\documentclass[a4paper,12pt]{article}
\input{../math/abbreviations}
\input{../math/format}
\input{../math/sectionless-theorems}
\newcommand{\todaysdate}{2005.04.28}
\title{Model-theory of elliptic curves}
\author{David Pierce}
\date{preliminary version: \todaysdate}

\newcommand{\dsets}[1]{\displaystyle\coprod_{I}D_I^{\str{#1}}}

\begin{document}
  \maketitle\thispagestyle{empty}
\setcounter{page}{0}
These notes are intended for a talk to be given at Bilgi University,
Istanbul, April 29, 2005.

\tableofcontents
\newpage
\setcounter{section}{-1}
\section{Introduction}

\defn{Model-theory}: mathematics done `self-consciously'---with an eye
on language and interconnections.

General question:  When are two \tech{structures} mathematically the same
(that is, \tech{isomorphic})?

Necessary model-theoretic condition:  when they are
\tech{elementarily equivalent}. 

When is this condition sufficient?

Is it sufficient when the structures are function-fields
of curves over an algebraically closed field?

---Yes, unless the curves are elliptic curves with complex
multiplication. 

\section{Powers of sets}

\begin{align*}
  \omega&=\{0,1,2,\dots\}\\
&= \text{closure of $\{\emptyset\}$ under $A\mapsto A\cup\{A\}$}\\
&=\{\emptyset,\{\emptyset\},\{\emptyset,\{\emptyset\}\},\dots\}.
\end{align*}
Then $0=\emptyset$ and $1=\{0\}=\{\emptyset\}$.

$n\in\omega$ means $n=\{0,1,2,\dots,n-1\}$, so $n\fsub\omega$.

Let $I\fsub\omega$,
let $M\neq\emptyset$, and define
\begin{equation*}
  M^I=\{\text{functions from $I$ to $M$}\}.
\end{equation*}
Typical element: $(a_i:i\in I)$ or $i\mapsto a_i$ or
$\tuple a$. 

Special case: Elements of $M^n$ are also
$(a_0,a_1,\dots,a_{n-1})$, and $M^n$ itself is
\begin{equation*}
  \underbrace{M\times\dotsb\times M}_n.
\end{equation*}

$M^0=\{0\}=1$.

$M^1=M$.

Each $\pow{M^I}$ is a Boolean algebra, equipped with: 
\begin{myenum}\setcounter{myenum}{-1}
  \item
the operations $\cap$, $\cup$ and ${}\comp$; 
\item
the distinguished elements $\emptyset$ and $M^I$; and 
\item
the relation $\included$.
\end{myenum}
Special case:
\begin{equation*}
  \pow{M^0}=\pow{\{0\}}=\{0,\{0\}\}=\{0,1\}=2.
\end{equation*}
  One can think of $2$
as $\{\False,\True\}$ and identify the study of $\pow{M^0}$ with
\defn{propositional logic}.

$\pow{M^2}$ contains $\{(a,a):a\in M\}$, the \defn{diagonal}
$\Delta_M$. 

Let also $J\fsub\omega$, and let $\alpha:I\to J$.  This induces 
\begin{align*}
M^J&\overset{\alpha^*}{\To} M^I\\
(b_j:j\in J)&\longmapsto (b_{\alpha(i)}:i\in I)
\end{align*}
and hence
\begin{align*}
  \pow{M^J}&\xrightarrow{\alpha^* \text{ or }(\alpha^*)\setim}\pow{M^I}\\
B&\longmapsto\{\alpha^*(\tuple b):\tuple b\in B\}
\end{align*}
as well as
\begin{align*}
  \pow{M^I}&\xrightarrow{\alpha_* \text{ or }(\alpha^*)\inv}\pow{M^J}\\
A&\longmapsto\{\tuple b:\alpha^*(\tuple b)\in A\}.
\end{align*}

For example, $\Delta_M=\alpha^*M$ when $\alpha:2\to 1$.

Also, let $\iota$ be the inclusion of $n$ in $n+1$.  Then
\begin{equation*}
  \iota^*(b_0,\dots,b_{n-1},b_n)=(b_0,\dots,b_{n-1}).
\end{equation*}
If $B\included
M^{n+1}$, then $\iota^*B=\{\tuple a\in M^n:(\tuple a,b)\in B \text{
  for some $b$ in $M$}\}$.

If $A\included M^n$, then $\iota_*A=A\times M$.

\section{Structures}

$M$ becomes a \defn{structure} $\str M$ when equipped with
some (or no):
\begin{myenum}\setcounter{myenum}{-1}
  \item
maps $f^{\str M}$ from $M^{n(f)}$ to $M$ for some
  $n(f)$ in $\omega\setminus\{0\}$; then $f^{\str M}$ is an
  \defn{$n(f)$-ary operation on $M$};
\item
distinguished elements $c^{\str M}$ of $M$;
\item
subsets $R^{\str M}$ of $M^{n(R)}$ for some $n(R)$
in $\omega\setminus\{0\}$; then $R^{\str M}$ is an \defn{$n(R)$-ary
  relation on $M$}. 
\end{myenum}
Then the \defn{signature} $\lang$ of $\str M$ consists of the various symbols
$f$, $c$ and $R$,---\defn{names} for the corresponding operations, 
elements and relations.  

$M$ is the \defn{universe} of $\str
M$, and $\str M$ is an \defn{$\lang$-structure}.

For example, $\R$ is a structure with signature
$\{+,-,\cdot,0,1,\leq\}$.

Structures with more than one universe are possible, \emph{e.g.}{}
vector-spaces.  

Different structures can have the same signature.  Any ordered field
is a structure with the same signature as $\R$.

Since $M=M^1$ and $1=M^0$, elements of $M$ are $0$-ary operations on
$M$. 

Any $n$-ary operation $f$ on $M$ is identified with the $(n+1)$-ary
relation 
\begin{equation*}
  \{(\tuple a,f(\tuple a)):\tuple a\in M^n\}. 
\end{equation*}
Hence the operations, distinguished elements and relations of $\str M$
correspond to certain elements of various $\pow{M^n}$: the
\defn{primitive relations} of $\str M$. 

Suppose $D_I^{\str M}\included\pow{M^I}$ and
$\dsets M$
is the smallest subset $X$ of  
$\displaystyle\coprod_{I\fsub\omega}\pow{M^I}$ such that:
\begin{myenum}\setcounter{myenum}{-1}
\item
$X$ contains $\Delta_M$ and each primitive relation of $\str M$;
\item
$X\cap\pow{M^I}$ is a sub-algebra of $\pow{M^I}$;
\item
if $\alpha:I\to J$, then
  $\alpha_*(X\cap\pow{M^I}),\alpha^*(X\cap\pow{M^J})\included X$.  
\end{myenum}
The elements of $\dsets M$ are the \defn{definable relations} of
$\str M$.

$\lang$-structures $\str M$ and $\str N$ are \defn{isomorphic},
\begin{equation*}
  \str M\cong\str N, 
\end{equation*}
if there is a
bijection from $M$ to $N$ taking each primitive relation of $\str
M$ to the corresponding relation of $\str N$.

$\str M$ and $\str N$ are \defn{elementarily equivalent},
\begin{equation*}
  \str M\equiv\str N,
\end{equation*}
if there is an isomorphism from
  $\dsets M$ to $\dsets N$ taking
 each primitive relation of $\str M$ to
the corresponding relation of $\str N$.  Then
\begin{equation*}
  \str M\cong\str N\implies\str M\equiv\str N.
\end{equation*}

\begin{example*}
  All algebraically closed fields of the same characteristic are
elementarily equivalent.  Their definable sets are the
constructible sets over the prime field.
\end{example*}

\section{Formulas}

Every definable relation $X$ of the $\lang$-structure $\str M$ has a
non-unique name $\phi$: a~string of symbols from
\begin{equation*}
  \lang\cup\{x_n:n\in\omega\}\cup\{=,\land,\lnot,\exists\}.
\end{equation*}
  Also symbols from $\{\lor,\to,\iff,\forall\}$ can be used.  Then
  $\phi$ is a \defn{formula} (of first-order 
logic), and $X$ is the \defn{interpretation}  
\begin{equation*}
  \phi^{\str M}
\end{equation*}
of $\phi$ in $\str M$.

Dictionary: (Here, $n=n(f)=n(R)$, and $\alpha:n\to I$.)
\begin{center}
  \begin{tabular}{|r|l|}\hline
$s$ & $s^{\str M}$\\ \hline
$x_k$ & $\tuple a\mapsto a_k=\iota^*(\tuple a)$, where
  $\iota:\{k\}\included I$\\ 
$fx_{\alpha(0)}\dotsb x_{\alpha(n-1)}$ & $\tuple a\mapsto
    f^{\str M}(\alpha^*(\tuple a))$ \\ 
$Rx_{\alpha(0)}\dotsb x_{\alpha(n-1)}$ & $\alpha_*(R^{\str M})$\\
$=$ & $\Delta_M$\\
$\land$ & $\cap$\\
$\lnot$ & ${}\comp$\\
$\Exists{x_k}\phi$ & $\iota^*(\phi^{\str M})$, where
    $\iota:I\setminus\{k\}\included I$\\ 
\hline
\end{tabular}
\end{center}


If $\phi^{\str M}\included M^0$, then $\phi$ is a \defn{sentence} $\sigma$.

If $\sigma^{\str M}=1$, then $\sigma$ is \defn{true in $\str M$}:
\begin{equation*}
  \str M\models\sigma.
\end{equation*}
So truth is a \emph{relation} between sentences and structures.

Let $\Th{\str M}=\{\sigma:\str M\models\sigma\}$, the \defn{theory of
  $\str M$}; then
\begin{equation*}
  \str M\equiv\str N\Iff\Th{\str M}=\Th{\str N}.
\end{equation*}
\newpage
\section{Curves}

Let $K=\alg K$.  (Perhaps $K=\C$.)  If $K\included L$, let
\begin{equation*}
  \Aff^n(L)=L^n.  
\end{equation*}
  Any irreducible $p$ in $K[X,Y]$
determines a \defn{curve} $C$ over $K$:
\begin{equation*}
C(L)=\{(x,y)\in \Aff^2(L):p(x,y)=0\}.
\end{equation*}
Let $(\alpha,\beta)\in C$ and $\{\alpha,\beta\}\not\included K$;
  then $(\alpha,\beta)$ is a \defn{generic point of $C$
  over~$K$}.

The \defn{field 
of rational functions on $C$ over $K$}, denoted
\begin{equation*}
  K(C),
\end{equation*}
is generated by 
\begin{equation*}
  \left.
  \begin{aligned}
    (x,y)&\longmapsto x\\
(x,y)&\longmapsto y 
  \end{aligned}
\right\}:C\To\Aff.
\end{equation*}
these are coordinates of a generic point of $C$;
hence
\begin{equation*}
  K(C)\cong K(\alpha,\beta).
\end{equation*}

Say also $D$ is a curve over $K$, with generic point
$(\gamma,\delta)$, and
\begin{equation*}
  h:K(\gamma,\delta)\To K(\alpha,\beta)
\end{equation*}
over $K$.  Let $h(\gamma)=f(\alpha,\beta)$ and
$h(\delta)=g(\alpha,\beta)$.  Then
\begin{equation*}
  (x,y)\longmapsto(f(x,y),g(x,y)):C\dashrightarrow D,
\end{equation*}
a \defn{dominant} rational map (its image contains a generic point).

Any such map $\phi$ induces the $K$-embedding $\phi^*$ of $K(D)$ in
$K(C)$ given by
\begin{equation*}
  \phi^*(f)=f\circ\phi.
\end{equation*}
Then
\begin{equation*}
  \deg\phi=[K(C):\phi^*K(D)].
\end{equation*}

\begin{example*}
$K(\Aff^1)\cong K(X)$.  Then
\begin{equation*}
  \deg(x\mapsto x^n:\Aff^1\to \Aff^1)=[K(X):K(X^n)]=n.
\end{equation*}
\end{example*}

\begin{example*}
  Let $C$ be given by $x^2+y^2=1$, and let
  \begin{equation*}
\phi:(x,y)\longmapsto \displaystyle\frac{y}{1+x}:C\To \Aff.
  \end{equation*}
  \begin{center}
    \setlength{\unitlength}{0.8pt}
  \begin{picture}(120,100)(-70,-50)
\put(0,0){\circle{50}}
\put(0,-50){\vector(0,1){100}}
\put(-50,0){\vector(1,0){100}}
\put(-25,0){\line(4,3){60}}
\put(-25,0){\line(-4,-3){25}}
\put(-25,0){\circle*{5}}
\put(-70,4){$(-1,0)$}
\put(7,24){\circle*{5}}
%\put(-15,-12){$1$}
\put(17,20){$(x,y)$}
  \end{picture}
  \end{center}
Let $(\alpha,\beta)$ be a generic point of $C$; then
\begin{gather*}
  \phi^*:f(X)\longmapsto f\left(\frac{\beta}{1+\alpha}\right):K(X)\To
  K(\alpha,\beta);\\
\deg\phi=
  \left[K(\alpha,\beta):K\left(\frac{\beta}{1+\alpha}\right)\right]=1  
\end{gather*}
since $\phi^*$ is invertible:  If $t=\beta/(1+\alpha)$, then
\begin{gather*}
  t^2=\frac{\beta^2}{(1+\alpha)^2}=\frac{1-\alpha^2}{(1+\alpha)^2}=
  \frac{1-\alpha}{1+\alpha};\qquad
\alpha=\frac{1-t^2}{1+t^2};\qquad
\beta=\frac{2t}{1+t^2}.
\end{gather*}
\end{example*}

Each curve $C$ has a \defn{genus} $\genus C$ in $\omega$.  (A curve
over $\C$ is a 
Riemann surface, hence an orientable surface over $\R$; its genus is
the number of holes.)  

If $\phi:C\dashrightarrow D$, dominant, then (by the Hurwitz formula)
\begin{myenum}\setcounter{myenum}{-1}
  \item
$\genus C>\genus D$, or
\item
$\genus C=\genus D\in\{0,1\}$, or
\item
$h$ is an isomorphism.
\end{myenum}
If $\genus C<\genus D$, then every point of $D(K(\alpha,\beta))$ has
coordinates in $K$.  

\begin{theorem*}
\mbox{}
  \begin{myitem}
  \item
$K$ is a definable subset of $K(C)$.
\item
If $\genus C\neq1$ or $\genus D\neq1$, then
\begin{equation*}
  K(C)\equiv K(D)\Iff K(C)\cong K(D).
\end{equation*}
\end{myitem}
\end{theorem*}

(Jean-Louis Duret proved this in case $\Char K=0$.)

\section{Function-fields (optional)}

A \defn{function-field} over $K$ is $K(\alpha_0,\dots,\alpha_n)$
(finitely generated).  If $L_i$ are such, then
\begin{equation*}
  L_0\equiv L_1\implies\trdeg(L_0/K)=\trdeg(L_1/K)
\end{equation*}
by the Tsen--Lang Theorem:

A \defn{quadratic form} over $K$ is a polynomial
\begin{equation*}
  \tuple x\cdot A\cdot\transp{\tuple x}
\end{equation*}
where $\transp A=A$ with entries from $K$.  Then $A$ is
diagonalizable, so by change of variables, the form becomes
\begin{equation*}
  \sum_{i<n}a_ix_i^2.
\end{equation*}
By Tsen and Lang, this form has a non-trivial zero from a
function-field $L$ over $K$ if and only if 
\begin{equation*}
  n>2^{\trdeg(L/K)}.
\end{equation*}
Every form $ax^2+by^2+cz^2$ has a non-trivial zero if and only if
every non-trivial equation $ax^2+by^2=1$ has a solution.  So a
function-field $L$ over $K$ is the
function-field of a curve if and only if
\begin{equation*}
  L\models\Forall z\Forall w\Exists x\Exists y(zw=0\lor zx^2+wy^2=1). 
\end{equation*}
Hence in particular
\begin{equation*}
  K(X)\not\equiv K(X,Y).
\end{equation*}
\newpage
\section{Elliptic curves}

A curve of genus $1$ is an \defn{elliptic curve}.  

A \defn{lattice} is a subgroup
$\gpgen{\omega_0,\omega_1}$
of $\C$, where
$\omega_0\omega_1\neq0$ and $\omega_0/\omega_1\notin\R$.  

Over $\C$, an elliptic curve is a
torus 
\begin{equation*}
  \C/\Lambda,
\end{equation*}
$\Lambda$ a lattice.
Then we may
assume $\Lambda=\gpgen{1,\tau}$ for some $\tau$ in $\uhp$.

How is $\C/\Lambda$ a curve?  The \defn{Weierstra\ss\ $\wp$-function
  for $\Lambda$} is given by
\begin{equation*}
\wp(z)=  \wp(z;\Lambda)=\frac 1{z^2}+\sum_{\omega\in
  \Lambda\setminus\{0\}}\left(\frac 1{(z-\omega)^2}-\frac
  1{\omega^2}\right).
\end{equation*}
$\wp$ is doubly periodic: 
\begin{equation*}
  \wp(z+\omega)=\wp(z)
\end{equation*}
if $\omega\in\Lambda$.  So $\wp$ is well-defined on $\C/\Lambda$.  
Now let 
\begin{equation*}
  G_k=G_k(\Lambda)=\sum_{\omega\in\Lambda\setminus\{0\}}\frac
1{\omega^{2k}},
\end{equation*}
and let $E$ be the curve given by
\begin{equation*}
  y^2= 4x^3 - 60 G_2x - 140 G_3.
\end{equation*}
Then $(\wp,\wp')\in E$, so $\C(E)=\C(\wp,\wp')$, and there is an
isomorphism 
\begin{equation*}
  z\longmapsto(\wp(z),\wp'(z)):\C/\Lambda\dashrightarrow E.
\end{equation*}
The induced group-structure of $E$ is given by polynomials:
\begin{equation*}
  4(\wp(a)+\wp(b)+\wp(a+b))=\lambda^2,
\end{equation*}
where
\begin{equation*}
  \lambda=
  \begin{cases}
    \displaystyle\frac{\wp'(b)-\wp'(a)}{\wp(b)-\wp(a)},& \text{ if
    }a\neq b;\\ 
\wp''(a),& \text{ if }a=b.
  \end{cases}
\end{equation*}

Let $E_i$ be $\C/\Lambda_i$.  A non-zero homomorphism from $E_0$ to
$E_1$ is an \defn{isogeny} and corresponds to $\alpha$ in $\unit{\C}$
such that
\begin{equation*}
  \alpha\Lambda_0\included\Lambda_1;
\end{equation*}
the {degree} of the isogeny is
$\size{\Lambda_1/\alpha\Lambda_0}$.

Any integer induces an endomorphism of $\C/\Lambda$; if any other
complex numbers do, then $\C/\Lambda$ has \defn{complex
  multiplication}. 

\begin{theorem*}
Let $E_i$ be elliptic curves over $K$ algebraically closed.
  The following are equivalent:
  \begin{myenum}\setcounter{myenum}{-1}
    \item
There are two isogenies from $E_0$ to $E_1$ of relatively prime
degrees.
%\item
%$K(E_0)\equiv K(E_1)$.
\item
$K(E_0)$ and $K(E_1)$ agree on all sentences
  \begin{equation*}
      \Forall{x_0}\forall{x_1}\dotsb\Forall{x_{n-1}}\Exists{x_n}\phi(\tuple
      x),
  \end{equation*}
  where $\phi$ is quantifier-free. 
\item
$K(E_0)$ and $K(E_1)$ agree on all $\forall\exists$ sentences.
  \end{myenum}
If one of the $E_i$ has no complex multiplication, then the following
is equivalent to the foregoing: 
  \begin{myenum}\setcounter{myenum}{2}
    \item
$K(E_0)\cong K(E_1)$.
  \end{myenum}
If one of the $E_i$ does have complex multiplication, and $\Char K=0$,
then the following is 
equivalent to \textnormal{(0)} \emph{et al.}:
  \begin{myenum}\setcounter{myenum}{3}
    \item
$\End{E_0}\cong\End{E_1}$.
 \end{myenum}
\end{theorem*}

 (Duret proved (1)$\Iff$(3) when $\Char K=0$.)

Relevant facts:  
\begin{myitem}
  \item
There are just 13 elliptic curves over $\C$ that are determined by
their endomorphism-rings.
\item
Say $E=\C/\gpgen{1,\tau}$.  Then 
\begin{equation*}
\End
E\cong\{\alpha\in\C:\alpha\gpgen{1,\tau}\included\gpgen{1,\tau}\}
\leq\gpgen{1,\tau}. 
\end{equation*}
If $E$ has complex multiplication, then $\tau$ is quadratic (and
conversely), since then
\begin{equation*}
  (x+A\tau)\tau\in\gpgen{1,\tau}
\end{equation*}
for some non-zero $A$.  If $\size A$ is minimal, then
\begin{equation*}
  \End E\cong\gpgen{1,A\bar\tau}.
\end{equation*}

\begin{example*}
$\End{\C/\gpgen{1,\tau}}\cong\gpgen{1,\tau}$ when $\tau$ is $i$ or
$(1+i\,\radix 3)/2$.
  \end{example*}
\item
Every isogeny $\alpha:E_0\to E_1$ has a \defn{dual} 
\begin{equation*}
  \widehat\alpha:E_1\to E_0
\end{equation*}
of the same degree $d$; then $\widehat\alpha\circ\alpha=[d]$
(multiplication by $d$).
\end{myitem}
\newpage
Ideas of proof:

(1)$\implies$(0).  If $p$ always divides $[K(E_0):\phi^*K(E_1)]$, then for
   some $E_2$, 
\begin{gather*}
  \phi^*K(E_1)\included \phi^*K(E_2)\included K(E_0);\\
[K(E_0):K(E_2)]=p.
\end{gather*}
Then $K(E_0)$ says---but $K(E_1)$ does not---that every point
of $E_1$ is the image of a point of some $E_2$ under a map of degree
$p$.

(0)$\implies$(2).  By (0), when $n>1$, some isogeny has degree prime
   to $n$.  Say $K(E_0)\models\Forall{\tuple x}\Exists{\tuple
   y}\phi(\tuple x,\tuple y)$, where $\phi$ is quantifier-free.
   Let~$n$ be the factorial of the degrees of the polynomials in
   $\phi$, and say 
   \begin{equation*}
            \gcd(n,[K(E_0):K(E_1)])=1.  
   \end{equation*}
If $\tuple a$ is from
   $K(E_1)$, then $\phi(\tuple a,\tuple y)$ must have a solution from
   $K(E_1)$. 







(0)$\implies$(4).  If $\alpha_i:E_0\to E_1$ and $\deg\alpha_i=d_i$ and
   $\sum a_id_i=1$, then
\begin{align*}
  \End{E_1}&\overset{\cong}{\To}\End{E_0}\\
\beta&\longmapsto\sum a_i\widehat\alpha_i\circ\beta\circ\alpha_i.
\end{align*}

(4)$\implies$(0).  Say $\End{E_0}\cong\End{E_1}$.  Then we may assume
\begin{gather*}
E_0=\C/\gpgen{1,\tau},\quad E_1=\C/\gpgen{1,n\tau},\\
A\tau^2+B\tau+C=0,\quad \gcd(A,B,C)=0,\quad n\divides A.
\end{gather*}
Hence
\begin{equation*}
  \Hom{E_0,E_1}\cong\gpgen {n,A\bar\tau}.
\end{equation*}
If $\alpha=nx+Ay\bar\tau$, then
\begin{gather*}
  \deg(z\mapsto\alpha z)= \frac 1n\abs{\alpha}= nx^2-Bxy+\frac{AC}ny^2,\\
\gcd\left(n,B,\frac{AC}n\right)=1,
\end{gather*}
so the degree takes two relatively prime values.



\end{document}


(0)$\implies$(2).  By (0), when $n>0$, some isogeny has degree prime
   to $n$.  \emph{Modulo} the theory of field-extensions of $K$, every
   sentence $\sigma$ can be 
   written 
   \begin{equation*}
\Quanti0{x_0}\Quanti1{x_1}\dotsb\Quanti{n-1}{x_{n-1}}\phi(\tuple x),
   \end{equation*}
where $\quant_i\in\{\exists,\forall\}$ and $\phi(\tuple x)$ is
\begin{equation*}
  \bigvee_i\bigwedge_jp_{i\;j}(\tuple x)=0
\end{equation*}
for some polynomials $p_{i\;j}$ over $K$.  Let $M$ be the product of
their degrees, let $N$ be prime to $M!$, and embed $K(E_1)$ in
$K(E_0)$ with index $N$.  Then
\begin{equation*}
  K(E_0)\models\sigma\implies K(E_1)\models\sigma.
\end{equation*}
