Sharedwww / Tables / computing_cp.texOpen in CoCalc
Author: William A. Stein
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% computing_cp.tex
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\documentclass[11pt]{article}
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\include{macros}
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\title{Component Groups of Optimal Quotients of $J_0(N)$}
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\begin{document}
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\maketitle
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\begin{abstract}
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\end{abstract}
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\section{Introduction}
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\comment{
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What are the simplest naturally occuring abelian varieties over $\Q$?
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In my opinion they are the ``optimal'' quotients of the Jacobian $J_0(N)$
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of the modular curve $X_0(N)=\overline{\sltwoz\backslash\h}$.
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Let $f\in S_2(\Gamma_0(N))$ be a newform. Then Shimura associated to
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$f$ a quotient $A_f$ of $J_0(N)$ having all of the right naturality
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properties. These quotients are optimal in the sense that the kernel
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inside $J_0(N)$ is itself an abelian variety.
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The Shimura-Taniyama conjecture, mostly proved by recent work
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of Wiles and Taylor-Wiles, asserts that every elliptic curve over
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$\Q$ is isogeneous to some $A_f$. Thus understanding the abelian varieties
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$A_f$ is paramount importance!
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Neron associated to any abelian variety a model of the integer ring
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which is an abelian scheme and whose points are just what one guesses
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by looking at the points over the base field. Let $\cA_f$ be the
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Neron model of $A_f$. Suppose $p$ is a prime which divides the level
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$N$, so that $\cA_f/\F_p$ isn't an abelian variety, and hence has no
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right to be connected. The group of components
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$\Phi_{A_f,p}$ is of important in many ways. For example,
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according to the Birch and Swinnerton-Dyer conjecture,
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the order of its rational subgroup appears in the numerator
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of a certain rational number associated to the $L$-function
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of $A_f$.
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}
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Let $f\in S_2(\Gamma_0(N))$ be a newform and $A_f$ the corresponding
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optimal quotient of $J_0(N)$.
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Let $\Phi_{A_f,p}$ be the special fiber of the Neron model of $A_f$ at $p$.
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The integer $\#\Phi_{A_f,p}$ is of great arithmetic interest. For example,
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it appears in the Birch and Swinnerton-Dyer conjecture (I lie: actually the
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cardinality of the rational component group appears there).
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\begin{center}
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How can we compute $\#\Phi_{A_f,p}$?
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\end{center}
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When $A_f=E$ is an elliptic curve the Riemann-Roch theorem gives us
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a powerful tool: we have a simple explicit Weierstrass equation which defines $E$.
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Using this Tate provided an efficient algorithm for computing the order
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of $\Phi_{E,p}$. I doubt that there is an analogue of the Weierstrass
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equation when $\dim A_f > 1$!
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Ribet, in a letter to Mestre, considered the case in which $A_f=E$
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is an elliptic curve which has multiplicative reduction at $p$,
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i.e., $\ord_p(N)=1$. Using Grothendieck's monodromy pairing
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on character groups of certain tori he was able to provide a
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``Weierstrass-free'' formula for $\Phi_{E,p}$.
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In this paper we generalize Ribet's method to $A_f$ of arbitrary
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dimension and thus obtain an explicit formula for $\#\Phi_{A_f,p}$.
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The formula is:
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$$\#\Phi_{A_f,p} =
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\sqrt{\deg(\delta)}
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\cdot \frac{\# \coker (X \ra \Hom(X[\p],\Z))^2}
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{\disc(X[\p] \cross X[\p] \ra \Z)}$$
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where
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\begin{itemize}
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\item $\delta:A_f^{\vee}\ra A_f$ is the modular polarization.
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\item $X$ is the character group of the torus associated to $J_0(N)/\Fp$.
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\item $\p_f=\Ann_{\T}(f)$, where $\T$ is the Hecke algebra.
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\item $X\cross X\ra \Z$ is the Monodromy pairing.
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\end{itemize}
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Each of these quantities can be explicitely computed. See
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\cite{kohel}, \cite{mestre}, and \cite{stein}.
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\section{The Monodromy Pairing}
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[to be written -- will contain the basic facts]
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\section{Ribet's Formulas}
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[to be written -- will contain Ribet's formulas as motivation.]
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\section{General Formula}
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In this section we prove our main theorem.
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Let $f\in S_2(\Gamma_0(N))$ be a newform and suppose $p|N$ but $p\nmid M=N/p$.
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The Hecke algebra $\T$ acts on the fundamental group $X_J$ of $J_0(N)$.
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The prime ideal
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$$\p_f=\Ann_\T(f) \subset \T$$
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cuts out a submodule
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$$V_f=X_J[\p_f] \subset X_J$$
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of $X_J$.
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\begin{theorem}
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$$\#\Phi_{A_f,p} =
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\sqrt{\deg(\delta)}
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\cdot \frac{\# \coker (X_J \ra \Hom(V_f,\Z))^2}
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{\disc(V_f, V_f)}$$
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\end{theorem}
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\vspace{.3in}
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We now begin the proof.
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\vspace{.3in}
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{\noindent \bf Fundamental diagrams:}
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\begin{center}
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\begin{picture}(310,150)
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\put(0,110){$A^{\vee}$}
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\put(70,110){$J$}
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\put(70,40){$A$}
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\put(75,100){\vector(0,-1){45}}
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\put(75,63){\vector(0,-1){3}}
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\put(13,105){\vector(1,-1){55}}
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\qbezier(65,45)(10,60)(5,103)\put(5,103){\vector(0,1){1}}
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\put(18,113){\vector(1,0){48}}
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\qbezier(18,113)(14,115)(18,117)
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\put(40,117){$\pi^{\vee}$}
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\put(40,80){$\delta$}
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\put(20,50){$\hat{\delta}$}
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\put(80,80){$\pi$}
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\put(230,110){$X_A$}
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\put(300,110){$X_J$}
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\put(300,40){$X_{A^{\vee}}$}
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\put(305,100){\vector(0,-1){45}}
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\put(305,63){\vector(0,-1){3}}
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\put(243,105){\vector(1,-1){55}}
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\qbezier(290,45)(240,60)(237,103)\put(237,103){\vector(0,1){1}}
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\put(248,113){\vector(1,0){48}}
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\qbezier(248,113)(244,115)(248,117)
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\put(270,117){$\pi^*$}
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\put(270,80){$\delta^*$}
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\put(250,50){$\hat{\delta}^*$}
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\put(310,80){$\pi_*$}
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\end{picture}
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\end{center}
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{\noindent \bf Monodromy pairings:}
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$$\langle \quad , \quad \rangle_J : X_J \cross X_J \ra \Z$$
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$$\langle \quad , \quad \rangle_A : X_A^{\vee} \cross X_A \ra \Z$$
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\vspace{.4in}
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{\noindent \bf Component groups:}
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$$0\ra X_J \ra \Hom(X_J,\Z)\ra \Phi_{J,p} \ra 0$$
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$$0\ra X_{A^{\vee}} \ra \Hom(X_A,\Z)\ra \Phi_{A,p} \ra 0$$
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\vspace{.4in}
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{\noindent \bf Step 1. Formulas:}
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\begin{lemma}
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Let $x\in X_{A^{\vee}}$, $y\in X_A$. Then
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$$\langle \pi^*\hat{\delta}^* x, \pi^* y\rangle_J =
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\deg(\delta) \langle x, y \rangle_A.$$
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\end{lemma}
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\begin{proof}
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This is a direct computation, using the functorial properties of
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the objects involved.
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\begin{eqnarray*}
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\langle \pi^*\hat{\delta}^* x, \pi^* y\rangle_J &=&
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\langle \pi_*\pi^*\hat{\delta}^* x, y\rangle_A \\
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&=& \langle \delta^* \hat{\delta}^* x, y\rangle_A \\
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&=& \langle (\hat{\delta} \circ \delta)^* x, y\rangle_A\\
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&=& \deg(\delta) \langle x, y \rangle_A
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\end{eqnarray*}
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\end{proof}
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\begin{proposition}
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$$\#\Phi_{A,p} = \frac{\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A)}{\deg(\delta)^d}.$$
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\end{proposition}
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\begin{proof}
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By the lemma we have
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$$\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A) = \deg(\delta)^d
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\cdot \disc(X_{A^{\vee}},X_A).$$
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Now use that
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$$\#\Phi_{A,p}=\#\coker(X_{A^{\vee}}\ra X_A)= \disc(X_{A^{\vee}},X_A).$$
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\end{proof}
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\begin{proposition}\label{discformula}
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$$\#\Phi_{A,p} = \frac{[X_A:\hat{\delta}^* X_{A^\vee}]\cdot
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\disc(\pi^* X_A,\pi^* X_A)}
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{\deg(\delta)^d}.$$
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\end{proposition}
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\begin{proof}
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Using the formula from the previous proposition we have
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\begin{eqnarray*}
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\#\Phi_{A,p} &=& \frac{\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A)}
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{\deg(\delta)^d} \\
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&=& \frac{[\pi^* X_A : \pi^*\hat{\delta}^*X_{A^{\vee}}]\cdot
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\disc (\pi^* X_A, \pi^* X_A)}
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{\deg(\delta)^d}
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\end{eqnarray*}
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Note that $\pi^*$ is an injection so it does not change
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the index:
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$$[\pi^*X_A : \pi^*\hat{\delta}^* X_{A^{\vee}}]
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= [X_A:\hat{\delta}^* X_{A^\vee}].$$
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\end{proof}
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{\noindent \bf Step 2. Identify $\Phi_{A,p}$ inside of $X_J$:}
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\begin{proposition}\label{snakecokernel}
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$$\Phi_{A,p} \isom \coker(X_J \ra \Hom(\pi^* X_A, \Z))$$
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\end{proposition}
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\begin{proof}
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Let $\tilde{X}_J$ denote the image of $X_J$ in $\Hom(\pi^*X_A,\Z)$.
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We have the following diagram with exact rows and columns:
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$$
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\begin{matrix}
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& & & & & & 0 & \\
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& & & & & & \da & \\
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& & & & 0 & \lra & K & \\
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& & & & \da & & \da & \\
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0&\lra& \tilde{X}_J& \lra & \Hom(\pi^*X_A,\Z) & \lra & W & \lra 0 \\
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& & \da \isom & & \da \isom & & \da & \\
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0&\lra& X_{A^{\vee}}& \lra & \Hom(X_A,\Z) & \lra & \Phi_{A,p} & \lra 0 \\
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& & \da & & \da & & \da & \\
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& & 0 & \lra & 0 & \lra & L & \\
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& & & & & & \da & \\
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& & & & & & 0 & \\
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\end{matrix}
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$$
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The snake lemma implies that the sequence
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$$0 \ra K \ra 0 \ra 0 \ra L \ra 0$$
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is exact,
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hence $K=L=0$ and so $W= \Phi_{A,p}$.
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\end{proof}
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{\noindent \bf Step 3. Relate $\pi^*X_A$ to $V_f$:}
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Let
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$$V_f = X_J[\p_f] = \bigcap_{t\in\Ann_{\T}(f)} \ker(t) \subset X_J.$$
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By multiplicity one and $\T$-invariance of $\pi^*$,
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$$\pi^* X_A \subset V_f,$$
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and the index is finite. Let
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$$ m = [V_f : \pi^* X_A].$$
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The discriminant expression of $\Phi_{A,p}$ is ``homogeneous of degree
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$2$ in $V_f$'':
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\begin{proposition}\label{homsquare}
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$$\#\Phi_{A,p} = \frac{[X_A:\hat{\delta}^* X_{A^\vee}]\cdot m^2 \cdot
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\disc(V_f, V_f)}
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{\deg(\delta)^d}.$$
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\end{proposition}
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\begin{proof}
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Note that
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$$\disc(\pi^* X_A, \pi^* X_A) = m^2 \disc(V_f,V_f).$$
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Now use (\ref{discformula}).
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\end{proof}
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The cokernel expression for $\Phi_{A,p}$ is ``homogeneous of degree
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$1$ in $V_f$'':
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\begin{proposition}\label{homunit}
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$$\#\Phi_{A,p} = m\cdot \#\coker( X_J\ra \Hom(V_f,\Z)).$$
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\end{proposition}
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\begin{proof}
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Use (\ref{snakecokernel}) and properties of cokernels.
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\end{proof}
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{\noindent \bf Step 4. Cancel $m$:}
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\begin{proposition}
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$$\#\Phi_{A,p}
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= \frac{\deg{\delta}^d}{[X_A:\hat{\delta}^* X_{A^{\vee}}]}
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\cdot \frac{\#\coker(X_J\ra \Hom(V_f,\Z))^2}{\disc(V_f,V_f)}$$
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\end{proposition}
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\begin{proof}
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Let $c=\#\Phi_{A,p}$. Using (\ref{homsquare}) and (\ref{homunit}) we obtain
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$$
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c = \frac{c^2}{c}
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= \frac{m^2\cdot \#\coker(X_J \ra \Hom(V_f,\Z))^2}
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{\frac{[X_A:\hat{\delta}^* X_{A^{\vee}}]\cdot m^2\cdot \disc(V_f,V_f)}
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{\deg(\delta)^d}}
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$$
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Cancelling the $m^2$'s gives the formula.
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\end{proof}
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{\noindent \bf Step 5. Analyze $\deg(\delta)$:}
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\begin{proposition}
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$$\frac{\deg(\delta)^d}{[X_A:\hat{\delta}^* X_{A^{\vee}}]}
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= [X_{A^{\vee}} : \delta^* X_{A}]$$
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\end{proposition}
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\begin{proof}
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We have
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\begin{eqnarray*}
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\deg(\delta)^d &=& [X_A:\deg(\delta) X_A] \\
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&=& [X_A : \hat{\delta}^* X_{A^{\vee}}]
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\cdot [\hat{\delta}^* X_{A^{\vee}} :
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\hat{\delta}^* \delta^* X_{A^{\vee}}] \\
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&=& [X_A : \hat{\delta}^* X_{A^{\vee}}]
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\cdot [X_{A^{\vee}} :
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\delta^* X_{A^{\vee}}]
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\end{eqnarray*}
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The last step uses that $\hat{\delta}^*$ is injective.
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Now divide both sides by $ [X_A : \hat{\delta}^* X_{A^{\vee}}]$.
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\end{proof}
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Putting everything together we obtain
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$$\#\Phi_{A,p}
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= \frac{\#\coker(X_A\ra X_{A^{\vee}}) \cdot \#\coker(X_J\ra \Hom(V,\Z))^2}
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{\disc(V_f, V_f)}.$$
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Thus to prove the theorem it remains only to show that
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$$\#\coker(X_A\ra X_{A^{\vee}}) = \sqrt{\deg(\delta)}.$$
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{\noindent \bf Step 6. Analyze cokernel of $\delta^*$:}
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\begin{theorem}
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$$\#\coker(X_A\ra X_{A^{\vee}}) = \sqrt{\deg(\delta)}.$$
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\end{theorem}
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\begin{proof}
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The construction of Neron models commutes with unramified base
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extension. Since $\ord_p(N)=1$ we can pass to an unramified
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extension of $\Q_p$ so that $A$ has universal cover $\Gm^d$.
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Applying the Raynaud-van der Put theorem we obtain
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an exact commutative diagram of rigid analytic spaces:
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$$\begin{matrix}
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& & & 0 & & 0 & & & \\
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& & & \da & & \da & & & \\
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0 \lra & 0 &\lra & X_A & \lra &X_{A^\vee} & \lra & L & \lra 0 \\
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& \da & & \da & & \da & & \da & \\
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0 \lra & \Lambda & \lra& \Gm^d & \lra & \Gm^d & \lra & 0 & \lra 0 \\
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& \da & & \da & & \da & & \da & \\
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0 \lra & K & \lra& A^{\vee}& \lra & A & \lra & 0 & \lra 0 \\
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& & & \da & & \da & & & \\
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& & & 0 & & 0 & & & \\
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\end{matrix}$$
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where all maps are induced by the map $\delta:A^{\vee}\ra A$, so
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$K=\Ker(\delta)$ and $L=\coker(X_A\ra X_{A^{\vee}})$.
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Applying the snake lemma, with the connecting map going from $K$ to $L$,
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we obtain the exact sequence
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$$0 \ra \Lambda \ra K \ra L \ra 0.$$
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Now compute the $1$-motive dual, i.e., $M'=\Hom(M,\Gm)$, of the above exact diagram to obtain:
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$$\begin{matrix}
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0 \lla & 0 & \lla& \Gm^d & \lla & \Gm^d & \lla & L' & \lla 0 \\
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& \ua & & \ua & & \ua & & \ua & \\
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0 \lla &\Lambda' &\lla & X_{A^\vee}& \lla &X_{A} & \lla & 0 & \lla 0 \\
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& \ua & & \ua & & \ua & & \ua & \\
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& K' & \lla & 0 & \lla & 0 & \lla & 0 & \\
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\end{matrix}$$
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Here $A'=A=0$, $(\Gm^d)'=X_A$ or $X_{A^{\vee}}$.
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Note that dualizing is not in general exact, though it is exact on the $X_A$ and $\Gm^d$ parts.
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Adding in the cokernels at the top gives:
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$$\begin{matrix}
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& & & 0 & & 0 & & & \\
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& & & \ua & & \ua & & & \\
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& 0 & \lla& A & \lla & A^{\vee} & \lla & K' & \lla 0 \\
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& \ua & & \ua & & \ua & & \ua & \\
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0 \lla & 0 & \lla& \Gm^d & \lla & \Gm^d & \lla & L' & \lla 0 \\
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& \ua & & \ua & & \ua & & \ua & \\
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0 \lla &\Lambda' &\lla & X_{A^\vee}& \lla &X_{A} & \lla & 0 & \lla 0 \\
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& \ua & & \ua & & \ua & & & \\
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& K' & \lla & 0 & \lla & 0 & & & \\
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\end{matrix}$$
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All of the maps in this diagram are induced by
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$$\delta^{\vee}:A^{\vee} \ra A.$$
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But $\delta$ is induced by the canonical polarization of the Jacobian $J_0(N)$,
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so $\delta^{\vee}=\delta$. This is the same diagram as before, but viewed
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from a different point of view! Thus:
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\begin{eqnarray*}
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K' &=& K\\
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L' &=& \Lambda\\
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\Lambda' &=& L\\
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\end{eqnarray*}
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If $G$ is a finite group scheme then $\#G = \#G'$.
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Thus the exact sequence $0\ra \Lambda \ra K \ra L \ra 0$ together with
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the above equalities yields
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$$\#K = \# L \cdot \# \Lambda = \# L \cdot \# L' = (\#L)^2.$$
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This completes the proof.
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\end{proof}
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\begin{question}
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In the proof we determined the structure of
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$\coker(X_A\ra X_{A^{\vee}})$
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in terms of the abelian group $\Ker(\delta)$. It is possible
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to compute $\Ker(\delta)$ explicitely hence it also possible
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to compute $\coker(X_A\ra X_{A^{\vee}})$. Is
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it possible to somehow use this to obtain the {\em structure}
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of $\Phi_{A,p}$, instead of just the order?
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\end{question}
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\section{Examples}
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[a couple of explicit worked examples illustrating various principles]
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\section{Tables}
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[table of all $c_p$ for level $\leq 100$.]
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[table of supringsly LARGE $c_p$.]
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[table of analytic orders of $\Sha$ for all $A_f$ of rank
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$0$ and prime level $\leq 1500$.]
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The table will essentially look like:
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\begin{center}
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\begin{tabular}{|l|c|c|c|c|}\hline
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$f$ & $\odd(L(A_f,1)/\Omega(A_f))$ & $\numer((N-1)/12)$ & $\deg(\delta)$ & $c_p$ \\\hline\hline
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{\bf 389E20} & $5^2/97$ & $97$ & $5^2\cdot 2^?$ & $97$\\
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{\bf 433D16}& $7^2/3^2$ & $2^2\cdot3^2$ & $3^2\cdot 7^2\cdot 37^2\cdot 2^?$ & \\
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{\bf 563E31}& $13^2/281$& $281$ & $13^2\cdot 2^?$ & \\
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{\bf 571D2}& $3^2/1$ & $5\cdot 19$ & $2^4\cdot 3^4\cdot 127^2$ & $1$\\
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{\bf 709C30}& $11^2/59$ & $59$ & $11^2\cdot 2^?$ & \\
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{\bf 997H42}& $3^4/83$ & $83$ & & \\
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{\bf 1061D46}& $151^2/(5\cdot53)$ & $5\cdot53$ & $61^2\cdot 151^2\cdot 179^2\cdot 2^?$ & \\
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{\bf 1091C62}& $7^2/(5\cdot109)$ & $5\cdot109$& $2^?$ & \\
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{\bf 1171D53}& $11^2/(3\cdot5\cdot13)$ & $3\cdot5\cdot13$& & \\
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{\bf 1283C62}& $5^2/641$ & $641$& & \\
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{\bf 1429B64}& $5^2/(7\cdot17)$ & $7\cdot17$& & \\
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{\bf 1481C71}& $13^2/(5\cdot37)$ & $2\cdot5\cdot37$& & \\
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{\bf 1483D67}& $3^2\cdot5^2/(13\cdot19)$ & $13\cdot19$& &\\\hline
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\end{tabular}
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\end{center}
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\comment{
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{\bf 389E20} & $\frac{5^2}{97}$ & $97$ & $5^2\cdot 2^?$ & $97$\\
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{\bf 433D16}& $\frac{7^2}{3^2}$ & $2^2\cdot3^2$ & $3^2\cdot 7^2\cdot 37^2\cdot 2^?$ & \\
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{\bf 563E31}& $\frac{13^2}{281}$& $281$ & $13^2\cdot 2^?$ & \\
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{\bf 571D2}& $\frac{3^2}{1}$ & $5\cdot 19$ & $2^4\cdot 3^4\cdot 127^2$ & $1$\\
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{\bf 709C30}& $\frac{11^2}{59}$ & $59$ & $11^2\cdot 2^?$ & \\
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{\bf 997H42}& $\frac{3^4}{83}$ & $83$ & & \\
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{\bf 1061D46}& $\frac{151^2}{5\cdot53}$ & $5\cdot53$ & $61^2\cdot 151^2\cdot 179^2\cdot 2^?$ & \\
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{\bf 1091C62}& $\frac{7^2}{5\cdot109}$ & $5\cdot109$& $2^?$ & \\
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{\bf 1171D53}& $\frac{11^2}{3\cdot5\cdot13}$ & $3\cdot5\cdot13$& & \\
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{\bf 1283C62}& $\frac{5^2}{641}$ & $641$& & \\
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{\bf 1429B64}& $\frac{5^2}{7\cdot17}$ & $7\cdot17$& & \\
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{\bf 1481C71}& $\frac{13^2}{5\cdot37}$ & $2\cdot5\cdot37$& & \\
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{\bf 1483D67}& $\frac{3^2\cdot5^2}{13\cdot19}$ & $13\cdot19$& &\\\hline
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}
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\begin{thebibliography}{HHHHHHH}
427
\bibitem[K]{kohel} D. Kohel, {\em Hecke module structure of quaternions},
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preprint, (1998).
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\bibitem[M]{mestre} J.F. Mestre,
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{\em La m\'{e}thode des graphs. Exemples et applications},
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Taniguchi Symp., Proceedings of the international
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conference on class numbers and fundamental units of
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algebraic number fields (Katata, 1986), 217--242, Nagoya Univ., Nagoya, (1986).
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\bibitem[S]{stein} W.A. Stein,
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{\em The Modular Polarization of a Quotient of $J_0(N)$},
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in preparation, (1999).
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