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Author: William A. Stein
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% congruences.tex
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\section{Congruences}
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\label{sec:congruences}
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Lo\"\i{}c's paper ``Arithmetic of elliptic curves and
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diophantine equations'' is very relevant to this
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section, esp. his section THREE. Do not forget
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to make proper attribution, etc.
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Let $f, g$ be nonconjugate newforms and $H=H_1(X_0(N),\Z)$.
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\begin{proposition}
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$(\Adual_f\intersect A_g^{\vee})[p]\neq 0$
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if and only if the mod $p$ rank of $H[I_f]+H[I_g]$
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is strictly less than $\rank H[I_f] + \rank H[I_g]$.
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\end{proposition}
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\begin{proof}
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By (\ref{Af}) $\Lambda_f=H[I_f]$ (resp., $\Lambda_g=H[I_g]$) is the submodule of
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$H$ which defines $A_f$ (resp., $A_g$). By reduction mod $p$ we mean
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the map $H\ra H\tensor \Fp$. Suppose
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$$\rank (\Lambda_f + \Lambda_g)\md p < \rank\Lambda_f + \rank \Lambda_g.$$
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Since $\Lambda_f$ (resp., $\Lambda_g$) is a kernel, it is saturated, so
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$\rank \Lambda_f \md p = \rank \Lambda_f$ (resp., for $\Lambda_g$).
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We conclude that the mod $p$ linear dependence must involve vectors
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from both $\Lambda_f$ and $\Lambda_g$; there is $v\in\Lambda_f$ and
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$w\in\Lambda_g$ so that $v, w\not\equiv 0\md p$ but $v+w\con 0 \md p$.
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Thus $\frac{v+w}{p}\in H$ is integral, i.e., in $J_0(N)(\C)$ we have
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$\frac{1}{p}v - (-\frac{1}{p} w)=0$. But $\frac{1}{p}v \not \in \Lambda_f$
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and $\frac{1}{p} w\not\in\Lambda_g$ (otherwise $v$ and $w$ would
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be $0\md p$), so
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$\frac{1}{p}v$ and $-\frac{1}{p}w$ are both nontrivial $p$-torsion in
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$\Adual_f$, $A_g^{\vee}$, resp. Conclusion:
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$0\neq \frac{1}{p}v = -\frac{1}{p} w \in (\Adual_f\intersect A_g^{\vee})[p]$.
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Conversely, suppose $0\neq x\in (\Adual_f\intersect A_g^{\vee})[p]$. Choose
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lifts modulo $H$ to $x_f\in\frac{1}{p}\Lambda_f$ and
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$x_g\in\frac{1}{p}\Lambda_g$.
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Then $px_f\in\Lambda_f$ (resp., $px_g\in\Lambda_g$), but
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$px_f\not\in p H$ (resp., $px_g\not\in pH$)
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because $x\neq 0$. Since $x_f-x_g\in H$,
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$px_f - px_g = p(x_f-x_g)\equiv 0\md p$.
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This is a nontrivial linear relation between
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$\Lambda_f$ and $\Lambda_g$.
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\end{proof}
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\begin{corollary}
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If $p>2$ and the sign of some Atkin-Lehner involution
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for $f$ is different than that for $g$ then
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$(\Adual_f\intersect A_g^{\vee})[p]=0$.
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\end{corollary}
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\begin{proof}
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Suppose $w_q(f) \neq w_q(g)$ and
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let $G=(\Adual_f\intersect A_g^{\vee})[p]$.
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Observe that $W_q$ acts as $w_q(f)\md p$ on $\Adual_f[p]$
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and as $w_q(g)\md p$ on $A_g^{\vee}[p]$. Hence $W_q$ acts
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as both $w_q(f)\md p$ and $w_q(g)\md p$ on $G$. Since
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$p>2$, this is not possible when $G\neq 0$.
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\end{proof}
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