head	1.1;
access;
symbols;
locks; strict;
comment	@% @;


1.1
date	2000.05.02.09.47.39;	author was;	state Exp;
branches;
next	;


desc
@Congruences
@


1.1
log
@Initial revision
@
text
@% $Header$

% congruences.tex

\section{Congruences}
\label{sec:congruences}

Lo\"\i{}c's paper ``Arithmetic of elliptic curves and
diophantine equations'' is very relevant to this
section, esp. his section THREE.  Do not forget
to make proper attribution, etc. 

Let $f, g$ be nonconjugate newforms and $H=H_1(X_0(N),\Z)$. 

\begin{proposition}
$(\Adual_f\intersect A_g^{\vee})[p]\neq 0$
if and only if the mod $p$ rank of $H[I_f]+H[I_g]$ 
is strictly less than $\rank H[I_f] + \rank H[I_g]$.
\end{proposition}
\begin{proof}
By (\ref{Af}) $\Lambda_f=H[I_f]$ (resp., $\Lambda_g=H[I_g]$) is the submodule of 
$H$ which defines $A_f$ (resp., $A_g$).  By reduction mod $p$ we mean
the map $H\ra H\tensor \Fp$. Suppose
  $$\rank (\Lambda_f + \Lambda_g)\md p < \rank\Lambda_f + \rank \Lambda_g.$$
Since $\Lambda_f$ (resp., $\Lambda_g$) is a kernel, it is saturated, so 
  $\rank \Lambda_f \md p = \rank \Lambda_f$ (resp., for $\Lambda_g$).  
We conclude that the mod $p$ linear dependence must involve vectors 
from both $\Lambda_f$ and $\Lambda_g$; there is $v\in\Lambda_f$ and
$w\in\Lambda_g$ so that $v, w\not\equiv 0\md p$ but $v+w\con 0 \md p$. 
Thus $\frac{v+w}{p}\in H$ is integral, i.e., in $J_0(N)(\C)$ we have
$\frac{1}{p}v - (-\frac{1}{p} w)=0$.  But $\frac{1}{p}v \not \in \Lambda_f$
and $\frac{1}{p} w\not\in\Lambda_g$ (otherwise $v$ and $w$ would 
be $0\md p$), so 
$\frac{1}{p}v$ and $-\frac{1}{p}w$ are both nontrivial $p$-torsion in
$\Adual_f$, $A_g^{\vee}$, resp.  Conclusion: 
$0\neq \frac{1}{p}v = -\frac{1}{p} w \in (\Adual_f\intersect A_g^{\vee})[p]$.

Conversely, suppose $0\neq x\in (\Adual_f\intersect A_g^{\vee})[p]$.  Choose
lifts modulo $H$ to $x_f\in\frac{1}{p}\Lambda_f$ and 
$x_g\in\frac{1}{p}\Lambda_g$.
Then $px_f\in\Lambda_f$  (resp., $px_g\in\Lambda_g$), but 
$px_f\not\in p H$ (resp., $px_g\not\in pH$)
because $x\neq 0$.  Since $x_f-x_g\in H$,
$px_f - px_g = p(x_f-x_g)\equiv 0\md p$.
This is a nontrivial linear relation between 
$\Lambda_f$ and $\Lambda_g$.
\end{proof}

\begin{corollary}
If $p>2$ and the sign of some Atkin-Lehner involution
for $f$ is different than that for $g$ then
$(\Adual_f\intersect A_g^{\vee})[p]=0$.
\end{corollary}
\begin{proof}
Suppose $w_q(f) \neq w_q(g)$ and 
let $G=(\Adual_f\intersect A_g^{\vee})[p]$. 
Observe that $W_q$ acts as $w_q(f)\md p$ on $\Adual_f[p]$
and as $w_q(g)\md p$ on $A_g^{\vee}[p]$.  Hence $W_q$ acts
as both $w_q(f)\md p$ and $w_q(g)\md p$ on $G$.  Since
$p>2$, this is not possible when $G\neq 0$.
\end{proof}

@