to 0
,\end{align*}
so
\begin{align*}
\# H^1(G; M) \cdot (\# M_G)^{-1}\cdot(\# M^G) \cdot (\# H^2(G; M))^{-1}= 1 = q(M)^{-1}
.\end{align*}

\end{proof}

\begin{lemma}[?]

If \(M, N\) are finitely generated in
\({\mathsf{G}{\hbox{-}}\mathsf{Mod}}\) and
\(M\otimes{\mathbb{R}}\cong N\otimes{\mathbb{R}}\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\),
then \(q(M) = q(N)\).

\end{lemma}

\begin{remark}

Analogy: Reidemeister torsion! Tensoring up to \({\mathbb{R}}\) somehow
doesn't lose all torsion information.

\end{remark}

\begin{proof}[of lemma]

We'll first show
\(M\otimes{\mathbb{Q}}{ { \, \xrightarrow{\sim}\, }}N\otimes{\mathbb{Q}}\)
implies \(q(M) = q(N)\).

\begin{claim}

There is a map \(M/{\operatorname{tors}}\to N/{\operatorname{tors}}\)
with finite kernel and cokernel.

\end{claim}

\begin{center}
\begin{tikzcd}[column sep=0.25in]
    0 && 0 \\
    \\
    {M_{\operatorname{tors}}} && {N_{\operatorname{tors}}} \\
    &&&& 0 & {M_{\operatorname{tors}}} & M & {N/N_{\operatorname{tors}}} & {\text{torsion!}} & 0 \\
    M && N \\
    \\
    {M\otimes{\mathbb{Q}}} && {N\otimes{\mathbb{Q}}}
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-1, to=5-1]
    \arrow[from=5-1, to=7-1]
    \arrow["{\sim, \quad n\cdot p}", from=7-1, to=7-3]
    \arrow[from=3-3, to=5-3]
    \arrow[from=5-3, to=7-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=5-1, to=5-3]
    \arrow[from=3-1, to=3-3]
    \arrow[dashed, from=5-1, to=7-3]
    \arrow[from=4-5, to=4-6]
    \arrow[from=4-6, to=4-7]
    \arrow[from=4-7, to=4-8]
    \arrow[from=4-8, to=4-9]
    \arrow[from=4-9, to=4-10]
\end{tikzcd}
\end{center}

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

\begin{claim}

\begin{align*}
M\otimes{\mathbb{R}}\cong N\otimes{\mathbb{R}}
\iff
M\otimes{\mathbb{Q}}\cong N \otimes{\mathbb{Q}}
.\end{align*}

\end{claim}

\begin{exercise}[?]

Prove this!

\end{exercise}

\end{proof}

\hypertarget{tuesday-september-28}{%
\section{Tuesday, September 28}\label{tuesday-september-28}}

\begin{quote}
See fppf cohomology. Note: statements of the form
\(A\otimes C \cong B\otimes C \implies A\cong B\) aren't quite descent!
There's no descent data or e.g.~Galois equivariance, and the downstairs
maps may not be related to the original map at all.
\end{quote}

\begin{theorem}[?]

For \(L_{/ {k}}\) cyclic of degree \(n\),
\begin{align*}
q(C_L)=n
.\end{align*}

\end{theorem}

\begin{remark}

Recall that \(q\) is multiplicative in exact sequences, equals 1 for
finite \(G{\hbox{-}}\)modules, and if \(M\otimes R \cong N\otimes R\)
then \(q(M)= q(N)\).

\end{remark}

\begin{proof}[of 3rd property]

It's enough to show this for \(M, N\) torsionfree, since
\(q(M) = q(M/M_{\operatorname{tors}})\). The claim is that for \(R\)
sufficiently divisible, letting
\(M\otimes{\mathbb{Q}}\xrightarrow{\phi} N \otimes{\mathbb{Q}}\),
\({ \left.{{\phi}} \right|_{{M}} }\) factors through \(N\) with torsion
kernel. Use that
\(M\otimes{\mathbb{R}}{ { \, \xrightarrow{\sim}\, }}N\otimes{\mathbb{R}}\)
implies
\(M\otimes{\mathbb{Q}}{ { \, \xrightarrow{\sim}\, }}N\otimes{\mathbb{Q}}\)
Now we claim that if \(G \in {\mathsf{Grp}}\) and
\(V_1, V_2\in {\mathsf{G}{\hbox{-}}\mathsf{Mod}}\) over a field \(k\)
and \(L_{/ {k}}\) is any extension, then
\(V_1\otimes L { { \, \xrightarrow{\sim}\, }}V_2\otimes L\) implies
\(V_1 { { \, \xrightarrow{\sim}\, }}V_2\).

\end{proof}

\begin{proof}[of claim]

1: Use that \(\mathop{\mathrm{Hom}}\) commutes with tensor products in
the following way:
\begin{align*}
\mathop{\mathrm{Hom}}_G(V_1\otimes L, V_2\otimes L) = \mathop{\mathrm{Hom}}_G(V_1, V_2)\otimes L
.\end{align*}
We can write the LHS as \((V_1 {}^{ \vee }\otimes V_2 \otimes_k L)^G\),
and the right-hand side as
\((V_1 {}^{ \vee }\otimes V_2)^G \otimes_k L\). It's enough to show that
for any \(G{\hbox{-}}\)representation \(V\), since
\(V^G \otimes L \cong (V\otimes L)^G\) where
\(V^G \coloneqq\ker( V{ {}^{ \scriptscriptstyle\oplus^{q} } } \xrightarrow{q} \bigoplus _{g\in G} V )\).
But now we're done since \(L_{/ {k}}\) is flat.

2: If both \(V_i\) are irreducible over \(L\), this follows from Schur.
For \(V_i\) irreducible over \(k\) an infinite field, then being an
isomorphism is a Zariski open condition, and any Zariski open subset of
\({\mathbb{A}}^{n}_{/ {k}}\) has infinitely many rational points.

\end{proof}

\begin{theorem}[?]

If \(L_{/ {k}}\) is cyclic and \(S\) is a set of primes of \(K\)
including all infinite primes, all primes that ramify, and all primues
under a set of generators of the class group of \(L\), letting \(T\) be
the set of primes of \(L\) over \(S\), we have

\begin{itemize}
\tightlist
\item
  \(q({\mathbb{I}}_{L, T}) = \prod_{v\in S} [{ L_{\widehat{v}} }: { k_{\widehat{v}} }]\)
\item
  \([L: k] q({\mathcal{O}}_{L, T}^{\times}) = \prod_{v} [{ L_{\widehat{v}} }: { k_{\widehat{v}} }]\)
\item
  \(q(C_L) = [L : k]\)
\end{itemize}

\end{theorem}

\begin{proof}[1 and 2 imply 3]

There is a SES
\begin{align*}
0 \to {\mathcal{O}}_{L, T}^{\times}\to {\mathbb{I}}_{L, T} \to C_L \to 1
,\end{align*}
where \({\mathcal{O}}_{L, T}^{\times}\) allows denominators in \(T\).
Then using (1) and (2),
\begin{align*}
q(C_L) = q({\mathbb{I}}_{L, T}) / q({\mathcal{O}}_{L, T}^{\times}) = [L: k]
.\end{align*}

\end{proof}

\begin{proof}[of 1]

Write
\({\mathbb{I}}_{L, T} = \prod_{v\in T} { L_{\widehat{v}} }^{\times}\times \prod_{v \not\in T} {\mathcal{O}}_{{ L_{\widehat{v}} }}^{\times}\),
so
\begin{align*}
q({\mathbb{I}}_{L, T})
&= \prod_{v\in S} q( \prod_{w\in  \mathrm{Pl}\qty{/L}  } L_w^{\times}) \\
&= \prod_{v\in S} { \# H^2({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} ; { L_{\widehat{v}} }^{\times}) \over \# H^2({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} ; { L_{\widehat{v}} }^{\times}) }\\
&= \prod_{v\in S} \# \mathop{\mathrm{Br}}({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} )\\
&= \prod_{v\in S} [{ L_{\widehat{v}} }: { k_{\widehat{v}} }]
.\end{align*}

\end{proof}

\begin{proof}[of 2]

Write
\(L_1 \coloneqq\mathop{\mathrm{Hom}}_{\mathsf{Set}}(T, {\mathbb{Z}})\)
and
\begin{align*}
L_2 \coloneqq m(\lambda: {\mathcal{O}}_{L, T}^{\times}&\to L_1\otimes{\mathbb{R}})
\alpha &\mapsto ( \log{\left\lvert { \alpha} \right\rvert}_w )_{w\in T} 
.\end{align*}
Dirichlet's unit theorem implies
\(L_2 \hookrightarrow L_1^0\otimes{\mathbb{R}}\coloneqq\left\{{\mathbf{x} {~\mathrel{\Big\vert}~}\sum x_i = 0}\right\}\)
is a lattice. We can write
\begin{align*}
L_1 
&= \bigoplus _{v\in S} \bigoplus _{w_{/ {v}} } {\mathbb{Z}}\\
&= \bigoplus _{v\in S} \operatorname{Ind}_{{ \mathsf{Gal}} ({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} )}^{ { \mathsf{Gal}} (L_{/ {k}} )} {\mathbb{Z}}
,\end{align*}
Thus
\begin{align*}
q(L_1)
&=
\prod _{v\in S} q\qty{ \operatorname{Ind}_{{ \mathsf{Gal}} ({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} )}^{ { \mathsf{Gal}} (L_{/ {k}} )} {\mathbb{Z}}} \\
&=
\prod_{v\in S} q({ L_{\widehat{v}} }_{/ {{ k_{\widehat{v}} }}} , {\mathbb{Z}}) \\
&= \prod_{v\in S} [{ L_{\widehat{v}} }: { k_{\widehat{v}} }]
.\end{align*}

To compute the other side, use that there is a SES
\(0\to L_1^0 \to L_1 \xrightarrow{\Sigma} {\mathbb{Z}}\to 0\). So
\begin{align*}
q(L_1^0) = q(L_1) / q(L_{/ {k}} ; {\mathbb{Z}}) = {\prod [{ L_{\widehat{v}} }: { k_{\widehat{v}} }] \over [L: k] } 
.\end{align*}
Now note \(q(L_k) = q({\mathcal{O}}_{L, T}^{\times})\) and there is a
SES
\begin{align*}
0 \to \mu(L) \to {\mathcal{O}}_{L, T}^{\times}\to L_k \to 0 \implies q({\mathcal{O}}_{L, T}^{\times}) = q(L_k)
,\end{align*}
where \(\mu(L)\) are the roots of unity in \(L\), which form a finite
group. Then
\begin{align*}
q({\mathcal{O}}_{L, T}^{\times}) = q(L_1^0) = {\prod [{ L_{\widehat{v}} }: { k_{\widehat{v}} }] \over [L:k] }
.\end{align*}

\end{proof}

\begin{fact}[from class field theory]

\begin{align*}
\#\qty{{\mathbb{I}}_K \over  k^{\times}\operatorname{Nm}_{L_{/ {k}} } {\mathbb{I}}_L } = [L: k]
.\end{align*}
How to prove: reduce to Kummer extensions, adjoin \(p\)th roots of
unity, etc.

\end{fact}

\begin{remark}

This fact implies \(H^1(L_{/ {k}} ; C_L) = 1\). The proof is that
\(\#\qty{H^2/H^1} = [L: k]\), which implies \(\# H^1 = 1\).

\end{remark}

\begin{theorem}[?]

Severi-Brauer varieties over \(k\) satisfy the Hasse principle, i.e.~the
following sequence is exact:
\begin{align*}
0 \to \mathop{\mathrm{Br}}(k) \to \bigoplus _{v\in  \mathrm{Pl}\qty{k} } \mathop{\mathrm{Br}}({ k_{\widehat{v}} })
.\end{align*}

\end{theorem}

\hypertarget{proof}{%
\subsection{Proof}\label{proof}}

\begin{theorem}[Hasse-Minkowski]

Let \(q\) be a quadratic form over a number field \(k\), then the
projective quadric
\(X\coloneqq\left\{{q=0}\right\} \subseteq {\mathbb{P}}^n_{/ {k}}\)
satisfies the Hasse principle: \(X\) has rational points over \(k\) iff
\(X\) has rational points over \({ k_{\widehat{v}} }\) for all
\(v\in \mathrm{Pl}\qty{k}\).

\end{theorem}

\begin{definition}[Quadratic forms representing elements]

Given \(q\) a quadratic form over \(k\) a field (e.g.~a number field or
a local field), then for \(a\in k\), we say \(q\) \textbf{represents}
\(a\) if there exist elements
\(\mathbf{x}\in k^n\setminus\left\{{0}\right\}\) such that
\(q(\mathbf{x}) = 0\).

\end{definition}

\begin{theorem}[a stronger one]

Given \(a\in k\), \(q\) represents \(a\) iff over \(k\) iff \(q\)
represents \(a\) over \({ k_{\widehat{v}} }\) for all
\(v\in \mathrm{Pl}\qty{k}\). Moreover, rational points are Zariski dense
on \(q(x) = a\).

\end{theorem}

\begin{remark}

That this implies the first theorem is easy, setting \(a=0\).
Conversely, consider
\(q'(\mathbf{x}, z) \coloneqq q(\mathbf{x}) - az^2\). Then \(q\)
represents \(a\) iff \(q'\) represents 0 -- however, this can go wrong
if \(z=0\)! Exercise: find a good proof.

\end{remark}

\begin{proof}[?]

Let \(n\) be the number of variables.

\begin{itemize}
\tightlist
\item
  For \(n=1\), we saw that \(x^2=a\) satisfies the Hasse principle in
  the first class. Moreover rational points are Zariski dense on the
  projective variety \(x^2 = ay^2\).
\item
  For \(n=2\), consider \(q(x_1, x_2) = a\). We'll pick this up next
  time!
\end{itemize}

\end{proof}

\hypertarget{tuesday-october-05}{%
\section{Tuesday, October 05}\label{tuesday-october-05}}

\begin{remark}

Goal: prove the Hasse-Minkowski theorem. We looked at \(n\leq 3\), so
today we'll look at \(n=4\).

\end{remark}

\begin{theorem}[?]

Let \(Q \subseteq {\mathbb{P}}^3_{/ {k}}\) be a smooth quadric, then
\begin{align*}
Q(k) \neq \emptyset \iff Q({ k_{\widehat{v}} }) \neq \emptyset
\end{align*}
for all places \(v\in \mathrm{Pl}\qty{k}\).

\end{theorem}

\begin{proof}[$n=4$ case]

Let \(X \subseteq {\operatorname{Gr}}_1({\mathbb{P}}^3)\) be the variety
of lines in \(Q\), and consider \({\mathcal{I}}\to X\) the universal
family. Then
\(X_{ { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } } = {\mathbb{P}}^1 \cup{\mathbb{P}}^1\)
since
\(Q_{ { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } } = {\mathbb{P}}^1 \times {\mathbb{P}}^1 \xhookrightarrow{{\mathcal{O}}(1, 1)} {\mathbb{P}}^3\).
Consider the case when \(X\) is not connected. Then

\begin{itemize}
\tightlist
\item
  \({\mathcal{U}}_c \to Q\) is an isomorphism, which can be checked over
  \({ \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\).
\item
  \(Q({ k_{\widehat{v}} }) \neq \emptyset\) implies
  \({\mathcal{U}}_c({ k_{\widehat{v}} }) \neq \emptyset\) and thus
  \(C({ k_{\widehat{v}} }) \neq \emptyset\) for all \(v\).
\item
  By the Hasse principle for Severi-Brauers, if \(C={\mathbb{P}}^1\)
  implies \(C(k) \neq \emptyset\).
\item
  Then \({\mathcal{U}}_c \to C\) is Zariski trivial, so
  \({\mathcal{U}}_c \cong Q\) has rational points.
\end{itemize}

\begin{center}
\begin{tikzcd}
    {{\mathcal{U}}_c} && {\mathcal{U}}&& Q \\
    \\
    C && X
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
    \arrow[hook, from=3-1, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-3, to=1-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJcXG1jdV9jIl0sWzAsMiwiQyJdLFsyLDIsIlgiXSxbMiwwLCJcXG1jdSJdLFs0LDAsIlEiXSxbMCwzXSxbMywyXSxbMSwyLCIiLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJob29rIiwic2lkZSI6InRvcCJ9fX1dLFswLDFdLFszLDRdXQ==}{Link
to Diagram}
\end{quote}

Now consider the case when \(X\) is connected. The claim is that there
exists a quadratic extensions \(k'/k\) where \(X_{k'}\) is not
connected:

\begin{itemize}
\item
  \(k' = {{\Gamma}\qty{X; {\mathcal{O}}_X} }\), which is a rank 2 vector
  bundle (which can be checked over
  \({ \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu }\)).
  So
  \begin{align*}
  {{\Gamma}\qty{X_{k'}; {\mathcal{O}}_{X_{k'}}} } = {{\Gamma}\qty{X; {\mathcal{O}}_X} }_{k'} = k' \otimes_k k' \cong k' \oplus k'
  ,\end{align*}
  so \(X_{k'}\) is disconnected.
\item
  We can take
  \([Q] \in H^1(k; {\operatorname{O}}_n) \xrightarrow{\operatorname{det}} H^1(k; \mu_2)\),
  and it maps to \([k']\).
\item
  \({ \mathsf{Gal}} ( { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } _{/ {k}} )\)
  acts on
  \({\operatorname{Pic}}(Q) \cong {\mathbb{Z}}{ {}^{ \scriptscriptstyle\times^{2} } }\),
  and this action factors through \(\left\{{\pm 1}\right\}\). Here
  \({\operatorname{O}}_n = \mathop{\mathrm{Aut}}(\sum x_i^2)\).

  \begin{itemize}
  \tightlist
  \item
    Why: this action preserves the \emph{effective cone} in
    \({\operatorname{Pic}}(Q_{ { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } })\)
    spanned by \(\pi_1^* {\mathcal{O}}(1)\) and
    \(\pi_2^* {\mathcal{O}}(1)\), which are those bundles with global
    sections (which is preserved by Galois).
  \end{itemize}
\end{itemize}

\begin{warnings}

Even if \([L] \in {\operatorname{Pic}}Q\) is Galois-invariant, this does
not imply that \(L\) is defined over \(k\)! This can be a common source
of errors.

\end{warnings}

By case 1, \(X(k')\neq \emptyset\), so there exists a rational line
\(L \subseteq Q\) contained in one connected component of \(X_{k'}\).
There is an action \({ \mathsf{Gal}} (k'_{/ {k}} )\curvearrowright L\),
so take \(\sigma \in { \mathsf{Gal}} (k'_{/ {k}} )\). Then \(\sigma(L)\)
is in the other component \(X_{k'}\), since Galois interchanges its
components pointwise. Then considering the two rulings of the quadric
yields the following picture, where they intersect at a point:

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/RationalPoints/sections/figures}{2021-10-05_10-08.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

But then \(\sigma(p) = p\) is Galois fixed, and is thus a
\(k{\hbox{-}}\)rational point.

\end{proof}

\begin{theorem}[?]

Let \(n\geq 5\) and \(Q = \sum_{l\leq i \leq n} a_i x_i^2\) be a
nondegenerate quadratic form over a number field \(k\). Then \(Q\)
satisfies the Hasse principle.

\end{theorem}

\begin{proof}[General case]

We'll proceed by induction on \(n\geq 5\). Write
\(Q = a_1 x_1^2 + a_2 x_2^2 + G(x_3,\cdots, x_n)\).

\begin{claim}

\(G\) represents \({ k_{\widehat{v}} }\) for almost all
\(v\in \mathrm{Pl}\qty{k}\).

\end{claim}

It's enough to show that
\(G'(x_3,\cdots, x_{n+1}) \coloneqq G(x_3, \cdots, x_n) + ax_{n+1}^2\)
represents 0 for all \(a\in { k_{\widehat{v}} }\) and almost all \(v\).
Without loss of generality, we can assume \(G\) is nondegenerate over
the residue field \(\kappa(v)\), by throwing out finitely many things.
Then \(G'\) has rank at least \(n-2\) over \(\kappa(v)\) for almost all
\(v\).

\begin{claim}

\(G'\) has a smooth rational point for for all

\end{claim}

Using the Lang-Weil estimates (using absolute irreducibility),
\(G'(\kappa(v))\) has about \((\# \kappa(v))^{n-3}\) rational points,
where the error term is uniform in \(v\). The singular locus is a
dimension smaller, so about \((\# \kappa(v))^{n-4}\), and for \(n\)
large enough for this to hold, the former is larger.

Now use Hensel's lemma, any smooth rational point on the special fiber
lifts to the generic fiber (i.e.~the infinitesimal smoothness criteria).
This proves the first claim that \(G\) represents
\({ k_{\widehat{v}} }\) for almost all \(v\).

\begin{fact}

For almost all \(v\), \(G(x_3,\cdots, x_n)\) represents \emph{every}
element of \(k\).

\end{fact}

Let
\(U \subseteq (\prod { k_{\widehat{v}} }) { {}^{ \scriptscriptstyle\times^{n-2} } } k[v]\)
be the set \(\left\{{ (x_3, v), \cdots, (x_n, v)}\right\}\) such that
there exists an \((x_1, v), (x_2, v)\) with \(Q(x_1, \cdots, x_n) = 0\).
Some claims:

\begin{itemize}
\tightlist
\item
  \(U\) is open, which follows from the fact above,
\item
  \(U\) is nonempty since \(Q\) represents 0 locally by hypothesis,
\item
  The set
  \(U' \subseteq (\prod { k_{\widehat{v}} }){ {}^{ \scriptscriptstyle\times^{2} } }\)
  of pairs \((x_1, v), (x_2, v)\) such that there exist
  \((x_3,v),\cdots,(x_n, v)\) with \(Q(x_1,\cdots, x_n)=0\) is also
  open.
\end{itemize}

Then by weak approximation, there exist \(x_1, x_2\in k\) such that
\((x_1, x_2)\in U\). So write \(c = a_1 x_1^2 + a_2 x_2^2\in k\) and
define \(Q'(z, x_3,\cdots, x_n) = -cz^2 + G(x_3,\cdots, x_n)\). This is
a quadratic form in \(n-1\) variables that represents 0 locally. Now by
induction, \(Q'\) represents zero globally.

\end{proof}

\hypertarget{tuesday-october-12}{%
\section{Tuesday, October 12}\label{tuesday-october-12}}

\begin{quote}
Reference: FGA Explained.
\end{quote}

\begin{proposition}[?]

For \(R\) a complete local ring with residue field \(\kappa\), there is
an isomorphism
\(\mathop{\mathrm{Br}}(R) { { \, \xrightarrow{\sim}\, }}\mathop{\mathrm{Br}}(\kappa)\).

\end{proposition}

\begin{remark}

We'll prove a stronger claim that there is a bijection
\(\mathsf{SB}{\mathsf{Sch}}_{/ {R}} /\sim \to \mathsf{SB}{\mathsf{Sch}}_{/ {k}} /\sim\),
which requires some deformation theory. A summary of obstruction theory
for schemes:

Let \(A \in \mathsf{CRing}\), \(I{~\trianglelefteq~}A\) is square zero
ideal, and \(X_{/ {A/I}}\) a smooth scheme. Then there exists a
functorial class
\(\operatorname{obs} (X) \in H^2(X; {\mathbf{T}}_X \otimes_{A/I} I)\)
such that \(X\) admits a flat lift to \(A\) iff
\(\operatorname{obs} (X) = 0\). If the obstruction vanishes, the set of
lifts is a torsor for \(H^1\), and the automorphisms of the lift are
given by \(H^0\). Here \({\mathbf{T}}_X\) is the tangent sheaf, and a
\emph{flat lift} is a flat scheme \(\tilde X_{/ {A}}\) equipped with an
isomorphism \(\tilde X\otimes(A/I) { { \, \xrightarrow{\sim}\, }}X\).

A word on this deformation-theoretic result is proved:

\begin{itemize}
\item
  Show affine schemes lift, e.g.~using Cohen structure theorem.
  Alternatively, something about being étale?
\item
  Try to glue, which may not satisfy the cocycle condition -- failure to
  glue will show up in this cohomology. Why the tangent sheaf: the
  difference between two gluing data is a derivation.
\end{itemize}

Note that for vector bundles \(E\to X\), the cohomology would be in
\(\mathop{\mathrm{End}}(E)\).

\begin{quote}
See also tangent/cotangent complex.
\end{quote}

\end{remark}

\begin{proof}[?]

We'll try to lift a Severi-Brauer over \(k\) to one over \(R\). Claim:
letting \(R_n \coloneqq R/{\mathfrak{m}}^n\), given a lift to \(R_n\),
there exists a unique lift to \(S_n \coloneqq R_{n+1}\). We have
\begin{align*}
\operatorname{obs} (S_n) \in H^2(S^n; {\mathbf{T}}_{S_n} \otimes{\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}) = H^2(S_n; {\mathbf{T}}_{S_n}) \otimes_k {\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}
,\end{align*}
which follows from base change in cohomology using
\({\mathbf{T}}_{S_n} \otimes_{R_n} k \otimes_k {\mathfrak{m}}^n/{\mathfrak{m}}^{n+1}\).
Here \(\operatorname{obs} (S_n) = 0\), since
\begin{align*}
H^2(S; {\mathbf{T}}_S) \otimes_k  { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } = H^2(S_{ { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } }; {\mathbf{T}}_{S,  { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } }) = H^2({\mathbb{P}}^n_{/ { { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } }} ; {\mathbf{T}}_{{\mathbb{P}}^n_{/ {k}} }) = 0
.\end{align*}

See Hartshorne, this uses the Euler exact sequence.

So a lift exists for each \(R_n\).

This lift is unique since lifts are torsors for
\(H^1(S_n; {\mathbf{T}}_{S_n} \otimes{\mathfrak{m}}^n/{\mathfrak{m}}^{n+1})\).

Why this lifts to \(R\): formal GAGA, which gives a way of going from
formal schemes to actual schemes. See ``FGA Explained'', Ch. 8. This is
because giving a scheme over \(R^n\) for all \(n\) amounts to giving a
formal scheme, since the underlying topological spaces are the same. The
input is an ample line bundle: here for \({\mathbb{P}}^n\) we can take
the dual of the dualizing sheaf \({\mathcal{O}}_{S_n} {}^{ \vee }\).

\end{proof}

\begin{remark}

Formal GAGA: one of the most useful techniques!

\end{remark}

\begin{proposition}[?]

Suppose \(X\in {\mathsf{Var}}_{/ {k}}\) and let
\(A\in \mathop{\mathrm{Br}}(X)\) (e.g.~represented by an Azumaya
algebra), then

\begin{itemize}
\item
  If \(k\) is a \(p{\hbox{-}}\)adic field, then there is a map
  \begin{align*}
  X(k) &\to \mathop{\mathrm{Br}}(X) \\
  x &\mapsto x^*(A)
  .\end{align*}
\item
  For \(k={\mathbb{R}}\), the map
  \(X({\mathbb{R}}) \to \mathop{\mathrm{Br}}(k) = {1\over 2}{\mathbb{Z}}/{\mathbb{Z}}\)
  is locally constant, i.e.~constant on connected components.
\end{itemize}

\end{proposition}

\begin{proof}[?]

For \(x\in X(k)\), \(\widehat{{\mathcal{O}}_{X, x}}\) is a complete
local \(k{\hbox{-}}\)algebra with residue field \(k\). Then for
\(A\in \mathop{\mathrm{Br}}(X)\), we have a map
\(\psi: { \left.{{A}} \right|_{{\widehat{ {\mathcal{O}}_{X, x} } }} } { { \, \xrightarrow{\sim}\, }}(A_x)\otimes_k \widehat{{\mathcal{O}}_{X, x}}\).
We want to spread \(\psi\) out to a \(p{\hbox{-}}\)adic neighborhood of
\(x\). In the analytic setting, this can be done using \textbf{Artin
approximation}, which will imply there exists an étale neighborhood
\(U\) of \(x\) and a map
\begin{align*}
U &\to X \\
y &\mapsto x
,\end{align*}
which extends (?) and induces an isomorphism on complete local rings.
Now applying the implicit function theorem, there exists a
\(p{\hbox{-}}\)adic neighborhood of \(x\) in any \(U(k)\).

\end{proof}

\begin{corollary}[?]

Let \(X_{/ {k}}\) for \(k\) a number field and
\(A\in \mathop{\mathrm{Br}}(X)\). Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  The following map on adeles is locally constant:
  \begin{align*}
  A^*: X({\mathbb{A}}_k) &\to {\mathbb{Q}}/{\mathbb{Z}}\\
  x &\mapsto \sum_{v\in  \mathrm{Pl}\qty{k} } m_{v_x}(x^* A)
  .\end{align*}
\item
  \(X({\mathbb{A}})^{{\mathbb{A}}} \coloneqq(A^*)^{-1}(0)\) is closed
  and open.
\item
  \(X({\mathbb{A}})^{\mathop{\mathrm{Br}}} = \displaystyle\bigcap_{A\in \mathop{\mathrm{Br}}(X)} X({\mathbb{A}})^{{\mathbb{A}}}\)
  is closed.
\item
  \(\mkern 1.5mu\overline{\mkern-1.5muX(k)\mkern-1.5mu}\mkern 1.5mu \subseteq X({\mathbb{A}})^{\mathop{\mathrm{Br}}}\).
\item
  If \(X\) is proper, then
  \(X({\mathbb{A}})^{\mathop{\mathrm{Br}}} \neq X({\mathbb{A}})\) and
  weak approximation does not hold.
\end{enumerate}

\end{corollary}

\begin{proof}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  Use the same Lang-Weil argument used previously, and that this is a
  sum of locally constant maps.
\item
  0 is closed and open in \({\mathbb{Q}}/{\mathbb{Z}}\) and \(A^*\) is
  continuous.
\item
  This is an intersection of closed sets.
\item
  We already know \(X(k)\) is contained in the RHS, and by (c) we know
  it's closed, so the RHS contains its closure.
\item
  Immediate from (d).
\end{enumerate}

\end{proof}

\begin{warnings}

The adelic topology is not the product topology.

\end{warnings}

\begin{definition}[Symbol Algebra]

For \(k\in \mathsf{Field}\) and let
\(\chi: { \mathsf{Gal}} ( { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } _{/ {k}} )\to C_n\)
and \(a\in k^{\times}/ (k^{\times})^n\), then recall that
\((\chi, a) \coloneqq L_\chi\left\langle{x}\right\rangle_{\sigma}/\left\langle{x^n-a}\right\rangle\)
where \(L_\chi\) is the fixed field of \(\chi\) and
\(L_\chi\left\langle{x}\right\rangle_\sigma\) is the twisted polynomial
ring where \(\ell x = x\sigma(\ell)\).

\end{definition}

\begin{example}[?]

Take a smooth proper model of
\(U = \left\{{y^2 + z^2 = (3-x^2)(x^2-2)}\right\}\) and the \emph{symbol
algebra} \(A = (3-x^2, -1)\).

\begin{exercise}[Homework]

Check that this has points locally!

\end{exercise}

Our goal is to show that \(X({\mathbb{A}})^A = \emptyset\). By Kummer
theory, choosing an isomorphism \(\mu_n(k) \to C_n\) induces a bijection
\begin{align*}
k^{\times}/(k^{\times})^n &{ { \, \xrightarrow{\sim}\, }}\left\{{\chi: { \mathsf{Gal}} ( { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } _{/ {k}} ) \to C_n}\right\} \\
a &\mapsto k[x] / \left\langle{x^n-a}\right\rangle
.\end{align*}

For \(n=2\) and \(\operatorname{ch}k \neq 2\), there is a canonical
isomorphism
\(\left\{{\pm 1}\right\} { { \, \xrightarrow{\sim}\, }}\mu_2(k) { { \, \xrightarrow{\sim}\, }}C_2\).
View \((\chi, a) \in H^2(k, \mu_n)\), and there is a cup product
\begin{align*}
H^1(k; C_n) \times H^1(k; \mu_n) &\to H^2(l; \mu_n) \\
\chi &\mapsto [\chi] \smile[a]
.\end{align*}

Another point of view: if \(L_{/ {k}}\) is Galois with Galois group
\(C_n\), it comes with a choice of generator \(\sigma\) and thus a
canonical element in
\([ \sigma] \in H^2(L_{/ {k}} ; {\mathbb{Z}}) { { \, \xrightarrow{\sim}\, }}C_n\).
Then there is another cup product
\begin{align*}
k^{\times}= H^0(L_{/ {k}} ; L^{\times}) \xrightarrow{({-}) \smile[\sigma]} H^2(L_{/ {k}} ; L^{\times}) = \mathop{\mathrm{Br}}(L_{/ {k}} ) = k^{\times}/\operatorname{Nm}_{L_{/ {k}} } k^{\times}
,\end{align*}
in which case \((\chi, a) = a \smile[\sigma]\).

\end{example}

\begin{corollary}[?]

\((\chi, a) =0 \iff a \in \operatorname{Nm}_{L_{/ {k}} } L^{\times}\).

\end{corollary}

\begin{remark}

For \(n=2\), one has
\((a, b) = k { \left[ {\sqrt{b}} \right] } \left\langle{x}\right\rangle_\sigma / \left\langle{x^n-a}\right\rangle\),
and this splits iff \(a\) is a norm from \(k(\sqrt{b})\) when this is a
field.

\end{remark}

\begin{exercise}[?]

What are the equations for the Severi-Brauer arising from \((a, b)\).

\end{exercise}

\hypertarget{tuesday-october-19}{%
\section{Tuesday, October 19}\label{tuesday-october-19}}

\todo[inline]{Missing some stuff! Find notes.}

\begin{remark}

Let \(k\) be a number field, \(X\in {\mathsf{Var}}_{/ {k}}\) a variety,
and \({\mathbb{A}}_{/ {k}}\) the adeles over \(k\). Let
\(F: {\mathsf{Sch}}^{\operatorname{op}}\to {\mathsf{Set}}\) be a
functor, we can then consider an \textbf{\(F{\hbox{-}}\)obstruction to
rational points}. A rational point is the data of a morphism
\(\operatorname{Spec}k\to X\), denoted
\(X(k) = {\mathsf{Sch}}(\operatorname{Spec}k, X)\), and adelic points
\(X({\mathbb{A}}_{/ {k}} ) = {\mathsf{Sch}}({\mathbb{A}}_{/ {k}} , X)\).
Since there is a morphism
\(k \to {\mathbb{A}}_{/ {k}} \in \mathsf{Ring}\), this yields a morphism
\(X(k)\to X({\mathbb{A}}_{/ {k}} )\) (noting contravariance). Note that
an adelic point is a lift:

\begin{center}
\begin{tikzcd}
    && {\operatorname{Spec}{\mathbb{A}}_{/ {k}} } \\
    \\
    X && {\operatorname{Spec}k}
    \arrow[from=3-3, to=3-1]
    \arrow[from=1-3, to=3-3]
    \arrow[dashed, from=1-3, to=3-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsMyxbMCwyLCJYIl0sWzIsMiwiXFxzcGVjIGsiXSxbMiwwLCJcXHNwZWMgXFxBQVxcc2xpY2UgayJdLFsxLDBdLFsyLDFdLFsyLDAsIiIsMix7InN0eWxlIjp7ImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

Moreover, \(F\) induces a diagram. Let \(i: \operatorname{Spec}k \to X\)
denote the inclusion, and \(x\in X(k)\) be a \(k{\hbox{-}}\)point. Then
writing \(F(k) \coloneqq F(\operatorname{Spec}k)\), we have the
following:

\begin{center}
\begin{tikzcd}
    {X(k)} && {X({\mathbb{A}}_{/ {k}} )} \\
    \\
    {F(k)} && {F({\mathbb{A}}_{/ {k}} )}
    \arrow[from=1-1, to=1-3]
    \arrow[from=3-1, to=3-3]
    \arrow["{x^* A \coloneqq Fi(a)}"', from=1-1, to=3-1]
    \arrow["{x_v^* A}"', from=1-3, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJYKGspIl0sWzAsMiwiRihrKSJdLFsyLDAsIlgoXFxBQVxcc2xpY2UgaykiXSxbMiwyLCJGKFxcQUFcXHNsaWNlIGspIl0sWzAsMl0sWzEsM10sWzAsMSwieF4qIEEgXFxkYSBGaShhKSIsMl0sWzIsMywieF92XiogQSIsMl1d}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{definition}[?]

Let \(A\in F(X)\), then
\begin{align*}
X({\mathbb{A}}_{/ {k}} )^A &\coloneqq\left\{{ (x_v) \in X({\mathbb{A}}_{/ {k}} ) {~\mathrel{\Big\vert}~}x_v^* A \in \operatorname{im}\qty{F(k) \to F({\mathbb{A}}_{/ {k}} )} }\right\} \\
X({\mathbb{A}}_{/ {k}} )^F &\coloneqq\displaystyle\bigcap_{A\in F(X)} X({\mathbb{A}}_{/ {k}} )^A
.\end{align*}

\end{definition}

\begin{example}[?]

For \(F({-}) \coloneqq\mathop{\mathrm{Br}}({-})\), the resulting
\(X({\mathbb{A}}_{/ {k}} )^F\) is the \textbf{Brauer-Manin obstruction}.
Consider \(A\in \mathop{\mathrm{Br}}(X)\) for
\(X = V(y^2 + z^2 = (x^2-2)(3-x^2))\), then any \(k{\hbox{-}}\)rational
point would yield \(x^* A\in \mathop{\mathrm{Br}}(k)\) and thus
\(x_v^* A \in \mathop{\mathrm{Br}}({\mathbb{A}}_{/ {k}} )\). The claim
is that \(\mathop{\mathrm{Br}}({\mathbb{A}}_{/ {k}} )^A = \emptyset\).
So consider
\(\left\{{ (x_v) \in \mathop{\mathrm{Br}}({\mathbb{A}}_{/ {k}} ) {~\mathrel{\Big\vert}~}x_v^* A \in \operatorname{im}\qty{\mathop{\mathrm{Br}}(k) \to \mathop{\mathrm{Br}}({\mathbb{A}}_{/ {k}} ) }}\right\}\).

\end{example}

\begin{exercise}[?]

Show this set is empty!

\end{exercise}

\hypertarget{descent-obstruction}{%
\subsection{Descent Obstruction}\label{descent-obstruction}}

\begin{definition}[fppf morphisms]

A morphism \(U\to X\in {\mathsf{Sch}}\) is \textbf{fppf} if

\begin{itemize}
\item
  \(f\) is flat\footnote{Flatness is a local condition:
    \(f_q: {\mathcal{O}}_{X, p}\to {\mathcal{O}}_(U, q)\) for
    \(q\in f^{-1}(p)\) should be a flat morphism of algebras, where
    \(A\to B\) is flat if \({-}\otimes_A B\) is flat as a functor on
    \({\mathsf{A}{\hbox{-}}\mathsf{Mod}}\).},
\item
  \(f\) is locally of finite presentation.\footnote{There exist affine
    open covers \({\mathcal{V}}\rightrightarrows X\) and
    \({\mathcal{U}}_i \rightrightarrows f^{-1}(V_i)\) such that
    \({ \left.{{f}} \right|_{{U_{ij}}} }: U_{ij} \to V_i\) is induced by
    \(B_i \to A_{ij} \in \mathsf{Ring}\) of finite presentation. So
    \(B\to A\in \mathsf{Ring}\) yields \(A = B[x_0, \cdots, x_n]/ I\)
    with \(I\) finitely generated.}
\end{itemize}

\end{definition}

\begin{remark}

Recall that \({\mathsf{Grp}}{\mathsf{Sch}}\leq {\mathsf{Sch}}\) is the
subcategory of group objects.

\end{remark}

\begin{definition}[?]

For \(G\in {\mathsf{Alg}}{\mathsf{Grp}}\) or
\({\mathsf{Grp}}{\mathsf{Sch}}_{/ {X}}\), and let
\begin{align*}
F({-}) = H^1({-}; G) = \left\{{G{\hbox{-}}\text{torsors over } X \text{ locally fppf trivial} }\right\}
.\end{align*}

\end{definition}

\begin{quote}
The fppf site is much more flexible than the etale site! Examples are
any morphisms \(X\to \operatorname{Hilb}({\mathbb{P}}^n)\).
\end{quote}

\begin{definition}[$G\dash$torsors]

A \textbf{\(G{\hbox{-}}\)torsor over \(X\)} is an object
\(T\in {\mathsf{Sch}}_{/ {X}}\) with a \(G{\hbox{-}}\)action
\(G\operatorname{fp}{X} T \xrightarrow{\sigma} T\) such that there is an
isomorphism
\begin{align*}
G\operatorname{fp}{X} T \underset{(\sigma, \pi_2)}{{ { \, \xrightarrow{\sim}\, }}} T\operatorname{fp}{X} T
.\end{align*}

\end{definition}

\begin{remark}

How to think about torsors: a family whose fibers are abstractly
isomorphic to \(G\), but not canonically.

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/RationalPoints/sections/figures}{2021-10-19_10-38.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

So the fibers are not canonically identified with \(G\), but \(G\) acts
naturally in a freely transitive way on them. This is essentially the
same data as a principal \(G{\hbox{-}}\)bundle.

\end{remark}

\begin{remark}

Let \(L\to X\) be a line bundle, so there exists a cover
\({\mathcal{U}}\rightrightarrows X\) with
\({ \left.{{L}} \right|_{{U_i}} } \cong {\mathbb{A}}^1\times U_i\) and
transition functions
\begin{align*}
{\mathbb{A}}^1 \times (U_i \times U_j) \xrightarrow{(t_{ij} , \operatorname{id})} {\mathbb{A}}^1 \times(U_i \times U_j)
\end{align*}
where \(t_{ij}: U_i \times U_j \to {\mathbb{G}}_m\). Then one can obtain
a \({\mathbb{G}}_m{\hbox{-}}\)torsor \(L\setminus\left\{{0}\right\}{0}\)
by removing the zero section, and in fact all such
\({\mathbb{G}}_m{\hbox{-}}\)torsors arise this way. One can check here
that
\({\mathbb{G}}_m(X)\operatorname{fp}{X} T { { \, \xrightarrow{\sim}\, }}T{ {}^{ \scriptscriptstyle{ \underset{\scriptscriptstyle {X} }{\times} }^{2} } }\).

\end{remark}

\begin{example}[of $\Grp\Sch\slice X$]

\envlist

\begin{itemize}
\item
  Take \(G\coloneqq{\mathbb{G}}_a, {\mathbb{G}}_m\) and consider
  \(G \times X\), then one can define a multiplication by multiplying
  the first factors. This generalizes to work for any
  \(G\in {\mathsf{Alg}}{\mathsf{Grp}}\).
\item
  An elliptic curve over \(\operatorname{Spec}{\mathbb{Z}}\setminus P\)
  for \(P\) a finite set of primes.
\item
  \({\mathbb{Z}}/n\) and \(\mu_n\).
\end{itemize}

\end{example}

\begin{remark}

Another formulation: \({\mathsf{Grp}}{\mathsf{Sch}}_{/ {X}}\) are
representable functors
\begin{align*}
F: {\mathsf{Sch}}_{/ {X}} ^{\operatorname{op}}&\to {\mathsf{Grp}}\\
Y &\mapsto {\mathsf{Sch}}_{/ {X}} (Y, G)
,\end{align*}
which composes with
\({\mathsf{Grp}}\xrightarrow{\mathop{\mathrm{Forget}}} {\mathsf{Set}}\)
in a certain way (?).

\end{remark}

\begin{remark}

Next time: descent obstructions.

\end{remark}

\hypertarget{descent-tuesday-october-26}{%
\section{Descent (Tuesday, October
26)}\label{descent-tuesday-october-26}}

\begin{quote}
Missed first few boards, find notes / see phone pics.
\end{quote}

\begin{remark}

If
\(G\in {\mathsf{sm}}{\mathsf{Aff}}{\mathsf{Grp}}{\mathsf{Sch}}_{/ {X}}\)
and \(T_{/ {X}}\) is a \href{./locally\%20fppf\%20trivial.html}{locally
fppf trivial} \(G{\hbox{-}}\)torsor, then \(T\) is
\href{./etale\%20locally\%20trivial.html}{etale locally trivial} by a
slicing argument.

Setup: let \(X\in {\mathsf{Var}}_{/ {k}}\) and
\(G\in{\mathsf{Grp}}{\mathsf{Sch}}_{/ {k}}\) with
\(T_{/ {X}} \in {\mathsf{G}{\hbox{-}}\mathsf{Torsors}}\). Partition
\(X(k)\) as
\begin{align*}
X(k) = \displaystyle\coprod_{\tau \in H^1(k; G)} \left\{{x\in X(k) {~\mathrel{\Big\vert}~}T_x \cong \tau }\right\}
.\end{align*}
Note that
\(H^1(k; G)\in {\mathsf{G}{\hbox{-}}\mathsf{Torsors}}_{/ {k}}\). Let
\(\tau\in H^1\) be such a torsor and \(G_\tau\) be the corresponding
\textbf{inner form}: note that \(G\) acts on itself by conjugation, so
inner forms are in the image of the induced map on \(H^1\):

\begin{center}
\begin{tikzcd}
    {\operatorname{im}\mathrm{conj}^*} && {\left\{{\text{forms of } G}\right\}} \\
    \\
    {H^1(k; G)} && { H^1(k; \mathop{\mathrm{Aut}}G)}
    \arrow["{\mathrm{conj}^*}", from=3-1, to=3-3]
    \arrow["{\mathrm{conj}^*}", from=3-1, to=1-1]
    \arrow[from=1-1, to=1-3]
    \arrow["\cong"', tail reversed, from=3-3, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwyLCJIXjEoazsgRykiXSxbMiwyLCJIXjEoazsgXFxBdXQgRykiXSxbMiwwLCJcXHRze1xcdGV4dHtmb3JtcyBvZiB9IEd9Il0sWzAsMCwiXFxpbSBcXGNvbmpeKiJdLFswLDEsIlxcY29ual4qIl0sWzAsMywiXFxjb25qXioiXSxbMywyXSxbMSwyLCJcXGNvbmciLDIseyJzdHlsZSI6eyJ0YWlsIjp7Im5hbWUiOiJhcnJvd2hlYWQifX19XV0=}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{exercise}[important: on what descent means and how to compute with it]

Prove the following claim: \(\tau\in H^1(k; G)\) is a left
\(G{\hbox{-}}\)torsor, and thus \(\tau\) is naturally a \emph{right}
\(G{\hbox{-}}\)torsor.

\end{exercise}

\begin{remark}

Let \(G\) be a discrete group and
\(T\in \mathsf{G}{\hbox{-}}\mathsf{Torsors}\), or equivalently a group
scheme over an algebraically closed field (in characteristic zero).
Write
\(T^r \coloneqq\left\{{T, T\times G\to T: tg\coloneqq g^{-1}t}\right\}\)
to exchange left and right actions, and apply this construction at the
level of points. Hint: you'll need to use \href{./descent.html}{descent}
to check this.

\end{remark}

\begin{remark}

On twisted \href{./torsors.html}{torsors} : define
\(T_\tau \coloneqq\leftquotient{G}{T{ \underset{\scriptscriptstyle {X} }{\times} }\tau}\),
and observe that \(T_\tau\) is a right \(G_\tau\) torsor via an action
on \(\tau\).

\end{remark}

\begin{proposition}[?]

There is an equality
\begin{align*}
\left\{{x\in X(k) {~\mathrel{\Big\vert}~}T_x \cong \tau}\right\} = \operatorname{im}\qty{T_\tau(k) \to X(k) }
.\end{align*}

\end{proposition}

\begin{proof}[?]

Note that
\(x\in \operatorname{im}\qty{T_\tau(k) \to X(k)} \iff { \left.{{T_\tau}} \right|_{{x}} }\)
is a trivial \(G_\tau{\hbox{-}}\)torsor, since having a rational point
implies being a trivial torsor since that point can be used to
translate. This happens iff
\(\leftquotient{G} T_x { \underset{\scriptscriptstyle {X} }{\times} } \tau\)
iff \(T_x\cong \tau\).

Conversely, left a rational point and check that you get the graph of an
isomorphisms (after base change).

\end{proof}

\begin{corollary}[?]

One can partition
\begin{align*}
X(k) = \displaystyle\coprod_{\tau \in H^1(k; G)} \operatorname{im}\qty{T_\tau(k) \to X(k)}
.\end{align*}

\end{corollary}

\begin{remark}

Everything up until now works in
\({\mathsf{Alg}}{\mathsf{Spaces}}\geq {\mathsf{Sch}}\), so it's a mostly
formal construction thus far.

\end{remark}

\begin{remark}

We showed the following proposition in the case of
\(\operatorname{PGL}_n\): assignments of points to Brauer classes were
locally constant.

\end{remark}

\begin{proposition}[Local constancy of evaluation]

Let \(k\in \mathsf{Local}\mathsf{Field}\) and
\(X\in {\mathsf{Sch}}_{/ {k}}\) proper. \footnote{Possibly not needed
  here, but included in Poonen's statement.}

For \(G\in {\mathsf{Grp}}{\mathsf{Sch}}_{/ {k}} ^\text{ét}\) and
\(T\in{\mathsf{G}{\hbox{-}}\mathsf{Torsors}}_{/ {X}}\), then the
following map is locally constant:
\begin{align*}
X(k) &\to H^1(k; G)\\
x &\to T_x
.\end{align*}

\end{proposition}

\begin{remark}

How did this proof go before? We showed something didn't deform, i.e.~an
argument in cohomology of the tangent bundle, and used Artin
approximation.

\end{remark}

\begin{proof}[?]

The point: étale sheaves don't deform,
i.e.~\(H^1({\mathbb{T}}_{T_{/ {X}} }) = 0\) for \({\mathbb{T}}\) the
relative cotangent bundle (i.e.~\(H^1\) with coefficients in the tangent
sheaf). See Poonen for a proof use Krasner's lemma. A word on this
proof: for a constant group scheme, this is a constant Galois cover and
the field extensions don't change under small perturbations (i.e.~of the
coefficients of the polynomial).

\end{proof}

\begin{corollary}[?]

The image \(\operatorname{im}\qty{X(k) \to H^1(k; G)}\) is finite.

\end{corollary}

\begin{proof}[?]

Use that the map is proper and \(X(k)\) is compact.

\end{proof}

\begin{remark}

\end{remark}

\hypertarget{selmer-sets}{%
\subsection{Selmer Sets}\label{selmer-sets}}

\begin{definition}[Selmer Sets]

Let \(k\in \mathsf{Number}\mathsf{Field}\),
\(X\in {\mathsf{Var}}_{/ {k}} , G\in {\mathsf{sm}}{\mathsf{Aff}}{\mathsf{Alg}}{\mathsf{Grp}}_{/ {k}} , T\in {\mathsf{G}{\hbox{-}}\mathsf{Torsors}}_{/ {X}}\).
Define
\begin{align*}
\operatorname{Sel}_T(k; G) \coloneqq\left\{{
\tau \in H^1(k; G) {~\mathrel{\Big\vert}~}
\tau_{{ k_{\widehat{v}} }} \in \operatorname{im}\qty{X({ k_{\widehat{v}} }) \to H^1({ k_{\widehat{v}} }; G) }
\text{for all }v\in  \mathrm{Pl}\qty{k} 
}\right\}
.\end{align*}

\end{definition}

\begin{example}[?]

If \(A_{/ {k}} \in {\mathsf{Ab}}{\mathsf{Var}}\) and
\(G \coloneqq A[n]\) with \(T: A \xrightarrow{[n]} A\), then
\(\operatorname{Sel}_T(k; G)\) is the
\href{./Selmer\%20group.html}{Selmer group} of \(A\).

\end{example}

\begin{remark}

Unpacking this definition, we can write this as
\begin{align*}
\operatorname{Sel}_T(k; G) = \left\{{\tau \in H^1 {~\mathrel{\Big\vert}~}T_\tau({ k_{\widehat{v}} }) \neq \emptyset}\right\} \supseteq \left\{{\tau {~\mathrel{\Big\vert}~}T_\tau(k) \neq \emptyset}\right\}
.\end{align*}
Then
\begin{align*}
X(k) = \displaystyle\coprod_{\tau\in \operatorname{Sel}_T(k; G)} \operatorname{im}\qty{T_\tau(k) \to X(k)}
.\end{align*}

\end{remark}

\hypertarget{thursday-october-28}{%
\section{Thursday, October 28}\label{thursday-october-28}}

\begin{remark}

Let
\(k\in\mathsf{Global}\mathsf{Field}, G\in {\mathsf{Grp}}{\mathsf{Sch}}_{/ {k}} , T\in {\mathsf{G}{\hbox{-}}\mathsf{Torsors}}_{/ {X}}\).
We defined the Selmer group as
\begin{align*}
\operatorname{Sel}_{T_{/ {X}} } (k; G) \coloneqq\left\{{ \tau \in H^1(k; G) {~\mathrel{\Big\vert}~}\tau_{{ k_{\widehat{v}} }} \in \operatorname{im}\qty{ X({ k_{\widehat{v}} }) \xrightarrow{x\to T_x} H^1({ k_{\widehat{v}} }; G)} }\right\} 
.\end{align*}

\end{remark}

\begin{theorem}[?]

For \(G\) smooth affine and \(X\) proper,
\(\operatorname{Sel}_{T_{/ {X}} }(k; G)\) is finite.

\end{theorem}

\begin{corollary}[?]

The following disjoint union is finite if \(X\) is proper:
\begin{align*}
X(k) = \displaystyle\coprod_{\operatorname{Sel}_{T_{/ {X}} } (k; G)} \left\{{x\in X({ k_{\widehat{v}} }) {~\mathrel{\Big\vert}~}\tau_x \cong \tau}\right\}
.\end{align*}

\end{corollary}

\begin{corollary}[Weak Mordell-Weil theorem]

For \(A\in {\mathsf{Ab}}{\mathsf{Var}}_{/ {k}}\), which is a smooth
proper group scheme, then \(A(k)/mA(k)\) is finite for any
\(m\in {\mathbb{Z}}_{\geq 0}\).

\end{corollary}

\begin{remark}

Note that Mordell-Weil is about finite generation of \(A(k)\), which
implies that the quotient is finite -- for counterexamples, take
\({\mathbb{Q}}\) and \({\mathbb{Q}}/m{\mathbb{Q}}= 0\). Existence of
\href{./variations\%20of\%20Hodge\%20structure.html}{variations of Hodge
structure}: follows from
\href{./nonabelian\%20Hodge\%20theory.html}{nonabelian Hodge theory}.

\end{remark}

\hypertarget{proof-of-finiteness-of-selmer-sets}{%
\subsection{Proof of finiteness of Selmer
Sets}\label{proof-of-finiteness-of-selmer-sets}}

\begin{remark}

Let \(T\) be defined as multiplication by \(n\) to get a SES
\begin{align*}
0 \to A[n] \to A \xrightarrow{x\mapsto nx} A \to 0
.\end{align*}

Then we get a diagram

\begin{center}
\begin{tikzcd}
    {A(k)} && {H^1(k; A[n])} \\
    \\
    {A(k) / nA(k)} && {\operatorname{Sel}_{T_{/ {X}} }(k; A[n])}
    \arrow["{x\mapsto T_x}", from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow[dashed, hook, from=3-1, to=1-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNCxbMCwwLCJBKGspIl0sWzIsMCwiSF4xKGs7IEFbbl0pIl0sWzAsMiwiQShrKSAvIG5BKGspIl0sWzIsMiwiXFxTZWxfe1RcXHNsaWNlIFh9KGs7IEFbbl0pIl0sWzAsMSwieFxcbWFwc3RvIFRfeCJdLFsxLDNdLFswLDJdLFsyLDNdLFsyLDEsIiIsMSx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn0sImJvZHkiOnsibmFtZSI6ImRhc2hlZCJ9fX1dXQ==}{Link
to Diagram}
\end{quote}

\end{remark}

\begin{remark}

We'll separately handle the ramified and unramified cases. We'll need to
set up some notation: let
\(k\in \mathsf{Global}\mathsf{Field}, S \subseteq \mathrm{Pl}\qty{(} k)\)
a finite subset of places,
\({\mathcal{G}}\in {\mathsf{Grp}}{\mathsf{Sch}}_{/ {{\mathcal{O}}_{k, S}}}\).
There is a SES
\begin{align*}
0 \to {\mathcal{G}}^0 \to {\mathcal{G}}\to {\mathcal{G}}/{\mathcal{F}}\to 0
,\end{align*}
where \({\mathcal{G}}_0\) is the component of the identity. Assume
\({\mathcal{F}}\) is finite étale, which can always be achieved by
enlarging \(S\) if necessary. We'll use Roman letters to denote fibers,
whence we get a SES over \(k\):
\begin{align*}
0 \to G^0 \to G\to F \to 0
.\end{align*}

\end{remark}

\begin{definition}[?]

Define the set of torsors unramified away from \(S\):

\begin{align*}
H^1_S(k; G) \coloneqq\left\{{ \tau\in H^1(k; G) {~\mathrel{\Big\vert}~}\tau_{{ k_{\widehat{v}} }} \in\operatorname{im}\qty{H^1({\mathcal{O}}_{{ k_{\widehat{v}} }};{\mathcal{G}}) \to H^1(k; {\mathcal{G}}) }\,\forall v\in S}\right\}
.\end{align*}
Note that this set depends on the model chosen for \({\mathcal{G}}\).

\end{definition}

\begin{theorem}[?]

The unramified case of Selmer finiteness

\begin{itemize}
\tightlist
\item
  \(H_S^1(k; G) \to H_S^1(k; G) \to \prod_{v\in S} H^1({ k_{\widehat{v}} }, F)\)
  has finite fibers.
\item
  If \(k\in \mathsf{Number}\mathsf{Field}\) then \(H^1_S(k; G)\) is
  finite.
\end{itemize}

\end{theorem}

\begin{remark}

This theorem is due to Lang, who was an AIDS denialist? \textbf{Yikes.}

\end{remark}

\begin{remark}

This \(H^1_S(k; G)\) is an approximation to something slightly more
natural, \(H^1_\text{ét}({\mathcal{O}}_{k, S}; {\mathcal{G}})\), the
\(G{\hbox{-}}\)torsors over \({\mathcal{O}}_{k, S}\).

\end{remark}

\begin{slogan}

Torsors for connected group schemes are not interesting?

\end{slogan}

\begin{theorem}[?]

Let
\(G\in {\mathsf{sm}}{\mathsf{Alg}}{\mathsf{Grp}}_{/ {{\mathbb{F}}_q}}\)
be connected, then
\(H^1({\mathbb{F}}_q; G) = \left\{{{\operatorname{pt}}}\right\}\).

\end{theorem}

\begin{exercise}[?]

Write down a variety over \({\mathbb{F}}_q\) with no rational points!

\end{exercise}

\begin{proof}[?]

Let \(T\in H^1({\mathbb{F}}_q; G)\), we want to show
\(T({\mathbb{F}}_q) \neq \emptyset\). Given a rational point, we want to
show \(G\cong T\). Take \(G\times T\to T\) and for any
\(x\in T({\mathbb{F}}_q)\) take the map
\begin{align*}
G &\to T \\
x&\mapsto gx
.\end{align*}

\begin{exercise}[?]

Prove this is an isomorphism. Hint: use the following diagram:

\begin{center}
\begin{tikzcd}
    G && T && {G\times T} && {T\times T} \\
    \\
    && X &&& T
    \arrow[from=1-1, to=1-3]
    \arrow[from=1-3, to=3-3]
    \arrow[from=1-5, to=3-6]
    \arrow[from=1-7, to=3-6]
    \arrow[from=3-3, to=3-6]
    \arrow[from=1-5, to=1-7]
    \arrow[from=1-1, to=3-3]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMCwwLCJHIl0sWzIsMCwiVCJdLFsyLDIsIlgiXSxbNCwwLCJHXFx0aW1lcyBUIl0sWzYsMCwiVFxcdGltZXMgVCJdLFs1LDIsIlQiXSxbMCwxXSxbMSwyXSxbMyw1XSxbNCw1XSxbMiw1XSxbMyw0XSxbMCwyXV0=}{Link
to Diagram}
\end{quote}

\end{exercise}

Let
\(t\in T(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\),
then any
\(t'\in T(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\)
satisfies \(t'=gt\) for a unique
\(g\in G(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\),
since this is a principal homogeneous space (so the action is free and
simply transitive). Consider Frobenius-fixed elements: take
\(\sigma\in{ \mathsf{Gal}} (\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu_{/ {{\mathbb{F}}_q}} )\),
then if \(t' = (t')^{ \sigma}\) then \(gt = g^{\sigma}t^{ \sigma}\) and
\((g^{ \sigma})^{-1}\tau= t^{ \sigma}\). Now \(t^{ \sigma} = bt\) for a
unique
\(b\in G(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\),
so to solve this equation it suffices to show that every
\(b \in G(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\)
can be written in the form \(b = (g^{ \sigma})^{-1}g\).

\begin{claim}

Consider the following funny action \(G\curvearrowright G\) by
\(g\cdot t \coloneqq(g^{ \sigma})^{-1}t g\), a twisted conjugation. This
acts transitively on
\(G(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\)
and thus has one orbit.

\end{claim}

Why this implies the theorem: take \(t=\operatorname{id}\), then any
\(b\in G(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\)
is in the orbit of \(t\) and \(b = (g^{\sigma})^{-1}g\).

\end{proof}

\begin{example}[?]

For \(G = {\mathbb{G}}_a\), this yields \(g\cdot t = tg - g^p\). For
\(G = {\mathbb{G}}_m\) it yields \(g\cdot t = t - g^{1-p}\).

\end{example}

\begin{observation}

For fixed
\(t \in G(\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu)\),
the following map is generically étale:
\begin{align*}
G_{\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu} &\to G_{\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu} \\
g &\mapsto (g^{ \sigma})^{-1}t g
,\end{align*}
and thus the image contains an open subset of
\(G_{\mkern 1.5mu\overline{\mkern-1.5mu\mathbb{F}_{q}\mkern-1.5mu}\mkern 1.5mu}\).
So any two orbits are open, and hence intersect since \(G\) is
connected. Since orbits are equal or disjoint, there is just one orbit.

Proof: continued next time!

\end{observation}

\begin{remark}

Some asides on philosophy: hyperbolicity should be related to having
rational points, and having ``big'' fundamental group and being general
type. A recent theorem by Ellenberg-Lawrence-Venketesh: for
\(X\in {\mathsf{Var}}_{/ {k}}\) for
\(k\in \mathsf{Number}\mathsf{Field}\) and \(\pi_1 X\) is big, then
\({\sharp}X({\mathcal{O}}_{k, S})\) has height bounded above by \(H\)
which grows like \(H^{{\varepsilon}}\) for any \({\varepsilon}\).

\begin{quote}
See paper: \url{https://arxiv.org/abs/2109.01043}
\end{quote}

Here ``big \(\pi_1\)'' means that for \(X_{\mathbb{C}}\), there is a
variation of Hodge structure on \(X\) such that the map
\(X\to {\mathcal{D}}/\Gamma\) into the period domain is quasi-finite.

\end{remark}

\hypertarget{tuesday-november-02}{%
\section{Tuesday, November 02}\label{tuesday-november-02}}

\begin{remark}

Setup: let \(k\in \mathsf{Global}\mathsf{Field}\),
\(S \subseteq \mathrm{Pl}\qty{k}\) a finite set of places,
\({\mathcal{G}}\in {\mathsf{sm}}{\mathsf{Grp}}{\mathsf{Sch}}_{/ {{\mathcal{O}}_{k, S}}}\)
yielding a SES
\begin{align*}
0\to {\mathcal{G}}^0 \to {\mathcal{G}}\to{\mathcal{F}}\to 0
,\end{align*}
where \({\mathcal{F}}\in {\mathsf{Grp}}{\mathsf{Sch}}\) is finite étale.
After base-changing to \(k\), this yields a SES of groups
\begin{align*}
0\to G^0\to G\to F\to 0
.\end{align*}
We defined
\(H^1_S(k; G) \coloneqq\left\{{\tau \in H^1(k; G) {~\mathrel{\Big\vert}~}\tau_{{ k_{\widehat{v}} }} \in \operatorname{im}\phi \forall v\not\in S}\right\}\)
where
\(\phi: H^1({\mathcal{O}}_{{ k_{\widehat{v}} }}; {\mathcal{G}})\to H^1(k;{\mathcal{G}})\).

\end{remark}

\begin{slogan}

General principal we'll use: there should only be finitely many things
over \({\mathcal{O}}_{k_S}\), see Shafarevich's conjecture.

\end{slogan}

\begin{theorem}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(H_S^1(k; G)\to H^!_S(k; F) \to \prod_{v\in S} H^1({ k_{\widehat{v}} }; F)\)
  has finite fibers, and
\item
  If \(k\) is a number field, then \(H_S^1(k; G)\) is finite.
\end{enumerate}

\end{theorem}

\hypertarget{proof-1}{%
\subsection{Proof}\label{proof-1}}

\hypertarget{proof-of-a}{%
\subsubsection{Proof of a}\label{proof-of-a}}

\begin{claim}[Step 1]

For \(v\not\in S\),
\(H^1({\mathcal{O}}_v; {\mathcal{F}}) \to H^1({ k_{\widehat{v}} }; F)\)
is injective.

\end{claim}

\begin{proof}[Sketch]

First consider a fiber that is a trivial torsor over
\(\left\{{{\operatorname{pt}}}\right\}\).
\({\mathcal{F}}_{/ {{\mathcal{O}}_{{ k_{\widehat{v}} }}}} \in{\mathsf{F}{\hbox{-}}\mathsf{Torsors}}\).
Why? Finite things are proper (finite iff proper and affine), so by the
valuative criterion of properness \({\mathcal{F}}({ k_{\widehat{v}} })\)
nonempty implies \({\mathcal{F}}({\mathcal{O}}_{{ k_{\widehat{v}} }})\)
nonempty.

Now one can use a twisting argument to show that every fiber is finite.

\end{proof}

\begin{claim}[Step 2]

\(H^1({\mathcal{O}}_{{ k_{\widehat{v}} }}; {\mathcal{G}}^0)\).

\end{claim}

\begin{proof}[of step 2]

Let \(\kappa(v)\) be the residue field.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  We know
  \(H^1(\kappa(v); { \left.{{ {\mathcal{G}}^0 }} \right|_{{\kappa(v)}} } ) = 1\)
  by Lang's theorem.
\item
  Let
  \(\tau\in H^1({\mathcal{O}}_{{ k_{\widehat{v}} }}; {\mathcal{G}}^0)\)
  be a torsor. Now by the infinitesimal lifting criterion for
  smoothness, i.e.~Hensel's lemma, if \(\tau(\kappa(v))\) is nonempty
  then \(\tau_{{\mathcal{O}}_{{ k_{\widehat{v}} }}}\) is nonempty.
\item
  \(H^1_S(k; G)\to H^1_S(k; F)\) has finite fibers. This uses input from
  a hard theorem which we'll black box here.
\end{enumerate}

The proof is that
\(H^1({\mathcal{O}}_{{ k_{\widehat{v}} }}; {\mathcal{G}})\to H^1({\mathcal{O}}_k; {\mathcal{F}}) \hookrightarrow H^1(k; F)\),
where the first map is injective by (2) above, and the second by (1).
Let \(\Sha\) denote the Tate-Shafarevich group, then
\(\ker h \subseteq \Sha^1_S(k; G) \coloneqq\ker \qty{H^1(k; C) \to \prod_{v\in S} H^1(k; G_{{ k_{\widehat{v}} }}) }\)
By Borel-Serre, for \({\mathcal{G}}\) affine\footnote{(conjecturally for
  \({\mathcal{G}}\) arbitrary, thought to be true but equivalent to
  BSD!),} the latter is finite.

Now a twisting argument shows this for every fiber.

\begin{claim}

\(H^1_S(k; {\mathcal{F}}) \to \prod_{v\in S} H^1({ k_{\widehat{v}} }; F_{{ k_{\widehat{v}} }})\)
has finite fibers.

\end{claim}

\begin{proof}[?]

Now use the source of all finiteness in arithmetic geometry:
Hermite-Minkowski's theorem, using that the previous information
determines the degree and discriminant.

\begin{itemize}
\item
  Choose
  \((\tau_v)\in \prod_{v} H^1({ k_{\widehat{v}} }; F_{{ k_{\widehat{v}} }})\),
  and suppose \(\tau \to (\tau_v)_{v\in S}\). As a scheme,
  \(\tau = \operatorname{Spec}R\) for \(R\in {\mathsf{Alg}_{/k} }\)
  finite étale, so \(R = \prod_i L_i\) with \(L_i\) unramified away from
  \(S\). This bounds the discriminant, so there are finitely many
  possibilities for \(L_i\).
\item
  For each \(R\) there are finitely many \(F\) actions making
  \(\operatorname{Spec}R\) into a torsor. Using representability, then
  \(F\to \underline{\mathop{\mathrm{Aut}}}_k(R)\) is a finite étale
  group scheme. Since both are now finite, there are only finitely many
  such maps.
\end{itemize}

\end{proof}

\end{proof}

\hypertarget{proof-of-b}{%
\subsubsection{Proof of b}\label{proof-of-b}}

\begin{remark}

For the proof of (b), it's enough to show that
\(\prod_v H^1({ k_{\widehat{v}} }; { \left.{{F}} \right|_{{{ k_{\widehat{v}} }}} } )\)
is finite. Krasner's lemma shows that are only finitely many degree
\(d\) extensions over a \(p{\hbox{-}}\)adic field (false for
\(\operatorname{fp}((t))\)! ), so \({ k_{\widehat{v}} }\) has finitely
many extensions. Then there are only finitely many extensions of \(k\)
of a given degree, so there are only finitely many possibility for
\(F{\hbox{-}}\)torsors.

\end{remark}

\begin{exercise}[?]

Produce infinitely many extensions of \(\operatorname{fp}((t))\).

\end{exercise}

\begin{remark}

Note: étale is stronger than smooth? In the following proof, we could
take everything to be étale to simplify things.

\end{remark}

\begin{remark}

Reminder of setup: let \(X\in {\mathsf{Var}}_{/ {k}}\) be proper,
\(G\in{\mathsf{sm}}{\mathsf{Grp}}{\mathsf{Sch}}_{/ {k}}\) finite ,
\(T\in {\mathsf{G}{\hbox{-}}\mathsf{Torsors}}_{/ {X}}\). We're trying to
show that \(\operatorname{Sel}_{T_{/ {X}} }(k; G)\) is finite.

\textbf{Step 1}: spread out. There are finitely many denominators and
thus finitely many places and \({\mathcal{G}}\) finite smooth and
\({\mathcal{O}}_{k, S'}\) a group scheme with \({\mathcal{X}}\) proper
over \({\mathcal{O}}_{k, S'}\). Then \({\mathcal{T}}_{/ {X}}\) is a
\({\mathcal{G}}{\hbox{-}}\)torsor with
\({\mathcal{G}}_k = G, {\mathcal{T}}_k = T, {\mathcal{X}}_k = X\). Let
\(\tau\in \operatorname{Sel}_{T_{/ {X}} }(k; G)\), then
\(\tau \in H^1_{S'}(k; G)\) since \(\tau_{{ k_{\widehat{v}} }}\) is in
the image of
\(X({ k_{\widehat{v}} }) \to H^1({ k_{\widehat{v}} }; G_{{ k_{\widehat{v}} }})\),
but
\(X({ k_{\widehat{v}} }) = {\mathcal{X}}({\mathcal{O}}_{{ k_{\widehat{v}} }})\)
by the valuative criterion of properness. So for a number field, the set
we want is contained in a finite set and we're done in this case.

For function fields, consider
\(\tau \in \operatorname{Sel}_{T_{/ {X}} } (k; G) \subseteq \operatorname{im}\qty{ H^1_{S'}(k; G) \to \prod_{v\in S'} H^1({ k_{\widehat{v}} }; F) }\),
and let \(\pi\) be the inclusion. The claim is that \(\pi\) has finite
image. The proof is by the local constancy lemma applied to
\(X({ k_{\widehat{v}} }) \to H^1({ k_{\widehat{v}} }; G/G^0)\), which is
essentially Krasner's lemma plus compactness -- here we used that
\(X({ k_{\widehat{v}} })\) is compact. Finite image plus finite fibers
implies finite, so we're done.

\end{remark}

\hypertarget{misc}{%
\subsection{Misc}\label{misc}}

\begin{theorem}[Minchev]

Let \(k \in \mathsf{Number}\mathsf{Field}\),
\(X,Y \in {\mathsf{Var}}_{/ {k}}\) geometrically integral, \(F:Y\to X\)
finite étale and not an isomorphism, and let
\(S \subseteq \mathrm{Pl}\qty{k}\) be a finite set of places and suppose
\(X({\mathbb{A}}^S) \neq \emptyset\). Then \(X\) does not satisfy strong
approximation, i.e.~\(X(k) \hookrightarrow X({\mathbb{A}}^S)\) is not
dense.

\end{theorem}

\begin{corollary}[?]

If \(X\) is geometrically integral and
\(\pi_1^\text{ét}(X_{ { \mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu } }) \neq 1\),
then \(X({\mathbb{A}}^S)\) nonempty implies \(X\) does not satisfy
strong approximation (since \(X\) admits no interesting étale covers)

\end{corollary}

\begin{remark}

Goal: find \(v\not\in S\) and \(U \in X({ k_{\widehat{v}} })\) open
where \(U\) does not contain any \(k{\hbox{-}}\)points.

\end{remark}

\hypertarget{tuesday-november-09}{%
\section{Tuesday, November 09}\label{tuesday-november-09}}

\begin{remark}

The descent obstruction does not suffice!

\end{remark}

\begin{theorem}[?]

There exists a nice \(X\in {\mathsf{Var}}_{/ {{\mathbb{Q}}}}\) with
\(X({\mathbb{A}})^{\text{ét}, \mathop{\mathrm{Br}}} \neq \emptyset\) but
\(X({\mathbb{Q}}) = \emptyset\).

\end{theorem}

\begin{remark}

\(X\) will be a quadric 3-fold bundle over a curve \(C\) with
\({\sharp}C({\mathbb{Q}}) = 1\) with 2 singular fibers. By the Lefschetz
hyperplane theorem, we'll be able to understand its \(\pi_1\), even
though this won't be a Serre fibration and thus we won't get a LES in
homotopy.

\end{remark}

\hypertarget{proof-2}{%
\subsection{Proof}\label{proof-2}}

\begin{remark}

First make \(Y\), a quadric 3-fold bundle over
\({\mathbb{P}}^1_{/ {{\mathbb{Q}}}}\) with 2 singular fibers. Since
we're fibering over \({\mathbb{P}}^1\), we'll define this in two affine
patches and then glue. Define
\begin{align*}
Y_t &\coloneqq\left\{{ t(t-1) x_0^2 + x_1^2 + \cdots + x_4^2 = 0}\right\} \subseteq {\mathbb{A}}^1_{/ {t}}  \times{\mathbb{P}}^1_{/ {[x_0,\cdots, x_4]}}  \\
Y_T &\coloneqq\left\{{ (1-T) X_0^2 + x_1^2 + \cdots + x_4^2 = 0}\right\} \subseteq {\mathbb{A}}^1_{/ {T}}  \times{\mathbb{P}}^1_{/ {[x_0,\cdots, x_4]}} 
.\end{align*}
Then define a gluing by
\begin{align*}
t &\mapsto {1\over T} \\
x_i &\mapsto x_i, \qquad i=1,\cdots, 4\\
x_0 &\mapsto {T \over X_0}
.\end{align*}

Now check that \(Y_0, Y_1\) are singular, and
\(Y_{\infty} = \left\{{\sum x_i^2 = 0}\right\}\):

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{42pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/RationalPoints/sections/figures}{2021-11-09_09-49.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\begin{quote}
Something about
\(({\mathbb{C}}^{\times}\times Q)/ \left\{{\pm 1}\right\} \to {\mathbb{C}}^{\times}\)?
This makes the geometry in this situation easy.
\end{quote}

Since \(Y_\infty({\mathbb{R}}) = \emptyset\) implies
\(Y_\infty(Q) = \emptyset\). Note that if \(t=0, 1\) this is a point,
for \(0< t< 1\) this is an ellipsoid, and for \(t>1\) or \(t<0\) there
are no real points.

\end{remark}

\begin{remark}

Issue: \({\mathbb{P}}^1\) has lots of rational points, so we'll make a
curve \(C\) with a map \(C\to {\mathbb{P}}^1\) and pull back. Let
\(C_{/ {{\mathbb{Q}}}}\) be a smooth proper curve such that

\begin{itemize}
\tightlist
\item
  \({\sharp}C({\mathbb{Q}}) = 1\), so write
  \(c \coloneqq C({\mathbb{Q}})\) as this single rational point. Write
  \(U \subseteq C({\mathbb{R}})\) as the connected component of the
  identity. The real points are compact 1-manifolds, so a disjoint union
  of circles whose number depends on \(g(C)\). Fun question: what is
  \(\pi_0 { \mathcal{M}_{g} }({\mathbb{R}})\)?
\item
  \(C \xrightarrow{f} {\mathbb{P}}^1\) such that

  \begin{itemize}
  \tightlist
  \item
    \(f(c) = \infty\) such that
  \item
    \(f\) is étale at \(c\) and also over
    \(0, 1 \in {\mathbb{P}}^1({\mathbb{Q}})\).
  \item
    \(f(U) \ni 1\).
  \end{itemize}
\end{itemize}

\end{remark}

\begin{example}[?]

\(y^2 = x^3-3\) is such a curve, see LMFDB.

\begin{itemize}
\item
  Probably the point is \(\infty\)!
\item
  Pick a uniformizer at \(\infty\), then compose with an affine map
  \(z\mapsto az+b\) to obtain the 2nd condition.
\item
  \(f(U)\) contains an open subset of \({\mathbb{P}}^1({\mathbb{R}})\),
  using that \(f\) is étale and thus a local homeomorphism, so apply the
  implicit/inverse function theorems. So compose with a generic
  \(z\mapsto z+1\) such that \(f(U) \ni 1\). This doesn't mess up the
  previous affine map, since it's generic.
\end{itemize}

Now define
\(X\coloneqq Y{ \underset{\scriptscriptstyle {{\mathbb{P}}^1} }{\times} } C\)
as the pullback.

\end{example}

\begin{claim}

\(X({\mathbb{Q}}) = \left\{{ \infty }\right\}\).

\end{claim}

\begin{proof}[?]

Note
\begin{align*}
X({\mathbb{Q}}) \xrightarrow{\pi} C({\mathbb{Q}}) 
,\end{align*}
and \(\pi^{-1}(c) \cong Y_\infty\) which has no rational points.

\end{proof}

\hypertarget{showing-xmathbbatextuxe9t-mathopmathrmbr}{%
\subsection{\texorpdfstring{Showing
\(X({\mathbb{A}})^{\text{ét}, \mathop{\mathrm{Br}}}\)}{Showing X(\{\textbackslash mathbb\{A\}\})\^{}\{\textbackslash text\{ét\}, \textbackslash mathop\{\textbackslash mathrm\{Br\}\}\}}}\label{showing-xmathbbatextuxe9t-mathopmathrmbr}}

\begin{remark}

Next goal: show the étale-Brauer set is empty.

\end{remark}

\begin{example}[?]

We can show \(X({\mathbb{A}})^{\mathop{\mathrm{Br}}} \neq \emptyset\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  An exercise using the Leray spectral sequence shows there is a
  surjection \(\mathop{\mathrm{Br}}(C) \to \mathop{\mathrm{Br}}(X)\).
\item
  Observe that \(x\in X({\mathbb{A}})\) is also in
  \(X({\mathbb{A}})^{\mathop{\mathrm{Br}}}\) if
\end{enumerate}

\begin{itemize}
\tightlist
\item
  \(\pi(x_{\infty}) \in C({\mathbb{Q}}_{v})\) is equal to \(C\) for all
  finite \(v\).
\item
  \(\pi(x_{\infty}) \in U\).
\end{itemize}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\tightlist
\item
  Find such a \((x_v) \in X({\mathbb{A}})\).
\end{enumerate}

\begin{itemize}
\tightlist
\item
  For \(v\) finite, choose \(x_v\in Y_\infty({\mathbb{Q}}_v)\). Then
  \(\left\{{ [x_0: \cdots : x_n] {~\mathrel{\Big\vert}~}\sum x_i = 0 }\right\} \neq \emptyset\).
\item
  For \(v\) infinite, choose any \(c'\) lying over
  \(1\in {\mathbb{P}}^1\), so \(c'\in U \cap f^{-1}(1)\). Then
  \(X_{c'} \cong Y_1\) but \(Y_1({\mathbb{R}}) \neq \emptyset\) since it
  contains \(p = [1:0:\cdots:0]\).
\end{itemize}

Why (2) is true: for \(v\) finite,
\(\mathrm{inv}_v(x_v^* \alpha) = \mathrm{inv}_v(\pi(x_v)^* \alpha') = \mathrm{inv}_v(c^* \alpha')\)
where
\(\alpha\in \mathop{\mathrm{Br}}(X), \alpha= \pi^* \alpha', \alpha'\in \mathop{\mathrm{Br}}(C)\),
and the same holds for \(v\) replaced by \(\infty\). Use that this
function is constant on \(U\) since it's connected. Now
\(\sum_v \mathrm{inv}_v( x_v^* \alpha) = \sum_v \mathrm{inv}_v(c^* \alpha') = 0\),
which shows it's in the Brauer set.

\end{example}

\begin{remark}

So the Brauer-Manin obstruction it not sufficient to obstruct rational
points.

\end{remark}

\hypertarget{uxe9tale-brauer-manin-set}{%
\subsection{Étale-Brauer-Manin Set}\label{uxe9tale-brauer-manin-set}}

\begin{remark}

Recall the definition:
\begin{align*}
X({\mathbb{A}})^{\text{ét}, \mathop{\mathrm{Br}}} \coloneqq
\displaystyle\bigcap_{\substack{ 
T \in H^1(X; G) = {\mathsf{G}{\hbox{-}}\mathsf{Torsors}} \\ G \text{ finite étale} 
}} 
\displaystyle\bigcup_{\tau \in H^1(k; G) } \operatorname{im}\qty{ T^\tau({\mathbb{A}})^{\mathop{\mathrm{Br}}} \to X({\mathbb{A}}) }
.\end{align*}
Let \((x_v) \in X({\mathbb{A}})\) as above.

\end{remark}

\begin{claim}

For each finite étale \(G\) and each \(T\in H^1(X; G)\) there exists a
\(\tau \in H^1(k; G)\) such that \((x_v)\) lifts to
\(T^\tau({\mathbb{A}})^{\mathop{\mathrm{Br}}}\).

\end{claim}

\begin{proof}[?]

In steps:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  (SGA1) \(T\) is a pullback of
  \(\tilde T\in {\mathsf{G}{\hbox{-}}\mathsf{Torsors}}_{/ {C}}\), using
  that \(\pi_1 X = \pi_1 C\).
\item
  There exists a \(\tau \in H^1(k; C)\) such that
  \(c\in C({\mathbb{Q}})\) lifts to
  \(\tilde c\in \tilde T^\tau({\mathbb{Q}})\).
\item
  For \(v\) finite, use the diagram:
\end{enumerate}

\begin{center}
\begin{tikzcd}
    & {T^\tau({\mathbb{Q}}_v)} && {X({\mathbb{Q}}_v)} & {\ni x_v} \\
    \\
    {\tilde c\in} & {\tilde T^\tau({\mathbb{Q}}_v)} && {C({\mathbb{Q}}_v)}
    \arrow[from=3-2, to=3-4]
    \arrow[from=1-4, to=3-4]
    \arrow[from=1-2, to=1-4]
    \arrow[from=1-2, to=3-2]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNixbMSwwLCJUXlxcdGF1KFxcUVFfdikiXSxbMywwLCJYKFxcUVFfdikiXSxbMSwyLCJcXHRpbGRlIFReXFx0YXUoXFxRUV92KSJdLFszLDIsIkMoXFxRUV92KSJdLFswLDIsIlxcdGlsZGUgY1xcaW4iXSxbNCwwLCJcXG5pIHhfdiJdLFsyLDNdLFsxLDNdLFswLDFdLFswLDJdXQ==}{Link
to Diagram}
\end{quote}

Then \(T^\tau_{\tilde c} \cong X_{c}\) implies \(x_v\) lifts.

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{3}
\tightlist
\item
  For \(v\) infinite, we have
  \(x_{\infty } \in Y_{1}({\mathbb{R}}) = X_{c'}({\mathbb{R}})\).
\end{enumerate}

\begin{itemize}
\tightlist
\item
  Check that \(c'\) lifts to \(c'' \in \tilde T^\tau({\mathbb{R}})\) in
  the same connected component of \(\tilde c\). This is because
  \(\tilde T^\tau({\mathbb{R}}) \to C({\mathbb{R}})\) is étale, and thus
  locally a covering map and thus surjective on any nonempty component.
  Thus \(T^\tau_{c''} \cong X_{c'}\) and thus \(x_\infty\) lifts.
\end{itemize}

\end{proof}

\hypertarget{chabauty-coleman-thursday-november-11}{%
\section{Chabauty-Coleman (Thursday, November
11)}\label{chabauty-coleman-thursday-november-11}}

\begin{theorem}[Coleman-Chabauty]

For \(X\in {\mathsf{Sch}}_{/ {{\mathbb{Q}}}}\) a proper curve with
\(g(X) \geq 2\) and
\(\operatorname{rank}_{\mathbb{Z}}\operatorname{Jac}(X)({\mathbb{Q}}) < g\)
where \(X\) has good reduction at \(p>2g\),
\({\sharp}X({\mathbb{Q}}) \leq {\sharp}X({\mathbb{F}}_p) + 2g-2\)

\end{theorem}

\begin{remark}

Idea: choose \(x\in X({\mathbb{Q}})\), define a map
\begin{align*}
\operatorname{AJ}_x: X &\xrightarrow{\text{Abel-Jacobi}} \operatorname{Jac}(X) \\
y &\mapsto [{\mathcal{O}}(y-x) ]
.\end{align*}
Then \(X^n\to \operatorname{Jac}(X)\) by
\(\mathbf{y} \mapsto \sum \operatorname{AJ}_x(y_i)\) is surjective for
\(n\geq g\). Note that
\begin{align*}
\mathop{\mathrm{Hom}}(T, \operatorname{Jac}(X)) = \left\{{ {\mathcal{L}}\in {\operatorname{Pic}}(X\times T) {~\mathrel{\Big\vert}~}{\mathcal{L}}\text{ has degree 0 on each fiber of } X\times T\to T}\right\}/{\operatorname{Pic}}(T)
\end{align*}
This equals
\(\left\{{{\mathcal{L}}\in {\operatorname{Pic}}X\times X {~\mathrel{\Big\vert}~}{\mathcal{L}}\text{ has fiberwise degree } 0}\right\}/{\operatorname{Pic}}(X)\).
Note that the LHS is
\({\mathcal{O}}(\left\{{x}\right\}\times X\to \Delta)\).

\end{remark}

\begin{claim}

For \(X^n \xrightarrow{\sum \operatorname{AJ}_x} \operatorname{Jac}(X)\)
is surjective for \(n\geq g\).

\end{claim}

\begin{proof}[?]

ETS that the following map is surjective:
\begin{align*}
X^n &\to {\operatorname{Pic}}^n(X) \\
\mathbf{y} &\mapsto {\mathcal{O}}\qty{ \sum y_i }
.\end{align*}
Given \([{\mathcal{L}}]\in {\operatorname{Pic}}^n(X)\), for
\({\mathcal{L}}= {\mathcal{O}}(D)\) for \(D\) effective iff
\(H^0({\mathcal{L}}) = 0\), so ETS \(H^0({\mathcal{L}}\neq 0)\) for
\(\deg {\mathcal{L}}\geq g\). By Riemann-Roch:
\begin{align*}
\dim H^0({\mathcal{L}}) - \dim H^1({\mathcal{L}}) = \deg {\mathcal{L}}+ 1 - g
.\end{align*}

The RHS is 1, so \(\dim H^0({\mathcal{L}}) \geq 1\).

\end{proof}

\begin{slogan}

\(X\) generates \(\operatorname{Jac}(X)\), and \(\operatorname{Jac}(X)\)
is the Albanese of \(X\).

\end{slogan}

\begin{remark}

Let
\(\Gamma = \operatorname{Jac}X({\mathbb{Q}}) \subseteq \operatorname{Jac}X({ {\mathbb{Q}}_p })\).
We have a factorization:

\begin{center}
\begin{tikzcd}
    {X({\mathbb{Q}})} && \Gamma \\
    \\
    {X({ {\mathbb{Q}}_p })} && \mkern 1.5mu\overline{\mkern-1.5mu\Gamma \mkern-1.5mu}\mkern 1.5mu&& {\operatorname{Jac}X({ {\mathbb{Q}}_p })}
    \arrow[from=1-1, to=3-1]
    \arrow[from=1-1, to=1-3]
    \arrow[curve={height=18pt}, from=3-1, to=3-5]
    \arrow[from=1-3, to=3-5]
    \arrow["\exists"', dashed, from=1-1, to=3-3]
    \arrow[from=3-3, to=3-5]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJYKFxcUVEpIl0sWzAsMiwiWChcXFFRcGFkaWMpIl0sWzIsMCwiXFxHYW1tYSJdLFsyLDIsIlxcYmFyXFxHYW1tYSJdLFs0LDIsIlxcSmFjIFgoXFxRUXBhZGljKSJdLFswLDFdLFswLDJdLFsxLDQsIiIsMCx7ImN1cnZlIjozfV0sWzIsNF0sWzAsMywiXFxleGlzdHMiLDIseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XSxbMyw0XV0=}{Link
to Diagram}
\end{quote}

Here
\(\mkern 1.5mu\overline{\mkern-1.5mu\Gamma\mkern-1.5mu}\mkern 1.5mu\)
has \(\dim n < g\) and \(\operatorname{Jac}X({ {\mathbb{Q}}_p })\) has
dimension ??. A fact from \(p{\hbox{-}}\)adic Lie groups: they're all
direct sums of balls!

Now construction functions on
\(\operatorname{Jac}X ({ {\mathbb{Q}}_p })\) vanishing on
\(\mkern 1.5mu\overline{\mkern-1.5mu\Gamma\mkern-1.5mu}\mkern 1.5mu\).
Then \({ \left.{{f_i}} \right|_{{X({\mathbb{Q}}_p)}} }\) vanish on
\(X({ {\mathbb{Q}}_p }) \cap\mkern 1.5mu\overline{\mkern-1.5mu\Gamma \mkern-1.5mu}\mkern 1.5mu\supseteq X({\mathbb{Q}})\).

This will show \(X({\mathbb{Q}})\) is finite. We'll show that not all of
the \(f_i\) are identically zero: ETS that
\(X({ {\mathbb{Q}}_p }) \not\subseteq \mkern 1.5mu\overline{\mkern-1.5mu\Gamma\mkern-1.5mu}\mkern 1.5mu\),
which will be true since \(X\) generates its Jacobian. Take whichever
\(f_i \not\equiv 0\), which is an \(p{\hbox{-}}\)adic analytic function,
then \(f_i\) will have finitely many zeros.

\end{remark}

\begin{remark}

There is a hypothesis that
\(\operatorname{rank}_{\mathbb{Z}}\operatorname{Jac}X({\mathbb{Q}}) < g\)
can be replaced by
\(\dim \mkern 1.5mu\overline{\mkern-1.5mu\Gamma \mkern-1.5mu}\mkern 1.5mu< g\).

\end{remark}

\hypertarget{proofs}{%
\subsection{Proofs}\label{proofs}}

\begin{lemma}[?]

Let \(\Gamma \leq A({ {\mathbb{Q}}_p })\) be a finitely-generated
abelian subgroup for \(A\in {\mathsf{Ab}}{\mathsf{Var}}\). Then
\(\dim_{{ {\mathbb{Q}}_p }} \mkern 1.5mu\overline{\mkern-1.5mu\Gamma \mkern-1.5mu}\mkern 1.5mu\leq \operatorname{rank}_{\mathbb{Z}}\Gamma\).

\end{lemma}

\begin{fact}

Any \(p{\hbox{-}}\)adic manifold \(X\) is locally isomorphic to
\({ {\mathbb{Z}}_p }{ {}^{ \scriptscriptstyle\times^{\dim X} } }\).

\end{fact}

\begin{proof}[?]

For
\(\Gamma' \subseteq { {\mathbb{Z}}_p }{ {}^{ \scriptscriptstyle\times^{n} } }\)
a finitely-generated abelian group,
\(\dim \mkern 1.5mu\overline{\mkern-1.5mu \Gamma\mkern-1.5mu}\mkern 1.5mu' \leq \operatorname{rank}_{\mathbb{Z}}\Gamma'\).
Pick generators \(g_1, \cdots, g_n\) generators, then write a linear map
\begin{align*}
\gamma: { {\mathbb{Z}}_p }^r &\to { {\mathbb{Z}}_p }^n \\
e_i &\mapsto g_i
.\end{align*}
The image is closed of dimension \(n = \operatorname{rank}\gamma\).
We'll now reduce the original lemma to proving this statement. This is
because any \emph{abelian} \(p{\hbox{-}}\)adic Lie group is of the form
\({ {\mathbb{Z}}_p }{ {}^{ \scriptscriptstyle\times^{n} } } \times \text{something finite}\)
for some \(n\). If \(A\) has good reduction at \(p\), then
\(A({ {\mathbb{Q}}_p }) \cong { {\mathbb{Z}}_p }\times A({\mathbb{F}}_p)\).
Recall that there are maps:
\begin{align*}
\exp: T_0 A({ {\mathbb{Z}}_p }) &\to A({ {\mathbb{Q}}_p }) \\
\log: \mathrm{sp}^{-1}(0) &\to A({ {\mathbb{Z}}_p })
,\end{align*}
where \(\mathrm{sp}\) is the specialization map, and this induces an
isomorphism of abelian groups.

The goal is to now construct a \(p{\hbox{-}}\)adic analytic function on
\(A({ {\mathbb{Q}}_p })\) vanishing on
\(\mkern 1.5mu\overline{\mkern-1.5mu\Gamma\mkern-1.5mu}\mkern 1.5mu\).
This construction: due to Coleman. Fix
\(\omega \in H^0(A_{{ {\mathbb{Q}}_p }}, \Omega^1_{A/{ {\mathbb{Q}}_p }})\),
then there exists a \(p{\hbox{-}}\)adic analytic function
\begin{align*}
A({ {\mathbb{Q}}_p }) &\to { {\mathbb{Q}}_p }\\
Q &\mapsto \int_0^Q \omega
.\end{align*}
This map is determined by

\begin{itemize}
\tightlist
\item
  \(Q\to \int_0^Q \omega\) is a morphism of topological groups,
\item
  FTC: locally near 0, if \(f\) is a \(p{\hbox{-}}\)adic analytic
  function with \(\omega = df\), then \(\int_0^Q \omega = f(Q) - f(0)\).
\end{itemize}

\end{proof}

\begin{remark}

Take
\(\mkern 1.5mu\overline{\mkern-1.5mu\Gamma \mkern-1.5mu}\mkern 1.5mu\subseteq { {\mathbb{Q}}_p }\),
then for
\(U \coloneqq\mkern 1.5mu\overline{\mkern-1.5mu\Gamma \mkern-1.5mu}\mkern 1.5mu\in B_0\)
a ball around the origin, consider
\(\log(U) \subseteq T_Q(A) = { {\mathbb{Q}}_p }{ {}^{ \scriptscriptstyle\times^{n} } }\).
Let \(v_1,\cdots, v_s \subseteq {\mathbf{T}}_0 {}^{ \vee }A\) be dual
vectors vanishing on
\(\log \mkern 1.5mu\overline{\mkern-1.5mu\Gamma\mkern-1.5mu}\mkern 1.5mu\).
Then

\begin{itemize}
\tightlist
\item
  \({\mathbf{T}}_0 {}^{ \vee }A \cong H^0(A, \Omega^1_{A})\)
\item
  \(f_i(Q) = \int_0^Q v_i\), viewing \(v_i\) as a 1-form.
\end{itemize}

\end{remark}

\hypertarget{the-mordell-conjecture-tuesday-november-23}{%
\section{The Mordell Conjecture (Tuesday, November
23)}\label{the-mordell-conjecture-tuesday-november-23}}

\begin{theorem}[Faltings]

For \(K\in\mathsf{Number}\mathsf{Field}\) and \(X_{/ {k}}\) a smooth
curve of genus \(g\geq 2\). Then
\begin{align*}
{\sharp}X(k) < \infty
.\end{align*}

\end{theorem}

\begin{example}[?]

Consider
\begin{align*}
x^n+y^n=1
.\end{align*}
Note that finiteness of rational points here is a weak form of Fermat:
scaling out the denominators yields a rational solution to FLT. This has
finitely many rational solutions for \(n\geq 4\), which ensures
\(g\geq 2\).

Another example is \(C \subseteq {\mathbb{P}}^2\) a smooth curve, then
we have finiteness when \(\deg C \geq 4\).

\end{example}

\begin{remark}

Strategy of proof: finite map with finite fibers, a standard way to
prove a set is finite. Note that Faltings proved a number of other
famous conjectures along the way to proving this:

\end{remark}

\begin{theorem}[Shafarevich for curves]

Let \(K\in \mathsf{Number}\mathsf{Field}\) and
\(S \subseteq \mathrm{Pl}\qty{k}\) a finite set of places. The set of
proper curves over \(k\) of genus \(g\) with good reduction outside of
\(S\) is finite after modding out by isomorphism.

\end{theorem}

\begin{remark}

Rephrasing: \({ \mathcal{M}_{g} }\) Deligne-Mumford stack, i.e.~the
functor sending a ring \(R\) to the groupoid of smooth proper curves
over \(R\), has finitely many \({\mathcal{O}}_{k, S}\) points:
\begin{align*}
{\sharp}{ \mathcal{M}_{g} }({\mathcal{O}}_{k, S}) < \infty
.\end{align*}
Note the stark contrast for \({\mathbb{Q}}\), where
\({\mathcal{M}}_2({\mathbb{Q}})\) has infinitely many points: take
\begin{align*}
E \coloneqq\left\{{y^2 = x(x-1) f(x)}\right\} && \deg f = 3 \text{ separable}
\end{align*}
where \(0, 1\) are not roots of \(f\). This produces an infinite family
of genus 2 curves.

\end{remark}

\begin{remark}

The Shafarevich conjecture for curves implies Mordell: fix \(S\) such
that \(X\) has a smooth proper model
\({\mathcal{X}}_{/ {{\mathcal{O}}_{k, S}}}\) with
\(X(k) = {\mathcal{X}}({\mathcal{O}}_{k, S})\). Then we win if we can
find a finite morphism
\({\mathcal{X}}\hookrightarrow{\mathcal{M}}_{g'}{}_{/ {{\mathcal{O}}_{k, S}}}\)
for some \(g'\). Note that if these were affine, this would be
impossible. The existence of such maps is the \textbf{Kodaira-Parshin
trick}:

\begin{itemize}
\tightlist
\item
  Over \({\mathbb{C}}\), existence was given by Kodaira,
\item
  Over \(k\), existence due to Parshin.
\end{itemize}

Faltings' (and others') proof pass though this trick.

\end{remark}

\begin{remark}

Note that a special case of Shafarevich was proved by Faltings,
Shafarevich proved it for all curves. Recent results along these lines:
proving \({\mathcal{O}}_{k, S}\) points are not Zariski-dense, or are
finite, for other interesting moduli spaces.

\end{remark}

\begin{theorem}[Shafarevich for abelian varieties]

For
\(k\in \mathsf{Number}\mathsf{Field}, S \subseteq \mathrm{Pl}\qty{k}\)
finite, and \(g,d > 0\), the set of isomorphism classes of abelian
varieties of dimension \(g\) with a polarization of degree \(d\) over
\(k\) with good reduction outside of \(S\), is finite. Then
\begin{align*}
{\sharp}{\mathcal{A}}_{g, d}({\mathcal{O}}_{k, S}) < \infty
.\end{align*}

\end{theorem}

\begin{remark}

Line bundles are maps to Picard, \(d\) is the degree. For \(d=1\), these
are principally polarized.

\end{remark}

\begin{remark}

It's enough to consider the case \(d=1\), which is not obvious -- this
is referred to as Zahrin's trick, and shows \(A\), \(A^8\) always has a
polarization of degree 1. Then Shafarevich for AVs implies Shafarevich
for curves, using the Torelli map:
\begin{align*}
{\mathcal{M}}_g &\to {\mathcal{A}}_g \\
[C] &\mapsto [\operatorname{Jac}(C), \Theta]
.\end{align*}
This is a finite map!

\end{remark}

\begin{definition}[Tate Modules]

\(A\in {\mathsf{Ab}}{\mathsf{Var}}_{/ {k}}\) and \(\ell\) some prime,
the \(\ell{\hbox{-}}\)adic Tate module of \(A\) is defined as
\begin{align*}
T_\ell(A) \coloneqq\cocolim_n A[\ell^n](\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu) \cong {\mathbb{Z}}_\ell{ {}^{ \scriptscriptstyle\times^{2g} }  }
,\end{align*}
and
\(V_\ell(A) = T_\ell(A) \otimes_{{\mathbb{Z}}_\ell} {\mathbb{Q}}_\ell\).

\end{definition}

\begin{remark}

Recall that finite generation of a field is being finitely generated
over a prime field.

\end{remark}

\begin{theorem}[Tate Conjecture]

For \(A_1, A_2 \in {\mathsf{Ab}}{\mathsf{Var}}_{/ {k}}\) with \(k\)
finitely generated,

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(V_\ell(A_i)\) are semisimple Galois representations.
\item
  After tensoring homs with \({\mathbb{Z}}_\ell\), we get an isomorphism
  of \({\mathbb{Z}}_\ell{\hbox{-}}\)modules:
  \begin{align*}
  \mathop{\mathrm{Hom}}_k(A_1, A_2)\otimes_{\mathbb{Z}}{\mathbb{Z}}_{\ell} { { \, \xrightarrow{\sim}\, }}\mathop{\mathrm{Hom}}(T_\ell(A_1), T_\ell(A_2)) \otimes{\mathbb{Z}}_\ell
  .\end{align*}
\end{enumerate}

\end{theorem}

\begin{remark}

Part 1 is already very deep, it's a special case of a conjecture we know
almost nothing about. In fact, it's false over \({ {\mathbb{Q}}_p }\):
take
\begin{align*}
E: \quad y^2=x(x-1)(x-p) \quad _{/ {{ {\mathbb{Q}}_p }}} 
.\end{align*}
This has multiplicative reduction. Similarly, take
\begin{align*}
E: \quad y^2=x(x-1)(x-t) \quad _{/ { {\mathbb{C}} { \left( {t} \right) }  }} 
,\end{align*}
since the monodromy matrix
\({ \begin{bmatrix}  {1} & {2} \\  {0} & {1} \end{bmatrix} }\) is not
diagonalizable. Note that
\(\mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{C}} { \left( {t} \right) } \mkern-1.5mu}\mkern 1.5mu\)
are the Puiseux series \(P\). Any topological generator of
\({ \mathsf{Gal}} (P/{\mathbb{C}} { \left( {t} \right) } ) \cong \widehat{{\mathbb{Z}}}(1)\)
acts on \(T_\ell(A)\) by a matrix conjugate to this.

\end{remark}

\begin{remark}

The general conjecture that we know almost nothing about:

\end{remark}

\begin{conjecture}

For \(X_{/ {k}}\) smooth proper,

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(G_k\curvearrowright H^i(X_{\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu}; {\mathbb{Q}}_\ell)\)
  semisimply.
\item
  \({\operatorname{CH}}^i(X) \otimes_{\mathbb{Z}}{\mathbb{Q}}\to H^{2i}(X_{\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu}; {\mathbb{Q}}_\ell(i))^{G_k}\)
  is surjective.
\end{enumerate}

\end{conjecture}

\begin{remark}

Extremely hard problems! Probably Fields material.

\end{remark}

\begin{theorem}[Main difficult ingredient]

For \(A\in {\mathsf{Ab}}{\mathsf{Var}}_{/ {k}}\), there exist only
finitely many isomorphism classes of abelian varieties over \(k\)
isogenous to \(A\).

\end{theorem}

\begin{proof}[Sketch/idea]

Take \(X_{/ {k}}\) smooth proper with \(g(X) \geq 2\) and good reduction
outside of \(S\). For \(\ell\) a prime, there is a map

Define

\begin{itemize}
\item
  \({\mathcal{M}}_{g'}({\mathcal{O}}_{k, S}) = \left\{{\text{Curves of genus $g'$ over $k$ with good reduction outside $S$}}\right\}/\sim\)
  and
  \begin{align*}
  \mathrm{KP}: X(k) \to {\mathcal{M}}_{g'}
  ,\end{align*}
  corresponding to the Kodaira-Parshin trick.
\item
  \({\mathcal{A}}_{g'}({\mathcal{O}}_{k, S})\) for the PPAVs of
  dimension \(g'\) with good reduction outside of \(S\) (mod
  isomorphism).
\item
  \({\mathsf{Rep}}_{2g'}(G_{k, S \cup\left\{{\ell}\right\}})\) for the
  set of semisimple \({\mathbb{Q}}_\ell\) Galois representations of rank
  \(2g'\) and \emph{weight} \(-1\) unramified outside of
  \(S\cup\left\{{l}\right\}\) such that the characteristic polynomials
  of \(\mathop{\mathrm{Frob}}\) (a conjugacy class) outside of
  \(S\cup\left\{{\ell}\right\}\) have integer coefficients.
\end{itemize}

Here \(\ell\) is the ambient prime. Weight \(-1\) means the eigenvalues
of \(\mathop{\mathrm{Frob}}\) are algebraic numbers \(\alpha\) where any
embedding \({\mathbb{Q}}(\alpha_j) \hookrightarrow{\mathbb{C}}\) and
\({\left\lvert {\alpha} \right\rvert} = {\left\lvert {k(v)} \right\rvert}^{1\over 2}\).
\(\mathop{\mathrm{Frob}}\) having integer coefficients comes from the
Weil conjectures.

Assembling these, we get a chain of maps
\begin{align*}
X(k) \xrightarrow{\mathrm{KP}} {\mathcal{A}}_{g'}({\mathcal{O}}_{k, S}) \xrightarrow{\operatorname{Jac}} {\mathcal{A}}_{g'}({\mathcal{O}}_{k, S}) \xrightarrow{V_\ell} {\mathsf{Rep}}_{2g'}(G_{k, S\cup\left\{{\ell}\right\}}; {\mathbb{Q}}_\ell)
,\end{align*}
which we claim all have finite fibers, and the last set is finite:

\begin{itemize}
\tightlist
\item
  \(\mathrm{KP}\) has finite fibers by construction
\item
  \(\operatorname{Jac}\) has finite fibers by the Torelli theorem
\item
  \(V_\ell\) has finite fibers, which is more difficult. By the Tate
  conjecture, the fiber over a representation in \(V\) are isogenous AVs
  over \(k\), but such sets are finite.
\end{itemize}

\end{proof}

\begin{theorem}[?]

\begin{align*}
{\sharp}{\mathsf{Rep}}_N(G_{k, S}; {\mathbb{Q}}_\ell) < \infty
.\end{align*}

\end{theorem}

\begin{remark}

We'll discuss this in detail next time, but the idea is that
representations are determined by traces of group elements. For
\(\Gamma \xrightarrow{\rho} \operatorname{GL}(V)\) a representation over
\(\operatorname{ch}k=0\), the semisimplification
\(\rho^{{\mathrm{ss}}}\) is determined by
\(\operatorname{Tr}\circ \rho\). We can use that Frobenii are dense, and
it's enough to determine this representation on a dense set, so we
consider \(\operatorname{Tr}(\rho(\mathop{\mathrm{Frob}}))\). This is
still an infinite set, so we have to argue that finitely many determine
it, but this comes from the bound on eigenvalues.

\end{remark}

\hypertarget{tuesday-november-30}{%
\section{Tuesday, November 30}\label{tuesday-november-30}}

\hypertarget{faltings-theorem}{%
\subsection{Faltings Theorem}\label{faltings-theorem}}

\begin{definition}[?]

Write \({\mathsf{Rep}}_N^*(G_{k, S}; { {\mathbb{Q}}_\ell })\) for
semisimple continuous representation
\(G_{k, S\cup\left\{{\ell}\right\}} \to \operatorname{GL}_n({\mathcal{O}}_\ell)\)
of weight -1 such that for all \(v\in S\cup\left\{{\ell}\right\}\), the
characteristic polynomial of \(\mathop{\mathrm{Frob}}_v\) has integer
coefficients. Note that the roots are
\(\alpha_i \in \mkern 1.5mu\overline{\mkern-1.5mu{\mathbb{Q}}\mkern-1.5mu}\mkern 1.5mu\)
such that \({\mathbb{Q}}(\alpha_i) \hookrightarrow{\mathbb{C}}\) and
\({\left\lvert {\alpha_i} \right\rvert} = q^{i\over 2}\).

\end{definition}

\begin{theorem}[Faltings]

\({\sharp}X(k) < \infty\).

\end{theorem}

\begin{theorem}[?]

\({\mathsf{Rep}}_N^*(G_{k, S}, {\mathcal{O}}_\ell)\) is finite.

\end{theorem}

\hypertarget{proof-3}{%
\subsubsection{Proof}\label{proof-3}}

\begin{lemma}[A very important fact]

If \(\Gamma\) is a group and
\(\Gamma \xrightarrow{\rho} \operatorname{GL}_n(L)\) is a semisimple
representation over \(L\) a field of characteristic zero. Then \(\rho\)
is determined by \(\operatorname{Tr}\circ \rho: \Gamma\to L\), i.e.~any
two representations for which these composites agree differ by
conjugacy.

\end{lemma}

\begin{fact}

\(\rho\in {\mathsf{Rep}}^*_N(G_{k, S}, {\mathcal{O}}_\ell)\) is
determined by
\(\left\{{\operatorname{Tr}\circ \rho \circ \mathop{\mathrm{Frob}}_v {~\mathrel{\Big\vert}~}v\in S \cup\left\{{\ell}\right\}}\right\}\)
since the Frobenii are dense in \(G_{k, S\cup\left\{{\ell}\right\}}\) by
Chebotarev density.

\end{fact}

\begin{fact}

Fix \(v\not\in S \cup\left\{{\ell}\right\}\), then there are only
finitely many possibilities for
\({\left\lvert {\operatorname{Tr}\circ \rho \mathop{\mathrm{Frob}}_v} \right\rvert} < N\cdot {\left\lvert {\kappa(v)} \right\rvert}^{1\over 2}\),
since traces of Frobenii are sums of eigenvalues.

\end{fact}

\begin{remark}

This provides a finite set of Frobenii whose traces determine \(\rho\).
Given \(\rho_1, \rho_2\), take the group ring of the Galois group and
consider the map
\begin{align*}
{\mathbb{Q}}_\ell  { \left[ { G_{k, S\cup\left\{{\ell}\right\}} } \right] }  \xrightarrow{\rho_1 \times \rho_2} \operatorname{Mat}(n\times n; { {\mathbb{Z}}_p }){ {}^{ \scriptscriptstyle\times^{2} }  }
,\end{align*}
where we used that
\(\operatorname{GL}_n({ {\mathbb{Z}}_p }) \leq \operatorname{GL}_n({ {\mathbb{Q}}_p })\)
is a maximal compact subgroup. The goal is to find a set of places \(T\)
such that
\(\left\{{ (\rho_1, \rho_2)(\mathop{\mathrm{Frob}}_v) {~\mathrel{\Big\vert}~}v\in T}\right\}\)
spans the image, which we can do since the RHS is a finite-dimensional
\({ {\mathbb{Z}}_p }\) module.

\end{remark}

\begin{remark}

Let \(\tilde k\) be the compositum of all extensions of \(k\) unramified
outside of the set of bad places \(S\cup\left\{{\ell}\right\}\) of
bounded degree \(\ell < 2n^2\), the dimension of the RHS above. This is
a finite extension by Hermite-Minkowski. Let
\(\tilde k^{ \operatorname{cl}}\) be the Galois closure, and by
Chebotarev choose \(T\) such that
\(\left\{{\mathop{\mathrm{Frob}}_v}\right\}_{v\in T}\) cover all
conjugacy classes of
\({ \mathsf{Gal}} (\tilde k^{ \operatorname{cl}} /k)\).

\end{remark}

\begin{claim}

For
\(\rho_1, \rho_2 \in {\mathsf{Rep}}_N^*(G_{k, S}; {\mathcal{O}}_\ell)\),
if
\(\operatorname{Tr}({ \left.{{\mathop{\mathrm{Frob}}_v}} \right|_{{\rho_1}} }) = \operatorname{Tr}({ \left.{{\mathop{\mathrm{Frob}}_v}} \right|_{{\rho_2}} })\)
for all \(v\in T\), then \(\rho_1 \cong \rho_2\).

\end{claim}

\begin{proof}[?]

Consider
\({ {\mathbb{Z}}_\ell }[ G_{k, S} ] \xrightarrow{(\rho_1, \rho_2)} \operatorname{Mat}_{N\times N}({ {\mathbb{Z}}_\ell }){ {}^{ \scriptscriptstyle\times^{2} } }\),
write \(M\coloneqq\operatorname{im}(\rho_1, \rho_2)\). ETS that the
images of \(\mathop{\mathrm{Frob}}_v\) generate this as a
\({ {\mathbb{Z}}_\ell }\) module. By Nakayama, it's enough to check that
they generate \(M/\ell M\). Note that
\(\# (M/\ell M)^{\times}< \ell^{2N^2}\), so write
\(\tau: G_{k, S\cup\left\{{\ell}\right\}} \to (M/\ell M)^{\times}\),
then \(\tau\) factors through
\({ \mathsf{Gal}} \qty{ \tilde k^{ \operatorname{cl}} / k}\). Then
\(\operatorname{im}\qty{ \left\{{\mathop{\mathrm{Frob}}_v {~\mathrel{\Big\vert}~}v\in T}\right\} } = \operatorname{im}\tau\).

\end{proof}

\hypertarget{the-kodaira-parshin-trick}{%
\subsection{The Kodaira-Parshin Trick}\label{the-kodaira-parshin-trick}}

\begin{theorem}[Parshin]

Let \(X\) be a curve over \({\mathcal{O}}_{k, S}\), \(g(X) \geq 2\).
After potentially increasing \(k\) and \(S\), there exists a finite map
of algebraic varieties
\begin{align*}
X' \to {\mathcal{M}}_{g'}/{\mathcal{O}}_{k, S}
\end{align*}
for some \(g'\) where \(X'\) is a finite étale cover of \(X\).

\end{theorem}

\begin{remark}

Note that \({\mathcal{M}}_2\) is affine and can't contain a smooth
proper curve. In general it isn't even proper, so it's difficult to map
a proper thing into a non-proper thing. Note that the RHS is a stack
over \({\mathcal{O}}_{k, S}\). Why compactify in general: argue
something is an open condition, degenerate to the boundary where the
objects are easier to work with and show it holds there and thus on an
open containing it.

\end{remark}

\begin{remark}

Note that \(\mathop{\mathrm{Hom}}(X, {\mathcal{M}}_{g'})\) is the of
smooth proper morphisms \(Y\to X\) over \({\mathcal{O}}_{k, S}\) such
that the geometric fibers are curves of genus \(g'\). Take
\({\mathcal{Y}}\xrightarrow{\pi} X\) and \(x\in X\) such that
\(\pi^{-1}(x)\) is a cover of \(X\) ramified only at \(x\).

\begin{slogan}

\(X\) is a moduli space of branched covers of \(X\) branched only at a
single point.

\end{slogan}

Idea: create a moduli space of such covers. Take
\(\eta: X{ {}^{ \scriptscriptstyle\times^{2} } }\setminus\Delta \to X\)
where the fiber over \(X\) is \(X\setminus\left\{{x}\right\}\).

\begin{figure}
\centering
\resizebox{\columnwidth}{!}{%
\begin{tikzpicture}
\fontsize{44pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/RationalPoints/sections/figures}{2021-11-30_10-31.pdf_tex} };
\end{tikzpicture}
}
\end{figure}

\end{remark}

\begin{remark}

Over \({\mathbb{C}}\), note that \(X\) is covered by \({\mathbb{H}}\) so
\(\pi_2 X =1\) and \(\eta\) is a fibration, so taking the LES in
homotopy yields a SES
\begin{align*}
1\to \pi_1(X\setminus\left\{{)}\right\} \to \pi_1(X{ {}^{ \scriptscriptstyle\times^{2} }  }\setminus\Delta) \to \pi_1(X) \to 1
.\end{align*}
Here
\(\pi_1(X) = \left\langle{a_1, b_1,\cdots, a_g, b_g {~\mathrel{\Big\vert}~}\prod [a_i b_i] = 1}\right\rangle\).
Pick any \(\gamma: \pi_1(X\setminus\left\{{)}\right\} \to S_3\) which
doesn't send the loop around the puncture to the identity,
i.e.~\(\gamma \qty{\prod [a_i b_i] } \neq \operatorname{id}\). This now
has ramification at a point.

Define
\(\Gamma \coloneqq\left\{{g{~\mathrel{\Big\vert}~}\pi_1(X{ {}^{ \scriptscriptstyle\times^{2} } } \setminus\Delta) {~\mathrel{\Big\vert}~}\gamma^g \text{ is conjugate to } g }\right\}\),
i.e.~there exists \(h_g \in S_3\) such that for all
\(x\in \pi_1(X\setminus\left\{{)}\right\}\) we have
\(\gamma(gxg^{-1}) = h_g \gamma(x) h_g^{-1}\) for all
\(x\in \pi_1(X\setminus\left\{{)}\right\}\).

\end{remark}

\begin{claim}

\([\Gamma: \pi_1(X{ {}^{ \scriptscriptstyle\times^{2} } }\setminus\Delta)] < \infty\)
has finite index and contains \(\pi_1(X\setminus\left\{{)}\right\}\).

\end{claim}

\begin{claim}

This yields a group morphism
\begin{align*}
\Gamma &\to S_g \\
g &\mapsto h_g
.\end{align*}

\end{claim}

\begin{remark}

Some words on why these are true: stabilizers of groups acting on finite
sets have finite index, and the subgroup claim comes from sending
\(h_g\to \gamma(g)\).

\end{remark}

\begin{remark}

The construction:

\begin{center}
\begin{tikzcd}
    {\mathcal{Y}}\\
    \\
    {X_\Gamma} && {X{ {}^{ \scriptscriptstyle\times^{2} }  }\setminus\Delta} \\
    \\
    {X'} && X
    \arrow[from=1-1, to=3-1]
    \arrow[from=3-1, to=3-3]
    \arrow[from=3-1, to=5-3]
    \arrow[from=3-3, to=5-3]
    \arrow["{\text{finite fibers}}"', from=5-1, to=5-3]
    \arrow[from=3-1, to=5-1]
    \arrow[curve={height=30pt}, from=1-1, to=5-1]
\end{tikzcd}
\end{center}

\begin{quote}
\href{https://q.uiver.app/?q=WzAsNSxbMCwwLCJcXG1jeSJdLFswLDIsIlhfXFxHYW1tYSJdLFsyLDIsIlhcXGNhcnRwb3dlcnsyfVxcc21cXERlbHRhIl0sWzIsNCwiWCJdLFswLDQsIlgnIl0sWzAsMV0sWzEsMl0sWzEsM10sWzIsM10sWzQsMywiXFx0ZXh0e2Zpbml0ZSBmaWJlcnN9IiwyXSxbMSw0XSxbMCw0LCIiLDEseyJjdXJ2ZSI6NX1dXQ==}{Link
to Diagram}
\end{quote}

The fibers of \({\mathcal{Y}}\to X\) are disjoint unions of covers of
\(X\) ramified at \(x\). Since we now have a family of curves,
\({\mathcal{Y}}\to X'\) yields a map
\(X'\to {\mathcal{M}}_{g'}{}_{/ {{\mathbb{C}}}}\).

\end{remark}

\begin{claim}

This map is not constant.

\end{claim}

\begin{remark}

Suppose all of the fibers are isomorphic to some fixed \(Y\) of a fixed
genus, then \(\mathop{\mathrm{Hom}}(Y, X)\) would be infinite. By
Hurwitz, there's a bound on the degree of such a map since \(g(Y)\) is
fixed, so there are only finitely many. \(\contradiction\)

\end{remark}

\begin{remark}

To do everything over \({\mathcal{O}}_{k, S}\), we use
\(\pi_1^\text{ét}\) instead of \(\pi_1\) and include \(2,3\in S\). Since
the map is unramified over \({\mathbb{C}}\), it's ramified at only
finitely many primes, so just add those to \(S\).

\end{remark}

\hypertarget{thursday-december-02}{%
\section{Thursday, December 02}\label{thursday-december-02}}

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

\begin{conjecture}[Tate conjecture, general]

Let \(X\in {\mathsf{sm}}\mathop{\mathrm{Proj}}{\mathsf{Var}}_{/ {k}}\)
for \(k\) a finitely generated field. Conjecturally the cycle class map
is surjective:
\begin{align*}
{\operatorname{CH}}^i(X)_{{ {\mathbb{Q}}_\ell }} \twoheadrightarrow H^{2i}(X_{\mkern 1.5mu\overline{\mkern-1.5muk\mkern-1.5mu}\mkern 1.5mu}; {\mathbb{Q}}_\ell(i) )^{G_k}
.\end{align*}

Replacing rational equivalence by \emph{homological equivalence} is
conjectured to yield a bijection, although this is closer to
tautological -- one quotients by the kernel of this map. A conjecture
with more content: replacing rational equivalence with \emph{numerical
equivalence} yields a bijection, where one essentially quotients by the
intersection pairing.

\end{conjecture}

\begin{remark}

Recall that we're proving \(\# X(k) < \infty\). Note
\begin{align*}
X(k) \xrightarrow{kP} {\mathcal{M}}_{g'}({\mathcal{O}}_{k, S}) \xrightarrow{\operatorname{Jac}} {\mathcal{A}}_{g'}({\mathcal{O}}_{k, S}) \xrightarrow{V_\rho} {\mathsf{Rep}}_{2g'}^*(G_{k, S}; { {\mathbb{Q}}_\ell })
.\end{align*}
Last time we constructed \(\mathrm{KP}\) and showed that
\({\mathsf{Rep}}_{2g'}\) above is finite. The strategy: proving some
finiteness statement, prove a special case of the Tate conjecture
(generally widely open), then get a stronger finiteness result.

\end{remark}

\begin{theorem}[Faltings, Tate conjecture for AVs over a number field]

For \(A, B \in {\mathsf{Ab}}{\mathsf{Var}}_{/ {k}}\),
\(k\in \mathsf{Number}\mathsf{Field}\),

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  The Tate modules \(V_\ell(A), V_\ell(B)\) are semisimple
  \(G_k{\hbox{-}}\)modules.
\item
  There is a bijection
  \begin{align*}
    \mathop{\mathrm{Hom}}(A, B)\otimes{ {\mathbb{Z}}_\ell }{ { \, \xrightarrow{\sim}\, }}\mathop{\mathrm{Hom}}_{G_k}(T_\ell(A), T_\ell(B))
    ,\end{align*}
  where the LHS are isogenies.
\end{enumerate}

\end{theorem}

\begin{remark}

This won't be true for \(p{\hbox{-}}\)adic fields, but there is a
version that works for finitely-generated fields, e.g.~function fields
of varieties defined over number fields.

\end{remark}

\begin{theorem}[Faltings, a finiteness result]

Isogeny classes over \(k\) are finite.

\end{theorem}

\begin{theorem}[Tate?]

Faltings' finiteness result implies the Tate conjecture

\end{theorem}

\begin{remark}

The proof idea: use a \textbf{height} defined by Faltings:
\begin{align*}
{\mathcal{A}}_g(k) \to {\mathbb{R}}
\end{align*}
satisfying

\begin{itemize}
\tightlist
\item
  For all \(N\),
  \({\sharp}\left\{{[A] \in {\mathcal{A}}_g(k) {~\mathrel{\Big\vert}~}h([A] ) < n }\right\}\)
  is finite.
\item
  If \(A \sim B\) are isogenous, then
  \({\left\lvert { h(A) - h(B) } \right\rvert} < C = C(A)\) a constant.
\end{itemize}

Note that these imply the finiteness result since
\(\left\{{h(B) {~\mathrel{\Big\vert}~}B\sim A}\right\} < \infty\). There
is a general height machine theory here, which extends this theory to
(nice, e.g.~DM) stacks. Recent work is going into extending height
theories to more complicated stacks, e.g.~algebraic spaces.

\end{remark}

\hypertarget{proof-finiteness-implies-the-tate-conjecture}{%
\subsection{Proof: Finiteness implies the Tate
conjecture}\label{proof-finiteness-implies-the-tate-conjecture}}

\begin{claim}

\(\mathop{\mathrm{Hom}}(A, B)\) is torsionfree.

\end{claim}

\begin{proof}[?]

This is because
\(\mathop{\mathrm{Hom}}(A, B) \hookrightarrow\mathop{\mathrm{Hom}}(T_\ell(A), T_\ell(B))\)
since the \(\ell^\infty{\hbox{-}}\)torsion points are dense, and the RHS
is torsionfree.

\end{proof}

\begin{claim}

There is a functor
\begin{align*}
T_\ell: {\mathsf{Ab}}{\mathsf{Var}}_{/ {k}}  \otimes{ {\mathbb{Z}}_\ell }\to {\mathsf{Rep}}_{G_k}
\end{align*}

which is

\begin{itemize}
\tightlist
\item
  Faithful (and we'll want to show it's full)
\item
  \(\mathop{\mathrm{Hom}}(A, B)\) is finitely generated.
\end{itemize}

\end{claim}

\begin{proof}[?]

The proof is essentially checking over \({\mathbb{C}}\).

\end{proof}

\begin{claim}

The following cokernel is torsionfree:
\begin{align*}
\operatorname{coker}\qty{
\mathop{\mathrm{Hom}}(A, B)\otimes{ {\mathbb{Z}}_\ell }\longrightarrow\mathop{\mathrm{Hom}}(T_\ell(A), T_\ell(B))
} 
.\end{align*}

\end{claim}

\begin{proof}[?]

Again, check over \({\mathbb{C}}\)!

\end{proof}

\begin{claim}

The truth of the Tate conjecture is preserved under base change: given
\(K/k\) a finite extension, the Tate conjecture for \(K\) implies the
Tate conjecture for \(k\).

\end{claim}

\begin{proof}[?]

Without loss of generality, let \(K'/k\) be finite Galois.

\begin{claim}

For semisimplicity, the claim is that if \(\Gamma' \leq \Gamma\) is a
subgroup of finite index and
\(\rho: \Gamma\to \operatorname{GL}_n({ {\mathbb{Q}}_\ell })\), then if
\({ \left.{{\rho}} \right|_{{\Gamma'}} }\) semisimple implies \(\rho\)
is semisimple.

\end{claim}

\begin{proof}[?]

Proof: get a splitting \(\rho \xrightarrow{s} \rho'\) and use an
averaging argument on \(\Gamma/\Gamma'\) (a la Maschke's theorem). This
works since \({ {\mathbb{Q}}_\ell }\) is positive characteristic.

\end{proof}

\begin{claim}

For the homs, the claim is that
\(\mathop{\mathrm{Hom}}(A_{K'}, B_{K'}) \otimes{ {\mathbb{Z}}_\ell }\to \mathop{\mathrm{Hom}}(T_\ell(A), T_\ell(B))\)
is bijective.

\end{claim}

\begin{proof}[?]

Noting that we already have injectivity over any field. Let
\(\sigma \in \mathop{\mathrm{Hom}}_{G_k}(T_\ell A, T_\ell B)\), then
there exists \(f: A\to B\) over \(k\) such that \(T_\ell(f) = \sigma\).
Since \(f\) is \(G_k{\hbox{-}}\)equivariant, we get density by Galois
descent.

\end{proof}

Reduce to \(A=B\) by using \(A\times B\). Reduce to proving the
following:

\begin{claim}

If \(W \subseteq V_\ell(A)\) is a Galois stable subspace, there exists a
\(u\in \mathop{\mathrm{End}}(A)\otimes{ {\mathbb{Q}}_\ell }\) with
\(u(V_l(A)) = W\).

\end{claim}

\begin{proof}[?]

For semisimplicity, use that subrepresentations are continuous? Then use
a double coset trick.

\end{proof}

Reduce to showing this for \(W\) a maximal isotropic subspace for the
Weil pairing (a symplectic form). This is Zahrin's trick: replace \(A\)
with \(A{ {}^{ \scriptscriptstyle\times^{4} } }\), gives a way to
``complete'' any subspace \(W \subseteq T_\ell(A)\) to a maximal
isotropic. The proof uses the Lagrange 4-squares theorem, how neat!

Write
\begin{align*}
G_n \coloneqq{W \cap T_\ell(A) + \ell^n T_\ell(A) \over \ell^n T_\ell(A) } \subseteq A[\ell^n], && B_n \coloneqq A/G_n
.\end{align*}
Then if \(W\) is maximal isotropic, \(B_n\) is a PPAV and thus so is
\(A\). The actual theorem Tate proves:

\begin{theorem}[?]

It suffices to show that \(\left\{{B_n}\right\}_{n\geq 1}\) is finite.

\end{theorem}

\begin{proof}[Idea]

Write \(A \xrightarrow{\psi_n} A/G_n = B_n\) and
\(\psi_n {}^{ \vee }: B_n\to A\), and from the construction of \(G_n\)
and unwinding definitions yields
\begin{align*}
\psi_n(T_\ell(B_n)) = W \cap T_\ell(A) \cdot \ell^n T_\ell(A)
.\end{align*}
So \(\operatorname{im}(\psi_n)\) is ``converging'' to
\(W \cap T_\ell(A)\). Use that eventually the \(B_n\) stabilize to
compose
\(\gamma_n: A \xrightarrow{\psi_n} B_n \xrightarrow{\psi_n {}^{ \vee }} A\)
and replace this with \(\gamma_n:A\to B\to A\). Use that
\(\mathop{\mathrm{End}}(A)\otimes{ {\mathbb{Z}}_\ell }\) is a compact
space to get a convergent subsequence.

\end{proof}

\end{proof}

\begin{remark}

The upshot: there exist finitely many PPAVs isogenous to \(A\).

\end{remark}

\hypertarget{faltings-heights}{%
\subsection{Faltings Heights}\label{faltings-heights}}

\begin{remark}

Write
\(\tilde {\mathcal{A}}\to {\mathcal{A}}\coloneqq\coloneqq{\mathcal{A}}_{\mathrm{univ}, g}\to {\mathcal{A}}_g\)
for the universal family, and define a line bundle
\({\mathcal{L}}\coloneqq\bigwedge\nolimits^{\text{top}} s^* \Omega^1_{\tilde {\mathcal{A}}/ {\mathcal{A}}}\).
A fact is that \({\mathcal{L}}\) is amply, since there exists a
compactification of \({\mathcal{A}}_g\) to which \({\mathcal{L}}\) is
ample. This is a hard theorem and involves Siegel modular forms.

\end{remark}

\begin{remark}

Define a height function associated to \({\mathcal{L}}\):
\begin{align*}
\tilde h: {\mathcal{A}}_g({\mathbb{Q}}) \xrightarrow{{\mathcal{L}}} {\mathbb{P}}^N({\mathbb{Q}}) &\to {\mathbb{R}}\\
{\left[ { {x_0 \over x_i}:\cdots : {x_n\over x_i} } \right]} &\mapsto \max({\left\lvert {x_i} \right\rvert}, {\left\lvert {x_i'} \right\rvert} )
.\end{align*}
This is the height machine, but it's fairly incomputable.

\end{remark}

\begin{remark}

A definition by Faltings: for \(A\) semisimple, write
\begin{align*}
H(A) = \prod_{k\hookrightarrow{\mathbb{C}}} \int_{A({\mathbb{C}})} \eta \wedge \mkern 1.5mu\overline{\mkern-1.5mu\wedge \mkern-1.5mu}\mkern 1.5mu&& \eta \in H^0(\mkern 1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu_{/ { {\mathcal{O}}_K}} ; \Omega_A^q ) 
,\end{align*}
taking a Néron model
\(\mkern 1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu\) for
\(A\). Then define
\begin{align*}
h(A) = {1\over [k: {\mathbb{Q}}]} \log(H(A))
.\end{align*}

\end{remark}

\begin{theorem}[Faltings]

There exist constant \(c_1, c_2\) such that
\begin{align*}
{\left\lvert {h(A) - \tilde c_1 h(A) } \right\rvert} < c_2
\end{align*}
where the \(c_i\) do not depend on \(A\).

\end{theorem}

\begin{proposition}[?]

For \(f: A\to B\) an isogeny induces
\(\mkern 1.5mu\overline{\mkern-1.5muA\mkern-1.5mu}\mkern 1.5mu\to \mkern 1.5mu\overline{\mkern-1.5muB\mkern-1.5mu}\mkern 1.5mu\)
on Néron models, and
\begin{align*}
h(B) - h(A) = {1\over 2} \log \deg(f)\cdot {1\over [K:{\mathbb{Q}}] } \log \mathop{\mathrm{length}}s^* \Omega^1_{/ {?}} 
.\end{align*}

\end{proposition}

\addsec{ToDos}
\listoftodos[List of Todos]
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% Hook into amsthm environments to list them.
\addsec{Definitions}
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\addsec{Theorems}
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\addsec{Exercises}
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\addsec{Figures}
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\printbibliography[title=Bibliography]


\end{document}
