% congruences.tex12\section{Congruences}3\label{sec:congruences}45Lo\"\i{}c's paper ``Arithmetic of elliptic curves and6diophantine equations'' is very relevant to this7section, esp. his section THREE. Do not forget8to make proper attribution, etc.910Let $f, g$ be nonconjugate newforms and $H=H_1(X_0(N),\Z)$.1112\begin{proposition}13$(\Adual_f\intersect A_g^{\vee})[p]\neq 0$14if and only if the mod $p$ rank of $H[I_f]+H[I_g]$15is strictly less than $\rank H[I_f] + \rank H[I_g]$.16\end{proposition}17\begin{proof}18By (\ref{Af}) $\Lambda_f=H[I_f]$ (resp., $\Lambda_g=H[I_g]$) is the submodule of19$H$ which defines $A_f$ (resp., $A_g$). By reduction mod $p$ we mean20the map $H\ra H\tensor \Fp$. Suppose21$$\rank (\Lambda_f + \Lambda_g)\md p < \rank\Lambda_f + \rank \Lambda_g.$$22Since $\Lambda_f$ (resp., $\Lambda_g$) is a kernel, it is saturated, so23$\rank \Lambda_f \md p = \rank \Lambda_f$ (resp., for $\Lambda_g$).24We conclude that the mod $p$ linear dependence must involve vectors25from both $\Lambda_f$ and $\Lambda_g$; there is $v\in\Lambda_f$ and26$w\in\Lambda_g$ so that $v, w\not\equiv 0\md p$ but $v+w\con 0 \md p$.27Thus $\frac{v+w}{p}\in H$ is integral, i.e., in $J_0(N)(\C)$ we have28$\frac{1}{p}v - (-\frac{1}{p} w)=0$. But $\frac{1}{p}v \not \in \Lambda_f$29and $\frac{1}{p} w\not\in\Lambda_g$ (otherwise $v$ and $w$ would30be $0\md p$), so31$\frac{1}{p}v$ and $-\frac{1}{p}w$ are both nontrivial $p$-torsion in32$\Adual_f$, $A_g^{\vee}$, resp. Conclusion:33$0\neq \frac{1}{p}v = -\frac{1}{p} w \in (\Adual_f\intersect A_g^{\vee})[p]$.3435Conversely, suppose $0\neq x\in (\Adual_f\intersect A_g^{\vee})[p]$. Choose36lifts modulo $H$ to $x_f\in\frac{1}{p}\Lambda_f$ and37$x_g\in\frac{1}{p}\Lambda_g$.38Then $px_f\in\Lambda_f$ (resp., $px_g\in\Lambda_g$), but39$px_f\not\in p H$ (resp., $px_g\not\in pH$)40because $x\neq 0$. Since $x_f-x_g\in H$,41$px_f - px_g = p(x_f-x_g)\equiv 0\md p$.42This is a nontrivial linear relation between43$\Lambda_f$ and $\Lambda_g$.44\end{proof}4546\begin{corollary}47If $p>2$ and the sign of some Atkin-Lehner involution48for $f$ is different than that for $g$ then49$(\Adual_f\intersect A_g^{\vee})[p]=0$.50\end{corollary}51\begin{proof}52Suppose $w_q(f) \neq w_q(g)$ and53let $G=(\Adual_f\intersect A_g^{\vee})[p]$.54Observe that $W_q$ acts as $w_q(f)\md p$ on $\Adual_f[p]$55and as $w_q(g)\md p$ on $A_g^{\vee}[p]$. Hence $W_q$ acts56as both $w_q(f)\md p$ and $w_q(g)\md p$ on $G$. Since57$p>2$, this is not possible when $G\neq 0$.58\end{proof}596061