Lecture 21: Proof of Mazur's theorem (part 2)

Proof. We have a containment $A[p^n]_{\fa} \subset A[\fa^m]$ for some $m$. Furthermore, we have an exact sequence

It thus suffices (by induction) to show that $A[\fa]$ satisfies $\JH(p)$. Let $V=A[\fa]$, regarded as an $\bF_p$-representation of the Galois group. Since $T_{\ell}$ acts by $\ell+1$ on $V$, Eichler--Shimura implies that the Frobenius $F_{\ell}$ satisfies $F_{\ell}^2-(\ell+1) F_{\ell}+\ell=0$ on $V$, for all $\ell \ne p,N$. We have a factorization $T^2-(\ell+1)T+\ell=(T-\ell)(T-1)$, and so all generalized eigenvalues of $F_{\ell}$ on $V$ are 1 or $\ell$. The result now follows from the following general lemma. ◾

Proof. Let $W=V \oplus V^*(1)$, where $(1)$ denotes tensoring with cyclotomic. If $m$ is the matrix for $F_{\ell}$ on $V$, then the matrix for $F_{\ell}$ on $V^*(1)$ is $\ell \cdot {}^tm^{-1}$. In particular, the dimension of the 1-generalized eigenspace of $F_{\ell}$ on $V$ is equal to the dimension of the $\ell$-generalized eigenspace of $F_{\ell}$ on $V^*(1)$. It follows that the dimension of the 1-generalized eigenspace of $F_{\ell}$ on $W$ is $n=\dim(V)$, and similarly for the $\ell$-generalized eigenspace. Thus the character of $W$ is the same as that of $n$ copies of the trivial plus $n$ copies of cyclotomic. It follows that every constituent of $W$ is trivial or cyclotomic, and so the same holds for $V$. ◾

Proof. Let $T$ be the integral Tate module of $G$. Since $T[1/p]$ is free of rank 2 over $\hat{\bT}_{\fa}[1/p]$, it follows that $T$ contains a free $\hat{\bT}_{\fa}$-module $T'$ of rank 2 with finite index. We have $\ell(G[p^n])=\len(T/p^nT)$. Of course, $\len(T/p^nT)=\len(T'/p^nT')+O(1)$, and $\len(T'/p^nT')=2\len(\hat{\bT}_{\fa}/p^n\hat{\bT}_{\fa})=2nd$. ◾

Proof. (For notational simplicity suppose $p \ne 2$, so that $\alpha$ counts the number of trivial constituents of $A[p^n]$ as a Galois representation.) Recall that the Hecke operators $T_{\ell}$ are self-adjoint under the Weil pairing on $J_0(N)[p^n]$. It follows that every element of $\hat{\bT}_p$ is also self-adjoint. Let $e$ be the idempotent of this ring that gives the projection onto the direct factor $\hat{\bT}_{\fa}$. Then $A[p^n]_{\fa}=J_0(N)[p^n]_{\fa}=eJ_0(N)[p^n]$. Now, $eJ_0(N)[p^n]$ and $(1-e)J_0(N)[p^n]$ are orthogonal under the Weil pairing, since $e$ is a self-adjoint idempotent. Since the Weil pairing on $J_0(N)[p^n]$ is perfect, it follows that it restricts to a perfect pairing on $A[p^n]_{\fa}$. Thus $A[p^n]_{\fa}$ is Cartier self-dual. Since it is made up of $\bZ/p\bZ$'s and $\mu_p$'s, the self-duality implies that there must be the same number of $\bZ/p\bZ$'s as $\mu_p$'s, and so the number of $\bZ/p\bZ$'s is half $\ell(A[p^n]_{\fa})$. This is $nd$, up to a bounded factor. (Note that $A[p^n]_{\fa}=G[p^n]$, where $G$ is the $p$-divisible group $A[p^{\infty}]_{\fa}=eA[p^{\infty}]$, and $V_p(G)$ is free of rank 2 over $\hat{\bT}_{\fa}[1/p]$.) ◾

Proof. Let $T=T_p(G)$ be the integral Tate module of $G$. Then $\cG_n(\ol{\bQ}_N)=T/p^nT$, and so

By definition, $\cG_n(\ol{\bF}_N)=\cG_n(\ol{\bQ}_N)^I=(T/p^nT)^I$, and so

We have an exact sequence

Since $\rH^1(I, T)$ is a finitely generated $\bZ_p$-module, its $p^n$-torsion is bounded independent of $n$. Thus

The result follows. ◾

Proof. Since $\cB[p^n]$ is étale over $\bZ_N$, every $p^n$-torsion point over $\ol{\bF}_N$ lifts to one over $\bZ^{\un}_N$. Furthermore, the Néron mapping property shows that every $\bQ^{\un}_N$ point of $B$ extends to one of $\cB$. We therefore have an isomorphism

Taking inverse limits gives the result. ◾

Proof. We have $\dim(V)=2 \dim(B)$. Let $\cB$ be Néron model of $B$. Then $V^I=V_p(\cB_{\bF_N})$. Since the identity component of $\cB_{\bF_N}$ is a torus of the same dimension as $B$, $V_p(\cB_{\bF_N})$ has dimension $\dim(B)$.  Thus $\dim(V)=2\dim(V^I)$.

Since $B$ has semi-stable reduction, Grothendieck's extension of Néron--Ogg--Shafarevich shows that $I$ acts unipotently on $V$. In fact, a stronger result is true: for any $g \in I$ we have $(g-1)^2=0$ on $V$. It follows that if $U$ is any Galois stable subspace of $V$ then $(g-1)^2=0$ on $U$ for any $g \in I$, and so $\dim(U^g) \ge \tfrac{1}{2} \dim(U)$. Since $I$ acts unipotently, its image is pro-$p$, and so the wild subgroup $I^w$ acts trivially. The group $I/I^w$ is pro-cyclic. If $g$ is a topologcal generator, then $U^I=U^g$. We thus see that $\dim(U^I) \ge \tfrac{1}{2} \dim(U)$.

Suppose now that $U$ is a summand of $V$, and write $V=U \oplus U'$. Then $V^I=U^I \oplus (U')^I$. By the above paragraph, $\dim(U^I) \ge \tfrac{1}{2} \dim(U)$ and $\dim((U')^I) \ge \tfrac{1}{2} \dim(U')$. Since $\dim(V^I)=\tfrac{1}{2} \dim(V)$, it follows that both inequalitities are equalities, which proves the result. ◾

Proof. Let $e \in \hat{\bT}_p$ be the idempotent projecting onto $\hat{\bT}_{\fa}$. Let $\cG$ be the $p$-divisible group $\cA[p^{\infty}]$ and let $V$ be the rational Tate module of $G=\cG_{\bQ}$. Then $\cG_{\fa}=e\cG$ is a $p$-divisible group and the rational Tate module of its generic fiber is $V_{\fa}=eV$. We have $\cG_{\fa}[p^n]=\cA[p^n]_{\fa}$, as both are just $\cA[\fa^{\infty}] \cap \cA[p^n]$. The group $\cA[p^n]$ is the maximal quasi-finite étale extension of its generic fiber, by the Néron mapping property, and the same is true for $\cA[p^n]_{\fa}$, since it is a summand. Thus $\delta(\cA[p^n]_{\fa})=n(\dim(V_{\fa})-\dim(V_{\fa}^I))+O(1)$. But $V_{\fa}$ is a summand of $V$ and $A$ has completely toric reduction, so $\dim(V_{\fa}^I)=\tfrac{1}{2} \dim(V_{\fa})$, and so $\delta(\cA[p^n]_{\fa})=\tfrac{1}{2} \dim(V_{\fa}) n+O(1)$. Since $V_{\fa}$ is a free $\hat{\bT}_{\fa}[1/p]$-module of rank 2, we have $\tfrac{1}{2} \dim(V_{\fa})=\rk(\hat{\bT}_{\fa})=d$. ◾

Proof. Let $\cA^{\circ}$ be the identity component of the Néron model of $\cA$ and let $\cG_n=\cA^{\circ}[p^n]_{\fa}$. We have

and

(the two $\delta$'s can only differ by the size of the component group of the special fiber of $\cA$ at $N$). Since the group $\cG_n$ is admissible, we have

Now, $\cG_n(\bZ) \subset \cA(\bZ)[p^n]=A(\bQ)[p^n]$ has bounded cardinality as $n$ varies, by the Mordell--Weil theorem. Thus $h^0(\cG_n)$ is bounded, and so $h^1(\cG_n)$ is bounded as well.

From the short exact sequence

we obtain an injection

Taking inverse limits over $n$, we have an injection

Applying the idempotent $e$, we obtain an injection

Since the group on the right is finite, so is the group on the left. The inclusion $\cA^{\circ}(\bZ) \subset \cA(\bZ) = A(\bQ)$ has finite index, and so we conclude that $A(\bQ) \otimes_{\bT} \hat{\bT}_{\fa}$ is finite as well. ◾

Proof. The map $M \to \bigoplus M/\fp M$ has finite kernel and cokernel, where the sum is over the minimal primes $\fp$ of $\bT$ contained in $\fa$. Thus the completion of $M/\fp M$ at $\fa$ is finite for each $\fp$. By the above lemma, each $M/\fp M$ is finite. Thus $M$ is finite as well. ◾

Proof. We have shown that $A(\bQ) \otimes_{\bT} \hat{\bT}_{\fa}$ is finite. Since $A(\bQ)$ is a $\bT/I$-module which is finitely generated as a $\bZ$-module, the result follows from the above lemma. ◾