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\title{
\rule{\linewidth}{1pt} \\
\textbf{
    Elliptic Curves
  }
    \\ {\normalsize University of Georgia, Spring 2020} \\
  \rule{\linewidth}{2pt}
}
\titlehead{
    \begin{center}
  \includegraphics[width=\linewidth,height=0.5\textheight,keepaspectratio]{figures/cover.png}
  \end{center}
       \begin{minipage}{.35\linewidth}
    \begin{flushleft}
      \vspace{2em}
      {\fontsize{6pt}{2pt} \textit{Notes: These are notes live-tex'd
from a graduate course in Elliptic Curves taught by Pete Clark at the
University of Georgia in Spring 2020. 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}
\textbf{D. Zack Garza} \\
\textit{University of Georgia} \\
\textit{dzackgarza@gmail.com} \\
{\tiny \textit{Last updated:} 2020-10-28 }
\end{flushleft}


\newpage
\tableofcontents

\hypertarget{wednesday-january-8}{%
\section{Wednesday January 8}\label{wednesday-january-8}}

Reference: \autocite{silverman_2009}

\hypertarget{summary}{%
\subsection{Summary}\label{summary}}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Mordell-Weil theorem
\end{enumerate}

\begin{itemize}
\item
  For elliptic curves over global fields

  \begin{itemize}
  \tightlist
  \item
    Number fields, function fields, finite fields, etc.
  \end{itemize}
\item
  Proof uses Galois cohomology and height functions. Essentially one
  proof!
\item
  Holds for abelian varieties, but more difficult.
\item
  Need an analog of height functions, i.e.~an \(x\dash\)coordinate).
\end{itemize}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\item
  Height functions.
\item
  Elliptic curves over \(\QQ_p\) or complete discrete valuation
  fields\footnote{See Silverman for basics, possibly Chapter 5},
  particularly Tate curves.
\item
  Weil-Chatelet groups \(E/k\) related to \(H^1(k; E)\) with
  coefficients in the elliptic curve
\item
  Galois representation of \(E/k\) for \(\ch k = 0\), for
  \begin{align*}
    \rho_n: g_k \to \GL(2, \ZZ/n\ZZ)
    \end{align*} which leads to
  \begin{align*}
    \hat \rho: g_k \to \GL(\hat \ZZ)
    \end{align*}
\end{enumerate}

\hypertarget{mordell-weil-groups}{%
\subsection{Mordell-Weil Groups}\label{mordell-weil-groups}}

Let \(E/k\) be an elliptic curve over a field \(k\)\footnote{Silverman
  is good for foundations, but assumes \(k\) is a perfect field. Here
  we'll let \(k\) be arbitrary.}, i.e.~a smooth, projective,
geometrically integral curve of genus 1 with a \(k\dash\)rational point
\(O\).

\begin{remark}

If \(k\) is not algebraically closed, such a point \(O\) may not exist.

By Riemann-Roch (easy computation) \(E\) embeds (non-canonically) into
\(\Bbb{P}^2/k\) as a Weierstrass cubic \begin{align*}
y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6 \quad \Delta \neq 0
.\end{align*}

This is a smoothness condition, and this equation has a
\(k\dash\)rational point at infinity \([0: 1: 0]\). The line at infinity
is a flex line (?), and so only intersects this curve at one point. If
\(\ch k \neq 2,3\) then \(y^2 = x^3 + Ax + B\). Every elliptic curve is
given by a Weierstrass equation, although not in a unique way.

\end{remark}

\begin{fact}[An amazing one!]

The set of \(k\dash\)rational points \(E(k)\) form an abelian group with
zero as the identity.

\end{fact}

\begin{proof}[?]

\envlist

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Given any plane cubic \(C/k\) and an origin \(O \in C(k)\), the chord
  and tangent process yields a group structure. Note that there is a
  symmetry in connecting rational points \(a, b\) with a line an
  intersecting at another rational point \(c\) which is not present in
  most groups, so an additional inversion about \(O\) is needed to
  actually make this into a group. Proving associativity: difficult!
\item
  Look at \(\mathrm{Pic}^0 E\), the degree 0 divisors on \(E\) mod
  birational equivalence (?), which is equal to the degree 0 line
  bundles on \(E\) mod bundle isomorphism.
\end{enumerate}

\begin{exercise}[?]

Show there is a map \(C(k) \to \mathrm{Pic}^1 C\) given by sending \(p\)
is its equivalence class; this is a bijection by Riemann-Roch
(straightforward exercise).

\end{exercise}

We can then compose this with a map
\(\mathrm{Pic}^1 \to \mathrm{Pic}^0 C\) given by \(D \mapsto D - [O]\),
which decreases the degree by 1. This gives a map
\(\Phi: C(k) \to \mathrm{Pic}^0 C\), just need to check that
\(\Phi(P \oplus Q) = \Phi(P) + \Phi(Q)\). Check that the groups are
independent of the \(k\dash\)rational point chosen, i.e.~changing
rational points yields isomorphic groups. So the group law itself
\textbf{does} actually depend on the rational point, although the
structure doesn't.

\end{proof}

\begin{exercise}[?]

Let \((E, O)/k\) be an elliptic curve and define
\(E^0 = E\setminus \theset{0}\) the (nonsingular, integral) affine curve
given by removing the point at infinity. Then the affine coordinate ring
\(k[E^0]\) is defined as \(k[x, y]/(y^2 -x^3 - Ax - B)\), which is a
Dedekind ring.

\begin{quote}
The interesting thing about Dedekind domains: the ideal class group!
(i.e.~the Picard group)
\end{quote}

This has ideal class group \(\mathrm{Pic} k[E^0]\), and one can show
that

\begin{align*}
\pic^0 E &\to \pic k[E^0] \\
\sum_p n_p \deg(p) [p] &\mapsto \sum_{p\neq 0} n_p [p] = \prod_p p^{n_p}
\end{align*}

with the sum ranging over all closed points is an isomorphism.

\begin{quote}
Just note that the RHS can't have a point at infinity, so we just forget
it. The isomorphism follows from some exact sequence with correction
terms that vanish.
\end{quote}

So the Mordell-Weil group of \(E(k)\) is isomorphic to \(\pic k[E^0]\),
the class group of a dedekind domain (?).

\end{exercise}

\begin{definition}[Class Group and the Mordell-Weil Group]

Let \(G\) be a commutative group.

\begin{itemize}
\tightlist
\item
  \(G\) is a \textbf{class group} iff there exists a dedekind domain
  \(R\) such that \(G \cong \pic R\).
\item
  \(G\) is an \textbf{(elliptic) Mordell-Weil group} iff there exists a
  field \(k\) and an elliptic curve \(E/k\) such that \(G \cong E(k)\).
\end{itemize}

\end{definition}

\textbf{Questions}:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Which \(G\) are class groups?
\item
  Which \(G\) are Mordell-Weil groups?
\end{enumerate}

\textbf{Answer 1:}

\begin{theorem}[Clayborn, 1966]

Every commutative \(G\) is a class group.\footnote{Subsequent proofs:
  Leetham-Green (1972) and Clark (2008) following Rosen, and uses
  elliptic curves. See the end of Pete's Commutative Algebra notes!}

\end{theorem}

\textbf{Answer 2}: Consider \(E/\CC\), then
\(E(\CC) \cong S^1 \cross S^1\), so the torsion subgroup is
\begin{align*}
T(1) \definedas (\QQ/\ZZ)^2 = \bigoplus_{\ell} (\QQ_\ell/\ZZ_\ell)^2
.\end{align*} This in fact holds for any algebraically closed field of
characteristic zero.

\begin{fact}

For any \(E/k\), the Mordell-Weil group \(E(k)\) is
``\(T(1)\)-constrained'', i.e.~\(E(k)[\mathrm{tors}] \injects T(1)\).

\end{fact}

\begin{theorem}[Clark, 2012]

\(G\) is a Mordell-Weil group \(\iff G\) is \(T(1)\dash\)constrained.

\end{theorem}

\begin{remark}[Some open problems.]

The analogous statement for abelian varieties, i.e being \(T(g)\)
constrained for some other genus \(g\neq 1\), is open. Fixing \(k= \QQ\)
still yields very interesting problems. Computing the rank and torsion
subgroups is currently open, and the subject of modern research.

\end{remark}

\hypertarget{monday-january-13th}{%
\section{Monday January 13th}\label{monday-january-13th}}

\hypertarget{every-abelian-group-is-a-class-group}{%
\subsection{Every Abelian Group is a Class
Group}\label{every-abelian-group-is-a-class-group}}

\begin{description}
\tightlist
\item[Theorem (Claborn - Leedham - Green - Clark)]
Any commutative group is the class group of some Dedekind domain.
\end{description}

Also see: partial re-proof by Rosen that uses elliptic curves. This
theorem: mostly a proof in commutative algebra, see end of Pete's
commutative algebra notes.

\hypertarget{proof-sketch}{%
\subsection{Proof Sketch}\label{proof-sketch}}

Let \(E/k\) be an elliptic curve over a field.

\hypertarget{step-1}{%
\subsubsection{Step 1}\label{step-1}}

Note that \(\mathrm{End}_k(E) \cong_\ZZ \ZZ^{a(E)}\) where
\(a(E) \in \theset{1,2,4}\).

\begin{quote}
Could be \(\ZZ\) as a \(\ZZ\dash\)module, could be an order in the
imaginary quadratic field (e.g.~a quaternion algebra)
\end{quote}

There is a short exact sequence
\begin{align*}0 \to E(k) \to E(k(E)) \to \mathrm{End}_K(E) \to 0.\end{align*}
This splits because (as seen above), the RHS term is free and thus
projective. So
\begin{align*}E / k(E) \cong E(k) \oplus \ZZ^{a(E)}.\end{align*}

Note that \(k(E)\) is an extension of \(E_k\) to \(E_{k(E)}\) the field
of rational functions over \(k\)? (function field). To simplify, take
\(a(E) = 1\) and \(E(k) = \theset{0}\).

\begin{quote}
Taking \(k=\QQ\), this happens (probably, asymptotically) half of the
time. It's easy to write down an elliptic curve that satisfies these
conditions
\end{quote}

Then \(E/k(E) \cong \ZZ\).

Now pass to the field of rational functions over this field, taking
\(E(~k(E) ~(E/k(E))~)\). Then
\begin{align*}
k^2(E) \definedas k(E) (E/k(E))
\end{align*} and inductively define \(k^n(E)\) by passing to function
fields. So \(E(k^n (E)) \cong \ZZ^n\).

So we can construct elliptic curves that have any free commutative group
as their Mordell-Weil group.

\hypertarget{step-2}{%
\subsubsection{Step 2}\label{step-2}}

Loosely speaking, we'll iterate this process transfinitely. Then for any
set \(S\), there exists a field \(k\) and an elliptic curve \(E/k\) such
that \(E(k) \cong \bigoplus_S \ZZ\). We now want to introduce a process
that allows passing to quotients. And \(R \definedas k[E^0]\) is the
affine coordinate ring of ?, remove the point at infinity (?).

\hypertarget{step-3}{%
\subsubsection{Step 3}\label{step-3}}

Let \(R\) be a Dedekind domain. Note it has a fraction field with a
certain ideal class group. Let \(W \subset \maxspec(R)\), then
\begin{align*}R^W \definedas \intersect_{\pr \in \maxspec R\setminus W} R_\pr.\end{align*}
Then \(R^W\) is Dedekind (and every overring of a Dedekind domain is of
this form) and \(\maxspec(R^W) = \maxspec(R\setminus W)\).

Then
\begin{align*}\pic R^W = \pic R / \generators{ [\pr] \suchthat \pr\in W }.\end{align*}
Note that if \((A, +)\) is a commutative group, writing
\(A = \bigoplus_S \ZZ/H\), we have a Dedekind domain \(R = k[E^0]\) such
that \(\pic R = \bigoplus_S \ZZ\).

\begin{quote}
Note: \(\pic R\) is the class group.
\end{quote}

\begin{description}
\tightlist
\item[Definition (Replete)]
A Dedekind domain \(R\) is \textbf{replete} iff every element of the
class group \(\pic R\) is the class group \([\pr]\) of some ideal
\(\pr \in \maxspec(R)\).
\end{description}

\begin{quote}
Is every ideal class the class of a prime ideal? For \(k\) a field,
\(R = \ZZ_k\). This follows from Chebotarev Density (most important
theorem for arithmetic geometers!)
\end{quote}

\begin{description}
\tightlist
\item[Definition (Weakly Replete)]
A Dedekind domain \(R\) is \textbf{weakly replete} iff every subgroup
\(H \subset \pic R\) is generated by classes of prime ideals.
\item[Exercise (Easy)]
\(K[E^0]\) is weakly replete, and an easy application of Riemann-Roch
shows that if \(0\neq p \in E(k) = \pic k[E^0]\), then
\([p] \in \pic k[E^0]\) is generated by a prime ideal.
\end{description}

Note: most applications of Riemann-Roch to elliptic curves are easy! In
this case, it gives you an identification \(E \cong \pic^1(E)\).

So there exists a subset \(W \subset \maxspec k[E^0]\) such that
\(\generators{[p] \suchthat p\in W} = H\). Then
\begin{align*}
\pic k[E^0]^W \cong \bigoplus_S \ZZ/H \cong A
.\end{align*}

\(\qed\)

Note that Dedekind domains don't have to be replete or even weakly
replete. The class group of a Dedekind domain could be \(\ZZ\), and the
class of every prime ideal could be \(1 \in \ZZ\)

\begin{description}
\item[Proof (Claborn)]
Start with an arbitrary Dedekind domain \(R\) and attach one that's
replete.

Can ask for a similar result for abelian varieties, there are
conjectures here, few clear results. Need to get
\(\ZZ/(m) \cross \ZZ/(n)\), since these occur as Mordell-Weil groups.
Take a modular curve and a generic point. Look at universal elliptic
curves over elliptic curves and take their Mordell-Weil groups (?)

If \(k\) is algebraically closed and \(\ch k = p\), can't have
\(\ZZ(p) \cross \ZZ/(p)\). Consider the \(p\dash\)primary torsion
\(E_k[p^\infty]\). It is zero iff \(E\) is supersingular (no points of
order \(p\)). It is \(\QQ_p/\ZZ_p = \lim_{\rightarrow_n} \ZZ/(p^n)\) iff
\(E\) is ordinary.

\begin{quote}
Can sometimes reduce to cases where \(k=\CC\) and do things
analytically.
\end{quote}
\end{description}

\hypertarget{mordell-weil}{%
\subsection{Mordell-Weil}\label{mordell-weil}}

\begin{description}
\item[Theorem (Mordell-Weil)]
Let \(k\) be a global field (extension of \(\QQ\) or function field over
\(\FF_p\)) and \(E/k\) and elliptic curve. Then
\(E(k) \cong \ZZ^r \oplus T\) (by classification of abelian groups)
where \(T\) is finite, and \(T \cong \ZZ/(m) \oplus \ZZ/(n)\) for
\(m\divides n\). So \(T\) is generated by at most two elements.
\item[Proof (3 steps)]
\textbf{Step 1:} Weak Mordell-Weil theorem.

Take any \(n\geq 2\) and \(\ch k\) not dividing \(n\). Show that
\(E(k)/n E(k)\) is finite.

\textbf{Step 2:} Define a height function \(h: E(k) \to \RR\) satisfying
3 properties (see next time). This is approximately a quadratic form.

\begin{quote}
Decompose at places of a number field, see Number Theory II.
\end{quote}

\textbf{Step 3:} For any commutative group \(A\), there is a notion of a
height function
\begin{align*}h: A \to \RR.\end{align*} Show the Height Descent Theorem:
if \(A\) admits a height function and \(A/nA\) is finite for some
\(n\geq 2\), then \(A\) is finitely generated.

\begin{quote}
Also how you'd prove this theorem for abelian varieties, more difficulty
defining \(h\).
\end{quote}
\end{description}

\hypertarget{wednesday-january-15th}{%
\section{Wednesday January 15th}\label{wednesday-january-15th}}

\hypertarget{proving-the-mordell-weil-theorem}{%
\subsection{Proving the Mordell-Weil
Theorem}\label{proving-the-mordell-weil-theorem}}

Recall that we're trying to prove the Mordell-Weil theorem. Let \(K\) be
a global field, so it's the field of functions over some nice curve.
Then the Mordell-Weil group \(E(K)\) is finitely generated.

\textbf{Step 1:} The weak Mordell-Weil theorem for all \(n\geq 2\) with
\(\ch k\) not dividing \(n\), \(E(k) / n E(k)\) is finite.

\textbf{Step 2:} Construction of a height function \(h: E(K) \to \RR\)
that is ``trying'' to be a quadratic form.

\textbf{Step 3 (Today):} The Height Descent Theorem, i.e.~if \((A, +)\)
is a commutative group such that \(A/nA\) is finite for some \(n\geq 2\)
and it admits a heigh function \(h: A\to \RR\), then \(A\) is finitely
generated.

\hypertarget{step-3-proving-the-height-descent-theorem}{%
\subsubsection{Step 3: Proving the Height Descent
Theorem}\label{step-3-proving-the-height-descent-theorem}}

\emph{Question:} What does the weak Mordell-Weil group \(E(K)/ nE(K)\)
tell us about \(E(K)\)?

Note that we'll inject this into a larger group, which we'll show is
finite, but this isn't great for learning about the size.

\begin{description}
\item[Example]
Consider \(E/\CC\), then \(E(\CC) = S^1 \cross S^1\) and
\(E(\CC)/nE(\CC) = 0\), so the map \(x\to nx\) is a surjective map and
\(E(K)\) is \(n\dash\)divisible here. In general, whenever
\(K = \bar K\) is algebraically closed, then \(x \mapsto nx\) is again
surjective and the weak Mordell-Weil group is trivial. So knowing this
is small doesn't tell us much about \(E(K)\) at all.
\item[Example]
For \(E/\RR\), \(E(\RR)\) is either \(S^1\) (cubic with one real root,
\(\Delta = 0\)) or \(S^1 \cross \ZZ/(2)\) (cubic with three real roots,
\(\Delta > 0\)) are the two possible group structure.

Then

\begin{align*}
? = 
\begin{cases}
0 & n \text{ odd } \\
0 & n \text{ even and } \Delta < 0 \\
\ZZ/(2) & n \text{ even and } \Delta > 0
\end{cases}
.\end{align*}
\item[Example]
Consider \(E/\QQ_p\), then for all \(\ell \gg 0\)
\(E(\QQ_p) \mapsvia{[\ell]} E(\QQ_p)\) with
\(E(\QQ_p)/\ell E(\QQ_p) = 0\) while \(E(\QQ_p) / p E(\QQ_p)\) is not
zero.
\end{description}

Note: here is an example of a Boolean space, that ends up being
homeomorphic to a Cantor set.

Suppose \(E(K)\) is finitely generated, so \(E(K) \cong \ZZ^r \oplus T\)
with \(T\) finite. Then knowing \(E(K)/ n E(K)\) gives an upper bound on
\(r\).

\begin{description}
\item[Example]
Take \(n=2\), then \(E(K) / nE(K) \cong (\ZZ/(2))^s\) for some
\(s\in \NN\). Then
\begin{align*}
(\ZZ^r \oplus T) / 2(\ZZ^r \oplus T) \cong (\ZZ/(2))^r \oplus T/2T
\end{align*} for \(r\leq s\). Then either

\begin{itemize}
\tightlist
\item
  \(r = 2\) and \(E(K[2]) = (0)\).
\item
  \(r=1\) and \(E(K[2]) \cong \ZZ/(2)\),
\item
  \(r = 0\) and \(E(K[2]) \cong (\ZZ/(2))^2\).
\end{itemize}
\end{description}

Note that we don't need the Mordell-Weil theorem to compute the torsion
subgroups of \(E(k)\). It is often easier to compute these directly. For
all non-archimedean places \(v\) of \(K\), \(E(K_v)[\text{tors}]\) is
finite (see Silverman?) and embeds into a number of finite things.

To compute \(E(K_v)[\text{tors}]\),

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Find \(N \in \ZZ^+\) such that \(E(k)[\text{tors}] \subset E[N]\).
\end{enumerate}

\begin{itemize}
\item
  Choose 2 different places \(v_0, v_1\) of good reduction (from
  Weierstrass equation) with different residue characteristics
  \(\ell_1 \neq \ell_2\)
\item
  Consider the map \(E(K_{v_i})[\text{tors}] \to E(\FF_{v_i})\)
\item
  The kernel is a finite \(p_i\dash\)primary group.
\item
  Comes down to torsion and formal groups, see first course.
\end{itemize}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  Compute \(E[N](K)\) (several algorithms, just checking for rational
  points on a zero-dimensional variety?)
\end{enumerate}

\begin{quote}
See division polynomials, can check for roots of polynomials over any
global field. Easy to check for rational points on finite fields.
\end{quote}

Suppose \(E(K) \cong \ZZ^r \oplus T\) is finitely generated and we know
\(E(K) / nE(K)\) for some \(n\) and we know \(T\). Then we explicitly
know \(r\).

\begin{quote}
See Tate Shafarevich group -- important! But difficult, a piece of
information that helps compute the rank (?).
\end{quote}

\begin{description}
\tightlist
\item[Definition (Height Functions)]
Fix \(n\geq 2\). An \(n\dash\)\textbf{height function} on \((A, +)\) is
a map \(h: A\to \RR\) satisfying
\end{description}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  For all \(R\geq 0\), the set \(h\inv(-\infty, R)\) is finite.
\item
  For all \(Q\in A\), there exists a \(C_2 = C_2(A, Q)\) such that for
  all \(P \in A\),
  \begin{align*}
    h(P + Q) \leq 2h(P) + C_2
    (?).\end{align*}
\item
  There exists a \(C_3 = C_3(A, n)\) such that for all \(P \in A\),
  \begin{align*}
    h(nP) \geq n^2 h(P) - C_3
    \end{align*}
\end{enumerate}

Note: (3) would be an equality for an honest quadratic function, so this
deviates in a controlled way.

\begin{description}
\item[Theorem (Height Descent)]
Let \((A, +)\) be a commutative group with an \(n\dash\)height function
\(h: (A, +) \to \RR\). If \(A/nA\) is finite, then \(A\) is finitely
generated.
\item[Proof]
Let \(r\) be the size of \(A/nA\). Choose coset representatives
\(Q_1, \cdots, Q_r\) of \(nA\) in \(A\). Let \(p\in A\) and define a
sequence \(\theset{P_k}_{k=0}^\infty\) in \(A\) by \(P_0 = P\) and for
\(k\geq 1\), choose \(P_k\) such that \(P_{k-1} = nP_k + Q_{i_k}\). Then
for all \(k\in \ZZ^+\), it's true that
\(P = n^k P_k + \sum_{j=1}^k n^{j-1} Q_{i_j}\).

\begin{description}
\tightlist
\item[Claim]
There exists a constant \(c > 0\) depending only on \(A, n\) such that
for all \(P \in A\), there exists a \(K = K(P\) such that for all
\(k \geq K\), we have \(h(P_k) \leq 0\).
\end{description}

Note that this is sufficient -- if so, \(A\) is generated by
\(\theset{Q_1, \cdots, Q_r} \union h\inv((-\infty, C])\), which are both
finite.
\end{description}

Next time: proof of claim.

\begin{quote}
Note: similar setup goes through for abelian varieties, see Néron--Tate
height canonical height, which yields an honest ``quadratic form''.
\end{quote}

\hypertarget{friday-january-17th}{%
\section{Friday January 17th}\label{friday-january-17th}}

\hypertarget{continuing-step-3}{%
\subsection{Continuing Step 3}\label{continuing-step-3}}

Recall the Height Descent Theorem (see previous notes). Most important
property of height function: the collection of elements under a given
height is finite.

\begin{quote}
Note that \(A/nA\) is the cokernel of multiplication by \(n\).
\end{quote}

\begin{description}
\tightlist
\item[Proof]
Let \(r\) be the size of \(A/nA\) and choose coset representatives
\(Q_1, \cdots, Q_r\). For \(P\in G\) (?) define \(P_0 = P\) and \(P_k\)
such that \(P_{k-1} = n P_k + Q_i\) for any \(i\).
\end{description}

For all positive \(k \in \ZZ\), we have \(P = n^k P_k + \sum n^j Q_i\).

\begin{description}
\item[Claim]
There exists a \(c> 0\) such that for all \(P \in A\) there exists a
\(K = K(P)\) such that for all \(k\geq K\), \(h(P_k) \leq C\). If this
holds, \(A\) is generated by
\(\theset{Q_i} \union h\inv((-\infty, C])\).
\item[Proof (of claim)]
Let \(c_2 = \max_{1\leq i \leq r} c_2(-Q_i)\).

Then

\begin{align*}
h(P_k) 
&\leq \frac 1 {n^2} \qty{ h(nP_k) + c_3  } \\
&= \frac 1 {n^2} \qty{ h(P_{k-1} - Q_i) + c_3  } \\
&\leq \frac 1 {n^2} \qty{ 2h(P_{k-1}) + c_2 + c_3  } \\
&\leq \frac 1 {n^2} \qty{ \frac 2 {n^2} \qty{ 2h(P_{k-1}) + c_2 + c_3  } + c_2 + c_3} \quad \text{by repeating} \\
&= \qty{ \frac{2}{n^2}  }^2 h(P_{k-2}) + (1 + \frac{2}{n^2})(c_2 + c_3) \\
&= \qty{\frac 2 {n^2}  }^k h(P) + \frac 1 {n^2} \qty{1 + 2/n^2 + (2/n^2)^2 + \cdots (2/n^2)^k   }(c_2 + c_3) \\
&\leq \qty{\frac{2}{n^2}  }^k h(P) + \qty{ \frac{1}{1 - \frac{2}{n^2}}  }(c_2 + c_3) 
,\end{align*}

where the last inequality follows because \(n \geq 2\) implies the
leading term is bounded by 1 and the middle term contains a convergent
series.

This proves the claim for any \(n\)?
\item[Definition (Linear and Quadratic Forms on Groups)]
A function \(h:A \to \RR\) from a commutative group is
\textbf{quadratic} if the associated function

\begin{align*}
B_h:A^2 &\to \RR \\
(x, y) &\mapsto h(x+y) - h(x) - h(y)
.\end{align*}

is bilinear. The function \(h\) is \textbf{linear} iff \(B_h\) is
constant.

The function \(h\) is a \textbf{quadratic form} iff \(h\) is quadratic
and for all \(m\in \ZZ\) and for all \(x\in A\), \(h(mx) = m^2 h(x)\),
i.e.~a degree 2 homogeneous function.
\item[Theorem (Canonical Height Descent)]
Suppose \((A, +)\) is commutative and \(h: A\to \RR\) is a quadratic
form. Suppose

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(A/nA\) is finite, and
\item
  \(h\inv((-\infty, R])\) is finite for all \(R\),
\end{enumerate}

then letting \(y_1, \cdots, y_r \in A/nA\) be coset representatives and
taking \(C = \max h(y_i)\), we can conclude that \(A\) is generated by
\(\theset{x\in A \suchthat h(x) \leq C}\).
\end{description}

\hypertarget{step-4}{%
\subsection{Step 4}\label{step-4}}

\begin{description}
\item[Theorem (Abstract Weak Mordell-Weil)]
Let \(k\) be a field, \(E/k\) an elliptic curve, choose \(n\) such that
\(\ch k\) doesn't divide \(n\), and let \(k' \definedas k(E[n])\) be
\(k\) with the \(n\dash\)torsion points of \(E\) adjoined. Note that
this adjoins finitely many algebraic points to \(k\).

Suppose there exists a Dedekind domain \(R\) with fraction field \(k'\)
with finite class group, so \(\Pic(R) < \infty\), and \(R\units\) is
finitely generated Then \(E(k) / n E(k)\) is \textbf{finite}.
\item[Corollary]
Let \(k\) be a global field \(n\geq 2\), then \(E(k)/ n E(k)\) is
finite.
\item[Proof (of Corollary)]
\(k\) is a number field, so is \(k'\). Pick \(k' = \ZZ_k\), which is a
Dedekind domain. By Number Theory I, the hypotheses above are satisfied.

If \(k\) is a function field, \(k/\FF_p(t)\) is finite and separable, so
\(k' / \FF_p(t)\) is finite and separable. For \(A = \FF_p(t)\),
\(A \subset \FF_q(t)\), then take \(R/A \subset k'/\FF_q(t)\) the
integral closure of \(A\) in \(k'\). By Number Theory I, \(R\) is a
Dedekind domain.

\includegraphics{figures/2020-01-17-12:57.png}\\

Then \(R = \FF_p[C^0]\), and by Number Theory II, \(\Pic(R)\) is finite.

\begin{quote}
Removing primes makes unit group larger and the class group smaller.
\end{quote}

\begin{quote}
Localizing at a prime ideal yields a DVR? This kills the Picard group
(since it's a PID?) but blows up the units group.
\end{quote}

\begin{quote}
Note that the proof for abelian varieties adapts very easily.
\end{quote}
\end{description}

\hypertarget{sketch-proof-of-abstract-mordell-weil}{%
\subsection{Sketch Proof of Abstract Mordell
Weil}\label{sketch-proof-of-abstract-mordell-weil}}

\textbf{Step 1:} Reduce to the case that \(E\) has full
\(n\dash\)torsion, i.e.~\(k' = k\). If \(L/k\) is finite Galois (as is
\(k'/k\)), and \(E(L)/nE(L)\) is finite, then \(E(k) / nE(k)\) is
finite.

\begin{description}
\tightlist
\item[Remark]
For any extension \(L/k\), there is an injection \(E(k) \injects E(L)\),
but \(E(k) / nE(k)\) need not inject into \(E(L)/n E(L)\). For
counterexamples, take \(k = \RR\) and \(\CC/k\), then
\(E(\CC) / nE(\CC)\) can be trivial.
\end{description}

\textbf{Step 2:} Let \(L\definedas k([n]\inv E(k))\) be the compositum
\(k[\theset{P}]\) over the \(P\in E/\bar k\) such that
\([n] P \in E(k)\) is \(k\dash\)rational. It's straightforward to show
that \(L\) is separable and Galois (it is an etale covering). That it's
galois: if \([n]P\) is rational, so is \([n] \sigma(P)\) for any
\(\sigma\) in the galois group. We'll show that this is a finite
extension.

\textbf{Step 3:} Construct a Kummer pairing to show that finiteness of
\([L: k]\) is equivalent to \(E(k)/nE(k)\) being finite.

\textbf{Step 4:} Reduce finiteness of \([L: k]\) to algebraic number
theory.

\hypertarget{wednesday-january-22nd}{%
\section{Wednesday January 22nd}\label{wednesday-january-22nd}}

\hypertarget{stronger-weak-mordell-weil}{%
\subsection{Stronger Weak Mordell
Weil}\label{stronger-weak-mordell-weil}}

\begin{description}
\item[Theorem (Stronger Weak Mordell-Weil)]
Let \(n\geq 2\), \(k\) a field, \(\ch(K)\) not dividing \(n\), \(E/k\)
an elliptic curve \(K \definedas K(E[n])\).

Suppose there exists a Dedekind domain \(R\) with fraction field \(K\)
such that

\begin{itemize}
\tightlist
\item
  \(\pic R[n]\) is finite
\item
  The cokernel of \(x\mapsto nx\) is finite
\end{itemize}

Then \(E(K)/ n E(K)\) is finite.
\item[Exercise]
Take \(k = \bar k\), \(C/k\) a ``nice'' affine curve, \(R = k[C]\),
\(K = k(C)\), \(\ch(k)\) not dividing \(n\). Show that if \(E/k\) is any
elliptic curve, then the hypotheses of Stronger Weak Mordell-Weil hold,
as does the conclusion, and in fact \(E(K)\) need not be finitely
generated.
\end{description}

\begin{quote}
Uses ANT II.
\end{quote}

\hypertarget{step-1-1}{%
\subsubsection{Step 1}\label{step-1-1}}

Let \(L/K\) be a Galois extension and consider
\begin{align*}
\iota: E(K)/n E(K) \to E(L)/ n E(L)
.\end{align*} This is not injective in general, and in fact
\begin{align*}
\ker(\iota) = (E(K) \intersect nE(L))/ nE(K)
.\end{align*}

\begin{description}
\item[Proposition]
\hfill

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  There exists a map \(\ker \iota \to \maps(g_{L/K}, E[n])\), where
  \(g_{L/K}\) is the Galois group of \(L/K\).
\item
  If \(L/K\) is finite, then \(\ker \iota\) is finite
\end{enumerate}
\end{description}

So if \(E(L) / nE(L)\) is finite, then \(E(K) / nE(K)\) is finite.

\begin{description}
\tightlist
\item[Proof]
Let \(p\in E(K) \intersect n E(K)\). Choose \(Q_p \in E(L)\) (only
determined up to an \(n\dash\)torsion point) such that \([n] Q_p = p\).
For all \(\sigma \in g_{L/K}\) we have
\end{description}

\begin{align*}
  [n] (\sigma(Q_p) - Q_p) = \sigma([n] Q_p) - [n] Q_p = \sigma(p) - p = 0
  .\end{align*}

Thus \(\sigma(Q_p) - Q_p \in E[n]\).

\begin{quote}
Note: this resembles a certain coboundary map in a cohomology theory.
\end{quote}

We then associate a map

\begin{align*}
  \lambda_p : g_{L/K} \to E[n] \\
  \sigma \mapsto \sigma(Q_p) - Q_p
  .\end{align*}

Suppose that for \(p, p' \in E(K) \intersect n E(L)\), so
\(\lambda_p = \lambda_{p'}\). Then for all \(\sigma \in g_{L/K}\), we
have

\begin{align*}
  \sigma(Q_p - Q_{p'}) &= \sigma(Q_p) - Q_p - (\sigma(Q_{p'}) - Q_{p'} ) + (Q_p - Q_{p'}) \\
  &= \lambda_p(\sigma) - \lambda_{p'}(\sigma) + Q_p - Q_{p'} \\
  &= Q_p - Q_{p'}
  .\end{align*}

So \(Q_p - Q_{p'} \in E(K)\), and thus
\(p - p' = [n] (Q_p - Q_{p'}) \in nE(K)\). Thus
\begin{align*}\ker \iota = \frac{E(K) \intersect nE(L)}{nE(K)}.\end{align*}

From now on, we'll assume \(K = K' = K(E[n])\).

\hypertarget{step-2-1}{%
\subsubsection{Step 2}\label{step-2-1}}

Define the Kummer pairing.

Let \(L = K\sep\) and take \(p \in E[K]\) such that \([n] Q_p = p\), and
define

\begin{align*}
\lambda_p : g_k \leq \aut(K\sep/K) &\to E[n] \\
\sigma &\mapsto \sigma(Q_p) - Q_p
.\end{align*}

Note that since \(E[n] = E[n](K)\), this no longer depends on the choice
of \(Q_p\).

Define a Kummer pairing

\begin{align*}
\kappa: E(k) \cross g_K &\to E[n] \\
(p, \sigma) &\mapsto \sigma Q_p - Q_p
.\end{align*}

\begin{description}
\tightlist
\item[Proposition]
This pairing satisfies the following properties:
\end{description}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  For all \(p, q\in E(K)\) and \(\sigma \in g_k\),
  \(\kappa(p+q, \sigma) = \kappa(p, \sigma) + \kappa(q, \sigma)\)
\item
  For all \(p \in E(K)\) and \(\sigma, \xi \in g_k\),
  \(\kappa(p, \sigma \xi) = \kappa(p, \sigma) + \kappa(p, \xi)\).
\item
  For all \(p\in E(K)\), we have \(\kappa(p, \sigma) = 0\) for all
  \(\sigma\) iff \(p\in nE(K)\).
\item
  For all \(\sigma \in g_K\), we have \(\kappa(p, \sigma) = 0\) for all
  \(p\in E(K)\) iff \(\sigma \in g_L\) where
  \(L\definedas K([n]\inv E(K)) = K(Q_p/P \in E(K))\).
\end{enumerate}

So \(\kappa\) induces group homomorphisms \begin{align*}
  g_K / g_L = g_{L/K} &\injects \hom(E(K) / nE(K), E[n]) \\
  E(K)/ nE(K) &\injects \hom(g_{L/K}, E[n])
  .\end{align*}

\begin{quote}
Thus \(E(K) / n E(K)\) is finite iff \(g_{L/K}\) is finite iff
\([L: K]\) is finite.
\end{quote}

Thus \(g_{L/K}\) is abelian of exponent dividing \(n\). So we can study
this using Kummer theory and class field theory.

\begin{description}
\tightlist
\item[Proof (of 1)]
Can take \(Q_{p + p'} = Q_p + Q_{p'}\), which is a fine choice, and then
\(\kappa(p + p', \sigma) = \kappa(p, \sigma) + \kappa(p', \sigma)\).
\item[Proof (of 2)]
\begin{align*}
  \kappa(p, \sigma \xi) 
  &= \sigma\xi Q_p - Q_p - \sigma \xi Q_p - \sigma Q_p + \sigma Q_p - Q_p \\
  &= \sigma(\xi Q_p - Q_p) + \kappa(p, \sigma) = \sigma(\kappa(p, \xi) + \kappa(p,\sigma)) \\
  &= \kappa(p, \xi) + \kappa(p, \sigma)
  .\end{align*}
\item[Proof (of 3)]
For \(\sigma \in g_K\), then \(\kappa(E(K), \sigma) = (0)\) iff
\(\sigma Q_p - Q_p = 0\) for all \(P \in E(K)\) iff \(\sigma\) pointwise
fixes \(L\) iff \(\sigma \in g_L\).
\item[Exercise]
Replace \(E/k\) with \(A/k\) a commutative group scheme such that
\([n]: A\to A\) is etale and \(A[n]\) is finite. The proof goes through
without modification if \(\ch k\) doesn't divide \(n\) and \(A/k\) is an
algebraic group, reduced, and of finite type.
\item[Exercise]
Take \(A = \GG_m\), the multiplicative group of \(K\). Then
\(A(K) / nA(K) = K\units/K^{\cross n}\). Suppose \(K\) contains \(n\)th
roots of unity, then regular Kummer theory gives a map
\(K\units/K^{\cross n} \cong \hom(g_K, \mu_n)\) where \(\mu_n\) are
\(n\)th roots of unity.
\end{description}

This says that \(K\units/K^{\times n} = \chi(g_{L/K})\) where \(L\) is
the maximal abelian extension of \(K\) of exponent dividing \(n\), and
\(\chi(\wait) = \hom(\wait, \ZZ/n\ZZ)\), i.e.~these are Pontryagin
duals.

So far, works for any algebraic group, but we'll need properness later.

\hypertarget{friday-january-24th}{%
\section{Friday January 24th}\label{friday-january-24th}}

\hypertarget{stronger-weak-mordell-weil-1}{%
\subsection{Stronger Weak
Mordell-Weil}\label{stronger-weak-mordell-weil-1}}

Let \(K\) be a field, \(n\geq 2\) with \(\ch k\) not dividing \(n\), and
\(A/K\) a commutative algebraic group (includes abelian varieties,
additive/multiplicative groups, etc). Assume that \(K\) contains all
\(n\)th roots of unity (we showed that this can be assumed).

Let
\begin{align*}L \definedas K([n] \inv A(K)) = K(\theset{Q \in A(K\sep) \suchthat [n] Q \in A(K)}).\end{align*}
We've shown that \(L/K\) is Galois, and moreover abelian of exponent
dividing \(n\) and \(L/K\) is finite iff \(A(K) / n A(K)\) is finite.

\begin{quote}
Take \(K = \QQ\), then \(\QQ[p]\) for every \(p\) gives infinite
extensions.
\end{quote}

We want to show that if \(A=E\) is an elliptic curve (or an abelian
variety) and \(K\) is the fraction field of some Dedekind domain \(R\)
with some finiteness condition on \(\pic(R)\) and \(R\units\), then
\(L/K\) is finite.

\hypertarget{step-3-1}{%
\subsubsection{Step 3}\label{step-3-1}}

Let \(\pr \in \maxspec(R)\) with \(\pr\) not dividing \(n\) such that
\(E\) has good reduction at \(\pr\).

\begin{quote}
Take an \(R\dash\)integral Weierstrass equation \(W/R\) for \(E\), then
the discriminant satisfies \(\Delta(W) \neq 0\). So just exclude the
(finitely many) primes where \(\pr \divides \Delta(W)\). For abelian
varieties, reducing the equations mod \(p\) can result in singularities
for only finitely many \(p\).
\end{quote}

Then \(L/K\) is unramified at \(\pr\) (i.e.~it's ramified at only
finitely many primes).

\emph{Proof:} We have
\begin{align*}
L = \prod_{Q \suchthat n[Q] = P \in E(K)} L(Q)
\end{align*} as a compositum of extensions, so it's enough to show that
\begin{align*}
L_p \definedas \prod_{[n] Q = p, ~p \in E(K)} K(Q)
\end{align*} is unramified over \(K\).

Take integral closures:

\begin{figure}
\centering
\includegraphics{figures/2020-01-24-12:34.png}
\caption{Image}
\end{figure}

Take the inertia group
\begin{align*}
I \definedas I(\pr \mid p) = \theset{\sigma \in g_{L_p/k} \sigma(\pr) = \pr, \sigma\actson S_p/\pr = \id} \in \aut(L_p/K)
.\end{align*}

We want to show that \(\forall \sigma \in I\), \(\sigma(Q) = Q\). We
have
\begin{align*}
0 = \sigma(P) - P = \sigma([n]Q) = [n]Q = [n](\sigma Q - Q)
\end{align*} and thus \(\sigma Q - Q \in E[n] = E[n](K)\).

We now introduce the reduction map

\begin{align*}
r: E(L_p) \to \tilde E(S/\pr)
,\end{align*}

where we use the fact that we can complete at \(\pr\) and then take a
reduction to obtain a map \(E(L_p) \to E(\hat L_p) \to \tilde E(S/p)\),
which is where we use the fact that the reduction is good.

We know \(\sigma Q - Q\) is \(n\dash\)torsion. Then \(r\) is a
homomorphism, so \(r(\sigma Q - Q) = r(\sigma Q) - r(Q) = 0\) by the
definition of the inertia group. So \(\sigma Q - Q\) is an
\(n\dash\)torsion point in the kernel of the reduction map.

Fact from elliptic curves I (Silverman Ch. 7), the only torsion in the
kernel of the reduction map is \(\ch (S/p)\dash\)primary torsion. Since
\(\ch(S/p)\) does not divide \(n\) here, \(\ker r = 0\), so
\(\sigma Q = Q\).

\(\qed\)

\begin{quote}
In the case of abelian varieties: The kernel of a good reduction is a
formal group law of \(g\) dimensions. Reference: Prop 3.1, Clark and
Xales (?) 2001. Here is where it doesn't work for \(\GG_m\). Projective
variety: clear denominators for the fraction field, now need to look at
integral points?
\end{quote}

\hypertarget{step-4-number-theory}{%
\subsubsection{Step 4: Number Theory}\label{step-4-number-theory}}

Let \(S \subset \maxspec(R)\) be the finite set of \(p\) such that
\(p \divides n\) (using global characteristic assumption here) or \(E\)
has bad reduction at \(p\). Then take \(L = K([n]\inv E(K))\). We know
\(L/K\) is abelian of exponent dividing \(n\), and is a compositum of
extensions \(L_Q\), each of bounded degree at most \(n^2\), each
unramified outside of \(S\).

\begin{quote}
\(Q\) is an \(n\) division point of something \(k\) rational.
Multiplication by \(n\) has degree \(n^2\), so \([K_Q : K] < n^2\).
\end{quote}

\begin{description}
\tightlist
\item[Theorem (Hermike - Minkowski)]
Let \(d \in \ZZ^+\), \(K\) a number field, and
\(s \subset \maxspec \ZZ_\ell\) finite. Then
\begin{align*}\#\theset{\text{ $L/K$ of degree $d$ unramified outside of $S$}} < \infty.\end{align*}
\end{description}

\begin{quote}
One of a few finiteness theorems in elliptic curves/NT, probably the
hardest one. Galois extension like branched cover, too few preimages.
Take surface with marked points (corresponding to prime) and take all
branched covering spaces that fix the degree and only ramify at those
points. I.e. take unramified coverings. Then \(\pi_1\) is free and f.g.,
which is the analogy here.
\end{quote}

\emph{Proof:} See Neukirch Ch. 3, 2.13, or Milne's ANT Theorem 8.42. Use
Hermite's theorem that any given \(\Delta \in \ZZ\) can only be the
discriminant \(\Delta(K)\) for finitely many \(K\).

If \(K\) is a number field, then take \(R = \ZZ_k\) and we're done.

Otherwise, let
\(R_s \definedas \intersect_{\pr \in \maxspec(R\setminus S)} R_{\pr}\).
Then \(R_s\) is a Dedekind domain,
\(\maxspec R_s = \maxspec(R\setminus S)\). Also \(\pic(R_s)\) is a
quotient \(\pic(R) / \generators{[\pr] \suchthat \pr \in S}\).

Assume \(\pic R\) is finitely generated, then so is \(\pic R_s\). If so,
by enlarging \(S\), we can make \(R_s\) a PID.

\begin{quote}
Note \(R_s\) is not necessarily a localization, although it could be, so
we expect the unit group to get bigger.
\end{quote}

\begin{description}
\item[Theorem (Pete's CA Notes Lemma 23.4)]
Let \(R\) be a Dedekind domain and \(R\subset T \subset K\). Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(T\units/R\units\) is torsion-free.
\item
  Let \(\pr \in \maxspec ?\) and define
  \(T \definedas \intersect_{q\neq \pr} R_q\). TFAE:
\end{enumerate}

\begin{itemize}
\tightlist
\item
  \(T\units / R\units = \ZZ\).
\item
  \(T\units \supsetneq R\units\).
\end{itemize}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{2}
\tightlist
\item
  \([p] \in \pic R[?]\).
\end{enumerate}
\end{description}

So killing off primes either adds factors of \(\ZZ\) or does nothing at
all.

By the lemma, passing from \(R\) to \(R_s\) by killing finitely many
maximal ideals, if \(\Pic R\) and \(R\units\) are both finitely
generated, then they are still finitely generated for \(R_s\). Thus we
can assume \(R_s\) is a PID.

\begin{quote}
Proof of theorem next time. Need ramification and taking \(n\)th roots
in local fields.
\end{quote}

\hypertarget{monday-january-27th}{%
\section{Monday January 27th}\label{monday-january-27th}}

\hypertarget{weak-mordell-weil-finishing-the-proof}{%
\subsection{Weak Mordell-Weil: Finishing the
Proof}\label{weak-mordell-weil-finishing-the-proof}}

Recall that \(n\geq 2\), \(K\) a field with \(\ch K\) not dividing
\(p\), \(E/K\) an elliptic curve, \(R\) a Dedekind domain with fraction
field \(K(E[n])\) such that \(\pic R\) and \(R\units\) are finitely
generated. Then \(E(K) / nE(K)\) is finite.

We've shown:

\begin{itemize}
\tightlist
\item
  May take \(K = K(E[n])\), so \(K\) contains the \(n\dash\)torsion and
  all \(n\)th roots of unity \(\mu_n \in K\).
\item
  \(R\) a PID with fraction field \(K\),
\item
  \(E(K) / n E(K)\) is finite iff \(L\definedas K([n]\inv E(K))\) is a
  finite degree extension over \(K\).
\item
  \(L/K\) is abelian of exponent dividing \(n\)
\item
  \(L/K\) is only ramified over \(\pr \in \spec(R)\) if \(p\divides n\)
  or \(E\) has bad reduction at \(\pr\) (so bad reduction at only
  finitely many primes).
\end{itemize}

Last step of the proof:

\hypertarget{applying-kummer-theory-to-show-lk-is-finite}{%
\subsubsection{\texorpdfstring{Applying Kummer theory to show \(L/K\) is
finite}{Applying Kummer theory to show L/K is finite}}\label{applying-kummer-theory-to-show-lk-is-finite}}

\begin{description}
\item[Lemma]
Let \(K\) be a field with a discrete valuation
\(v: K \surjects \ZZ \union \theset{\infty}\) such that \(v(n) = 0\).
When are Kummer extensions unramified? Every Kummer extension \(L\) is a
compositum \(L \definedas K(a^{1/n})\). Then \(L/K\) is unramified at
\(v\) iff \(n\) divides \(v(a)\).
\item[Proof (of Lemma)]
\hfill

\(\implies\): We can extend \(v\) to \(L\), and \(v\) is unramified iff
\(v(L\units) = \ZZ\). Noting that \(v(a^{1/n}) = \frac 1 n v(a)\), if we
suppose \(v(L\units) = \ZZ\), we're done.

\(\impliedby\):

\begin{quote}
Can check being unramified by passing to completion.
\end{quote}

Pass from \(K\) to the completion \(K_v\) and let \(\pi\) be a
uniformizer, so \(v(\pi) = 1\). Then define \(a' = a/\pi^{v(a)}\), so
\(v(a') = v(a) - v(a) = 0\) and thus \(a \in R_v\units\). Writing
\begin{align*}
a/a' = \pi^{v(a)} \in (K\units)^n
,\end{align*} so \(K(a^{1/n}) = K((a')^{1/n})\). So we reduce to the
case \(v(a) = 0\).

Since \(R_v\) is integrally closed, we have
\(x\in K_v^{\times n} \iff x\in R_v^{\times n}\), so there is an
isomorphism

\begin{align*}
R_v\units / R_v^{\times n} \cong k_v\units / k_v^{\times n}
\end{align*}

which follows from Hensel's Lemma.

Thus \(K_v(a^{1/n})/K\) is unramified.

\begin{quote}
Standard argument from NT II.
\end{quote}
\end{description}

Now let \(S\subset \spec R\) be the set
\begin{align*}
S = \theset{\pr \in \spec R \suchthat \pr \divides n \text{ or } E  \text{ has bad reduction at } \pr }
,\end{align*} and
\begin{align*}
T_S =  \theset{ a\in K\units / K^{\times n} \suchthat \forall \pr\in \spec(R\setminus S),~ n \divides \ord_p(a)}
.\end{align*}

We will be done if \(T_S\) is finite.

\begin{description}
\tightlist
\item[Claim]
There exists a surjective group homomorphism \(\psi: R\units \to T_S\).
\end{description}

Then since \(R\units\) is finitely generated, so is its image, but if
\(T_S = T_S[n]\) is an \(n\dash\)torsion group, this forces \(T_S\) to
be finite.

\begin{center}\includesvg[width=0.7\linewidth]{823d6ace849775220ab7535acc8a77289de9acd0}\end{center}

Take \(\bar a \in T_s\) and lift it to \(a\in K\units\), and consider
\((a)\) as a fractional \(R_S\) ideal; it is an \(n\)th power. So write
\((a) = I^n\) for \(I\) another fractional \(R_S\) ideal.

\begin{quote}
Really only need to assume that the class group of \(R_S\) does not have
\(n\dash\)torsion.
\end{quote}

Since \(R_S\) is a PID, \(I = (b)\), so \((a) = (b)^n = (b^n)\) so there
is some \(u\in R_S\units\) with \(a = ub^n\). But then
\begin{align*}
\bar a = \psi(a) = \psi(u) \psi(b^n)
 = \psi(u)
 ,\end{align*} so \(\psi\) is surjective.

This finishes the proof of weak Mordell-Weil. \(\qed\)

\begin{description}
\tightlist
\item[Remark]
We could weaken the assumptions to \(R\) with fraction field \(K(E[n])\)
such that \((\pic R)[n]\) is finite and \(R\units / R^{\times n}\) is
finite.
\item[Exercise]
Let \(k = \bar k\) and \(C/k\) be a ``nice'' affine curve. Let \(R\) be
the affine coordinate ring of \(C\), i.e.~\(R = k[C]\), and \(K = k(C)\)
with \(\ch(k)\) not dividing \(n\). Show that if \(E/K\) is an elliptic
curve then \(E(K) / n E(K)\) is finite.
\end{description}

\(n\dash\)torsion on an abelian group is finite. Note that the
Mordell-Weil group here may not itself be finite! Also note that proof
almost goes through for abelian varieties, since we didn't use anything
particular to elliptic curves.

\hypertarget{height-functions}{%
\subsection{Height Functions}\label{height-functions}}

\begin{description}
\tightlist
\item[Definition (Product Formula Field)]
A \textbf{product formula field} is a triple \((K, \Sigma_K, A)\) where
\(K\) is a field, \(\Sigma_K\) is a set of places on \(K\), and \(A\) is
a set of normalizing constants.
\item[Definition (Place)]
A \textbf{place} \(v\) on a field \(K\) is an equivalence class of
absolute values \(\abs{\wait}: K \to [-, \infty)\subset \RR\), where
e.g.~the equivalence is given by absolute values inducing the same
topology.
\end{description}

Here we don't mind if raising to some power invalidates the triangle
inequality, as long as raising to \emph{some} power preserves it.

\begin{quote}
See ultrametric, Archimedean.
\end{quote}

\(\Sigma_K\) is \textbf{a} set of equivalence classes of absolute values
(not necessarily all) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \textbf{(PF1)}
  \begin{align*}
    \Sigma_K^{\text{Arch}}  \subset \Sigma_K
    .\end{align*}
\item
  \textbf{(PF2)} For all \(x\in K\units\),
  \begin{align*}
    \# \theset{ v\in \Sigma_K  \suchthat \abs{x}_v \neq 1} < \infty.
    \end{align*}

  A non-Archimedean valuation in \(\Sigma_K\) satisfies
  \(\abs{x}_v = c_v^{-v(x)}\) for some rank 1 valuation
  \(v: K \to \RR\union\infty\) with \(c_v > 1\), and \(A\) is that data
  of a choice of \(c_v\) for each non-Archimedean \(v\).
\item
  \textbf{(PF3)} For all \(x\in K\units\), we have
  \begin{align*}
    \prod_{v\in \Sigma_K} \abs{x}_v = 1
    .\end{align*}
\end{enumerate}

Note that we don't care about the \(c_v\)s necessarily, but we do need
to choose them to preserve this product formula.

All global fields are PFFs, so we'll use this to define height functions
on projective spaces.

\hypertarget{wednesday-january-29th}{%
\section{Wednesday January 29th}\label{wednesday-january-29th}}

\hypertarget{remarks-on-weak-mordell-weil}{%
\subsection{Remarks on Weak
Mordell-Weil}\label{remarks-on-weak-mordell-weil}}

\begin{description}
\tightlist
\item[Definition (Small Groups)]
A profinite group (e.g.~group occurring as an absolute Galois group) is
\textbf{small} if for all \(n \in \ZZ^+\), the set of open subgroups of
\(g\) of index \(n\) is finite.
\item[Proposition]
For any \(g = g_K = \Aut(K\sep/K)\), \(g\) is small iff \(K\) has only
finitely many separable extensions of any fixed degree \(d\).
\item[Definition (Big Fields)]
A field \(K\) is \emph{big} iff its Galois group \(g_K\) is small.
\end{description}

\textbf{Warning}: In general not related to cardinality.

If \(K\) is big and contains \(n\)th roots of unity \(\mu_n \in K\),
then for all \(n\geq 2\), the maximal abelian extension of exponent
dividing \(n\) is finite (by Kummer theory) Therefore, if \(A/K\) is any
commutative algebraic group (plus conditions?), then \(A(K)/ nA(K)\) is
finite.

Fields that are big:

\begin{itemize}
\tightlist
\item
  Finite fields
\item
  Algebraically closed fields
\item
  \(\RR\) and any real closed field \(\iff\)
\item
  \(p\dash\)adic fields
\end{itemize}

But \(\FF_q((A))\) is \emph{not} big.

\begin{quote}
The Weak Mordell Weil proof goes through if you look at integral points
instead of just rational points. Dirichlet unit theorem gives WMW for
the multiplicative group.
\end{quote}

\hypertarget{height-functions-1}{%
\subsection{Height Functions}\label{height-functions-1}}

Let \((K, \Sigma_K, A)\) be a product formula field where \(K\) is a
number field and \(K = k(C)\) for \(C/K\) is a nice curve for \(k\) any
field.

\begin{quote}
Don't need to assume separability, but we will!
\end{quote}

A height function is a function \(H: \PP^n(K) \to \RR\). We assume the
product formula field \(K\) has the \emph{Northcott property},
i.e.~\(\theset{ p\in \PP^n(K) \suchthat H(P) \leq R }\) is finite for
ever \(n\) and every \(R\).

\begin{description}
\tightlist
\item[Remark]
A number field \(K\) has the Northcott property, and \(k(C)\) has this
property iff \(k\) is finite.
\end{description}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  This gives an estimate for change in \(H\) under finite morphisms
  \(f: \PP^n \to \PP^m\).
\item
  For \(E/K\) with \(y^2 = P_S(x)\), then \(H: E(K) \to \RR\) given by
  \(p\mapsto H([X(p): 1])\).
\end{enumerate}

View \(X: E \to \PP^1\) is a 2:1 map given by taking \(x\) coordinates.

Sticking point: The ``near-quadraticity'' of \(H\), i.e.~the second
property of height functions. We won't have Weierstrass equations for
abelian varieties, so we don't want to do this. Instead, we'll develop

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Weil's height machine
\item
  Néron-Tate canonical heights on abelian varieties \(A/K\).
\end{enumerate}

Note that if \((x_0, \cdots, x_n) \in \PP^n(\QQ)\), where \(\QQ\) is the
fraction field of \(\ZZ\), a UFD, so this is a UFD. Appropriately
clearing denominators, we can get \(\gcd(x_0, \cdots, x_n)\). So
\(H(x_0, \cdots, x_n) = \max_{0\leq i \leq n} \abs{x_i}\), and (check!)
the Northcott property holds.

Let \((K, \Sigma_K, A)\) where \(\Sigma_K\) is a set of inequivalent
places \(\abs{\wait}_v\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  PFF1: The Archimedean places \(\Sigma_K^{\text{Arch}}\) have
  valuations \(v: K \to \RR \union \infty\) where
  \(\abs{\wait}_v = c_v > 1\).
\item
  PFF2: For all \(x\in k\units\), \(\abs{x}_v = 1\) for all \(v\).
\item
  PFF3: Height is invariant under scaling, i.e.~for all
  \(x\in k\units\), \(\prod_{v\in\Sigma_K} \abs{x}_v = 1\).
\end{enumerate}

\begin{description}
\tightlist
\item[Example]
Take \(K = \QQ\), then
\end{description}

\begin{itemize}
\tightlist
\item
  \(\abs{x}_p = p^{-\ord_p(x)}\)
\item
  \(\abs{x}_\infty = \abs{x}\) the standard real absolute value
\item
  \(\abs{x}_\infty = \qty{\prod_p \abs{x}_p }\inv\) for all
  \(x\in \ZZ^\bullet\).
\end{itemize}

\begin{description}
\item[Example]
\(k\) any field, \(R = k[t]\), and \(K = k(t)\). Then it's not quite
true that
\(\Sigma_K = \theset{v_p, ~p \text{ monic irreducible polynomials}}\).
Question: what is \(C_{v_p}\).

For \(c>1\), we can write
\begin{align*}
\abs{\wait}_p \definedas c^{-v_p}
.\end{align*} Write \(x = \eps p_1^{a_1} \cdots p_{r}^{a_r}\) with the
\(p_i\) monic irreducibles and \(\eps \in k\units\). Then
\begin{align*}
\prod_p \abs{x}_p = \prod_p c^{-v_p(x)} = c^{-\sum^r a_i}
.\end{align*}

But this isn't one! We don't have the ``counterbalance'' here, so we may
want to add an infinite place to compensate.
\item[Remark]
If \(k\) is algebraically closed, then \(p_i = t - x_i\) and
\(\sum a_i = \deg v\).
\end{description}

Adding a place: Define \(v_\infty(f/g) = v_\infty(f) - v_\infty(g)\),
and
\begin{align*}
\abs{\wait}_{v_\infty} = c^{-v_\infty(f/g)} = c^{\deg f - \deg g}
.\end{align*} This yields

\begin{align*}
\abs{x}_\infty \prod_p \abs{x}_p 
&= \prod_p C^{-v_p(x)} C^{\deg x} \\
&= c^{-\sum^r a_i} C^{\deg x} \\
&= C^{-\deg x}C^{\deg x}
.\end{align*}

In general,
\begin{align*}
\deg\qty{\prod_{i=1}^r p_1^{a_1} \cdots p_r^{a_r}} = \sum_{i=1}^r a_i \deg(p_i)
.\end{align*}

Then
\begin{align*}
k_v \definedas \deg p_v \cdot c \quad\text{and}\quad \abs{x}_p \definedas C^{-\deg(p) v_p(x)}
.\end{align*}

\hypertarget{friday-january-31st}{%
\section{Friday January 31st}\label{friday-january-31st}}

\hypertarget{finite-separable-extensions-are-pffs}{%
\subsection{Finite Separable Extensions are PFFs
?}\label{finite-separable-extensions-are-pffs}}

Given a product formula field (PFF) \((K, \Sigma_K, A)\), recall that we
want the product formula:

\begin{align*}
\forall x\in K\units,\quad \prod_{v\in \Sigma_K} \abs{x}_v = 1
.\end{align*}

We want to show that given any finite separable extension \(L/K\), endow
it with a PF structure where the places of \(L\) are given by
\(\Sigma_L =\) all extensions of \(v\in L/K\) to \(L\) (?)

\begin{quote}
Note that separability is not necessary but does simplify things, c.f.
Serre.
\end{quote}

The structure for \(A\): to be decided.

A technical remark: A global field is given by \(K = \FF_q(C)\) where
\(C/\FF_q\) is a nice curve.

Then \(\QQ\) has a PF structure, \(K(t)\) has a PF structure. Note that
the extensions over \(\QQ\) will be separable, since we're in
characteristic zero, so the question is about separable extensions over
\(K(t)\).

If \(C/K\) is a nice curve over any field, then there exists a finite
degree extension \(K(C) / K(t)\) iff \(C \mapsvia{f} \PP^1\).

\begin{quote}
Reminder: Noether's normalization, integral affine variety over any
field, can geometrically map it down to affine d-space via an algebraic
(finite) map. So we end up with a finitely generated module over a
polynomial ring.
\end{quote}

We want a separable extension, and from field theory, since \(\FF_q\) is
perfect we can find \(\FF_q(t) \subset \FF_q(C)\) which is separable.

If \(k\) is any perfect field, then we can find a separable Noether
normalization. If \(K/k\) is a finitely generated field extension of
transcendence degree \(d\). Moreover, we can find an extension
\begin{align*}
K \mapsvia{\text{finite, separable}} k(f_1, \cdots, f_d) \to k
.\end{align*}

So let \(L/K\) be degree \(d\) and separable. There is a surjective map
\(\Sigma_L \surjects \Sigma_K\) with fibers of size \(\leq d\). For
\(v\in \Sigma_k\), this is defined by \(L_v\definedas L \tensor_k K_v\),
and if \(L/K\) is separable then \(L \cong k[t]/(f)\) where \(f\) is
separable and irreducible. Thus \(L_v \cong K_v[t] / (f)\) where
\(f = f_1 \cdots f_r\) which are irreducible over \(K_v\). By the
Chinese Remainder Theorem,
\begin{align*}L_v \cong \prod_{i=1}^r K_v[t] / (f_i) \cong \prod_{w \mid v} L_w.\end{align*}

\begin{quote}
Primitive element theorem: finite separable extensions of fields are
generated by one element.
\end{quote}

How do we extend the norm? For \(x\in C_w\), define
\begin{align*}
\abs{x}_w \definedas \abs{ N_{L_w}/ K_v (x) }_v^{\frac 1 d}
\end{align*} where \(d\) is the global degree \(d = [L : K]\).

\begin{quote}
Note that the exponent here is not necessary to get the product formula,
but it works anyway and is useful when we later look at heights.
\end{quote}

\begin{description}
\tightlist
\item[Lemma]
For \(x\in C\),
\begin{align*}
\prod_{w \mid v} \abs{x}_w = \abs{N_{L/K}(x)}_v^{\frac 1 d}
.\end{align*}
\end{description}

This yields the normalization \(A_L\).

\begin{description}
\tightlist
\item[Theorem (Finite Separable Extensions are PFFs)]
The product formula holds, i.e.~\((L, \Sigma_K, A)\) is a PFF.
\end{description}

For all \(x\in L\dual\), we have
\begin{align*}
\prod_{w\in \Sigma_L} \abs{x}_w = \prod_{v\in \Sigma_K} \prod_{w\mid v} \abs{x}_w = \prod_{v\in \Sigma_K} \abs{ N_{L/K}(x)  }_v^{\frac 1 d} = 1
\end{align*} by the product formula on \(K\).

\hypertarget{height-functions-2}{%
\subsection{Height Functions}\label{height-functions-2}}

For \((K, \Sigma_K, A_K)\) a PFF, define
\(x\definedas (x_0, \cdots, x_n) \in \AA^{n+1}(K)\) and
\begin{align*}
H(x) \definedas \prod_{v\in \Sigma_K} \max_{0\leq i \leq n} \abs{x_i}_v
.\end{align*} Then \(H(x) = 0\) iff \(x = (0, 0, \cdots, 0)\).

\begin{description}
\tightlist
\item[Lemma]
For all \(\lambda \in K\units\),
\(H(\lambda x_1, \cdots, \lambda x_n) = H(x_1, \cdots, x_n)\), so \(H\)
descends to \(H: \PP^n(K) \to \RR\).
\end{description}

Thus for any \(p \in \PP^n(K)\), can write \(p = [x_0, \cdots, x_n]\)
where some \(x_i = 1\). So each term appearing in the product is at
least 1.

For \(x\in K\), we define \(H(x) = H([x: 1])\).

\begin{description}
\item[Example]
Take \(K = \QQ\) and \(n=1\), we can compute \(H([x: y]) = H([a, b])\)
where \(\gcd(a, b) = 1\). This equals
\(\prod_{p\leq \infty} \max(\abs{a}_p, \abs{b}_p)\), the \(p\dash\)adic
norms. This equals
\begin{align*}
\qty{\max(\abs a, \abs b)} \qty{ \prod_p \max(\abs{a}_p, \abs{b}_p)  }
,\end{align*} where the second term collapses to 1 because every term is
1, because no \(p\) can divide both \(a\) and \(b\). So
\(H([a, b]) = \max(\abs a, \abs b)\).
\item[Example]
Take \(K=\QQ, n\) arbitrary. WLOG, we can consider \(\theset{x_i}\) with
\(\gcd(x_i) = 1\), then
\(H([x_0: \cdots : x_n]) = \max(\abs{x_0}, \cdots, \abs{x_n})\).

We in fact have a finite bound

\begin{align*}
\#\theset{P \in \PP^n(\QQ) \suchthat H(P) \leq R \in \ZZ} \leq (2R + 1)^{n+1} 
.\end{align*}
\end{description}

Note that the probability that two numbers are prime is \(1/\zeta(2)\);
look at Euler product expansion.

\begin{quote}
Idea of proof: don't want both to be divisible by 2, by 3, by 5, etc.
\end{quote}

\begin{description}
\item[Example]
Let \(K = k(t) \supset k[t]\) for \(k\) arbitrary. Write
\(H([x_0: x_1]) = H([a:b])\) where \(a, b\in k[t]\) with
\(\gcd(a, b) = 1\). Recall that the infinite-adic norm is given by
\(\abs{a}_\infty = C^{\deg(a)}\). Then

\begin{align*}
H([a: b]) &= \prod_{p\leq \infty} \max(\abs{a}_p, \abs{b}_p) \\
&= \qty{ \max(\abs{a}_\infty,\abs{b}_\infty)} \prod_p \max(\abs{a}_p, \abs{b}_p) \\
&= \max(C^{\deg a}, C^{\deg b})
,\end{align*}

where the same argument goes through. Thus
\(\log_C H([a: b]) = \max(\deg a, \deg b)\), literally the maximum
degree of these two polynomials.
\end{description}

Note that there are only finitely many polynomials of a given degree
\(d\) iff the field is finite. So the PFF formalism doesn't care about
the field, but the Northcott property depends on the cardinality of the
field in a key way.

\begin{description}
\item[Fact]
The Northcott property holds for all \(R> 0\) and for all \(n\), i.e.~
\begin{align*}
\#\theset{P\in \PP^n \suchthat H(P) \leq R} < \infty
\end{align*} iff \(K\) is finite.

This holds iff
\begin{align*}\#\theset{P\in \PP^n \suchthat \log_C H(P) \leq R}\end{align*}
is finite. Note that if \(K = \FF_q\), then for \(R\in \ZZ^+\),
\begin{align*}
\#\theset{P \in \PP^n(\FF_q) \suchthat \log_C H(P) \leq R} = (q^{r+1})^{n+1}
\end{align*} by counting choices for coefficients.
\end{description}

We'll show that the Northcott property passes to finite extensions.

\begin{description}
\tightlist
\item[Theorem (Finiteness for Northcott PFFs)]
Let \(K\) be a Northcott PFF, then for all \(n\in \ZZ^+\), for all
\(R\in \RR\), for all \(d\in \ZZ^+\),
\begin{align*}
\#\theset{ P \in \PP^n(K\sep) \suchthat [K(P): K] \leq d,~H(P) \leq R  }
\end{align*} is finite.
\end{description}

\hypertarget{monday-february-3rd}{%
\section{Monday February 3rd}\label{monday-february-3rd}}

\hypertarget{extending-height-functions-to-separable-closure}{%
\subsection{Extending Height Functions to Separable
Closure}\label{extending-height-functions-to-separable-closure}}

We started with a product formula field \((K, \Sigma_K, A)\), then took
\(L/K\) finite and separable and equipped it with a PFF structure:
\(\Sigma_L\) is equal to the places extending \(v\in \Sigma_K\), and for
\(v\in \Sigma_L\), we have \(\abs{x}_w = \abs{L_w/K_v (x) }_v^{1/d}\).

We defined a height function \begin{align*}
H: \PP^n(K) &\to [1,\infty) \\
[x_0, \cdots, x_n] &\mapsto \prod_{v\in \Sigma_K} \max(\abs{x_0}_v, \cdots)
.\end{align*}

If \(L/K\) is separable of degree \(d\), we can consider
\(p\in \PP^n(K) \subset \PP^n(L)\) and define \(H_K, H_L\) respectively.
It is then a fact that \(H_L(p) = H_K(p)\). For \(\alpha \in \AA^1\), we
have
\begin{align*}
H_L(\alpha) = H_K(N_{L/K}(\alpha))^{1/d} = H_K(\alpha^d)^{1/d} = H(\alpha)
.\end{align*}

This allows us to extend height functions to the separable closure,
\(H: \PP^n(K\sep) \to \RR\).

\begin{description}
\tightlist
\item[Lemma]
If \(p\in \PP^n(K\sep)\) and \(\sigma \in g_K = \aut(K\sep/K)\), then
\(\sigma(p) \definedas [\sigma(x_0), \cdots, \sigma(x_n)]\) does not
change the height.
\item[Proof]
Exercise. Check that the norms of points are the same as the norms of
their Galois conjugates.
\end{description}

So the height is Galois-equivariant.

For \(v\in \Sigma_K\), let \(\eps_v\) be the \textbf{Artin constant}:
the smallest \(\eps_v \in \RR\) such that
\(\abs{x + y}_v \leq \eps_v \max(\abs{x}_v, \abs{y}_v)\).

\begin{quote}
Artin constant: measures how much better a norm does than the triangle
inequality. Note that \(\eps_v = 1\) for ?
\end{quote}

\begin{description}
\item[Lemma]
Let \(K\) be a PFF and \(d\in \ZZ^+\). If there exists an
\(M(K, d) > 0\) such that for all separable
\begin{align*}
f(T) = T^d + a_{d-1}T^{d-1} + \cdots a_1 T + a_0 \in K[T]
,\end{align*} i.e.~\(f(T) = \prod_{i=1}^\alpha (T-\alpha_i)\) with
\(\alpha_i \in K\sep\), then
\begin{align*}
H(a_0, \cdots, a_{n-1}, 1) \leq M(K, d) \prod_{i=1}^d H(\alpha_i)
.\end{align*}
\item[Proof]
\textbf{Step 1:} Suppose \(\alpha_i \in K\) for all \(i\). If so, we can
take \(M(K, d) = \prod_{v\in \Sigma_K} \eps_v^{d-1}\).

Why? We need to show that for all \(v\in \Sigma_K\), we have
\begin{align*}
\max \abs{a_i}_v \leq \eps_v^{d-1} \prod_{i=1}^d \max(\abs{\alpha_i}_v)
.\end{align*} If so, take the product over all \(v\).

Proceeding by induction, where the \(d=1\) is clear, suppose \(d\geq 2\)
so result holds for \(d-1\). Choose \(k\) such that
\(\abs{\alpha_k}_v \geq \abs{\alpha_i}_v\) for all \(i\). We can then
write
\begin{align*}
f(T) = (T-\alpha_k)( T^{d-1} + b_{d-1}T^{d-2} + \cdots + b_1 T + b_0 )
\end{align*} where \(b_{d-1} = 1\) and \(b_{-1} = b_d = 0\). Then for
all \(i\), we have \(\alpha_i b_{i-1} - \alpha_k b_j\).

We then apply the triangle inequality:

\begin{align*}
\abs{a_i}_v 
&= \max_i \abs{b_{i-1} - \alpha_j b_i} \\
&\leq \eps_v  \max \abs{b_{i-1}}_v \abs{\alpha_k b_i}_v \\
&\leq \eps_v (\max_n \abs{b_i}_v) \max(\abs{\alpha_k}_v, 1) \\
&\leq_{IH} \eps_v^{d-1} \max_i \abs(\abs{a_i}_v, 1)
.\end{align*}

\textbf{Step 2:} Now suppose \(\alpha, \cdots, \alpha_d \in K\sep\).

Replace \(K\) with \(K(\alpha_1, \cdots, \alpha_d) \definedas L\). This
is an extension of \(K\) of degree at most \(d!\).

Note that if we define \(\eps = \max \eps_v\), and
\(\abs{\Sigma_K^{Arch}} = A\), then
\begin{align*}
\prod_{v\in \Sigma_K} \eps_v^{d-1} \leq (\eps^A)^{d-1} = \eps^{A(d-1)}
.\end{align*} We also have
\begin{align*}
\abs{\Sigma_L^{Arch}} \leq d! A
\end{align*} and for the argument over \(K\), we can take
\(M(K, d) = \eps^{A d!(d-1)}\).
\item[Theorem (Northcott implies Strong Northcott)]
Let \(K\) be a PFF satisfying the Northcott property. Then for all
\(d, n \in \ZZ^+\) and for all \(R\in \RR\), the set
\begin{align*}\theset{ P \in \PP^n(K\sep) \suchthat [K(P): K] \leq d,~ H(P) \leq R  }\end{align*}
is finite.
\end{description}

I.e. passing to finite separable extensions preserves the Northcott
property, although this statement is stronger, since this is a
\emph{uniform} bound (hence the term ``strong'').

Stated scheme-theoretically, we can write \(P = [x_0, \cdots, x_n]\)
where some \(x_i = 1\) be rescaling. Then \(K(P) = K(x_0, \cdots, x_n)\)
is a finite separable extension.

\begin{description}
\item[Proof]
Let \(P \in \PP^n(K\sep)\) with \(\deg P \leq d\), where
\(P = [x_0, \cdots, x_n]\) with one \(x_i = 1\).

\begin{align*}
H(P) 
&= \prod_{v\in \Sigma_K} \max_i \abs{x_i}_v \\
&= \prod_{v\in \Sigma_K} \max_i (\abs{x_i}_v  , 1) \\
&\geq \max_i \prod_{v\in \Sigma_K} \max(\abs{x_i}_v  , 1) \\
&= \max_i H(x_i)
.\end{align*}

So if \(H(P) \leq C\) and \([K(P):K] \leq d\), then
\(\max_i H(x_i) \leq C\) and \(\max_i [K(x_i): K] \leq d\).

Define
\begin{align*}X(K, C, d) = \theset{ x \in K\sep \suchthat H(x) \leq C,~ [K(x): K] \leq d  },\end{align*}
we'll show this is a finite set. Let \(x \in X(K, C, d)\) and
\(d' \definedas [K(x): K] \leq d\), and
\begin{align*}f(T) \definedas \minpoly x = \prod_{i=1}^{d'} (T - x_i) = T^{d'} + \sum_{i=0}^{d'-1} a_i T^i.\end{align*}
Take \(M\definedas \max_{1\leq i \leq d} M(K, i)\) from the previous
lemma.

By the lemma, we have
\begin{align*}H(a_0, \cdots, a_{d'-1}, 1) \leq M \prod{i=1}^{d'} H(x_i).\end{align*}
But Galois conjugate points have the same height, so this is equal to
\(M H(x)^{d'} \leq MC^d\). By the Northcott property, there exist only
finitely many such tuples \((a_0, \cdots, a_{d'-1}, 1)\), and thus only
finitely many possibilities for \(x\).
\end{description}

\hypertarget{key-theorem-bounds-on-heights-of-morphisms-for-pffs}{%
\subsection{Key Theorem: Bounds on Heights of Morphisms for
PFFs}\label{key-theorem-bounds-on-heights-of-morphisms-for-pffs}}

\begin{description}
\item[Theorem (The Key Theorem)]
Let \(K\) be a PFF, and \(f: \PP^n \to \PP^m\) be a degree \(d\)
morphism over \(K\) (given by a collection of homogeneous polynomials of
degree at most \(d\)). Then there exist constant \(C_1, C_2 > 0\)
depending only on \(K\) such that

\begin{align*}
C_1 H(P)^d \leq H(f(P)) \leq C_2 H(P)^d
.\end{align*}
\item[Proof]
See Theorem 8.5.6 in Silverman I.
\end{description}

\hypertarget{friday-february-7th}{%
\section{Friday February 7th}\label{friday-february-7th}}

\hypertarget{weil-height-machine}{%
\subsection{Weil Height Machine}\label{weil-height-machine}}

What is the analog of the \(x\dash\)coordinate of an elliptic curve for
a general abelian variety? Answer: an ample base point divisor.

Let \(K\) be an NPFF (Northcott Product Formula Field), \(V/k\) a
projective variety and \(\phi: V \to \PP^n\), then define

\begin{align*}
h_\phi: V(K\sep) &\to [0, \infty) \\
p &\mapsto \log H(\phi(p))
.\end{align*}

\emph{Key property:} Take hyperplanes \(\psi: V \to \PP^m \supset H'\)
and \(\phi: V\to \PP^n \supset H\), noting that the Picard group of
\(\PP^n\) is \(\ZZ\).

Suppose \(\phi^* \sim \psi^* H'\), then \(h_\phi = h_\psi + O(1)\).
Generally, we'll want to identify functions whose difference is a
bounded function.

\begin{quote}
Highly recommended: Diophantine Geometry, Handry-Silvermann GTM
\end{quote}

\begin{description}
\tightlist
\item[Theorem (Weil Height)]
Let \(K\) be a NPFF (e.g.~a global field) and \(V/K\) a smooth
projective variety. There exists a map
\(h_{V, \wait}: \div V \to \RR^{V(K\sep)}\) from divisors on \(V\) to ?
which is unique modulo bounded functions such that
\end{description}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  For \(H \subset \PP^n\) a hyperplane, the divisor
  \(h_{\PP_1^n H} = h + O(1)\) (for the previously defined \(h\)).
\item
  Pullbacks: For \(\phi: V\to W\) a smooth morphism and \(D\in \div W\),
  we have \(h_{V, \phi^* D} = h_{W, D}\circ \phi + O(1)\).
\item
  Linearity: For all divisors \(D_1, D_2 \in \div V\),
  \(h_{V, D_1 + D_2} = h_{V, D_1} + h_{V, D_2} = O(1)\).
\item
  If \(D_1 \sim D_2\), then \(h_{V, D_1} = h_{V, D_2} + O(1)\).
\item
  Positivity: Let \(D\subset \div V\) be effective and \(B\) the base
  locus (common intersection of linearly equivalent effective divisors),
  then\(h_{V, D}(p) \geq O(1)\) for all \(P \in (V\setminus B)(K\sep)\).
\item
  Let \(D, E. \in \div V\) with \(D\) ample and \(E\) algebraically
  equivalent to \(0\). Then
  \begin{align*}
    \lim_{P \in V(K\sep), h_{V,D}(p) \to \infty} h_{V, E}(p) / h_{V,D}(p) = 0
    .\end{align*}
\end{enumerate}

\begin{quote}
Ample: Chern class of line bundle is zero.
\end{quote}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{6}
\tightlist
\item
  Finiteness: \(D \in \div V\) is ample. For all map into \(L/K\) and
  \(r\in \RR\), the set
  \(\theset{p\in V(L) \suchthat h_{V, 0}(p) \leq R }\) is finite.
\end{enumerate}

Note every ample divisor has the Northcott property. Every abelian
variety has an ample \emph{symmetric} divisor.

Let \(V = A/K\) be an abelian variety and \(D\in \div A\) a divisor, see
section A.7 in Silverman.

\emph{Some properties of such divisors}:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  (A.7.25) For all \(n\in \ZZ\),
  \begin{align*}
    [n]^* D \sim \frac{n^2 + n}{2}D + \frac{n^2 - n}{2} [-1]^*D
    \end{align*} (standard fact).
\end{enumerate}

If \([D]\) is symmetric, i.e.~ \([-1]D \sim D\), then
\begin{align*}
  [n]^* D \sim n^2 D
  \end{align*}

If \([D]\) anti-symmetric, i.e.~ \([-1]^* D \sim -D\), then
\begin{align*}
  [n]^*D \sim nD \text{ .i.e. } [-1]^* D \sim -D
  ,\end{align*} then
\begin{align*}
  [n]^*D \sim nD \text{ i.e. } [-1]^* D \sim -D
  ,\end{align*} then \([n]^*D \sim nD\).

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  (A.7.210?) Let \(D\) be effective, then \(2D\) is basepoint free and
  TFAE:
\end{enumerate}

\begin{itemize}
\tightlist
\item
  \(D\) is ample.
\item
  \(\psi_\ell: A \to A\dual\) where \(x \mapsto \tau_x^*D - D\) (where
  \(\tau_x\) is translation by \(x\)) has finite kernel.
\item
  \(A \to \PP(L(2D))\) is finite.
\end{itemize}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{2}
\tightlist
\item
  \textbf{(A.7.33)} Define 4 maps,
  \(\sigma, \delta, \pi_1, \pi_2: A\cross A \to A\), \begin{align*}
  \sigma(P, Q) = P + Q \quad \delta(P, Q) = P - Q \quad \pi_1(P, Q) = P \quad \pi_2(P, Q) = Q
  ,\end{align*}
\end{enumerate}

Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  For \(D\in \div A\), \([D]\) is symmetric iff
  \begin{align*}
    \sigma^* D + \delta^* D \sim 2\pi_1^* D + 2\pi_2^* D
    .\end{align*}
\item
  {[}D{]} is antisymmetric iff
  \begin{align*}
    \sigma^* \sim \pi_1^* D + \pi_2^* D \iff [D] \in \pic^0 A
    .\end{align*}
\end{enumerate}

\begin{quote}
Proof: see Silverman, uses ``Theorem of the Square'' and the ``Seesaw
Principle''.
\end{quote}

\begin{description}
\tightlist
\item[Corollary]
For \(A/K\) an abelian variety over \(K\) a NPFF and \(D\in \div A\),
then
\end{description}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  \(p\in A(K\sep)\) and
  \begin{align*}
    h_{A, 0}([n]p) = \frac{n^2 + n}{2} h_{A, D}(p) + \frac{n^2 - n}{2} h_{A, D}(-p) + O(1)
    .\end{align*} If \(D\) is symmetric, then
  \begin{align*}
      h_{A, D}([n]p) = n^2 h_{A, D}(p) + O(1)
      .\end{align*} If \(D\) is antisymmetric, then
  \begin{align*}
      h_{A, D}([n]p) = nh_{A, D}(p) + O(1)
      .\end{align*}
\end{enumerate}

\begin{quote}
Note: This follows from properties of the Weil height machine.
\end{quote}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  If \([D]\) is symmetric, than we get the \textbf{quadraticity}
  relation: for all \(P, Q \in A(K\sep)\),
  \begin{align*}
    h_{A, D}(P+Q) + h_{A, D}(P-Q) = 2h_{A, D}(p) + 2h_{A, D}(Q) + O(1)
    .\end{align*}
\end{enumerate}

\begin{description}
\tightlist
\item[Proof]
We have
\begin{align*}
  h_{A\cross A, D}(\sigma(P, Q)) + h_{A\cross A, D}(\delta(P,Q)) = 2h_{A\cross A, D}(\pi_1^* D) + 2h_{A\cross A, D}(\pi_2^* D) + O(1)
  .\end{align*}
\end{description}

For \(\phi: V \to W\) and \(D\in \div W\), we have
\begin{align*}
  h_{C, \phi^*(D)}(p) = h_{W, D}(\phi(p)) + O(1)
  .\end{align*} So
\begin{align*}
  h_{A\cross A, D}(\sigma(P, Q)) = h_{A, D}(\sigma(P, Q)) = h_{A, D}(P + Q)
  ,\end{align*} and applying it to the other terms works the same way.

Abelian variety has attached \emph{polarization}, and are projective
group varieties. Every projective variety has an ample divisor, by
definition?

\hypertarget{sketch-of-proof-of-mordell-weil-for-ak-a-global-field}{%
\subsection{\texorpdfstring{Sketch of Proof of Mordell-Weil for \(A/K\)
a Global
Field}{Sketch of Proof of Mordell-Weil for A/K a Global Field}}\label{sketch-of-proof-of-mordell-weil-for-ak-a-global-field}}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  Weak MW
\item
  Choose an ample symmetric \(D \in \div A\), by definition \(A\)
  carries an ample divisor \(D_0\); then
  \(D \definedas D_0 + [-1]^* D_0\) is ample and symmetric.
\item
  Work with \(h_{A, D}\) instead of \(h_X\) for elliptic curves
\end{enumerate}

Then the proof goes through as before.

Note \(h\) is trying to be a quadratic form (we'll review such forms on
groups) except for the \(O(1)\). Maybe we can modify \(h\) by a bounded
function to make it quadratic -- this is the Neron-Tate canonical
height.

\begin{quote}
Regulator: defined in terms of the Neron-Tate canonical height on the
Mordell-Weil group of an elliptic curve.
\end{quote}

\hypertarget{wednesday-february-12th}{%
\section{Wednesday February 12th}\label{wednesday-february-12th}}

\hypertarget{correction-to-weils-height-machine}{%
\subsection{Correction to Weil's Height
Machine}\label{correction-to-weils-height-machine}}

Let \(K\) be a NPFF (from now onwards: just a global field) and \(X/K\)
a nice variety. Let \(D\in \div(X)\) be ample and consider the points of
bounded height \(\theset{ p\in V(K) \suchthat h(p) \leq R }\); by the
Northcott property this is finite for every \(R \in \RR\).

Counterexample: Let \(X = E\) an elliptic curve and \(D = [0]\). Then
\(\ell(D) = \dim{f\in K(X) \suchthat \div f\geq -D}\), the rational
functions with poles no worse than order \(D\). Then \(\ell[0] = 1\).

Then there is a corresponding morphism to projective space
\begin{align*}
\phi_D: X \to \PP(L(D)) = \PP^{\ell(d) - 1}
.\end{align*}

Here
\begin{align*}
\phi_D: E \to \PP^{\ell[0] - 1} = \PP^0 = \pt
.\end{align*}

Thus if \(E(K)\) is infinite, which is possible, then
\(\forall R\geq 1\) the set
\(\theset{p\in E(K) \suchthat h(p) \leq R} = E(K)\) is infinite as well.

How to fix this? Replace ``\(D\) is ample'' with ``\(D\) is ample and
\(\ell(D) \geq 2\)''.

\begin{quote}
Note: not listed as a correction in Silverman's book.
\end{quote}

\begin{description}
\item[Theorem (C.1.9 Of Hindry-Silverman)]
Let \(K\) be a number field (perhaps works with global function fields)
and \(n\geq 2\), with \(A/K\) an abelian variety of dimension \(g\).
Suppose \(A[n] = A[n](K)\) (so a large extension of \(K\)), and let
\(S\) be a set of finite non-Archimedean places of \(K\) containing
places dividing \(n\) and places of bad reduction.

Suppose all that the ring of \(S\dash\)integers, \(\ZZ_{K, S}\), is a
PID.

Then
\begin{align*}
\rk A[k] \leq 2g \cdot \rk \ZZ_{K, S}\units
.\end{align*} We can then apply the Dirichlet unit theorem, the RHS is
\(2g(r_1 + r_2 + \# S - 1)\) where \(r_1\) is the number of real places
and \(r_2\) is the number of complex places.
\end{description}

This is an explicit upper bound, but a pretty bad one!

\hypertarget{nuxe9ron-tate-canonical-height}{%
\subsection{Néron-Tate Canonical
Height}\label{nuxe9ron-tate-canonical-height}}

Let \(K\) be a global field. We'll get a good canonical representative
of the height functions studied previously, provided we get one
additional piece of data.

\begin{description}
\item[Theorem (Néron-Tate)]
Let \(V/K\) be a nice variety and \(D \in \div V\), and suppose there
exists a morphism \(\phi: V\to V\) such that there exists an
\(a\in \QQ^{> 1}\) with the induced map on \(\pic V\) satisfies
\(\phi^* D \sim \alpha D\).

\begin{quote}
So \(D\) is an ``eigendivisor''.
\end{quote}

Then there exists a unique function

\begin{align*}
\hat{h}_{V, \phi, D}: V(K\sep) \to \RR
\end{align*}

such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  (CH1) \(\hat h_{V, \phi, D} = h_{V, D} + O(1)\)
\item
  (CH2) \(\hat h_{V, \phi, D} \circ \phi = \alpha \hat h_{V, D}\)
\end{enumerate}

Moreover, for all \(p\in K\sep\), there is a stabilization

\begin{align*}
\hat h_{\phi, D}(p) = \lim_n \frac{1}{\alpha^n} h_{V, D}(\phi^n(p))
.\end{align*}
\item[Proof]
We first need to show that the stated limit exists. Applying Weil's
Height Machine to \(\phi^* D \sim \alpha D\), there exists a \(c\) such
that
\end{description}

\begin{align*}
  \abs{ h_{V, D} \circ \phi  - \alpha h_{V, D}} \leq c
  .\end{align*}

So take a separable point \(p\in V(K\sep)\).

\begin{description}
\item[Claim]
\(\alpha^{-n} h_{V, D}(\phi^n(p))\) is Cauchy in \(\RR\) and thus
convergent.
\item[Proof]
\textbf{Step 1}:

Let \(m\leq n\), then we want to show

\begin{align*}
\abs{ \alpha^{-n} h_{V, D}( \phi^n (p) ) - \alpha^{-m} h_{V, D}( \phi^n (p) ) } 
&=
\abs{ \sum_{i=m+1}^n \alpha^{-i} \qty{ h_{V, D} (\phi^i(p)) - \alpha h_{V, D}( \phi^{i-1}(p)  )  }  } \\
&=
\sum_{i=m+1}^n \alpha^{-i} \abs{ \qty{ h_{V, D} (\phi^i(p)) - \alpha h_{V, D}( \phi^{i-1}(p)  )  }  } \\
&\leq 
\sum_{i=m+1}^n \alpha^{-i} C \\
&<
C \sum_{i=m+1}^\infty \alpha^{-i}\\
&= \frac{\alpha^{-m-1}{1-\alpha}} C \\
&\converges{m\to\infty}\to 0
.\end{align*}

\textbf{Step 2}: Write

\begin{align*}
\hat h_{V, D}(\phi(p)) 
&=
\lim_{n\to\infty} \alpha^{-n} h_{V, D} ( \phi^n(\phi(p))  ) \\
&=
\alpha \lim_{n\to\infty} \frac{1}{\alpha^{n+1}} h_{V, D} ( \phi^n(\phi(p))  ) \\
&=
\alpha \hat h_{V, \phi, D}(p)
.\end{align*}

\textbf{Step 3}: Suppose \(\hat h\) and \(\hat{h'}\) both satisfy CH1
and CH2. Let \(g\definedas \hat h - \hat{h'}\). By CH1,
\(\abs g \leq C\) for some \(C\). Then by CH2,
\(g\circ \phi = \alpha g\), and for all \(n\in \ZZ^+\),
\(g\circ \phi^n = \alpha^n g\). We thus get

\begin{align*}
g = \frac{g\circ \phi^n}{\alpha^n} \implies \abs{g} \leq \frac{C}{\alpha^n} \converges{n\to\infty}\to 0
.\end{align*}
\end{description}

\hypertarget{cents-of-dynamics}{%
\subsection{10 Cents of Dynamics}\label{cents-of-dynamics}}

Anytime we have an endomorphism, we can iterate it and consider the
associated dynamical system.

Let \(S\) be a set, \(\phi: S\to S\) a function,
\(\phi^n = \phi \circ \phi \circ \cdots \circ \phi\) with
\(\phi^0 = \id_S\). Then \(p\in S\) is \(\phi\dash\)periodic if there
exists an \(n\in \ZZ^+\) such that \(\phi^n(p) = p\). We say \(p\) is
\(\phi\dash\)preperiodic iff the set
\(\theset{\phi^n(p) \suchthat n\in \NN}\) is finite. In this case, there
exists some \(n\in \NN\) such that \(\phi^n(p)\) is
\(\phi\dash\)periodic:

\includegraphics{figures/2020-02-12-13:01.png}\\

\textbf{Proposition:} Setup as in the last result, \(\phi: V\to V\),
\(D\in \div V\) and \(\phi^*D \sim \alpha D\) for
\(\alpha \in \QQ^{\geq 1}\). For \(\ell(D) \geq 2\) and \(D\) ample, we
have

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  For all \(p\in V(K\sep)\), \(\hat h_{V, D, \phi}(p) \geq 0\) Moreover,
  \(\hat h_{V, D, \phi}(p) = 0\) off \(p\) is \(\phi\dash\)preperiodic.
\item
  The set
  \(\theset{p\in V(K) \suchthat p \text{ is $\phi\dash$ preperiodic} }\)
  is finite.
\end{enumerate}

\emph{Proof of (a):}

We know positivity by the Weil Height Machine. Let \(p\in A(K\sep)\),
then \(K \to K(p)\) for ?. Suppose \(\hat h_{V, D, \phi}(p) = 0\), then

\begin{align*}
h_{V, D}( \phi^n(p)  ) 
&= \hat h_{V, \phi, D}(\phi^n(p)) + O(1) \\
&= \alpha^n \hat h_{V, \phi, D}(p) + O(1) \\
&= O(1)
.\end{align*}

Then the set of forward orbits \(\theset{\phi^n(p) \suchthat n\in \NN}\)
has bounded height. By the finiteness property of the Weil Height
Machine, this set is finite and \(p\) is thus \(\phi\dash\)preperiodic.

If \(p\) is \(\phi\dash\)preperiodic, then \(\theset{h_{V, D}(p)}\)
bounded, so

\begin{align*}
\hat h_{V, \phi, D} (p) = \lim_n \frac{1}{\alpha^n} h_{V, D}(\phi^n(p)) = 0
.\end{align*}

\emph{Proof of (b):}

By part (a), if \(p\in V(K)\) then \(p\) is \(\phi\dash\)preperiodic, so
\(\hat h_{V, \phi, D} = 0\) and thus \(h_{V, D}(p) = O(1)\). By the Weil
Height Machine, this set of points is finite.

\(\qed\)

\hypertarget{important-examples}{%
\subsubsection{Important Examples}\label{important-examples}}

\textbf{Example 1:} Let \(\phi: \PP^n \to \PP^n\) of degree \(d\geq 2\)
and \(H\subset \PP^n\) the hyperplane divisor, \(\phi^* H \sim d H\). So
take \(\alpha = d\), this is an eigendivisor, now applying this theory,
a consequence is that
\(\theset{p\in V(K) \suchthat p \text{ is } \phi\dash\text{preperiodic}}\)
is finite. (Theorem by Northcott, properties of heights were established
to show this statement!)

\textbf{Example 2:} Let \(A/K\) an abelian variety with \(D\in \div A\)
ample, \(\ell(D) \geq 2\) with \([-1]^* D \sim D\). Take \(\phi = [n]\)
for some \(n\geq 2\), then \(\phi^*(D) \sim n^2 D\) where
\(\alpha = n^2 > 1\). The \(\phi\dash\)preperiodic points of
\(A(K\sep)\) are precisely the torsion points.

\begin{quote}
Note that this proves that the set of torsion points on an abelian
variety is finite.
\end{quote}

\hypertarget{friday-february-14th}{%
\section{Friday February 14th}\label{friday-february-14th}}

\hypertarget{canonical-heights}{%
\subsection{Canonical Heights}\label{canonical-heights}}

Last time: the Néron-Tate Canonical Height. Simple if you buy the Weil
Height Machine! Recall the statement of the theorem:

\begin{description}
\item[Theorem (Existence of Canonical Height Functions)]
Let \(K\) be a global field, \(V/K\) a nice variety, \(D\in \div(V)\).
Suppose there exists \(\phi:V \to V\) such that \(D\) is an
``eigendivisor'', i.e.~\(\phi^* D \sim \alpha D\) for some
\(\alpha \in \QQ\).

Then there exists a unique function
\(\hat h_{V, \phi, D}: V(K\sep) \to \RR\) such that

\begin{itemize}
\tightlist
\item
  (CH1) \(\hat h_{V, \phi, D} = h_{V, D} O(1)\)
\item
  (CH2)
  \(\hat{h_{V, \phi, D}} \circ \phi = \alpha \hat{h_{V, \phi, D}}\),
  which is constructed in the following way: For all \(p\in K\sep\),
  \(\hat h_{V, \phi, D}(p) = \lim_n \frac{1}{\alpha^n} h_{V, D}(\phi^n(p))\).
\end{itemize}
\end{description}

Two important examples:

\begin{description}
\item[Example]
Morphism from \(\PP \to \PP\), then \(\pic \PP = \ZZ\) so everything's
an eigendivisor. So just take a degree \(\geq 2\) map and a hyperplane
divisor.
\item[Example (Key example)]
\(A/K\) an abelian variety, \(D\) ample, \(\ell(D) \geq 2\) (so at least
2 sections of the line bundle), and symmetric in the sense that
\([-1]^* D \sim D\). Then for all \(n\in \ZZ\), \([n]^* D \sim n^2 D\),
so take \(\phi\) to be multiplication by \(n\) for \(\abs{n} \geq 2\).

In this case \(\hat h_{A, D} = \hat h_{A, [n], D}\). It agrees with the
usual Weil height, and moreover
\(\hat h_{A, D}([n]p) = n^2 \hat h_{A, D}(p)\). Note that this is at
least one property we'd want from a quadratic form.
\end{description}

The geometry associated to the height function on other types of
varieties can become more interesting, e.g.~\(K_3\) surfaces and
Calabi-Yau.

Next up: Showing \(\hat h_{A, D}: A(K\sep) \to \RR\) is a ``quadratic
form'', but what is a quadratic form on an abelian group?

\hypertarget{quadratic-functions}{%
\subsection{Quadratic Functions}\label{quadratic-functions}}

Let \(A, B\) be abelian groups written additively. Then a map
\(f: A\to B\) is \emph{quadratic} if the associated function

\begin{align*}
B_f: A\cross A &\to B \\
(x, y) &\mapsto f(x+y) - f(x) - f(y) + f(0)
\end{align*} is bilinear.

A function is \emph{homogeneous quadratic} iff \(f\) is quadratic and
\(f(0)= 0\). A \emph{quadratic form} is homogeneous, quadratic, and
symmetric in the sense that \(f(-x) = f(x)\) for all \(x\in A\).

\begin{quote}
Note: multiplication on \(\RR\) by 2 is an isomorphism, i.e.~\(\RR\) is
\emph{uniquely 2-divisible}. If \(B\) doesn't have this property, things
can get slightly more complicated.
\end{quote}

\begin{description}
\item[Lemma]
\(f: A\to B\) is quadratic iff for all \(x, y, z\in A\) we have

\begin{align*}
f(x + y + z) - f(x + y) - f(x + z) - f(y+z) + f(x) + f(y) + f(z) - f(0) = 0
.\end{align*}
\end{description}

In other words, knowing what it does on values and pairs determines what
it does on triples.

\begin{description}
\item[Proof]
Straightforward computation.
\item[Lemma]
Let \(f:A\to B\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(f\) is a quadratic form.
\item
  \(f\) satisfies the \emph{parallelogram law} (also: quadraticity)
  \(f(x + y) + f(x-y) = 2f(x) + 2f(y)\).
\end{enumerate}

Then 1 implies 2, and if the 2-torsion \(B[2]\) is trivial, then 2
implies 1.
\item[Lemma]
Suppose \(f\) is a quadratic function, then \(f\) is a quadratic form
iff \(\forall n\in \ZZ,~\forall x\in A\), \(f(nx) = n^2f(x)\).
\end{description}

Note that \(n=0\) implies homogeneity, and \(n=1\) implies symmetric.
Proceed by strong induction.

\begin{description}
\item[Lemma]
Let \(f: A\to B\) be a quadratic function. Define the even and odd parts
of \(f\) in the following way:

\begin{align*}
q: x &\mapsto \qty{ f(x) - f(0) } + \qty{ f(-x) - f(0)  } \\
l: x &\mapsto \qty{ f(x) - f(0) } - \qty{ f(-x) - f(0)  } \\
.\end{align*}

Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(q\) is a quadratic form, \(l\) is a group morphism, and
  \(2f = q + l + 2f(0)\).
\item
  If \(B\) is uniquely 2-divisible, then dividing by 2 yields
  \(f = \frac 1 2 q + \frac 1 2 l + f(0)\) where \(q\) is a quadratic
  form, \(l\) is a linear morphism, and \(f(0)\) is constant. This
  decomposition is unique.
\end{enumerate}
\item[Exercise]
Show that if \(f: A\to B\) a quadratic function,

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  \(f(A[\tors]) \subseteq B[\tors]\).
\item
  For \(n\) odd, \(f(A[n]) \subset B[n]\).
\item
  The map \(f: \ZZ/2\ZZ \to \ZZ/4\ZZ\) where \(0\mapsto 0, 1\mapsto 1\)
  is a quadratic form. Thus the \(n\) odd condition above is necessary.
\end{enumerate}
\end{description}

This comes up in Galois cohomology of abelian varieties? See Pete's
paper. It takes an \(n\) torsion to an element that is no worse than
\(2n\) torsion.

From now on, take \(B = \RR\). Note that if \(B=\RR\), height functions
kill all torsion.

Define a pairing

\begin{align*}
\inner{\wait}{\wait}: A\cross A &\to B \\
\inner{x}{y} &= \frac 1 2 \qty{ f(x+y) - f(x) - f(y)  }
.\end{align*}

If \(f\) is a quadratic form, \(\inner{x}{x} = f(x)\), so this reduces
to the usual bilinear form associated to a quadratic form.

\begin{description}
\item[Exercise]
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item[]
\item
  Show that for all \(x\in A, y\in A[\tors]\), we have
  \(f(x+y) = f(x)\). Thus \(f\) factors through a morphism
  \(\hat f: A/A[\tors] \to \RR\) through the torsion-free quotient.
\end{enumerate}

\begin{quote}
This is stronger than just killing torsion, since \(f\) isn't
necessarily linear.
\end{quote}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  If \(f\) is a quadratic form, \(\hat f\) is as well.
\end{enumerate}
\item[Theorem (Canonical Height Descent)]
Let \(h: A\to \RR\) be a quadratic form and suppose \(\exists n\geq 2\)
such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(\cok{A \mapsvia{\times 2} A} = A/nA < \infty\),
\item
  The Northcott property holds,
  i.e.~\(\theset{x\in A \suchthat h(x) \leq R} < \infty\) for each
  \(R\in \RR\),
\end{enumerate}

Letting \(y_1, \cdots, y_r\) be coset representatives for
\(nA \subset A\) and \(C_0 \definedas \max h(y_i)\), then \(A\) is
generated by the finite set \(\theset{x\in A \suchthat h(x) \leq C_0}\).
\item[Proof:]
Omitted.
\item[Theorem (The Néron-Tate Height is a Quadratic Form)]
Let \(K\) be a global field, \(A/K\) an abelian variety, \(D\) ample and
\(\ell(D) \geq 2\) with \([D]\) symmetric. Then there exists a unique
function \(\hat h_{A, D}: A(K\sep) \to \RR\) such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  \(\hat h_{A, D} = h_{A, D} + O(1)\)
\item
  \(\hat h_{A, D}\) is a \emph{quadratic form} (stronger conclusion),
  and in particular \(\hat h_{A, D}([n]p) = n^2 \hat h_{A, D}(p)\).
\end{enumerate}
\item[Proof]
We have 1, we'll show 2 by establishing the parallelogram law and use a
previous lemma.

\begin{quote}
Note \(\phi\) is suppressed here, since by uniqueness, multiplication by
\(n\) always yields the same function. So we'll take multiplication by
2.
\end{quote}

For all \(P, Q \in A(K\sep)\), we can compute

\begin{align*}
\hat h_{A, D}(P + Q) + \hat h_{A, D}(P-Q) 
&=
\lim 4^{-n} \qty{ h_{A,D}([2^n]P + [2^n]Q) + h_{A, D}\qty{ [2^n]P - [2^n]Q  }  } \\
&=
\lim 4^{-n} \qty{ 2 h_{A,D}([2^n]P) + 2 h_{A, D}([2^n]Q) + O(1) } \\
&=
2 \hat h_{A, D}(P) + 2\hat h_{A, D}(Q)
.\end{align*}
\end{description}

\hypertarget{monday-february-17th}{%
\section{Monday February 17th}\label{monday-february-17th}}

\hypertarget{canonical-heights-for-abelian-varieties}{%
\subsection{Canonical Heights for Abelian
Varieties}\label{canonical-heights-for-abelian-varieties}}

\begin{description}
\item[Theorem (Canonical Heights for Abelian Varieties)]
If \(A/K\) is an abelian variety, \(D\in \div(A)\), ample,
\(\ell(D) \geq 2\), \([D]\) symmetric, then there exists a canonical
height
\begin{align*}\hat h_{A, D}: A(K\sep) \to \RR\end{align*} such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\hat h_{A, D} = h_{A, D} + O(1)\)
\item
  \(\hat h_{A, D}\) is a \emph{quadratic form} (defined last time).
\end{enumerate}
\end{description}

In particular, \(\hat h_{A, D} \circ [n] = n^2 \hat h_{A, D}\). We can
define an associated bilinear form

\begin{align*}
\inner{P}{Q} = \frac 1 2 \qty { \hat h_{A,D}(P+Q) - \hat h_{A, D}(P) - \hat h_{A, D}(Q) }
,\end{align*}

so consider the \textbf{linear extension}

\begin{align*}
\hat h_{A, D}: A(K\sep) \tensor \RR &\to \RR \\
\sum_{i=1}^n [P]\tensor \alpha_i &\mapsto \sum_{i=1}^n \alpha_I \hat h_{A, D}
.\end{align*}

\begin{description}
\tightlist
\item[Exercise]
This yields a quadratic form on the domain.
\end{description}

\emph{Question:} What is \(\ker(A(K\sep) \to A(K\sep) \tensor \RR)\)?

\emph{Answer:} It contains the torsion, but turns out to be precisely
this. Note that \(\ker(M \to M\tensor \QQ) = M[\tors]\), and we can
tensor up in two steps:
\(A(K\sep) \tensor_\ZZ \RR = (A(K\sep) \tensor_\ZZ \QQ) \tensor_\QQ \RR = A(K\sep)[\tors]\).

\begin{description}
\tightlist
\item[Theorem (Linear Extension is Positive Definite)]
With the same setup as the previous theorem, the linear extension is
\emph{positive definite}, i.e.~
\begin{align*}
q(x) \geq 0 ~\forall x \quad\text{and}\quad q(x) = 0 \iff x=0
\end{align*}
\item[Definition (Regulator)]
If \(P_1, \cdots, P_n \in A(K\sep) \tensor \RR\) are
\(\RR\dash\)linearly independent, then we can define the
\textbf{regulator}
\begin{align*}\reg(P_1, \cdots, P_2) \definedas \det(\inner{P_i}{P_j}) > 0.\end{align*}
\end{description}

Note that this is the determinant of the Gram matrix, forms can be
diagonalized into \(\sum a_i x_i^2\), and positive-definite implies
\(a_i \geq 0\)?

If \(\theset{P_i}\) is a \(\ZZ\dash\)basis for
\(A(K\sep) / A(K\sep) [\tors]\), then we define
\(\reg(A, K, D) \definedas \det(\inner{P_i}{P_j}) > 0\).

\begin{quote}
This is well-defined for exactly the same reason that the discriminant
of a number field is well-defined.
\end{quote}

\begin{description}
\item[Proof]
Mostly algebra, mostly omitted here. Uses the following lemma.
\item[Lemma]
Let \(A \cong \ZZ^n\) and \(q:A\to \RR\) be a quadratic form such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  For all \(x\in A\), \(q(x) \geq 0\) and \(q(x) = 0 \iff x=0\),
\item
  The Northcott Property holds,
  i.e.~\(\forall R\in \RR, \# \theset{x\in A \suchthat q(x) \leq R} < \infty\).
\end{enumerate}

Then \(q\tensor \RR: A\tensor \RR \to \RR\) is still positive-definite.
\item[Example]
See Cassels-Silverman. Let \(K\) be a real quadratic field,
e.g.~\(K = \ZZ[\sqrt 2] \subset \RR\). Define
\(q:A \to \RR, x\mapsto x^2\); this is positive-definite but for
\(q\tensor \RR\), there exist \(x\neq 0\) such that \(q(x) = 0\).
\end{description}

\emph{Note:} There exists a \textbf{canonical} canonical height which is
used for the BSD conjecture.

\hypertarget{constructing-canonical-heights-for-abelian-varieties}{%
\subsubsection{Constructing Canonical Heights for Abelian
Varieties}\label{constructing-canonical-heights-for-abelian-varieties}}

\textbf{Step 1:} For elliptic curves \(E/K\), \(D = 2[O] \iff D_x\)
where \(x: E\to \PP^1\) is taking the \(x\dash\)coordinate, note that if
\(y^2 = f(x)\), then \(x([-P]) = x([P])\) making this an even function.
So taking the quotient \(E \mapsvia{q} E/[-1]\) yields a 2-to-1 map,
which factors through an isomorphism \(E/[-1] \cong \PP^1\). However,
\(2[-O] = [-1]^* D = D\), but we also have \(2[O] = x^*([\infty])\).
Then \(K(x) \subset K(E)\) contains all of the even functions,
i.e.~every even function is a function of the \(x\dash\)coordinate. So
take any rational function \(f\in K(x) \setminus K\) yields a
composition \(F: E\mapsvia{x} \PP^1 \mapsvia{f} \PP^1\) where
\(\deg(F) = 2\deg(f)\).

Define \(D_f = F^*([\infty])\), which is a pullback of a symmetric
function (since \(x\) is only a function of one coordinate), so
\([-1]^* D_f = D_f\), so \(D_f\) is a symmetric divisor. Moreover,
\(\deg D_f = 2\deg f\). Take \(\hat h_{E, f} = \hat h_{E, D_f}\), since
by Riemann-Roch the number of sections is at least 2. How does this
depend on \(f\)? Does the regulator depend on \(f\)?

\begin{quote}
There is a dependence, but it's understandable enough to be removed at
the end.
\end{quote}

We can apply the Weil Height Machine, and a comparison shows
\(h_{E, D_f} = \qty{\deg f} h_{E, D_x} + O(1)\). Therefore
\(\hat h_{E, D_f} = \qty{\deg f} \hat h_E, D_x\) on the nose. By a
computation, \(\reg(E, D_f) = \deg(f)^{n} \reg(E, D_x)\) where
\(n = \rank(E(K))\). So there is a dependence, but nothing too scary.

Silverman defines the canonical height in the curve case as

\begin{align*}
\hat h_E \definedas \frac{\hat h_{E, D_f}}{2\deg f}
.\end{align*}

\textbf{Step 2:} For abelian varieties, define a height pairing
\(\hat h: A(K\sep) \cross A\dual(K\sep) \to \RR\), where the dual
variety is defined such that the functor of points
\(A\dual \iff \pic^0(A)\), where \(A\dual(K) = \pic^0(A)\).

There is a (well-defined up to linear equivalence) \emph{Poincare
divisor} \(\mcp \in \div(A\cross A\dual)\) which is ample and symmetric.
We can then define \(\hat h_{A\cross A\dual, \mcp}\) and take the
associated bilinear form

\begin{align*}
\inner{\wait}{\wait}_{A\cross A\dual, \mcp}:\qty{A(K) \cross A(K\sep)}^2 \to \RR
.\end{align*}

So choose bases \(P_1, \cdots, P_R\) for the Mordell-Weil group mod
torsion \(A(K)/\tors\), and by isogeny, a basis \(Q_1, \cdots, Q_R\) of
\(A\dual(K)/\tors\). Then define
\(\reg(A, K) \definedas \abs{\det{\inner{P_i}{Q_j}}} > 0\).

\begin{quote}
Note that swapping signs in the basis changes the determinant by
\(\pm 1\), hence the absolute value. This is the regulator that appears
in the BSD conjecture.
\end{quote}

Recall Mumford's formula: If \(D\in \div(A)\), then

\begin{align*}
[n]^* D \sim \left[ \frac{n^2 + n}{2}\right] D + \left[ \frac{n^2-n}{2}\right] [-1]^* D
.\end{align*}

If \(D\) is symmetric, then \([n]^* D \sim n^2 D\), i.e.~the quadratic
eigendivisor condition. In this case \(\hat h_{A, D}\) is a quadratic
\emph{form}.

If \(D\) is skew-symmetric, i.e.~\([-1]^* D = -D \iff D \in \pic^0(A)\),
then \([n]^* D = nD\) and we obtain a \emph{linear} eigendivisor
condition.

\begin{quote}
Note: \(D\in \pic^0(A)\) means algebraically equivalent to zero.
\end{quote}

If \(D\in \pic^0(A)\), then \(\hat h_{A, D}: A(K\sep)\to \RR\) is a
linear form. By the Weil Height Machine, up to \(O(1)\) we have
\(\hat h_{A, D_1 + D_2} = \hat h_{A, D_1} + \hat h_{A, D_2}\). So we do
have a bilinear pairing

\begin{align*}
\inner{\wait}{\wait}: A(K\sep)^2 &\to \RR \\
(P, Q\iff D\in \div(A) \text{ after extending bases  }) &\mapsto \hat h_{A, D}(P) \in \RR
.\end{align*}

\begin{quote}
Note: we could equivalently define it on \(A(L)\) for every finite
separable extension \(L/K\).
\end{quote}

This explains the regulator appearing in BSD. This concludes the
discussion of height functions and the Mordell-Weil group.

\hypertarget{friday-february-21st}{%
\section{Friday February 21st}\label{friday-february-21st}}

Question: how do we define \(h_{V, D}\)?

Answer: write \(D = D_1 - D_2\) which are (very) ample divisors and
basepoint free. We then obtain embeddings

\begin{align*}
\varphi_1: V &\injects \PP_K^{n_1} \\
\varphi_2: V &\injects \PP_K^{n_2}
.\end{align*}

So write
\begin{align*}
h_{V, D}(p) = h(\varphi_1(p)) - h(\varphi_2(p)) + O(1)
\end{align*}

\begin{description}
\item[Example]
For \(E/K\) an elliptic curve,

\begin{itemize}
\tightlist
\item
  \(2[0]\) is an ample divisor
\item
  \(3[0]\) is a very ample divisor.
\end{itemize}
\end{description}

Let \(K\) be a local field (i.e.~\(\CC, \RR,\) a \(p\dash\)adic field,
or \(\FF_q((t))\) formal Laurent series) and \(A/K\) be an abelian
variety; we want to understand \(A(K)\). We know this has the structure
of compact abelian \(K\dash\)analytic Lie group.

\begin{itemize}
\item
  Question 1: What does Lie theory say?
\item
  Question 2: What extra information comes from \(A/K\) being a
  \(g\dash\)dimensional abelian variety?
\end{itemize}

If \(K = \CC\), then \(A(K) \cong (\RR/\ZZ)^{2g}\). If \(K = \RR\), then
\(A(K) \cong (\RR/\ZZ)^g \oplus \prod_{i=1}^d \ZZ/2\ZZ\) where
\(0\leq d \leq g\).

Fix \(d\), then

\begin{itemize}
\item
  Let \(E_1/\RR\) with \(\Delta > 0\) (and thus 3 real roots), then
  \(E_1(\RR)[2] = \qty{\ZZ/2\ZZ}^2\).
\item
  Let \(E_2/\RR\) with \(\Delta < 0\) (and 1 real root), then
  \(E_2(\RR)[2] = \ZZ/2\ZZ\).
\end{itemize}

By taking products of \(E_1\) and \(E_2\),
i.e.~\(A =(E_1)^{d} \cross (E_2)^{g-d}\).

\begin{quote}
Todo: find reference in Silverman?
\end{quote}

\begin{description}
\item[Fact]
\(A(K)\) is totally disconnected and homeomorphic to a Cantor set.
\item[Fact (From Lie Theory, Serre p.116)]
If \(G\) is an abelian compact \(K\dash\)analytic Lie group, then there
exists a filtration by open finite index subgroups

\begin{align*}
G = G^0 \supset G^1 \supset \cdots \supset G^n \supset \cdots
\end{align*}

such that

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  The successive quotients are finite, and each \(G^i\) is
  \emph{standard}, i.e.~obtained by evaluating a formal group law on
  \(\qty{\mfm^i}^g\).
\item
  \(\intersect_i G^i = (0)\).
\item
  \(G^i / G^{i+1}\) has exponent \(p\), i.e.~it is a finite dimensional
  \(\ZZ/p\ZZ\dash\)vector space.
\item
  \(G'[\tors] = G'[p^\infty]\), all of the prime-to-\(p\) torsion is
  \(p\dash\)primary.
\end{enumerate}
\end{description}

\begin{quote}
Todo: define \(p\dash\)primary torsion, \(\QQ_p\).
\end{quote}

What structure theorem does this give?

\begin{description}
\item[Theorem (C-Lacy)]
Let \(G\) be a compact, second countable, \(K\dash\)analytic abelian Lie
group of dimension \(g\geq 1\). Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  If \(\ch K = 0\) and \(d = [K: \QQ_p]\), then
  \begin{align*}
  G \cong_{\text{TopGrp}} \ZZ_p^{dg} \oplus G[\tors]
  \end{align*} where \(G[\tors]\) is finite.
\item
  If \(\ch K = p\), i.e.~\(K = \FF_q((t))\), and if it is true that
  \(G[\tors]\) is finite iff \(G[p]\) finite, then
  \begin{align*}
  G \cong_{\text{TopGrp}} \prod_{i=1}^\infty \ZZ_p \oplus G[\tors]
  \end{align*}
\end{enumerate}
\end{description}

For any \(g\geq 1\), \((T, +)\) a finite discrete abelian group
\((R, +) \cong (\ZZ_p^d, +)\) and \(R^g \oplus T\) is a compact abelian
\(K\dash\)analytic Lie group isomorphic to \(\ZZ_p^{dg} \oplus T\) (?).

\begin{quote}
Question: what does this mean for \(G = S^1\)? Ask Pete!
\end{quote}

\begin{description}
\tightlist
\item[Theorem (Cartan)]
Let \(K\) be a local field, \(\QQ\injects K\) dense (so
\(K = \RR, \QQ_p\)). Then if \(G_1, G_2\) are \(K\dash\)analytic, and
\(\varphi \in \hom_{\text{TopGrp}}(G_1, G_2)\), then
\(\varphi \in \hom_{k\dash\text{analytic}}(G_1, G_2)\).
\item[Example]
For \(R = \FF_q[[t]]\), \((R, +)^g [p] = (R, +)^g\).
\item[Example]
Take \(G = \GG_a^g(K)\) the additive group or \(A/K\) a
\(g\dash\)dimensional abelian variety (i.e.~\(G = A(K)\)) then
\(G[p] \subsetneq (\ZZ/p\ZZ)^{2g}\) and is finite.
\end{description}

\hypertarget{proof-of-cartans-theorem}{%
\subsection{Proof of Cartan's Theorem}\label{proof-of-cartans-theorem}}

\hypertarget{step-1-2}{%
\subsubsection{Step 1}\label{step-1-2}}

We want to show that \(G[p] < \infty\), then \(G[\tors] < \infty\).
We'll use the filtration in Serre's result; then for \(i \gg 1\), we'll
have \(G^i[p] = 0\). Thus for \(i \gg 1\), we'll have
\(G^i[p^\infty] = 0\); but this is the only torsion that can appear.
We'll then obtain an injection \(G[\tors] \injects G/G^i < \infty\).

\begin{description}
\item[Lemma]
Let \(H\) be an abelian torsionfree pro\(\dash p\) group
(e.g.~\(\prod \ZZ_p\)). Then \(H \cong \prod_{i\in I} \ZZ_p\), and if
\(H\) is second-countable, then \(I\) is countable.
\item[Proof]
Omitted. Idea: use Pontrayagin duality
\(G\dual \definedas \hom_{\text{Top}}(G, \RR/\ZZ)\) for locally compact
abelian groups.

\hfill\break

Use the fact that this is naturally an exact contravariant functor, this
lets you trade in profinite groups for discrete torsion abelian groups.
\end{description}

\begin{quote}
Note: look up pro-p groups. Is the Pontrayagin dual a cohomotopy group?
\end{quote}

\hypertarget{monday-february-24th}{%
\section{Monday February 24th}\label{monday-february-24th}}

\hypertarget{classification-of-locally-kdashanalytic-lie-groups}{%
\subsection{\texorpdfstring{Classification of Locally \(K\dash\)Analytic
Lie
Groups}{Classification of Locally K\textbackslash dashAnalytic Lie Groups}}\label{classification-of-locally-kdashanalytic-lie-groups}}

Let \(K\) be a locally compact field with discrete valuation, \(R\) a
valuation ring with \(\mfm = (\pi)\) its maximal ideal where
\(R/\mfm \cong \FF_q = \FF_{p^a}\). There are two cases:

\begin{itemize}
\tightlist
\item
  If \(\ch K = 0\), then \(\QQ_p \subset K\) and \(d = [L: \QQ_p]\)
\item
  If \(\ch K = p > 0\), then \(K \cong \FF_q((t))\).
\end{itemize}

\begin{description}
\item[Fact (Lie Theoretic)]
Let \(G\) be a compact commutative \(g\dash\)dimensional
\(K\dash\)analytic lie group (i.e.~locally looks like \(k^n\) with
transition harts given by convergent power series) which is 2nd
countable.

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  There exists a filtration by open subgroups
  \(G = G^0 \supset G^1 \supset \cdots \supset G^n \supset \cdots\) such
  that
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\setcounter{enumi}{-1}
\tightlist
\item
  \(G^0/G^i\) is finite and discrete for all \(i\),
\item
  \(G^i \cong (\mfm^i)^g\), with addition given by a
  \(g\dash\)dimensional formal group law,
\item
  \(\intersect G^i = (0)\), so the filtration is exhaustive,
\item
  \(G/G^{i+1}\) is \(p\dash\)torsion,
\item
  \(G^1[\tors] = G^1[p^\infty]\)
\end{enumerate}

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\setcounter{enumi}{1}
\tightlist
\item
  If \(\ch(K) = 0\), then there exists an open subgroup \(U\) of \(G\)
  such that \(U \cong (R^g, +)\) as \(K\dash\)analytic Lie groups.
\end{enumerate}
\end{description}

\begin{quote}
Analog in Lie theory: Lie groups with isomorphic Lie algebras yield
isomorphic universal covers. Can then recover the formal group from the
Lie algebra. Wildly false in characteristic \(p\), since we lose
information about the height of the formal group.
\end{quote}

\begin{description}
\item[Theorem (C-Lacy)]
\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item[]
\item
  If \(\ch(K) = 0\) (i.e.~in a \(p\dash\)adic field), then
  \(G \cong (R, +)^g \oplus G[\tors]\) as topological groups, where
  \(G[\tors]\) is finite, which is in turn isomorphic to
  \(\ZZ_p^{dg} \oplus G[\tors]\).
\item
  If \(\ch(K) = p\), then there exists a countable set \(I\) such that
  \(G \cong \prod_{i\in I} \ZZ_p \oplus G[\tors]\) as topological
  groups, where each of the groups on the RHS are closed subgroups.
  Moreover, \(G[\tors] < \infty \iff G[p] < \infty\), and when these
  conditions hold, \(I\) is infinite.
\end{enumerate}
\item[Lemma]
If \(H\) is a a commutatve torsionfree pro-\(p\) group, then
\(H \cong \prod_{i\in I} \ZZ_p\). If \(H\) is 2nd countable, then \(I\)
is countable.
\end{description}

\begin{proof}[of lemma, sketch]

We'll take \emph{Pontryagin duals}. Recall that if \(G\) is an locally
compact abelian (LCA) group, then \(G\dual \definedas \hom(G, \RR/\ZZ)\)
is an LCA group. Note that the dual of a profinite group (inverse limit)
is an ind-finite group (direct limit), which are discrete torsion
groups.

\(H\dual\) is a discrete \(p\dash\)primary torsion group. Example:
\(\ZZ_p\dual = \QQ_p/\ZZ_p\), which flips the following exact sequence:

\begin{center}\includesvg[width=0.7\linewidth]{e20a263933086c813ea017a6de84fafdca606ac7}\end{center}

Then \(H\dual = [p] H\dual\), and thus \(H\dual\) is divisible. We then
apply the structure theory of divisible group to get a direct sum, then
applying duality again yields a direct product, which proves the lemma.

\end{proof}

\begin{description}
\item[Proof (of Theorem)]
Assume that \(G[\tors]\) is finite. We have a SES of commutative
profinite groups:

\begin{align*}
0 \to G[\tors] \to G \to G/G[\tors] \to 0
,\end{align*}

and taking Pontryagin duals yields

\begin{align*}
0 \to (G/G[\tors])\dual \to G\dual \to G[\tors]\dual
.\end{align*}

Then \(G/G[\tors]\) is torsionfree and has a finite index pro-\(p\)
subgroup, so \(G/G[\tors]\) is itself a pro-\(p\) group. By the lemma,
\(G/G[\tors] \cong \prod_{i\in I} \ZZ_p\), so
\begin{align*}
\qty{G/G[\tors]}\dual \cong \bigoplus_{i\in I} \QQ_p/\ZZ_p
.\end{align*} But this is divisible, and hence injective since we're
over a PID, so the dual sequence above splits. So the original sequence
splits.

We thus have an isomorphism of topological groups
\begin{align*}
G \cong G/G[\tors] \oplus G[\tors] \cong \prod_{i\in I} \ZZ_p \oplus G[\tors]
,\end{align*} where \(G[\tors]\) was assumed finite.

Suppose \(\ch(K) = 0\). We have two open subgroups of \(G\),
\(\prod_{i\in I} \ZZ_P \leq G\) (open since its complement is finite)
and \((R, +)^g \cong \ZZ_p^{dg}\) by Serre. It follows that
\(\abs{I} = dg\).

Suppose instead that \(\ch(K) > 0\). The claim is that \([G: pG]\) is
infinite and thus \(\abs{I}\) is infinite. This is because the cokernel
of multiplication by \(p\) on \(\prod \ZZ_p \oplus G[\tors]\) is
infinite iff \(I\) is infinite, so it suffices to check the size of this
cokernel.

Consider the formal group law in characteristic \(p\) given by

\begin{align*}
[p] \in R[[X_1^p, \cdots, X_g^p]]^g
.\end{align*}

It suffices to restrict to \(G^1 = (t \FF_q[[t]]^g, \text{fgl} )\). Then
\(pG_1 \subseteq t \FF_q[[t]]^g \intersect \FF_q[[t^p]]^g\). But
\([\FF_q[[t]]: \FF_q[[t^p]] ]\) is infinite, so \([G: pG_1]\) is
infinite, so \([G: pG]\) is infinite and thus \(I\) is infinite.
\end{description}

If we know the torsion is finite, can we find bounds on their size?
We'll need to revisit Néron models (as covered in the abelian varieties
course), and introduce Tate curves.

\hypertarget{wednesday-february-26th}{%
\section{Wednesday February 26th}\label{wednesday-february-26th}}

\begin{description}
\tightlist
\item[Example (regarding generalization from last time)]
Let \(G = \prod^\infty \ZZ/p^n\ZZ\), which is a commutative pro-\(p\)
group. Then
\begin{align*}\oplus \ZZ_p^n \ZZ ~(?)~\subsetneq G[\tors] \subsetneq G
  \end{align*} with proper containments, but the direct sum is dense in
the direct product, so \(G[\tors]\) is proper and dense and thus not
closed.
\end{description}

\hypertarget{clark-xarles-local-bounds-for-torsion-points-on-abelian-varieties}{%
\subsection{(Clark-Xarles) Local Bounds for Torsion Points on Abelian
Varieties}\label{clark-xarles-local-bounds-for-torsion-points-on-abelian-varieties}}

\begin{quote}
Recommended reading! Introduces a lot of tools in their most basic form.
\end{quote}

Let \(K/\QQ_p\) be a \(p\dash\)adic field with \([K: \QQ_p] = d\), \(R\)
a valuation ring, \(\mfm\) a maximal ideal, \(\FF_q = R/\mfm\).

\begin{quote}
Recall previously: finiteness on torsion subgroups in certain cases.
Mordell-Weil works for global fields, but what is that finite group? No
positive results bounding the torsion for a Laurent series field, can be
arbitrary large.
\end{quote}

Let \(A/K\) be a \(g\dash\)dimensional abelian variety and recall

\begin{align*}
A(K) &\cong_{\mathrm{TopGrp}} (R, +)^g \oplus T \cong \ZZ_p^{dg} \oplus T
,\end{align*} where \(\abs{T} < \infty\)

We'll prove bounds in three steps:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Good reduction
\item
  Potentially good reduction
\item
  General reduction and results in the case of \emph{anisotropic
  reduction}.
\end{enumerate}

\hypertarget{good-reduction}{%
\subsection{Good Reduction}\label{good-reduction}}

\begin{description}
\tightlist
\item[Definition (Good Reduction for Abelian Varieties)]
\(A/K\) has \emph{good reduction} if there exists a smooth group scheme
\(A?R\) with generic fiber \(A/K\).
\end{description}

Noting that \(\spec(R) = \theset{(0), (\pi)}\) is a two point space, we
have the fibers

\begin{center}\includesvg[width=0.7\linewidth]{dab7abc6ece599cf6ef6d59720bc7973678a4415}\end{center}

We get a SES
\begin{align*}
0 \to A' \to A(R) \mapsvia{\text{reduction}} A(\FF_q) \to 0
\end{align*} and the fact that it is surjective is due to
\textbf{Hensel's lemma}. Here \(A' = \mfm^g\) endowed with a formal
group law. Since we have a projective variety, we can clear denominators
to replace \(A(R)\) with \(A(K)\) in the above sequence. Note that the
quotient is precisely a Mordell-Weil group, so using the Weil
conjectures, we can say quite a bit about its size.

Thus \(A' \injects A(K)\) is an open finite-index subgroup. Moreover,
all of the torsion is \(p\dash\)primary, and
\(A'[\tors] = A'[p^\infty]\). Define \(G[\tors]'\) to be the torsion
that is prime to \(p\). If the kernel is \(p\dash\)primary torsion, then
the reduction map induces an isomorphism

\begin{align*}
r: A(K)[\tors]' \mapsvia{\cong} A(\FF_q)[\tors]'
.\end{align*}

\begin{description}
\tightlist
\item[Theorem (Weil, Serre. Improved Weil Bounds)]
\(\# A(\FF_q) \leq \floor{ (1 + \sqrt q)^2}^g\). (Note that Weil's
original bound is just this without the floor function.)
\end{description}

Therefore, for \(A/K\) and \(K/\QQ\) a number field of degree \(d\),
choose \(\mfp_1, \mfp_2\) in \(\spec \ZZ_k\) points of good reduction.
Then

\begin{align*}
\# A(K)[\tors] \leq \floor{ \qty{1+p_1^{\frac d 2}   }^2  }^g
\floor{ \qty{ 1 + p_2^{\frac d 2}  }^2  }
.\end{align*}

\begin{quote}
Note: potentially ``blah'' always means ``blah'' after some base
extension.
\end{quote}

\hypertarget{potentially-good-reduction}{%
\subsection{Potentially Good
Reduction}\label{potentially-good-reduction}}

\begin{description}
\tightlist
\item[Definition (Potentially Good Reduction)]
Let \(K\) be a \(p\dash\)adic field, then \(A/K\) has \emph{potentially
good reduction} iff there exists a finite degree extension \(L/K\) such
that base extending to \(L\), \(A/L\), has good reduction.
\item[Proposition]
\(\# A(K)[\tors]' \leq \floor{\qty{1 + \sqrt q }^2 }^g\).
\item[Remark]
If \(\endo(A)\) is ``large enough'', e.g.~if \(A\) has
\(X\dash\)multiplication for \(X = \CC, \HH\) (complex or quaternionic)
then \(A\) must have potentially good reduction (see Pete's thesis).
\item[Proof]
Serre-Tate observed that if you have potentially good reduction, it can
always be obtained via a totally ramified extension \(L/K\). That is, if
\(S\) is the valuation ring of \(L\) and \(\mfm_S\) is its maximal
ideal, then \(S/\mfm_S \cong \FF_q\), so it doesn't change the size of
the residue field. Applying step 1 over \(L\) yields a bound.
\item[Application]
If \(E/\QQ\) is an elliptic curve with potentially good reduction over
\(2, 3\) (i.e.~the \(j\dash\)invariant doesn't have 2 or 3 in the
denominator). Applying the bound yields
\end{description}

\begin{align*}
  \# E(\QQ)[\tors] 
  &\leq \floor{(1 + \sqrt 2)^2  } \floor{ (1 + \sqrt 3)^2 } \\
  &= (5)(7) \\
  &= 35
  .\end{align*}

Note that a much harder theorem yields a tighter bound at \(15\), but
this is much easier and not much worse of a bound.

Taking now \(A/\QQ\) with \(\dim A = 2\) with the same conditions yields
\(\# A(\QQ)[\tors]' \leq (35^2) = 1225\).

Note that without this minor observation, the known bound is
astronomically larger!

\begin{description}
\tightlist
\item[Theorem (C-Xarles)]
Fix \(A/\QQ\) with \(\dim A = 2\) and everywhere potentially good
reduction. Then
\begin{align*}
  \# A(\QQ)[\tors] \in [1, 16] \union [18, 20], \union \theset{22, 24, 25, 28, 30, 36, 48, 60, 72}
  \end{align*}
\end{description}

Conversely, \([1, 10] \union \theset{12, 16, 18, 19 ,20, 24, 36}\) are
known to occur.

\begin{quote}
Note that for elliptic curves over \(\QQ\), it's \(\theset{1,2,4,6}\).
\end{quote}

\begin{description}
\tightlist
\item[Proof (Sketch)]
For \(K\) a \(p\dash\)adic field, the absolute unramified extension has
galois group \(\hat \ZZ\), then extend to \(\bar K\) with galois group
\(H\) to obtain a SES
\begin{align*}
  1 \to H \to g_K \to \hat \ZZ \to 0
  \end{align*} Since \(\hat \ZZ\) is a free profinite group, thus a
projective object in that category of profinite groups, so this sequence
splits and \(g_K \cong \hat \ZZ \oplus H\).
\end{description}

What does this buy us? We can think of the subgroup \(H\) now as a
quotient, so there is a \(K^{\mathrm{tr}} \leq \bar K\) such that
\(\Aut(K\tr/K) = H\), i.e.~we have a compositum

\begin{center}\includesvg[width=0.7\linewidth]{a64f46bbd5fee8b0c2b4fd7e801b521ad3425de1}\end{center}

Thus \(\bar K = K^{\text{unramified}} \cross K^{\text{tr}}\).

\hypertarget{general-reduction}{%
\subsection{General Reduction}\label{general-reduction}}

Let \(A/K\) be an abelian variety, \(A/R\) the Néron model, which is a
smooth finite type group \(R\dash\)group scheme with generic fiber
\(A/K\) and \(A(K) = A(R)\). There is a reduction map
\(A' \to A(R) \mapsvia{r} A(\FF_1) \to 0\) given by reduction, which is
again surjective by Hensel's lemma.

Here the kernel \(A'\) is exactly as before, obtained by evaluating a
\(g\dash\)dimensional formal group law on \(\mfm^g\). The difference
here is that \(A(\FF_q)\) is no longer an abelian variety, and instead
the group of \(\FF_q\) rational points of a commutative algebraic group
over \(\FF_q\).

There is structure theory of commutative algebraic groups, which may be
disconnected, so we'll need to consider things like the component group
-- but this can be arbitrarily large, which can be an issue.

\hypertarget{friday-february-28th}{%
\section{Friday February 28th}\label{friday-february-28th}}

\hypertarget{clark-xarles-paper}{%
\subsection{Clark-Xarles Paper}\label{clark-xarles-paper}}

Continuing Clark-Xarles paper on torsion points of abelian varieties
\(A/k\), where \([k: \QQ_p] = ef\) and \(q = p^f\). If \(A/k\) has
potentially good reduction, the prime to \(p\) torsion injects into the
torsion of the residue field, which gave an estimate
\begin{align*}
\# A(k)[\tors]' \text{(prime to $p$)}
\leq 
\floor{ \qty{1 + \sqrt q  }^2  }^ g
= 
\floor{ q + 2\sqrt q + 1  }^ g
\end{align*}

Can also get a bound on \(\# A(k)[p^\infty]\) in terms of \(p, g, e\),
and if \(p \gg e\), the bound is precisely 1.

We want to generalize this: let \(R\) be a valuation ring, then there
exists a finite type smooth commutative (not necessarily proper)
\(R\dash\)group scheme \(A/R\) with generic fiber \(A/k\), the Néron
model, such that \(A(R) = A(K)\) and
\begin{align*}
0 \to A'(R) \to A(R) = A(k) \mapsvia{r} A(\FF_q) \to 0
.\end{align*} Note that \(A'(R)\) here comes from a formal group law.

For \(A/\FF_q\) a smooth commutative algebraic group, and \(A/k\) has
good reduction \(\iff A/\FF_q\) is an abelian variety.

We obtain
\begin{align*}
0 \to A^0/\FF_q \to A/\FF_q \to \Phi \to 0
,\end{align*} where the first entry is the component group and the third
is a finite etale \(\FF_q\dash\)group scheme. Identify \(\Phi\) with
\(\Phi(\bar \FF_q)\) as a \(g_{\FF_q}\dash\)module, then
\(\Phi(\FF_q) = \Phi^{g_{\FF_q}}\) coming from a Galois action.

Note that \(A^0/\FF_q\) is a connected commutative smooth algebraic
group over the perfect field \(\FF_q\), so we can apply some
classification theory.

\hypertarget{the-chevalley-decomposition}{%
\subsection{The Chevalley
Decomposition}\label{the-chevalley-decomposition}}

Let \(k\) be any perfect field and \(G/k\) be a commutative connected
algebraic group, then there exists a unique SES of such groups
\begin{align*}
0 \to T\oplus U\to G \to B \to 0
.\end{align*} Plugging in \(\bar k\) points would yield a SES of Galois
modules. Plugging in \(k\) points may not work, because a surjection of
elliptic curves (isogeny) is not necessarily surjective on
\(k\dash\)points. Say \(\dim G= g, \dim B = \beta\) be the abelian
ranks.

Since \(B/k\) is an abelian variety, \(T/k\) is a linear torus and
\(T/\bar k = \GG_m^\mu\), so define \(\dim T = \mu\) the ``toric rank''.
Then \(U/k\) is a commutative unipotent \emph{linear algebraic group},
so it embeds in some \(\GL(n, k)\) with eigenvalues only 1. Note that
\(\GG_a = \theset{ [1, x; 0, 1] \suchthat x\in k}\), and anything here
will be a repeated extension of the additive group. Let
\(\alpha = \dim U\) be the additive rank.

If \(A/\FF_q\) is an abelian variety of dimension \(\beta\), we have the
bound \(\# A(\FF_q) \leq \floor{ q + 2\sqrt q + 1 }^\beta\), if
\(T_{\FF_q}\) is a torus of dimension \(u\), then
\(\# T(\FF_q) \leq \qty{q+1}^\mu\). Note that if \(\dim T = 1\), then
\(T\) is isomorphic to either \(\GG_m\) or
\begin{align*}N = \ker( \GG_m/\FF_{q^2} \to \GG_m/\FF_q  )\end{align*}
(Weyl restriction) over \(\FF_q\). Moreover,
\(N(\FF_q) = \ker(\FF_{q^2}\units \to \FF_q\units)\) which is cyclic of
order \(q+1\).

Note that if we have a nodal cubic, and the group splits to \(\GG_m\) if
the slopes of the two tangents at the node do (??) lie in \(\FF_q\), and
lands in the second case if the slopes are Galois conjugate.

If \(U/\FF_q\) is a commutative unipotent algebraic group of dimension
\(\alpha\), then \(\# U(\FF_q) = q^[\alpha]\), the number of points of
\(\GG_m^\alpha\) (?).

We'll find a filtration
\begin{align*}
0 = G^3 \subset G^2 \subset G^1 \subset G^0 = A(k)
.\end{align*} Take \(G^1\) to be the points reducing into
\(A^0(\FF_q)\), i.e.~
\begin{align*}\ker( A(k) \mapsvia{r} A(\FF_q) \mapsvia{?} \Phi(\FF_q) ).\end{align*}
Then \(G^0 / G^1 \injects \Phi(\FF_q)\). Take
\begin{align*}G^2 = \ker(A(k) \mapsvia{r} A(\FF_q)),\end{align*} then
\(G^1/ G^2 \injects A^0(\FF_q)\). Note that \(G^2\) comes from the
formal group law.

Now let \(T^i = G^i \intersect A(k)[\tors]\), then
\begin{align*}
\# A(k)[\tors]' \leq \# \Phi(\FF_Q) \# A^0(\FF_q),
,\end{align*} and since we know everything about \(A^0(\FF_q)\), we have

\begin{align*}
\#A(k)[\tors]' \leq \# \Phi(\FF_q) \floor{ q + 2\sqrt q + 1  }^\beta \qty{q + 1}^\mu
.\end{align*}

Note that \(q^\alpha\) is \(p\dash\)torsion, so we don't need the
unipotent piece here.

\begin{description}
\tightlist
\item[Theorem (Lestra-Oort)]
Suppose \(\mu = 0\), so no toric part. Then
\(\# \Phi' \leq 2^{2\alpha} \leq 2^{2g}\).
\end{description}

Thus if \(\mu = 0\), we have

\begin{align*}
\#A(k)[\tors]' \leq \# \Phi(\FF_q) ~\floor{ q + 2\sqrt q + 1  }^\beta ~\qty{q + 1}^\mu ~2^{2\alpha}
.\end{align*}

Noting that \(\alpha + \mu + \alpha = g\), this can give a better bound
than in the case of good reduction.

\begin{description}
\item[Definition (Anisotropic Torii)]
Let \(T/k\) be a torus, then \(T\) is \emph{anisotropic} iff there does
not exist an embedding \(\GG_m \injects T\) over \(k\).
\item[Definition (Anisotropic Reduction)]
\(A/k\) has \emph{anisotropic reduction} (or AR) iff the torus part is
anisotropic.
\item[Theorem (Clark-Xarles)]
If \(A/k\) has anisotropic reduction, then

\begin{align*}
\#A(k)[\tors]' 
&\leq \# \Phi(\FF_q) \floor{ q + 2\sqrt q + 1  }^\beta \qty{q + 1}^\mu 2^{\alpha} \\
&\leq \floor{ q + 2\sqrt q + 1 }^g
.\end{align*}
\end{description}

This generalizes the potentially good reduction case, since p.g.r.
implies anisotropic.

Next up: need to understand \(p\dash\)adic uniformization of abelian
varieties, and particularly for abelian varieties.

\hypertarget{friday-march-6th}{%
\section{Friday March 6th}\label{friday-march-6th}}

\hypertarget{analytic-uniformization-and-tate-curves}{%
\subsection{Analytic Uniformization and Tate
Curves}\label{analytic-uniformization-and-tate-curves}}

Gives us a way to use analysis to understand a variety.

Tate curves: let \((K, \abs{\wait}) = \RR, \CC\) or a complete
nonarchimedean field. Recall that if \(K\) is \(p\dash\)adic, then
\((K, +)\) has no nontrivial lattices (closed, discrete subgroups). For
\(\CC\) we have an infinite-dimensional covering map
\begin{align*}
\operatorname{exp}: \CC/\ZZ \mapsvia{z\mapsto e^{2\pi i z}} (\CC\units, \cdot)
\end{align*} which is an isomorphism of \(\CC\dash\)lie groups.

If \(\tau \in \HH\) and \(\Lambda_s = \generators{1, \tau}\), then
taking the exponential yields \begin{align*}
C/\Lambda_S \mapsvia{\operatorname{exp}} \CC\units/\generators{q_\tau}
\end{align*} where \(q_\tau = e^{2\pi i q}\) and
\(0 < \abs{q_\tau} < 1\), which is a type of ``additive
uniformization''.

Thus if \(E/\CC\) is any elliptic curve, then there exists such a
\(q\in \CC\) such that
\begin{align*}
E(\CC) \cong C\units/\generators{q}
.\end{align*}

\begin{remark}

The map \(q \mapsto j(q)\) is generally not injective. Note that \(j\)
factors through the quotient

\begin{center}\includesvg[width=0.7\linewidth]{1cdd8df2d9daf9c66c6313a3f07e5e43f62dcad2}\end{center}

In other words, for \((a, b; c, d) \in \liesl(2, \ZZ)\) we have
\(j(\tau) = j\qty{\frac{a\tau + b}{c\tau + d}}\). The indeterminacy here
is given by \(e^{2\pi i \tau_1} = e^{2\pi i \tau_2}\) iff
\(j_2 = j_1 + b\) for some \(b\in \ZZ\) and
\((1, b; 0, 1) \xi = \xi + b\).

\end{remark}

Last time we saw that \((K, \abs{\wait})\) is locally compact and
\(q\in K\) with \(0< \abs q < 1\), then \(\generators{q} = q^\ZZ\) is a
full lattice in \(K\units\).

\begin{quote}
Proof last time: a one-liner using compactness of the unit ball.
\end{quote}

\begin{description}
\tightlist
\item[Exercise]
If \(\Lambda \in K\units\) is a full lattice (discrete, closed, compact
quotient/cocompact), then \(\Lambda = q^\ZZ\) for a unique \(q\) with
\(0 < \abs q < 1\). The content: can't take \(\abs{q} = 1\). Think about
what happens if you take a \(p\dash\)adic field and start taking powers,
see what it is dense in.
\end{description}

\begin{quote}
Modding out by finite group doesn't lose compactness.
\end{quote}

\begin{description}
\tightlist
\item[Exercise]
What are the full lattices in \((K\units)^g\)? Higher dimensional
abelian varieties are also uniformized by lattices in \(\CC^g\). Claim:
all isomorphic to \(\ZZ^g\).
\item[Exercise]
For \(K = \RR\), \(0 < \abs{q} < 1\), then
\end{description}

\begin{align*}
R\units /\generators{q} \cong 
\begin{cases}
\RR/\ZZ \cross \ZZ/2\ZZ \quad q > 0 \\
\RR/\ZZ \quad q < 0
\end{cases}
.\end{align*}

\begin{quote}
Primary source material for Tate Curves: Silverman Chapter 5
\end{quote}

In fact, for all \(E/\RR\) there exists a \(0 < \abs q < 1\) such that
\(E(\RR) \cong E_q\) as \(\RR\dash\)lie groups, where
\(E_q \cong \RR\units / \generators{q}\). So the Tate parameter \(q\) is
telling you whether or not 2-torsion exists. Also relates sign of \(q\)
to sign of discriminant. Can take \(q\) series in the sense of modular
forms, starting next week.

\begin{quote}
Note: these \(q\) series will still converge in \(K\).
\end{quote}

For our purposes, take \((K, \abs{\wait})\) a complete nonarchimedean
field, \(0< \abs{q} < 1\) in \(K\),
\(E_q \definedas K\units/\generators{q}\) is a compact commutative
1-dimensional \(K\dash\)analytic lie group.

How all of this will end up working:

\begin{description}
\item[Theorem (Tate, Part A)]
For \((K, \abs{\wait})\) a complete nonarchimedean field and \(q\) a
Tate parameter, there exists an elliptic curve \(E_q/K\) and a
\(K\dash\)analytic group isomorphism
\begin{align*}
  \phi: K\units/\generators{q} \mapsvia{\cong} E_q(K)
  \end{align*} Moreover, \(E_q\) has split multiplicative reduction.
\item[Theorem (Tate, Part B)]
Moreover, for all finite extensions \(L/K\), note that we can extend the
norm to the algebraic closure uniquely (by Number Theory II), and the
following diagram commutes

\begin{center}\includesvg[width=0.7\linewidth]{ee905e5f6388f7472cd557fa3a2a2d3be4e60658}\end{center}
\end{description}

Taking the \(\directlim\) over finite \(L/K\),
\begin{align*}
\phi: \bar K\units /\generators{q} \mapsvia{\cong} E_q(\bar k)
.\end{align*} Moreover, \(\phi\) is equivariant for \(\aut(\bar K/K)\),
i.e.~for all

\begin{align*}
\forall \sigma \in \aut(\bar k/k),~ x\in \bar K\units/\generators{q}, \quad\quad \phi(\sigma x) = \sigma(\phi x)
.\end{align*}

For \(N \in \ZZ^+\) with \(\ch K \notdivides N\), we have
\begin{align*}
E_q[N] \cong (K\sep)\units / \generators{q} [N]
\end{align*} as \(g_K\dash\)modules. Note that \(\mu_N\) is a subgroup
of the RHS, which is fixed under the galois action.

Punch line: every elliptic curve comes with a unique order \(N\)
subgroup which is galois-invariant. Thus we reduce the question of
computing the galois module structure of torsion points to a qual-level
algebra problem of computing galois extensions.

\begin{description}
\tightlist
\item[Theorem (Tate, Part C)]
Suppose \(K\) is discretely valued, \(\abs{x} = C^{-v(x)}\),
\(v(K^x) = \ZZ\). Then \(E_q\) has \textbf{split} multiplicative
reduction (by A), and
\begin{align*}
  v(j(E_q)) = v\qty{\frac 1 q} = -v(q) \in \ZZ^+
  .\end{align*} This implies that the component group of the Néron
special fiber is \emph{constant} and isomorphic to \(\ZZ/v(q)\ZZ\).
\end{description}

\begin{quote}
See Kodaira-Néron classification
\end{quote}

Major difference: only uniformizing curves with split multiplicative
reduction, as opposed to all curves in \(\RR, \CC\) case.

\begin{description}
\tightlist
\item[Theorem (Tate, Part D)]
For all \(j\in K\) with \(\abs{j} > 1\), there exists a unique
\(q\in K\) such that \(0 \abs{q} < 1\) and \(j(E_q) = j\). So the Tate
parameter is uniquely recovered.
\end{description}

For every \(q\), we build a curve \(E_q\), and get a galois structure on
its torsion points. They must have split multiplicative reduction, and
every such curve is isomorphic to \(E_q\) for a unique \(q\). Pretty
strong result!

\begin{description}
\tightlist
\item[Theorem (Tate, Part E)]
Assume \(\ch K \neq 2\). If \(E/K\) is any elliptic curve with
\(\abs{j(E)} > 1\) (so not integral), then there exists a unique \(q\)
such that \(E/K\) is a \emph{quadratic twist} of \(E_q/ K\). For
\(K(\sqrt \alpha) / K\), we have
\begin{align*}
  E/K(\sqrt \alpha) \cong E_q / K(\sqrt \alpha)
  .\end{align*} Moreover,
\end{description}

\begin{align*}
  E/K \text{ has } 
  \begin{cases}
  \text{split mult. reduction} & K(\sqrt \alpha) = K \quad \sim \GG_m\\
  \text{non-split mult. reduction} & K(\sqrt \alpha)/K \text{ is unramified } \quad\sim T\text{ a nonsplit torus} \\
  \text{additive reduction} & K(\sqrt \alpha) / K \text{ is ramified }
  \end{cases}
  .\end{align*}

Remark: the torsion can be arbitrarily large.

\begin{description}
\tightlist
\item[Example]
Take \(K = \QQ_p\) and \(q = p^n\) so \(\abs{p^n} < 1\). Then
\begin{align*}
  E_q(\QQ_p) \cong \QQ_p\units / \generators{p^n}
  .\end{align*} What is the torsion subgroup?
\end{description}

Note that roots of unity in \(\QQ_p\) inject into the quotient, and
\(\mu(\QQ_p) \cong \FF_p\units \cong \ZZ/(p-1)\) (by Number Theory II).
The class of \(p\) is order \(n\), and thus also torsion, so
\begin{align*}
  T \cong \ZZ/(p) \times \ZZ/(p-1)
  ,\end{align*} where the containment is obvious and staring at this
calmly will show that this in fact the entire torsion subgroup.

In general, for \((K, \abs{\wait})\) discretely valued, it's easy to
give an upper bounds. We have
\begin{align*}
E_q(K)[N] \injects \mu_N(K) \times Z
,\end{align*} where \(Z\) is cyclic of order \(v(q)\). Equality occurs
if (not iff) \(q\) is a perfect \(v(q)\)th power.

Taking \(E_q(\bar K)[N]\) is generated by \(\zeta_n\), a primitive
\(n\)th root of unity, and any \(n\)th root of \(q\), \(q^{\frac 1 n}\).
Thus this is literally a Kummer extension, and this is no big deal to
work out.

The culprit: the component group can be arbitrarily large!

\hypertarget{monday-march-30th}{%
\section{Monday March 30th}\label{monday-march-30th}}

We'll begin by recalling some properties that Tate curves have.

\hypertarget{tate-curves}{%
\subsection{Tate Curves}\label{tate-curves}}

Let \((K, \abs{\wait})\) be a complete nonarchimedean normal field and
\(q\in K\) with \(0 < \abs{q} < 1\) so
\(\abs{j(E_q)} = {1 \over \abs q} > 1\). Let \(R\) be its valuation
ring. Then

\textbf{Remark (a)}:

There exists an elliptic curve \(E_1/K\) and a \(K\dash\)analytic group
isomorphism \begin{align*}
\phi_K: K\units/\gens{q} &\mapsvia{\cong} E_q(K)
,\end{align*} a \emph{multiplicative uniformization}.

For all finite \(L/K\), the norm extends uniquely over \(L\), and there
is a compatibility in the following sense

\begin{center}\includesvg[width=0.7\linewidth]{e1426af6a5ceaeb90d38edb4401d0f8951a36fb6}\end{center}

And taking the \(\directlim\) over separable \(L/K\) yields
\begin{align*}
\phi: (K\sep)\units/\gens{q} \mapsvia{\cong} E_q(K\sep)
,\end{align*} which is an isomorphism of \(g_K\dash\)modules where
\(g_K = \aut(K\sep/K)\).

Note that in this case, \(E_q(K)\) will not have potentially good
reduction, and in fact we precisely get all curves with split
multiplicative reduction.

\textbf{Remark (b)}:

Suppose \(*K, \cdot)\) is discretely valued. Then \(E_1\) has split
multiplicative reduction and the component group is cyclic of order
\(v(q)\).

\begin{description}
\tightlist
\item[Example]
If \(\pi\) is a uniformizing element for \(K\) and \(q = \pi^n\) then
\(E_q(K)[\tors] \cong K\units / (\pi^n)\) and the class of \(\pi\) has
order \(n\). So the torsion can be arbitrary large, but we can
understand this quotient using Kummer theory.
\end{description}

Suppose that \(K\) is a complete discrete valuation field with perfect
residue field \(k\).

\begin{itemize}
\tightlist
\item
  If \(\abs{j(E)} \leq 1\) then \(E\) has potentially good reduction.

  \begin{itemize}
  \tightlist
  \item
    Note that we just need to pass to an extension of degree 2 or 4.
  \end{itemize}
\item
  If \(\abs{j(E)} > 1\) then \(E\) has \textbf{no} potentially good
  reduction and is a unique quadratic twist of a \emph{unique} Tate
  curve.
\end{itemize}

The latter is a strong statement: we'll attached a unique \(q\) and
\(E_q\) to \(E\) such that \(j(E) = j(E_q)\). Thus these \(E\) will have
explicit identifications, and since we understand the former case of
potentially good reduction somewhat well, we'll be in good shape.

\begin{description}
\item[Theorem]
If \((K, v)\) is a complete discrete valuation field with perfect
residue field \(k\). Let \(E/K\) be an elliptic curve suchj that
\(\abs{j(E)} < 0\) (so it's a quadratic twist of a Tate curve). Let
\(\ell \geq 3\) where \(\ell\) does not divide \(\ch K v(j(E))\).

\begin{quote}
If \(\ch K = 0\) and \(\ch k = p\), then \(\ell = p\) is allowed.
\end{quote}

Let \(I_K\) be the inertia group of \(K\) (noting that we're over a
local field so \(g_K\) is local), a subgroup of the absolute Galois
group of \(K\), i.e. \(1 \to I_K \to g_K \to g_k \to 1\).

Then there exists a basis \(e_1, e_2\) of \(E[\ell](K\sep)\) and
\(\sigma \in I_K\) such that \begin{align*}
\phi_\ell(\sigma) =
\begin{bmatrix}
1 & 1 \
0 & 1
\end{bmatrix}
,\end{align*} i.e.~\(e_1 \mapsto e_1\) and \(e_2 \mapsto e_1 + e_2\).
\end{description}

Note that this matrix is in Jordan form. It is unipotent, and every
unipotent matrix in a field of \(\ch = \ell\) has order \(\ell\). Thus
we have a very particular element of order \(\ell\) in the Galois group.

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

\begin{description}
\item[Proof]
We can replace \(K\) with any algebraic extension \(L/K\) such that
\(\phi(L/K)\) is finite and prime to \(\ell\). In particular, we can
assume that \(L\) contains the maximal unramified extension
\(L \supset K^\text{unr}\), which occurs iff \(k = \bar k\) is
algebraically closed. Thus we can take Galois groups instead of inertia
groups.

We may also assume that \(E\) is isomorphic to a Tate curve over \(L\),
i.e.~\(E \cong_L E_q\) by replacing \(L\) with a quadratic extension,
and that \(L\) has a primitive \(\ell\)th root of unit \(\zeta_\ell\).
This is because the ramification index is equal to the degree \(\ell-1\)
which is still prime to \(\ell\). Thus we can use the Tate
parametrization and Kummer theory.

Let \(Q \in L\sep\) such that \(Q^\ell = q\), i.e.~take an \(\ell\)th
root of the Tate parameter. By classical Kummer theory, \(L(Q) / L\) is
cyclic of degree \(\ell\), so let \(\sigma \in \aut(L(Q) / K)\) be a
generator. Then \(\zeta \definedas \sigma(Q) / Q\) is nontrivial and
thus a primitive \(\ell\)th root of unity.

Note that \(E_1(L\sep) \cong (L\sep)\units / \gens{q}\). So we can
construct a basis by taking \(e_1 = \zeta\) and \(e_2 = Q\).
\item[Corollary]
For \(K, E/K, \ell\) as in the theorem, so \(\abs{j(E)} > 1\) and
\(\ell\) does not divide \(\ch K v(j(E))\) Then \(E/K\) has an
\textbf{unique} \(K\dash\)rational \(\ell\dash\)isogeny.
\item[Proof]
There exist a quadratic twist \(E_q\) of \(E\) that is a Tate curve
since the reduction is not potentially good, so \(j\neq 0, 1728\), so
the only twists are quadratic. Thus WLOG we can assume \(E \cong E_q\).

\begin{quote}
Next time: to each quadratic twist we'll associate a quadratic character
\(\chi\), to which we associate a Galois representation.
\end{quote}

If we take \(H \definedas \theset{\mu_\ell \gens{q}} < \GG_m/\gens{q}\)
the group of roots of \(\ell\)th roots of unity, which is a
\(g_K\dash\)stable subgroup. We also have \(\sigma \in I_K\) with
\(?\phi(\sigma) = \begin{bmatrix} 1 & 1 \
0 & 1 \end{bmatrix}\) and the cyclic subgroups of \(E[\ell]\) are
generated by \([1, 0]\) and \([x, 1]\) where \(x\in ?\).
\end{description}

\hypertarget{monday-april-6th}{%
\section{Monday April 6th}\label{monday-april-6th}}

We've been talking about quadratic twists of Tate curves. Let \(K\) be a
field, \(\ch K \neq 2\), \(V/K\) a nice variety, \(\iota: \aut(V)\) an
involution, \(L = K(\sqrt d)/K\), then \(J(V, \iota, L) = V^d/K\) is a
quadratic twist. We have \(V^d / K(\sqrt d) = V_{K(\sqrt d)}\), and
\(\Aut K(\sqrt d) / K = \theset{1, \sigma}\) is a two element set, so
\begin{align*}
V^d(K) \definedas \theset{ p \in V(K(\sqrt d)) \suchthat \sigma(p) = \iota(p) }
.\end{align*}

Example: For \(E/k\) an elliptic curve with \(\iota = [-1]\), if
\(E: y^2 =x^3 + Ax + B\) yields
\(E^d: dy^2 = x^3 + Ax + B \sim y^2 = x^3 + d^2Ax + d^2 B\). This works
out similarly for hyperelliptic curves \(C\): for \(y^2 = f(x)\), we
obtain \(C^d: dy^3 = f(x)\). Note that there is not such a description
in equations for abelian varieties.

\begin{description}
\item[Theorem (C- Stankewicz, 2018)]
Let \(K\) be a number field and \(C/K\) a ``nice'' curve (conditions to
be described) with \(\iota\) an involution on \(C\). Then

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\tightlist
\item
  For all but finitely many \(d \in K\units / (K\units)^2\),
  \(C^d(K) = \emptyset\).
\item
  There exist infinitely many \(d\) such that \(C^d\) has
  \(K_v\dash\)rational points for all places \(v\).
\item
  If \(K = \QQ\), then as \(x\to\infty\) the number of \(d\) with
  \(\abs{d} \leq x\) such that \(C^d\) violates the Hasse principle is
  \(\gg x/\log x\)
\end{enumerate}
\end{description}

Punchline: under some conditions, you get a lot of Hasse principle
violations.

Conditions:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  \(\theset{p\in C(K) \suchthat \iota(p) = p} = \emptyset\)
\item
  \(\theset{p \in C(\bar K) \suchthat \iota(p) = p} \neq \emptyset\).
\item
  There exists some \(d\in K\units / \qty{K\units}^2\) such that \(C^d\)
  has points everywhere locally.
\item
  \((C/\iota)(K)\) is finite.
\end{enumerate}

This works for hyperelliptic curves, conditional on the abc conjecture.

For condition 4, we can just take the genus \(>2\) and apply Falting's
theorem. We can apply this to Atkin-Laner (?) twists of elliptic curves.

\begin{quote}
Note that there is a lot of work being done on quadratic twists.
\end{quote}

For \(A/K\) an abelian variety, take \(\iota = [-1]\) and
\(K(\sqrt{d}) / K\) with \(\ch K \neq 2\). What is the quadratic twist
\(A^d/K\)? For \(N\geq 3 \in \ZZ^+\) and \(\dim A = g\), there is a
representation
\(\rho_N: g_K \to \Gl(2g, \ZZ/n\ZZ) = \Aut(A[N]) =\Aut(\qty{\ZZ/N\ZZ}^{2g})\).
This is the ``mod \(N\) galois representation''.

What is the galois representation on the quadratic twist
\(\rho^d_N: g_K \to \GL(2g, \ZZ/N\ZZ)\)? They both restrict to the same
function on \(g(K(\sqrt d))\). The succinct answer is that
\(\phi^d_N = \phi_N \tensor \chi_d\), where
\(\chi_d: g_k \to \theset{\pm 1}\) where \(\sigma \mapsto 1\) iff
\(\restrictionof{\sigma}{K(\sqrt d)} = 1\) and -1 otherwise. What is
this tensor? We can think of
\(\theset{\pm 1}\in \qty{\ZZ/N\ZZ}\units = \Gl(1, \ZZ/N\ZZ)\). In other
words, for all \(\sigma \in g_k\), we have
\(\rho_N^d = \chi_d(\sigma) \rho_N(\sigma) = \pm \rho_N(\sigma)\), where
we identify \(\pm 1\) with the scalar matrices \(\pm I\) in
\(\Gl(2g, \ZZ/N\ZZ)\). Thus it's a twist by a quadratic character.

Now let \(G \subset A[n] \subset A\) where \(G = \gens{p}\) is cyclic of
order \(N\) and \(p\in A(K\sep)\). Then \(G\) is \(g_k\dash\)stable, so
we can choose a basis \(p = p_1, \cdots, p_{2g}\) of
\(A[n] / \ZZ/N\ZZ\). We can then write \begin{align*}
\phi_N(\sigma) =
\begin{bmatrix}
\chi_G(\sigma) & \ast \\
0 & \ast
\end{bmatrix}
\end{align*}

where \(\sigma(p) = \chi_G(\sigma) p\), and thus \begin{align*}
\begin{bmatrix}
\chi_d(\sigma) \cdot \chi_G(\sigma) & \ast \\
0 & \ast
\end{bmatrix}
.\end{align*}

We then have a diagram of inclusions

\begin{center}\includesvg[width=0.7\linewidth]{1ffc9bd8376a37db29d5d29d4ede4367dede4538}\end{center}

What if we now want an actual \(K\dash\)rational point of order \(N\)?
Cyclic order \(N\) \(K\dash\)rational subgroups of \(A\) are equal to
cyclic order \(N\) \(K\dash\)rational subgroups of \(A^d\), so

For \(E/K\), we proved that it has a unique cyclic order \(\ell\)
subgroup corresponding to \(\mu_\ell\). So the same holds for every
quadratic twist, and every rational point has to come from the original
\(\mu_\ell\) since cyclic subgroups aren't changing under quadratic
twists. Thus only the isogeny character can change.

So when does quadratic twisting gain or lose a rational point? The image
of the isogeny character must be order 1 or 2, corresponding to a
quadratic extension, and the isogeny character is trivialized. We'll
eventually conclude that none of the quadratic twists will have a
rational point of order \(N\).

\hypertarget{wednesday-april-8th}{%
\section{Wednesday April 8th}\label{wednesday-april-8th}}

\begin{description}
\tightlist
\item[Exercise]
If \(\ch K \neq 2,3\) and \(E_1, E_2/ K\) are elliptic curves, set
\(L = K(\sqrt{d})\). If \(E_1 \not\cong_K E_2\) but
\(E_1 \cong_{K(\QQ)} E_2\), then \(E_2 \cong E_1^d\) (so \(E_2\) is a
quadratic twist of \(E_1\)).
\end{description}

\begin{quote}
Note that we take the twist with respect to the inversion map.
\end{quote}

This is not generally true for quadratic twists of \((V, \iota)\). It
depends on \(\aut(V)\). If \(\aut(E) \theset{\pm 1}\), this holds for
\(j = 0 ,1728\). Then for \(E/K\) an elliptic curve,
\(J(E) = \theset{ E'_K \suchthat E' \cong_{K(\sqrt d)} E}\), which is in
correspondence with
\(H^1_{\text{gal}} (\Aut(K(\sqrt d) / K), \aut(E) )\). This is given by
\(H^1(\theset{1, \sigma}, I)\), which reduces to a hom set.

Last time we justified the argument that every ramified quadratic twist
of a Tate curve, resulting in an elliptic curve with additive reduction.

Question:

Given \(E_1, E_2/K\), how we can decide if \(E_1 \cong_K E_2\) if
\(j\neq 0,1728\)?

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  Check if they are isomorphic over \(\bar K \iff j(E_1) = j(E_2)\). If
  not, stop, otherwise if so, they are quadratic twists of each other.
\item
  Use an invariant (see Silverman) \(\gamma(E)\); then
  \(E_1 \cong E_2 \iff j(E_1) = j(E_2)\) and
  \(\gamma(E_1) = \gamma(E_2)\).
\end{enumerate}

\begin{description}
\tightlist
\item[Proposition (An Application)]
Let \(F\) be a field, \(F \injects \CC\), \(E/F\) an elliptic curve with
complex multiplication. Then \(\endo(E) = \OO\) (taken over the
algebraic closure) is some order in an imaginary quadratic field
\(K = \QQ(\sqrt{\Delta})\), and \(j(E) \in \bar{\ZZ}\), the algebraic
integers. Then \(\QQ(j(E))\) is a number field \(L\), and for all
\(\mfp \normal \ZZ_\ell\), \(E\) has potential good reduction at
\(\mfp\) (i.e.~the \(j\dash\)invariant is \(p\dash\)adically integral).
\end{description}

\begin{quote}
We more or less need to show that complex multiplication and ``being a
Tate curve'' are mutually exclusive.
\end{quote}

\begin{description}
\item[Proof (Sketch)]
Let \(K = \endo(E) \tensor_\ZZ \QQ\) and \(L\definedas K(j(E))\). Then
there are two cases:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\tightlist
\item
  If \(j(E)\) is algebraic then \(L\) is a number field.
\item
  Otherwise \(j(E)\) is transcendental, and thinking of \(j\) as an
  indeterminate we have \(L \cong K(j)\), a rational function field.
\end{enumerate}

\textbf{Case 1:} Take \(\mfp \in \ZZ_L\) and \(E/L_\mfp\) a
\(p\dash\)adic field. We want to show that \(v_\mfp (j(E)) \geq 0\).
Assume the proposition doesn't hold, toward a contradiction. Then after
a quadratic base extension \(M_\mfp / L_\mfp\), the curve \(E/M_\mfp\)
is a Tate curve.

\hfill\break

For \(\ell \gg 0\), define \(N_\ell \definedas M_\mfp (E[\ell])\) and
there is a map \(\theta_\ell: g_{M_\mfp} \to \GL(2, \ZZ/\ell\ZZ)\). Note
that \(N_\ell / M_\mfp\) is Galois, and
\(\theta_\ell \definedas \Aut(N_\ell / M_\mfp) \cong \rho_\ell(g_{M_\mfp})\).

\hfill\break

Because \(E / M_\mfp\) has \(\mu_\mfp\dash\) central (?) complex
multiplication, \(G_\ell\) is commutative and
\(\rho_\ell: g_{M_\mfp} \to \GL(1, \OO/\ell\OO)\) which is commutative.
Since it's a Tate curve, we have an explicit description of the
\(\ell\dash\)torsion coming from Kummer theory. This buys us the fact
that \(\mu_\mfp(E[\ell]) = M_\mfp(\mu_\ell, q^{1/\ell})\), which is just
the splitting field of \(t^\ell - q\) over \(M_\mfp\) by ``non-abelian
Kummer Theory''.

\begin{quote}
Qual alert: computing with these splitting fields shows up often on
Algebra quals!
\end{quote}

Choosing \(\ell \gg 0, \ell > 2\), and \(\ell\) not dividing
\(v_\mfp(q)\) makes this splitting field irreducible. Thus the galois
group is given by
\(G_\ell \cong \ZZ/\ell \ZZ \semidirect \chi_\ell(g_{M_\mfp})\) where
the RHS is the cyclotomic character sitting inside of
\(\qty{\ZZ/\ell\ZZ}\units\), with the action given by multiplication.
(See Lang's Algebra or Pete's Field Theory notes.)

\hfill\break

What is the abelianization? We can compute
\(\#[G_\mfp: G_\ell] = \# \chi_\rho(g_{M_\mfp})\), and eventually this
reduces to the case of a finite field \(\FF_q\) and considering
\([\FF(?): \FF_q]\).

\textbf{Case 2}: Suppose \(j(E)\) is transcendental over \(K\), then
\(K(j)\) is a rational function field equal to \(ff(K[j])\) with
\(v_\infty \definedas -\deg(P(j))\) and \(v_\infty(j) = -1\). Letting
\(K_\infty\) be the completion of \(K(j)\) with respect to \(v_\infty\),
then \(K_\infty \cong K((1/j))\) with residue field \(K = \QQ(\Delta)\).
Now \(E/K_\infty\) has complex multiplication, \(v_\infty(j(E)) < 0\),
and we again have a quadratic extension and \(E\) is a Tate curve with
complex multiplication. Using the fact that a number field only contains
finitely many roots of unity, the same argument works.
\end{description}

There is a higher-genus notion of a Mumford curve which are analogously
\(p\dash\)adically uniformized (much in the same way \(\HH\)
uniformizes) by lattices in \(\GG_m^g\).

\hypertarget{friday-april-10th}{%
\section{Friday April 10th}\label{friday-april-10th}}

Let \((K, \abs{\wait})\) be complete and \(q\in K\units\) be the Tate
parameter with \(\abs{q} < 1\). The \emph{Tate curve} is given in
equations by \begin{align*}
E_q: y^ + xy = x^3 + a_1(q) x + a_2(q)
.\end{align*}

For all \(k\geq 1\), we define
\(S_k(q) = \sum_{n=1}^\infty {n^k q^n \over 1 - q^n}\) for which
convergence in \(K\) can easily be checked since in the non-Archimedean
settings, series converge iff the terms go to zero.

Fact: \begin{align*}
a_4(q) &= -S_3(q) \\
a_6(q) &= {-5 S_3(q) + 7 S_5(q) \over 12}
.\end{align*}

This is some multiplicative analog of exponentiating the Weierstrass
elliptic equations in \(\CC\units\). The discriminant is given as in the
theory of modular forms as \begin{align*}
\Delta (E_q) = q \prod_{n=1}^\infty \qty{1 - q^n}^{?} \neq 0
,\end{align*}

so \(E_q\) is an elliptic curve.

The \(j\dash\)invariant is given by \begin{align*}
j(E_q) &= {1 \over q} + 744 +  196844q + \cdots \\
& = \frac 1 q \sum_{n=1}^\infty c(n) q^n,\quad c(n) \in \ZZ
.\end{align*}

\begin{quote}
Connection to irreducible representations of the Monster group.
\end{quote}

Note that \(\abs{j(E_q)} = {1 \over \abs q} > 1\) and (claim) the norms
of \(S_k, a_4, a_q\) are less than 1.

After reducing mod \(n\), we obtain \(\ts{x\in K \st \abs x < 1}\) and
\begin{align*}
\tilde E_q:  y^2 + xy = x^3
\end{align*} which becomes singular at \((0, 0)\).

Taking the lowest degree part yields \(0 = y^3 + xy = y(y+x)\), meaning
we have split multiplicative reduction. We now want the Tate
parameterization.

For \(u \in K\units \sm q^\ZZ\), define \begin{align*}
X(u, q) &\definedas \sum_{n\in \ZZ} {q^n u \over \qty{1 - q^n u}^2} - 2S_1(q)
Y(u, q) &\definedas \sum_{n\in \ZZ} {\qty{q^n u}^2 \over \qty{1 - q^n u}^2} - 2S_1(q)
.\end{align*}

\begin{quote}
Note the similar formulas involving \(g_2, g_3\) in the case of modular
forms.
\end{quote}

We have a uniformizer map \begin{align*}
\phi: K\units &\to E_q(K) \\
u &\mapsto
\begin{cases}
(X(u, q), Y(u, q)) & u\in q^\ZZ \\
0 & u\in q^\ZZ
\end{cases}
.\end{align*}

\begin{quote}
Usual group symmetry trick, just add up over orbit of the group.
\end{quote}

This immediately yields a map \(K\units / \gens{q^\ZZ} \to E_q(K)\), and
the work goes into showing that \(\phi\) is a \(K\dash\)analytic group
isomorphism. The hard part here is showing surjectivity, see Silverman
II, Chapter 5 (4 page proof).

Since \((K, v)\) is complete and discretely valued (with multiplicative
reduction), we have \(v(j(E_q)) = -v(q)\). See Kodaira-Neron. Moreover,
the identity component of the generic fiber is cyclic of order \(v(q)\).

Question: is every curve with an integral \(j\dash\)invariant a Tate
curve?

\begin{description}
\tightlist
\item[Lemma]
Let \(\alpha \in K\) with \(\abs \alpha > 1\). Then there exists a
unique \(q \in K(\alpha)\) (a finite extension) such that \(\abs q < 1\)
and \(j(E_q) = \alpha\).
\end{description}

Thus the function \(j\) which maps \(q \mapsto j(E_q)\) is bijective, so
you can recover the Tate parameter \(q\) from either the elliptic curve
or the \(j\dash\)invariant, which is not true over \(K = \CC\).

\begin{description}
\item[Proof]
Define
\(f(q) \definedas {1 \over j(q)} = {1 \over 1 + ?q + ? q^2 + \cdots}\).
By inverting formal power series, this is equal to
\(q - 744 q^2 + 35665 q^3 + \cdots \in \ZZ[[q]]\). To get a functional
inverse, for any ring \(R\), any \(f\in R[[q]]\) such that
\(g \circ f = f\ \circ g = \id\) iff \(a_1(f)\) is a unit. Then there is
a \(\beta \in E, \abs \beta < 1\) such that
\(\abs{g(\beta)} = \abs{\beta}\). Defininng
\(q \definedas g(1/\alpha) \in K(\alpha)\), we have
\(0 < \abs q = \abs{1 \over \alpha} < 1\), and inverting yields
\({1 \over J(q)} = f(q) = f(g(1/\alpha)) = 1\alpha\), so
\(\alpha = j(q)\). This shows existence, so the \(j\) function is
surjective.

\hfill\break

For uniqueness, suppose \(j(q) = j(q')\) with \(\abs{q}, \abs{q'} < 1\).
Then \(f(q) = f(q')\) and so
\(0 = \abs{f(q) - f(q')} = \abs{q-q'} \abs{1 - 744(q+q') + 35665 \qty{q^2 + qq' + (q')^2} + \cdots}\).
The second term has norm 1, so this is equal to \(\abs{q - q'}\). So the
\(j\) function is injective.
\end{description}

Note that this proof doesn't go through for \(\CC\), and the injectivity
is false. Over \(\CC\), the function \begin{align*}
j: \HH/ \SL(2, \ZZ) &\mapsvia{\cong} \CC
j(q) = j(q') &\iff q' = {aq + b \over cq + d},
\begin{bmatrix}
a & b \
c & d
\end{bmatrix}
\in \SL(2, \ZZ)
\end{align*}

where \(q = e^{2\pi i j}, q' = e^{2\pi i j'}\), we have
\(j(q) = j(q') \notimplies q = q'\). E.g. for curves over \(\CC\), the
\(\tau\) parameter can be associated to multiple curves.

Next time: non-Archimedean uniformization of higher dimensional
varieties, particularly analytic torii. Any compact connected
\(\CC\dash\)analytic Lie group is a torus. Note that the moduli space of
elliptic curves of dimension ? is greater than the corresponding moduli
space of abelian varieties, so additional conditions are needed to
ensure they are torii. The abelian varieties that can be uniformized are
exactly those with split multiplicative reduction.

\hypertarget{monday-april-13th}{%
\section{Monday April 13th}\label{monday-april-13th}}

\hypertarget{non-archimedean-uniformization-of-abelian-varieties}{%
\subsection{Non-Archimedean Uniformization of Abelian
Varieties}\label{non-archimedean-uniformization-of-abelian-varieties}}

We'll start by recalling the Archimedean uniformization theory,
i.e.~over \(\CC\): every compact connected \(\CC\dash\)analytic (thus
commutative) Lie group is isomorphic to \(\CC^g / \Lambda\) a discrete
cocompact lattice isomorphic as a \(\ZZ\dash\)module to \(\ZZ^{2g}\).
Note that \(\Lambda\) is a complex torus, and not a linear torus
\(\GG_m^n\).

\begin{quote}
Thus all complex torii are isomorphic as topological (even real Lie)
groups.
\end{quote}

Note that if \((K, \abs{\wait})\) is a complete non-Archimedean field,
then full lattices in \(k\) do not exist in characteristic zero, so we
want a multiplicative version. The right notion of isomorphism to use
here is homothety (scaling by a complex unit) and thus
\(\Lambda \subset \CC^g\) has an action of \(\GL_n(\CC)\) so
\(\CC^g/\Lambda \cong \CC^g/m \Lambda\), so we may choose \(\Lambda\) to
have a \(\ZZ\dash\)basis given by \(\lambda_1 = e_1, \lambda_g = c_g\)
generalizing the usual basis of \(1, \tau\) in the 1-dimensional case.

We have an exponential map
\(\operatorname{exp}: \CC^g \to (\CC\units)^g\) where
\(\vector z \mapsto e^{2\pi i \vector z}\), so pushing \(\Lambda\)
through \(\operatorname{exp}\) yields a full lattice
\(\Lambda' \subset (\CC\units)^g\) where
\(\CC^g / \Lambda \cong (\CC\units)^g/\Lambda'\).

In \(\dim g > 1\), not every \(\CC\dash\)torus corresponds to an abelian
variety (contrasting the \(g=1\) case). Changing basis for \(\Lambda\)
yields a matrix \((I \mid Z)\) where \(Z \in \Mat_{g, g}(\CC)\), and
Riemann showed that \(\CC^g / \Lambda\) determines an abelian variety
iff this matrix is symmetric with positive definite imaginary part. This
occurs iff \(\Lambda\) admits a \emph{Riemann form}, which is an analog
of the polarization form, and is given by
\(H: \CC^g \cross \CC^g \to \CC\) that is Hermitian and positive
definite such that \(H(\Lambda) \subset \ZZ\).

\begin{quote}
Equivalently, having an ample line bundle, some multiple of which embeds
it into some projective space.
\end{quote}

By counting parameters, we can determine the dimension of the moduli
space of \(g\dash\)dimensional complex torii: since we have a
\(g\times g\) nonzero matrix, this yields \(g^2\). Similarly, counting
moduli for abelian varieties yields \(g(g+1)/2\). Note that these
coincide for \(g=1\) but differ for \(g>1\), so there's an entire
dimension of torii that aren't abelian varieties.

We define a \(K\dash\)analytic torus as
\(T(\Lambda) = (K\units)^g / \Lambda\). This is a \(g\dash\)dimensional
\(K\dash\)analytic manifold which is compact iff \(K\) is locally
compact.

Question: For which \(\Lambda\) is there an abelian variety \(A/K\) such
that \(A \cong T(A)\)? This should mean that for all
\(K \subset L \subset \bar K\), we have
\(A(L) \cong (L\units)^g / \Lambda\). But which category should we take
this in? We need to ensure that the notion of isomorphism isn't so weak
that these isomorphism occur for trivial reasons.

We'll work in the category of rigid \(K\dash\)analytic spaces: locally
ringed spaces (with a sheaf of functions) equipped with a Grothendieck
topology, which are locally isomorphic to \emph{affinoid} spaces. The
most basic example of an affinoid space: \(\ZZ_p \injects \QQ_p\),
locally compact but not compact, taking the \(p\dash\)adic disc (defined
by an inequality in the norm) is affinoid. In general, points in a space
satisfying norm inequalities and thus bounded, generalizing what happens
in \(p\dash\)adic fields.

\begin{quote}
Affinoid spaces will be analogous to CW complexes.
\end{quote}

A Grothendieck topology (like the etale topology) will be a
generalization of open sets to arbitrary categories by saying what it
means for an object to cover another. These restrict the coverings
allowed by the general topology to those that appear in algebraic
geometry.

Remark: A connected rigid \(K\dash\)analytic space \(X\) has a in fact
has a \emph{field} of meromorphic functions \(K(X)\), which is an
important invariant.

Example: Take a connected complex manifold, we can consider its
meromorphic functions which is a sheaf that is locally given by
quotients of holomorphic functions.

For \(\Lambda \subset (K\units)^g\), the transcendence degree of
\(K(T(\Lambda))\) over \(K\) is finite and \(\leq g\) because
\(T(\Lambda)\) is a proper (similar to compact) \(K\dash\)analytic
spaces. There is a correspondence between curves and function fields,
which we get whenever the transcendence degree is 1. So this yields a
good analog for complex torii.

Recall that if \(T(\Lambda)\) is a complex torus implies that
\(0 \leq \deg_\CC^{\text{transc.}} \CC(T(\Lambda)) \leq g\), then there
exists a universal quotient \(A_n\) with \(T\surjects A_n\).

Let \(\Lambda = (K\units)^g\) and
\(\lambda\dual = \hom_\ZZ(\Lambda, \ZZ)\), then \(T(\Lambda)\) is an
abelian variety iff there exists a map
\(\sigma: \Lambda \to \Lambda\dual\) such that
\(\sigma(\gamma_1)(\gamma_2) = \sigma(\gamma_2)(\gamma_1)\) and
\((\gamma_1, \gamma_2) \definedas - \log \abs{ \sigma(\gamma_2)(\gamma_1)}\)
is positive definite.

Assume no that \(K\) is a discretely valued valuation field, and
abbreviate the identity component of the Néron special fiber as the
Néron group (\(g\dash\)dimensional algebraic group over the special
fiber which is an abelian variety iff good reduction)

\begin{description}
\tightlist
\item[Theorem]
\hfill
1. If \(T(\Lambda)\) is an abelian variety, this it has split
multiplicative reduction and the Neron group is isomorphic to
\(\GG_m^g / \Lambda\) 2. Conversely, if \(A/K\) has split multiplicative
reduction, then \(A\) is isomorphic to an analytic torus.
\end{description}

We define \(S/K\) to be a \emph{semi-abelian} variety if
\(0 \to T \to S \to B \to 0\) where \(T\) is a linear torus of dimension
\(u(s)\) and \(B\) is an abelian variety of dimension \(\beta(s)\).

\begin{description}
\tightlist
\item[Theorem]
If \((K, \abs{\wait})\) is a complete non-Archimedean field and \(A/K\)
a \(g\dash\)dimensional abelian variety then there exists a semi-abelian
variety \(0\to T\to S\to B \to 0\) of dimension \(g\) which has
potentially good reduction and a \(g_K\dash\)module
\(M \cong_\ZZ \ZZ^{u(s)}\) and
\(0 \to M \to S(K\sep) \to A(K\sep) \to 0\) where \(\rk_\ZZ M^{g_k}\)
(the invariants) is the split toruc ? of the Neron group of \(A\) and
for all \(L/K\) the Néron groups of \(S/L\) and \(A/K\) are isomorphic.
\end{description}

\begin{quote}
Note that we're taking the Néron group of a semi-abelian variety here,
which is a generalization.
\end{quote}

\hypertarget{wednesday-april-15th-non-archimedean-uniformization-of-algebraic-curves}{%
\section{Wednesday April 15th: Non-Archimedean Uniformization of
Algebraic
Curves}\label{wednesday-april-15th-non-archimedean-uniformization-of-algebraic-curves}}

See Mumford 1972, or the Springer monograph ``? Groups and Mumford
Curves'' which we'll attempt to summarize today.

Let \((K, \abs{\wait})\) be a complete discretely valued field with
valuation ring \(R\) and residue field \(k\).

\begin{description}
\tightlist
\item[Definition]
A \emph{Mumford curve} is a curve \(C/R\) that is proper, flat, normal,
and nice generic fibers \(C/K\) where \(C/k\) is semistable and totally
degenerate: each component \(u\) is isomorphic to \(\PP^1_k\), and the
intersections of fibers are only transverse double points.
\end{description}

To such a curve we attach a \emph{dual graph} \(G(C)\) with vertices
corresponding to components and edges corresponding to intersections
between components. We thus think of the special fiber as
combinatorially object.

Blowing up such a curve stays within this class of curves, and for genus
\(>1\), there is a unique regular minimal model. For an elliptic curve
\(E\), this says that \(E\) has split multiplicative reduction.

For \(C/k\) has genus \(g\geq 1\), then the Euler characteristic of the
dual graph \(G\) is given by \(\chi(G(C)) = 1-g\), the number of
vertices minus the number of edges.

Example: \(E/k\) with split multiplicative reduction, given a minimal
regular model, the dual graph is \(C_n\).

For \(C/k\) a genus \(g\) Mumford curve, for \(\Jac(C)\) the Néron group
is \(\GG_m^g/k\). Thus the Jacobian of a Mumford curve has an analytic
uniformization \((K\units)^g/\Lambda\)

\begin{description}
\tightlist
\item[Theorem (Main Result)]
A curve \(C/k\) admits an analytic uniformization iff it is a Mumford
curve.
\end{description}

Similar to how abelian varieties admit uniformizations iff split
multiplicative reduction.

\begin{description}
\tightlist
\item[Example]
If \(B/\QQ\) is a nonsplit indefinite quaternion algebra, which are
classified by their discriminants \(\disc B = D = p_1 \cdots p_n\). Then
\(X^D/\QQ\) is a Shimura curve, and for each \(p\) dividing \(D\),
\(X^D/\QQ_p\) is a Mumford curve and has an explicit \(p\dash\)adic
uniformization (given by Cerdnik-Drinfeld?).
\end{description}

A quick opaque way is to define \(\mch\) as the \(p\dash\)adic upper
half plane for some (complicated) ``connected'' rigid analytic space,
then \(\mch(K) = \PP^1(\bar K) \setminus \PP^1(K)\). There is a faithful
action by some discrete subgroup \(\Gamma \subset \PGL(2, K)\) on
\(\mch(K)\) by linear fractional transformations. Every Mumford curve
can be realized as \(\Gamma \textbackslash \mch\).

Note that \(\Gamma\) acts on \(\PP \definedas \PP^1(E)\).

\begin{description}
\tightlist
\item[Definition]
A point \(x\in \PP\) is a \emph{limit point} of \(\Gamma\) if there
exists a \(y\in \PP\) and distinct \(\gamma_n\) in \(\Gamma\) such that
\(\gamma_n y \to x\).
\end{description}

These are essentially the accumulation points of the orbits of
\(\Gamma\). Define \(\Lambda(\Gamma)\) be the set of all limit points,
which turns out to be closed in \(\PP\), and
\(U(\Gamma) - \PP\setminus \Lambda(\Gamma)\) the regular points. Say
that \(\Gamma\) is \emph{discontinuous} if \(U(\Lambda) \neq \emptyset\)
and every neighborhood of \(\Gamma\) in \(\PP\) has compact closure.

\begin{description}
\tightlist
\item[Definition]
A \emph{Schotky group} is a discrete \(\Gamma \subset \PGL(2, K)\) that
is finitely generated and torsionfree.
\item[Theorem]
\hfill 
1. Every Schotky group is free on \(g\) generators, each being a
hyperbolic element of \(\PGL(2, K)\), i.e.~the two eigenvalues have
distinct norms. 2. Every discontinuous group has a finite index normal
Schotky subgroup.
\end{description}

Here \(\Gamma\) is a Schotky group of genus \(g\) and \(\Gamma\) acts
freely on \(U(\Gamma)\), which is open in \(\PP^1\). Since \(\PP^1\)
naturally has the structure of a \(k\dash\)analytic space, the space
\(\Gamma \textbackslash U(\Gamma)\) exists in the category of rigid
\(k\dash\)analytic spaces, and is the analytic space associated to a
genus \(g\) Mumford curve \(X/k\). Conversely, every such \(X/k\) arises
from such a \(\Gamma\).

\begin{quote}
Analogy: properly discontinuous group actions. For any point, taking a
neighborhood and translating it by the group action yields only finitely
many intersecting with the original neighborhood.
\end{quote}

Example: Take \(K = \CC\) and \(0 < \abs{q} < 1\). Define
\(\Gamma = \gens{\gamma_q}\) where \begin{align*}
\gamma_q =
\begin{bmatrix}
q & 0 \
0 & 1
\end{bmatrix}
.\end{align*} What is the subset of \(\CC\) on which \(\Gamma\) acts
freely? We see \(0, \infty \in \PP^1(\CC)\) are the only fixed points,
and at every other point \(\Gamma\) acts discontinuously. So
\(\Lambda(\Gamma) = \theset{0, \infty}\) and \(U(\Gamma) = C\units\),
and thus
\(\Gamma \textbackslash U(\Gamma) = q^\ZZ \textbackslash \CC\units\).
What is the fundamental domain for the \(\Gamma\dash\)action on
\(U(\Gamma)\)? It will be given by an annulus with the boundaries
identified, which yields a torus.

In general, there will be \(2g\) discs given by
\(A_1, B_1, \cdots, A_g, B_g\), so choose \(\gamma_i \in \PGL(2 ,K)\)
such that \(\gamma_i\) maps \(A_i^\circ \mapsvia{\cong} B_i^\circ\).
Then \(\Gamma = \gens{\gamma_i}\) is a free group of rank \(g\), and in
fact a Schotky group. Then
\(F(\Gamma) \definedas \PP^1 \setminus \union^{2g} D_i\) for \(D_i\)
open discs, then \(F(\Gamma)\) is a fundamental domain for the action of
\(\Gamma\) on \(U(\Gamma)\) and is a genus \(g\) torus.

\includegraphics{figures/image_2020-04-15-13-20-43.png}\\

Note that this yields an affinoid space? We can take the \(\Gamma\)
invariant meromorphic functions on \(U(\Gamma)\), this will yield an
extension of transcendence degree 1 that will be the function field of
the curve.

\hypertarget{friday-april-17th}{%
\section{Friday April 17th}\label{friday-april-17th}}

Let \(X\) be locally compact and \(\Gamma \actson X\) by homeomorphisms.
Recall that \(\Gamma\) acts properly discontinuously at \(x\in X\) iff
there exists a neighborhood \(N_x\) such that
\(\theset{\gamma\in \Gamma \suchthat \gamma\cdot N_x \cdot N_x \neq \emptyset}\)
is finite. Equivalently, th e orbit \(\Gamma\cdot x\) is discrete in
\(X\) and the stabilizer \(\Gamma^x\) is finite. Moreover \(\Gamma\)
acts freely iff \(\Gamma^x\) is trivial for all \(x\).

\begin{description}
\tightlist
\item[Proposition]
For such \(X, \Gamma\) as above, the map \(X \to \Gamma/X\) is a galois
covering iff \(\Gamma\) acts freely and properly discontinuously.
\end{description}

\begin{quote}
Note: Galois covering here is just the usual notion involving deck
transformations.
\end{quote}

This requires \(\Gamma\) to be discrete, otherwise a sequence converging
to the identity would have infinitely many neighborhoods converging on a
neighborhood of the identity.

A \emph{Fuchsian} group is a discrete subgroup of automorphisms of
\(\HH \subset \CC\),
i.e.~\(\Gamma \subset \aut_\CC (\HH) = \PSL(2, \RR)\).

\begin{description}
\tightlist
\item[Fact]
Every Fuchsian group acts properly discontinuously on all of \(\HH\).
Thus the discreteness which is necessary is also sufficient in this
case. Then \(\HH \to \Gamma \textbackslash \HH\) is a covering map and
the action is free \(\iff \Gamma\) has no elements of finite order.
\end{description}

Thus the complex structure (sheaf of holomorphic functions?) pushes
forward to the quotient, giving it the structure of a Riemann surface.

\begin{description}
\tightlist
\item[Definition]
A \emph{Kleinian} group is a discrete subgroup of
\(\PSL(2, \CC) = \PGL(2, \CC) = \aut_\CC(\PP^1)\).
\end{description}

Note that \(\HH\) carries a hyperbolic structure, and this group
precisely preserves it.

Define the regular set as
\(\ell(\Gamma) = \theset{x\in \PP^1 \suchthat x \text{ acts freely and PDC}}\)

\begin{description}
\tightlist
\item[Fact]
A Kleinian group is conjugate in \(\PSL(2, \CC)\) to a Fuschian group if
it stabilizes a circle in \(\PP^1(\CC)\).
\item[Exercise]
Let \(\Gamma = \gens{[q, 0; 0, 1]} \subset \PSL(2, \CC)\) and
\(\Lambda(\Gamma) = \PP^1\setminus \theset{\vector 0}\). Show
\(\Lambda(\Gamma)\) is open but could be empty, has either 0,1,2, or
infinitely many connected components.
\item[Exercise]
Show that for \(\Gamma = \PSL(2, \ZZ)\),
\(\Lambda(\Gamma) = \HH^+ \union \HH^-\).
\end{description}

If \(U(\Gamma) \neq \emptyset\), then
\(\Lambda(\Gamma) \to \Lambda(\Gamma)/\Gamma\) is a galois covering that
endows the quotient with the structure of a Riemann surface.

For \(G\) a group and \(X\) a set with a map \(M: G\cross X \to X\), we
say \(X\) is a \(G\dash\)set. We can make a category from this: given
\((X, \mu_x)\) and \((Y, \mu_y)\), a \(G\dash\)map is given by a diagram

\begin{center}\includesvg[width=0.7\linewidth]{050cdb0072f7af841e577f10f8a166a33d60280b}\end{center}

The set \(X\) is transitive if for all \(x, y\in X\), there is a
\(g\in G\) such that \(gx = y\). \(X\) is \emph{simply transitive} if
this \(g\) is unique.

\begin{description}
\tightlist
\item[Exercise]
\(G\cross G \mapsvia{\mu} G\) given by group multiplication is simply
transitive. Conversely, any simply transitive \(G\dash\)set \(X\) is
isomorphic to \(G\). Choosing \(p_0 \in X\), consider the map
\(\phi: X\to G\) where \(p \mapsto p - p_0\).
\end{description}

This is a torsor, i.e.~a principal homogeneous space?

Example: ODEs, the solutions for the inhomogeneous equation are torsors
for the homogeneous solutions (?).

Let \(k\) be a field and \(G/k\) a commutative algebraic group. A torsor
(or principal homogeneous space) under \(G\) is a nonempty
\(k\dash\)variety \(X/A\) equipped with a morphism
\(\mu: G\cross X \to X\) such that the following holds.

1st try: define
\(\mu(\bar k): G(\bar k) \cross X(\bar k) \to X(\bar k)\) to be a simply
transitive action. This is a good definition in characteristic zero. In
general, we have an isomorphism \begin{align*}
G\cross X &\to X
(g, x) &\mapsto (x, gx)
.\end{align*} This is trivial iff \((X, \mu) \cong (G, \cdot)\) with its
group structure.

\begin{description}
\tightlist
\item[Claim]
\(X\) is trivial \(\iff X(k) \neq \emptyset\)
\item[Proof]
\(\implies: X \cong_K G\) and \(G(k) \neq \emptyset\), so
\(X(k)\neq \emptyset\).\\
\(\impliedby:\) Let \(p_0 \in X(k)\), then the map \begin{align*}
\mu(\wait, p_0): G \to X
\end{align*} is defined over \(k\) and is a bijection on \(X(\bar k)\),
and thus an isomorphism in characteristic zero.
\end{description}

This implies that every \(X\) under \(G\) is a twisted form of \(G\),
i.e.~\(X/\bar k \cong G/\bar k\).

There's a group law on the torsors called the Baer sum, and a group
structure yielding the Weil-Chevallet group. Every torsor will be given
by a galois cohomological cocycle.

\hypertarget{monday-april-20th}{%
\section{Monday April 20th}\label{monday-april-20th}}

Today: torsors.

Let \(k\) be a (perfect, separable) field and \(G/k\) a commutative
algebraic group (a finite type reduced group ?).

\begin{description}
\item[Definition]
A variety \(X/k\) is a \emph{torsor under \(G\)} is
\(\mu: G\cross X \to X\) a group action such that the map

\begin{align*}
G\cross X \to X \cross X \\
(g, x) \mapsto (\mu(gx), x)
\end{align*} is an isomorphism.
\end{description}

For \(\ell/k\) any field extension, the base change to \(X/\ell\)
induces \(\mu_\ell\) making \(X/\ell\) a \(G/\ell\) torsor. \(X\) is
trivial iff it is isomorphic to \begin{align*}
\mu: G\cross G &\to G \\
(g, hg) &\mapsto gh
.\end{align*}

\begin{description}
\tightlist
\item[Claim]
\(X\) is trivial iff \(X(k) \neq \emptyset\).
\end{description}

First ``proof'': for \(p\in X(k)\), define \(\mu(\wait, p): G \to X\).
Want to get a map \(G(k\sep) \to X(k\sep)\), when does this happen? In
characteristic zero, we have some map \(G\to X\) (???) which is
surjective with trivial kernel and thus an isogeny but has not \(k\sep\)
points. But this doesn't work in positive characteristic.

Second proof: the map \(G\cross X \to X \cross X\) being an isomorphism
says that upon base change on \(X \to \spec k\), \(X\) becomes
isomorphic to \(G\). But then it also becomes isomorphic over base
change for which \(X\) is intermediate. So if we have

\begin{center}\includesvg[width=0.7\linewidth]{a98a2166d6350a203f08350519e6e08615ad4aee}\end{center}

which factors through \(Y\), if \(p\in X(k)\) then \(\spec k\to X\) and
thus \(X/k \cong G/k\).

The form of the assumed isomorphic implies that the base change of the
\(G\dash\)torsor \(X\) from \(\spec k\) to \(X\) is trivial as a
\(G\cross X\) torsor over \(X\).

\(\qed\)

For \(k\) a field, \(G/k\), an equivalent definition would be that a
\(G\) torsor is \(X/k\) with a \(G\) action that becomes trivial over
\(k\sep\). Therefore A \(G\) torsor \(X\) is a \(k\sep /k\) twisted form
of \(X\) where \(X/k\sep \cong G/k\sep\).

Example: Let \(G = E\) an elliptic curve, and \(X/k\) is a nice curve of
genus 1, but \(X(k)\) is likely empty. Conversely, given such a curve of
genus 1, we can take the Picard variety \(\underline{\Pic}^0 X\),
i.e.~the Jacobian. Then there is an isomorphism \begin{align*}
X &\mapsvia{\cong} \underline{\Pic}^1 X \\
p &\mapsto [p]
.\end{align*}

So every nice curve is a torsor for its Jacobian (?). Note that in
higher dimensions, we'd need to take the albanese, and the same
statement would work: every abelian variety is a torsor over its
albanese.

For \(G/k\) commutative, we can make the set of torsors \(X\) for
\(G/k\) modulo equivalence into a commutative group. We define the
Weil-Chatelet group of \(G/k\) as \(WC(k, G)\). For two torsors, we can
define the \emph{Baer sum} \(X_1 \oplus X_2\) by first defining a map
\begin{align*}
\mu_{\pm}: G\cross \qty{X_1 \cross X_2} &\to X_1 \cross X_2 \\
(g, x_1, x_2) &\mapsto (\mu_1(g, x_1), \mu_2([-1]g, x_2)  )
\end{align*} and defining
\(X_1 \oplus X_2 = \qty{X_1 \cross X_2} / \mu_{\pm}\). Then the action
\(\mu_{\pm}\) on \(X_1 \oplus X_2\) is a \(G\) torsor.

This makes \(WC(k , G)\) into a commutative group where
\(\mu: G\cross G\to G\) defines
\([-1](X, \mu) \definedas (X, \mu([-1]\wait ))\).

\begin{description}
\tightlist
\item[Exercise]
For \(C/k\) a nice genus one curve, \(G = E = \underline{\Pic}^0 C\) and
\(C = \underline{\Pic}^1 C\). Show that \(n[C] \underline{\Pic}^n C\).
\end{description}

\begin{quote}
Note that by adding divisor classes, there is a map
\(\underline{\Pic}^1 C \cross \underline{\Pic}^1 C \to \underline{\Pic}^2 E\).
\end{quote}

\begin{description}
\tightlist
\item[Corollary]
For \(E/k\) an elliptic curve, \(WC(k, E)\) is a torsion abelian group
iff for all genus 1 curves \(C\), there exists an \(n\in \ZZ^{\geq 0}\)
such that \(\qty{\underline{\Pic}^n C}(k) \neq \emptyset\).
\end{description}

We can define the \emph{period} of an elliptic curve as the least \(n\)
for which the torsor becomes trivial, this is an interesting numerical
invariant.

Next up: cocycles and descent.

\hypertarget{wednesday-april-22nd}{%
\section{Wednesday April 22nd}\label{wednesday-april-22nd}}

Remarks:

\begin{enumerate}
\def\labelenumi{\arabic{enumi}.}
\item
  In the category of \(G\dash\)sets, we have
  \(\aut_{G\dash\text{set}}(G) = G\) where \(G\) acts on itself by
  translation.
\item
  For \(k\) a field, \(G/k\) a smooth algebraic group, \(X/k\) a
  variety, \(\mu: G\cross X \to X\) is a torsor iff the map
  \begin{align*}
    \phi: G \cross X &\to X\cross X \\
    (g, x) &\mapsto (\mu(g, x), x)
    \end{align*} is an isomorphism. Letting
  \(G_X \definedas G\cross X, X_X \definedas X\cross X\) be the base
  changes, this asks for a commuting diagram
\end{enumerate}

\begin{center}\includesvg[width=0.7\linewidth]{f2a14cedc862405a2a181f9416fa8b133adf4b36}\end{center}

thus the base change to \(X\) is the trivial \(G\dash\)torsor on \(X\).

Suppose \(G\) is commutative and recall we have a Weil-Chevalley group
\(WC(k, G)\). Question: What is the difference between a twisted form
\(X/k\) of \(G/k\) and a torsor under \(fG/k\)?

Better question: Suppose \((X, \mu) \in WC(k, G)\). How many other
elements of \(WC(k, G) \ni (X', \mu)\) have \(X' \cong_k X\)?

Recall that \(\wait [(X, \mu)] = (X, \mu([-1]\circ \wait, \wait) )\).
Letting \(G = E\) an elliptic curve, we can consider the subtraction map
\begin{align*}
v: X\cross X &\to E \\
(p, g) &\mapsto g
\end{align*}

with \(p= g+q\) iff

\begin{center}\includesvg[width=0.7\linewidth]{37bce190caed6c35d21428a65a43d1b99769c676}\end{center}

For \(X/k\) a nice genus one curve, \(E/k\) an elliptic curve,
\(\mu: E\cross X \to X\) is a torsor iff the map
\(\underline{\Pic}^0 X \to E\) is an isomorphism. Therefore two elements
\(X_1, X_2 \in WC(k, E)\) are isomorphic iff \(X_1, X_2\) lie in the
same \(\Aut(E)\dash\)orbit of \(WC(k, E)\).

Remarks: Thus the torsors over \(E\) aren't much more interesting than
\(E\) itself. E.g. characteristic zero, \(j\neq 0, 1728\), you just mod
out by \(\pm 1\). There is a version of this for abelian varieties.

\begin{itemize}
\tightlist
\item
  \(WC(k, E) = (0) \iff\) every genus 1 curve with Jacobian \(E\) has a
  \(k\dash\)rational point.
\item
  \(\forall E/k WC(k, E) = (0) \iff\) every genus 1 \(C/k\) has
  \(C(k) \neq\emptyset\).
\end{itemize}

\begin{description}
\item[Example]
For \(k = \bar k\), all nonempty \(V/k\) have \(V(k) \neq \emptyset\).
\item[Example]
Say a \(k\) is \emph{pseudo algebraically closed} iff every
geometrically integral \(V/k\) has a \(k\dash\)rational point,
i.e.~\(V(k) \neq \emptyset\). E.g., if \(k\) is \emph{separably closed}
it is pseudo algebraically closed.
\item[Example]
For \(k = \FF_q\) a finite field, if \(X/k\) is nice of genus \(g\),
then
\begin{align*}\abs{\# X(\FF_q) - (q + 1)} \leq 2g \sqrt q.\end{align*}
Thus for \(g=1\), for elliptic curves, we get
\begin{align*}q+1 - 2\sqrt q \leq \# X(\FF_q) \leq q+1 + 2\sqrt{q}\end{align*}
and since \(q>2\), the number of points is strictly positive.
\item[Example (Non-Example)]
Take \(C: y^2 = P_4(x) \in k[x]\) for \(P_4\) a separable degree 4
polynomial. Look at \(C \mapsvia{x} \PP^1\), and define the \emph{index}
\(I(C)\) of a genus 1 curve \(C/k\) to be the least positive degree of a
\(k\dash\)rational divisor on \(C\), equivalently the gcd of degrees of
closed points on \(C\).
\item[Exercise]
If \(C\) is a genus 1 curve, then \(C\) is given by \(y^2 = P_4(x)\) iff
\(C\) has a \(k\dash\)rational divisor of degree 2 iff
\(I(C) \in \theset{1, 2}\).
\item[Exercise]
If \(C: y^2 = ax^4 + bx^3 + cx^2 + dx + e\), then \(C(k)\) is ? iff
there exists \(x, y\in k\) such that \(y^2 = P_4(x)\) or
\(a\in k\units\) is a square.
\item[Example]
Take \(k=\RR\) and \(C: y^2 = - \qty{x^4 + 1}\). The leading term is
negative, and not a square, and the point at \(\infty\) doesn't need to
be check (this would yield exactly 2 real points, thus not a
1-dimensional real manifold). Thus \(C\) is a nontrivial element of
\(WC(\PP, \underline{\Pic}^0 C)\).
\item[Exercise]
Let \(p\) be a prime number and find \(P_4(x) \in \QQ_p[x]\) such that
\(y^2 = P_4(x)\) has no \(\QQ_p\dash\)points.

\begin{quote}
Try not choosing \(p=2\), and try polynomials in \(\ZZ[x]\) and apply
Hensel's lemma.
\end{quote}
\item[Exercise]
If \(C: F(x, y,z) = 0\) is a nice plane cubic curve over \(k\)

\begin{enumerate}
\def\labelenumi{\alph{enumi}.}
\item
  Show that \(C/k\) admits such a defining equation iff it has a
  rational divisor of degree 3 iff \(I(c) \in \theset{1, 3}\).
\item
  Take \(k= \QQ_p\) and find \(C/k\) with no \(k\dash\)rational points.
\end{enumerate}
\end{description}

For \(G/k\) a smooth (commutative, but not necessary) group, \(X/k\) a
\(G\dash\)torso, choose \(p\in k\sep\). Then defining
\(g\definedas \aut(k\sep / k)\), then \(X(k\sep)\) has two actions: a
galois action \(g\actson\) the left, and a \(G(k\sep)\) action on the
right. For all \(\sigma \in g\), there exists a unique
\(a_\sigma \in G(k\sep)\) such that \(\sigma p = p a_{\sigma}\).

This defines a map \(a_\bullet: g\to G(k\sep)\) -- however, this is not
a group morphism, it is a ``twisted'' version. For
\(\sigma, \tau \in g\), by definition we have
\(p_{a_{\sigma \tau}} = \sigma \tau p = \sigma(\tau p) = \sigma (P a_\tau) = \cdots\)
and we can conclude
\begin{align*} a_{\sigma \tau} = a_\sigma^\sigma a_\tau, \end{align*}
which is in fact a 1-cocycle.

\newpage

\newpage
\section{Indices}
\listoftodos[List of Todos]

% Hook into amsthm environments to list them.
\renewcommand{\listtheoremname}{Definitions}
\listoftheorems[ignoreall,show={definition}, numwidth=3.5em]

\renewcommand{\listtheoremname}{Theorems}
\listoftheorems[ignoreall,show={theorem,proposition}, numwidth=3.5em]

\renewcommand{\listtheoremname}{Exercises}
\listoftheorems[ignoreall,show={exercise}, numwidth=3.5em]

\listoffigures


\printbibliography[title=Bibliography]


\end{document}
