\documentclass[a4paper,twoside,11pt]{amsart}
\title{Logical classification of curves}
\author{David Pierce}
\date{\today}
\address{Mathematics Dept, Middle East Technical University, Ankara 06531, Turkey}
\email{dpierce@metu.edu.tr}
\urladdr{http://metu.edu.tr/~dpierce/}
\thanks{Notes prepared for the algebra seminar, November 20, 2009}

\usepackage{hfoldsty}
\usepackage{typearea}
%\usepackage{parskip}
\usepackage{pstricks}
\usepackage[all]{xy}

%\newcommand{\alg}[1]{#1^{\mathrm{alg}}}
\newcommand{\included}{\subseteq}
\newcommand{\pincluded}{\subset}
\newcommand{\genus}[1]{\gamma(#1)}
\newcommand{\Exists}[1]{\exists#1\;}
\newcommand{\Forall}[1]{\forall#1\;}
\newcommand{\lto}{\Rightarrow}
\newcommand{\End}[1]{\operatorname{End}(#1)}
\newcommand{\Hom}[1]{\operatorname{Hom}(#1)}
\newcommand{\C}{\mathbb C}
\newcommand{\Z}{\mathbb Z}
\newcommand{\R}{\mathbb R}
\newcommand{\Lat}[1]{\langle#1\rangle}
\newcommand{\lat}[1]{\Lat{1,\tau_{#1}}}
\newcommand{\Lam}[1]{\Lambda_{#1}}
\newcommand{\size}[1]{\lvert#1\rvert}
\newcommand{\dee}{\operatorname d}
\newcommand{\Proj}[1]{\mathbb P(#1)}

\newtheorem*{myrule}{Rule}
\newtheorem*{theorem}{Theorem}

\usepackage{amssymb}
\renewcommand{\setminus}{\smallsetminus}
\renewcommand{\leq}{\leqslant}
\renewcommand{\phi}{\varphi}

\begin{document}
\maketitle
\tableofcontents

\section{Ellipses and elliptic curves}

An \textbf{ellipse} is given by an equation
\begin{equation*}
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
\end{equation*}
In general, length along a curve from $P$ to $Q$ is given by
$\int_P^Q\sqrt{\dee x^2+\dee y^2}$.
For the ellipse, we compute
\begin{align*}
\frac{2x\dee x}{a^2}+\frac{2y\dee y}{b^2}&=0,&
\dee y^2&=\frac{b^4x^2}{a^4y^2}\dee x^2
=\frac{b^2x^2}{a^2(a^2-x^2)}\dee x^2,
\end{align*}
so
\begin{multline*}
\int\sqrt{\dee x^2+\dee y^2}
=\int\sqrt{\rule[-1.5ex]{0ex}{4ex}}\frac{a^2(a^2-x^2)+b^2x^2}{a^2(a^2-x^2)}\dee x\\
%=\int\sqrt{1+\frac{b^4x^2}{a^4y^2}}\dee x\\
%=\int\sqrt{1+\frac{b^2x^2}{a^2(a^2-x^2)}}\dee x
=\frac1a\int\sqrt{\rule[-1.5ex]{0ex}{4ex}}\frac{a^4-c^2x^2}{a^2-x^2}\dee x
=\frac1a\int\frac y{a^2-x^2}\dee x,
\end{multline*}
where $b^2+c^2=a^2$ and
\begin{equation*}
y^2=(a^2-x^2)(a^4-c^2x^2).
\end{equation*}
Assuming $c\neq0$, the last equation defines an \textbf{elliptic curve} and is equivalent to:
\begin{gather*}
y^2=(x^2-a^2)(c^2x^2-a^4),\\
\Bigl(\frac y{(x+a)^2}\Bigr)^2
=\Bigl(\frac{x-a}{x+a}\Bigr)
\Bigl(\frac{cx+a^2}{x+a}\Bigr)
\Bigl(\frac{cx-a^2}{x+a}\Bigr).
\end{gather*}
We rewrite this as
\begin{equation*}
v^2=\beta u(u-\mu)(u-\rho),
\end{equation*}
where
\begin{align*}
v&=\frac y{(x+a)^2},&
u&=\frac{x-a}{x+a},
\end{align*}
and $\beta$, $\mu$, and $\rho$ are such that
\begin{gather*}
\Bigl(\frac{cx+a^2}{x+a}\Bigr)
\Bigl(\frac{cx-a^2}{x+a}\Bigr)
=\beta(u-\mu)(u-\rho),\\
\begin{aligned}
c^2\Bigl(x-\frac{a^2}c\Bigr)\Bigl(x+\frac{a^2}c\Bigr)
&=\beta(x-a-\mu(x+a))(x-a-\rho(x+a))\\
&=\beta\bigl((1-\mu)x-(1+\mu)\bigr)\bigl((1-\rho)x-(1+\rho)a\bigr)\\
&=\beta(1-\mu)(1-\rho)\Bigl(x-\frac{1+\mu}{1-\mu}\Bigr)
\Bigl(x-\frac{1+\rho}{1-\rho}\Bigr).
\end{aligned}
\end{gather*}
So it suffices if
\begin{align*}
c^2&=\beta(1-\mu)(1-\rho),&
\frac{a^2}c&=\frac{1+\mu}{1-\mu},&
-\frac{a^2}c&=\frac{1+\rho}{1-\rho},
\end{align*}
that is,
\begin{align*}
\mu&=\frac{a^2-c}{a^2+c},&
\rho&=\frac1{\mu},&
\beta&=-\frac{c^2\mu}{(1+\mu)^2}.
\end{align*}
After another change of variables, the equation becomes
\begin{equation*}
y^2=x(x-1)(x-\lambda)
\end{equation*}
(where $\lambda=\rho/\mu$).  On this curve, the differential form
$\dee x/y$ is holomorphic.  But
\begin{equation*}
Q\mapsto\int_P^Q\frac{\dee x}y
\end{equation*}
is well defined, not on $\Proj{\C}$ (that is, $\C\cup\{\infty\}$), but rather on the Riemann surface got by cutting and gluing two copies of this along lines from $0$ to $\infty$ and $1$ to $\lambda$: the surface is then a \textbf{torus.}  This then is the elliptic curve, and the function above is an analytic bijection onto $\C/\Lambda$ for some lattice $\Lambda$.

\section{Curves and function fields}

Let $K$ and $L$ be algebraically closed fields, with
$K\pincluded L$ and $\operatorname{tr-deg}(L/K)=\infty$.

An irreducible $f$ in $K[X,Y]$ defines a \textbf{curve} $C$ over $K$, namely
\begin{equation*}
C=\{(x,y)\in L^2\colon f(x,y)=0\}.
\end{equation*}
We define
\begin{align*}
K[C]&=K[X,Y]/(f),\\
K(C)&=\text{ fraction field of }K[C];
\end{align*}
this is the field of \textbf{rational functions} on $C$ over $K$.  Then
						\begin{align*}
					K[C]&= K[a,b]\\		
					K(C)&= K(a,b),
\end{align*}
where
\begin{equation*}
\left.
						\begin{aligned}
					a&=((x,y)\mapsto x)\\		
					b&=((x,y)\mapsto y)
\end{aligned}
\right\}\text{ on }C,
\end{equation*}
so that
$f(a,b)=0$
and $(a,b)$ is a \textbf{generic point} of $C$ over $K$; we may assume $(a,b)\in L^2$.

Say also
\begin{equation*}
D=\{(x,y)\in L^2\colon g(x,y)=0\},
\end{equation*}
and $\phi^*$ is an embedding of $K(C)$ in $K(D)$ over $K$.  Then
\begin{equation*}
0=\phi^*(f(a,b))=f(\phi^*(a),\phi^*(b)),
\end{equation*}
so $(\phi^*(a),\phi^*(b))$ is a generic point of $C$ and is also a \textbf{dominant rational map} $\phi$ from $D$ onto $C$.  We recover $\phi^*$ by
\begin{equation*}
\phi^*(h)=h\circ\phi.
\end{equation*}
Indeed, 
\begin{equation*}
\phi^*(a)=a(\phi^*(a),\phi^*(b))=a\circ(\phi^*(a),\phi^*(b))=a\circ\phi,
\end{equation*}
and likewise for $b$.

\begin{myrule}
The $K$-algebra $K(C)$ embeds in $K(D)$ if and only if $C$ has a generic point with coordinates from $K(D)$.
\end{myrule}

We also have
\begin{equation*}
K(C)\cong K(D)\iff\text{ $D$ and $C$ are \textbf{birationally equivalent.}}
\end{equation*}
For example, the function 
\begin{equation*}
(u,v)\mapsto\Bigl(\frac{x-a}{x+a},\frac y{(x+a)^2}\Bigr)
\end{equation*}
determines a birational equivalence between the elliptic curves above.

Or let $f=X^2+Y^2$ and $g=X$.  See Figure~\ref{fig:1}.
Then $\phi\colon C\to D$, where
\begin{align*}
\phi(x,y)&=\frac y{1+x},&
\phi^{-1}(t)&=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1-t^2}\right),
\end{align*}
so $C$ and $D$ are birationally equivalent, and
\begin{align*}
K(D)\cong K(e)&\cong K(a,b)\cong K(C)\\
e&\mapsto\frac b{1+a}\\
\frac{1-e^2}{1+e^2}&\gets a\\
\frac{2e}{1-e^2}&\gets b
\end{align*}

\begin{figure}
\begin{center}
\begin{pspicture}(-3,-3)(5,3)
%\psgrid
\psset{unit=5mm}
\psline{->}(-6,0)(6,0)
\psline{->}(0,-6)(0,6)
\pscircle(0,0){5}
\psline(-5,0)(3,4)
\psdots(-5,0)(3,4)(0,2.5)
\uput[ur](3,4){$(x,y)=\displaystyle\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right)$}
\uput[dr](0,2.5){$t=\displaystyle\frac y{1+x}$}
\end{pspicture}
\caption{Birational equivalence of circle and straight line}\label{fig:1}
\end{center}
\end{figure}

Every curve $C$ has a \textbf{genus} $\genus C$ in $\mathbb N$.  
If $K(C)$ embeds in $K(D)$ over $K$, then
\begin{equation*}
\genus C\leq \genus D.
\end{equation*}
If the embedding is \emph{proper,} then
	either $\genus C<\genus D$ or 
	\begin{equation*}
	0\leq \genus C\leq \genus D\leq 1.
\end{equation*}
If $\genus C=0$, then $K(C)\cong K(X)$.

\section{Logic and elliptic curves}

Suppose $K(C)\ncong K(D)$.  We may assume $\genus C\leq\genus D<\genus E$ for some curve $E$.  Then the formula
\begin{equation*}
\Exists y(x,y)\in E%\land\Exists z(z,x)\in E)
\end{equation*}
\textbf{defines} $K$ in $K(C)$ and $K(D)$.  
If $\genus C<\genus D$ or $1<\genus C=\genus D$, then
the sentence
\begin{equation*}
\Forall x\Forall y\Exists z((x,y)\in D\lto (x,z)\in E)
\end{equation*}
is true in $K(C)$, but not $K(D)$, so these algebras have different \textbf{theories;} we say they are not \textbf{elementarily equivalent,} and we write
\begin{equation*}
K(C)\not\equiv K(D).
\end{equation*}
We cannot then have $0=\genus C=\genus D$.
The remaining possibility is $1=\genus C=\genus D$, that is, $C$ and $D$ are \textbf{elliptic curves.}

An elliptic curve $E$ is also an abelian group; the curve has \textbf{complex multiplication} if $\End E\ncong\mathbb Z$.

\begin{theorem}[Jean-Louis Duret (1992); D.P. (1998)]
If $C$ and $D$ are curves over $K$, and $C$ is not an elliptic curve with complex multiplication, then
\begin{equation*}
K(C)\ncong K(D)\implies K(C)\not\equiv K(D).
\end{equation*}
\end{theorem}

In general, if $\phi\colon D\to C$, then
\begin{equation*}
\deg(\phi)=[K(D):K(C)]
\end{equation*}

\begin{theorem}[D.P. (1998)]
Suppose $C$ and $D$ are elliptic curves over $K$ with complex multiplication.  The following are equivalent.
\begin{enumerate}
	\item 
	There are $\phi$ and $\phi'$ from $C$ onto $D$ with 
	\begin{equation*}
	\gcd(\deg(\phi),\deg(\phi'))=1.
	\end{equation*}
	\item
	$K(C)$ and $K(D)$ agree on all sentences
	\begin{equation*}
	\Forall{(x_0,\dots,x_{n-1})}\Exists y\psi(x_0,\dots,x_{n-1},y), 
	\end{equation*}
	where $\psi$ is quantifier-free.
\end{enumerate}
If $\operatorname{char}(K)=0$, then the foregoing are equivalent to the following.
\begin{enumerate}\setcounter{enumi}2
	\item 
	$\End C\cong\End D$.
\end{enumerate}
\end{theorem}

Say $E_0$ and $E_1$ are elliptic curves over $\C$.  For each $i$ in $\{0,1\}$ there are $A_i$ and $B_i$ in $\C$ such that $E_i$ is birationally equivalent to the curve defined by
\begin{equation*}
y^2=4x^3-A_ix-B_i.
\end{equation*}
So we may assume $E_i$ is this curve.  There is a lattice $\Lam i$, namely $\lat i$, where $\Im(\tau_i)>0$, and there is a function $\wp_i$, namely
\begin{equation*}
z\mapsto
\frac 1{z^2}+\sum_{\omega\in
  \Lam i\setminus\{0\}}\left(\frac 1{(z-\omega)^2}-\frac
  1{\omega^2}\right),
\end{equation*}
such that $(\wp_i,\wp_i{}')$ is a generic point of $E_i$ and is a bijection from $\C/\Lam i$ to $E_i$.  Say $\phi\colon E_0\to E_1$.  There are $\alpha$ and $\omega$ in $\C$ such that the following commutes.
\begin{equation*}
\xymatrix{
\C/\Lam0 \ar[d]_{z\mapsto\alpha z} \ar[rr]^{(\wp_0,\wp_0{}')} && E_0 \ar[dd]^{\phi}\\
\C/\Lam1 \ar[d]_{z\mapsto z+\omega} &&\\
\C/\Lam1 \ar[rr]_{(\wp_1,\wp_1{}')} && E_1
}
\end{equation*}
We may assume $\omega=0$, so $\phi$ is an \textbf{isogeny} and, in particular, a homomorphism.  We must have 
\begin{equation*}
\alpha\Lam0\included\Lam1,
\end{equation*}  
and then
\begin{equation*}
\deg(\phi)=[\Lam1:\alpha\Lam0].
\end{equation*}
Also, if $\alpha\neq0$, there is a matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ or $M$ in $\operatorname{M}_n(\Z)$ such that
\begin{equation*}
\alpha\begin{pmatrix}1\\\tau_0\end{pmatrix}
=\begin{pmatrix}
a+b\tau_1\\c+d\tau_1
\end{pmatrix}
=
\begin{pmatrix}
a&b\\c&d
\end{pmatrix}
\begin{pmatrix}
1\\\tau_1
\end{pmatrix}
=
M
\begin{pmatrix}
1\\\tau_1
\end{pmatrix},
\end{equation*}
and then
\begin{equation*}
\deg(\phi)=\det(M).
\end{equation*}
Also
\begin{equation*}
\begin{pmatrix}
d&-b\\-c&a
\end{pmatrix}
\begin{pmatrix}1\\\tau_0\end{pmatrix}
=\alpha^{-1}
\det(M)
\begin{pmatrix}
1\\\tau_1
\end{pmatrix}=
\alpha^{-1}
\deg(\phi)
\begin{pmatrix}
1\\\tau_1
\end{pmatrix},
\end{equation*}
so
\begin{equation*}
z\mapsto\alpha^{-1}\deg(\phi)z\colon\C/\Lam1\to\C/\Lam0
\end{equation*}
corresponding to an isogeny $\hat{\phi}$ from $E_1$ to $E_0$.  Then
\begin{gather*}
\deg(\hat{\phi})=\deg(\phi),\\
\hat{\phi}\phi=[\deg(\phi)]
\end{gather*}
where $[n]$ is multiplication by $n$. 

If $E$ corresponds to $\C/\Lam{}$, then
\begin{equation*}
\End E\cong\{z\in\C\colon z\Lam{}\included\Lam{}\}.
\end{equation*}
For example, if 
\begin{equation*}
\tau=\frac{-1+\sqrt{{-7}}}4.
\end{equation*}
then (see Figure~\ref{fig:lat-end})
\begin{equation*}
  \End E=\Lat{1,2\tau}.
\end{equation*}
\begin{figure}
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%
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\uput[ur](1,0){$1$}
\uput[dl](-0.25,0.661){$\tau$}
  \end{pspicture}
\mbox{$\qquad\qquad$}
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%\psgrid
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\psset{fillstyle=solid}
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%
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\uput[ur](1,0){$1$}
\uput[dl](-0.50,1.323){$2\tau$}
  \end{pspicture}
\caption{A lattice and its endomorphisms}\label{fig:lat-end}
\end{figure}%
In general, if $E$ has complex multiplication, this means, for some $\alpha$ in $\C\setminus\R$, we have
\begin{equation*}
\alpha
\begin{pmatrix}1\\\tau\end{pmatrix}
=\begin{pmatrix}
a+b\tau\\c+d\tau
\end{pmatrix},
\end{equation*}
so
\begin{gather*}
\alpha=a+b\tau,\\
c+d\tau=\alpha\tau=(a+b\tau)\tau,\\
b\tau^2+(a-d)\tau-c=0.
\end{gather*}
So $E$ has complex multiplication if and only if $\tau$ is quadratic.  If indeed
\begin{equation*}
b\tau^2+a\tau-c=0
\end{equation*}
in lowest terms,
then one shows
\begin{equation*}
\End E\cong\langle1,b\bar{\tau}\rangle;
\end{equation*}
in any case, $\End E$ embeds in $\Lambda$.

In general, since $\End E$ embeds in $\C$, it is commutative.  Suppose $\phi$ and $\psi$ are isogenies from $E_0$ to $E_1$ of relatively prime degrees.  There are integers $m$ and $n$ such that
\begin{equation*}
m\deg(\phi)+n\deg(\psi)=1.
\end{equation*}
Then $\End{E_1}\cong\End{E_0}$ by
\begin{equation*}
\alpha\mapsto m\hat{\phi}\alpha\phi+n\hat{\psi}\alpha\psi.
\end{equation*}

Now suppose conversely $\End{E_1}\cong\End{E_0}$, and each curve has complex multiplication.  Then $\Lam0$ and $\Lam1$ have a common sublattice, so by linear algebra we may assume $\tau_1=n\tau_0$ for some $n$.  

\begin{theorem}[D.P.]
Say $\End{E_1}\cong\End{E_0}\ncong\Z$, and
\begin{equation*}
b\tau_0{}^2+a\tau_0{}-c=0
\end{equation*}
in lowest terms, and $\tau_1=n\tau_0$.  Then
\begin{equation*}
\Hom{E_0,E_1}\cong\langle n,b\bar{\tau}\rangle.  
\end{equation*}
If this takes $\phi$ to $nx+by\bar{\tau}$, then
\begin{equation*}
\deg(\phi)=nx^2-axy-\frac{bc}ny^2,
\end{equation*}
a quadratic form with relatively prime coefficients, so it represents coprime numbers.
\end{theorem}

Suppose now $p$ divides the degree of every isogeny from $E_0$ to $E_1$.  Then there is a finite set $\mathcal L$ of lattices, each having index $p$ in $\Lam1$, such that, if
\begin{equation*}
\alpha\Lam0\included\Lam1,
\end{equation*}
then, for some $\Lam{}$ in $\mathcal L$,
\begin{equation*}
\alpha\Lam0\included\Lam{}\pincluded\Lam1.
\end{equation*}
Hence
\begin{equation*}
K(E_0)\ncong K(E_1),
\end{equation*}
because $K(E_0)$ but not $K(E_1)$ is a field $L$ such that, if 
\begin{equation*}
\phi^*[K(E_1)]\included L,
\end{equation*}
then
\begin{equation*}
\phi^*[K(E_1)]\pincluded F\included L,
\end{equation*}
where the isomorphism-class of $F$ over $\phi^*[K(E_1)]$ has
finitely many possibilities.


\end{document}
