Lecture 21: Proof of Mazur's theorem (part 2)
Today we finish the proof Mazur's theorem. In the previous lecture, we constructed a quotient of $A$ of $J_0(N)$, and showed that to prove the theorem it was enough to establish that $A(\bQ)$ has rank 0. Unfortunately, our criterion for rank 0 does not apply directly to $A$. However, the proof of the criterion almost applies, and with a little extra work we show what we want.
Review
From the previous lecture
We are in the process of proving Mazur's theorem:
Theorem. Let $N$ be a prime $\gt 7$ and not 13. Then no elliptic curve over $\bQ$ has a rational point of order $N$.
Let's recall where we left off. We picked a prime $p$ dividing the order of $[0][\infty]$ in $J_0(N)$, and defined $\fa$ to be the $p$Eisenstein ideal, that is, the ideal of $\bT$ generated by $p$ and $T_{\ell}(\ell+1)$ for all $\ell$. We showed that $\fa$ is not the unit ideal of $\bT$, and thus a maximal ideal with $\bT/\fa=\bF_p$. We defined $I$ to be the intersection of the minimal primes of $\bT$ contained in $\fa$, and $A$ to be the quotient $J_0(N)/IJ_0(N)$. To complete the proof, we need to show that $A(\bQ)$ has rank 0. That is what we'll do in this lecture.
From Lecture 11
To show that $A(\bQ)$ has rank 0, we're going to adapt the proof of Theorem B from [PP69BM]. We now briefly recall how that went.
A $p$power order group scheme over $\bZ$ or $\bZ[1/N]$ is called admissible if: (1) it's finite flat away from $N$; (2) it's quasifinite étale away from $p$; and (3) over $\bZ[1/N]$, admits a filtration with quotients $\mu_p$ or $\bZ/p\bZ$. We showed that a group scheme $G$ satisfying (1) and (2) is admissible if and only if the Galois module $G(\ol{\bQ})$ satisfies $\JH(p)$, i.e., it admits a filtration whose successive quotients are trivial or cyclotomic.
Let $G$ be an admissible group over $\bZ$. We defined several numerical invariants:

$\ell(G)=\log_p(\# G)$. (We'll use this notation for any group of $p$power order.)

$\alpha(G)$ is the number of $\bZ/p\bZ$'s occurring in $G$.

$\delta(G)=\ell(G_{\bQ})\ell(G_{\bF_N})$.

$h^i(G)$ is the length of $\rH^i_{\rm fppf}(\Spec(\bZ), G)$.
We proved the following inequality:
$$ h^1(G)h^0(G) \le \delta(G)  \alpha(G). $$Suppose now that $A/\bQ$ is an abelian variety satisfying the hypotheses of Theorem B, i.e., $A$ has good reduction away from $N$, completely toric reduction at $N$, and $A[p](\ol{\bQ})$ satisfies $\JH(p)$. Let $\cA/\bZ$ be the Néron model of $A$. Then $\cA[p^n]$ is admissible. Using the hypotheses, we computed $\alpha(\cA[p^n])$ and $\delta(\cA[p^n])$, and found them to be approximately equal, from which we concluded that $h^1(\cA[p^n])h^0(\cA[p^n])$ is bounded as $n \to \infty$. Since $h^0(\cA[p^n])$ is bounded (by the MordellWeil theorem), it follows that $h^1(\cA[p^n])$ is bounded. By Kummer theory, $A(\bQ)$ injects into the inverse limit of $\rH^1_{\rm fppf}(\bZ, \cA[p^n])$, which shows that $A(\bQ)$ is finite. (Actually, one must use $\cA^{\circ}$ here, but that does not cause a problem.)
Proof of rank 0
Notation
The quantities $N$, $p$, $\fa$, $I$, and $A$ are as above. We write $d$ for the $\bZ_p$rank of $\hat{\bT}_{\fa}$ and $e$ for the idempotent of $\hat{\bT}_p$ which projects onto $\hat{\bT}_{\fa}$. We let $\cA$ for the Néron model of $A$ over $\bZ$.
Admissibility
Proposition. $\cA[p^n]_{\fa}$ is admissible.
By our previous work, it is enough to show the following:
Lemma. $A[p^n]_{\fa}$ satisfies $\JH(p)$.
Proof. We have a containment $A[p^n]_{\fa} \subset A[\fa^m]$ for some $m$. Furthermore, we have an exact sequence
$$ 0 \to A[\fa] \to A[\fa^m] \to A[\fa^{m1}]. $$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 $V$, EichlerShimura 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 $T^2(\ell+1)T+\ell=(T\ell)(T1)$, and so all generalized eigenvalues of $F_{\ell}$ on $V$ are 1 or $\ell$. The result now follows from the following general lemma. ◾
Lemma. Let $V$ be a finite dimensional representation of $\Gal(\ol{\bQ}/\bQ)$ over $\bF_p$ such that 1 and $\ell$ are the only generalized eigenvalues of $F_{\ell}$, for all but finitely many $\ell$. Then $V$ satisfies $\JH(p)$.
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 1generalized 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 1generalized 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 $V$. ◾
Computation of α
Lemma. Let $G$ be a $p$divisible group with an action of $\hat{\bT}_{\fa}$ such that the rational Tate module $V_p(G)$ is free of rank 2 as a $\hat{\bT}_{\fa}[1/p]$module. Then $\ell(G[p^n])=2nd+O(1)$.
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 $\len(T'/p^nT')=2\len(\hat{\bT}_{\fa}/p^n\hat{\bT}_{\fa})=2nd$. ◾
Proposition. $\alpha(\cA[p^n]_{\fa})=nd+O(1)$.
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 selfadjoint under the Weil pairing on $J_0(N)[p^n]$. It follows that every element of $\hat{\bT}_p$ is also selfadjoint. 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 $(1e)J_0(N)[p^n]$ are orthogonal under the Weil pairing, since $e$ is a selfadjoint 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 selfdual. Since it is made up of $\bZ/p\bZ$'s and $\mu_p$'s, the selfduality 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 $\hat{\bT}_{\fa}[1/p]$.) ◾
Computation of δ
Lemma. Let $G/\bQ_N$ be a $p$divisible group, let $V=V_p(G)$ be its rational Tate module, and let $\cG_n/\bZ_p$ be the maximal extension of $G[p^n]$ to an étale quasifinite group (see <a title="Lecture 9: Néron models" href="http://asnowden.com/679/lecture9/">Lecture 9</a>). Then $\delta(\cG_n)=(\dim(V)\dim(V^I)) n + O(1)$, where $I \subset \Gal(\ol{\bQ}_N/\bQ_N)$ is the inertia subgroup.
Proof. Let $T=T_p(G)$ be the integral Tate module of $G$. Then $\cG_n(\ol{\bQ}_N)=T/p^nT$, and so
$$ \ell((\cG_n)_{\bQ_N})=\len(T/p^nT)=n \dim(V). $$By definition, $\cG_n(\ol{\bF}_N)=\cG_n(\ol{\bQ}_N)^I=(T/p^nT)^I$, and so
$$ \ell((\cG_n)_{\bF_N})=\len((T/p^nT)^I). $$We have an exact sequence
$$ 0 \to T^I/p^n T^I \to (T/p^nT)^I \to \rH^1(I, T)[p^n] \to 0. $$Since $\rH^1(I, T)$ is a finitely generated $\bZ_p$module, its $p^n$torsion is bounded independent of $n$. Thus
$$ \len((T/p^nT)^I)=\dim(T^I/p^n T^I)+O(1)=n\dim(V^I)+O(1). $$The result follows. ◾
Lemma. Let $B/\bQ_N$ be an abelian variety and let $\cB/\bZ_N$ be its Néron model. Then $V_p(\cB_{\bF_N})=V_p(B)^I$, where $I$ is the inertia group.
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
$$ \cB[p^n](\ol{\bF}_N) = \cB[p^n](\bQ_N^{\un}) = \cB[p^n](\ol{\bQ}_N)^I. $$Taking inverse limits gives the result. ◾
Lemma. Let $B/\bQ_N$ be an abelian variety with completely toric reduction and let $U$ be a summand of $V=V_p(B)$ as a Galois representation. Then $\dim(U^I)=\tfrac{1}{2} \dim(U)$.
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 semistable reduction, Grothendieck's extension of NéronOggShafarevich shows that $I$ acts unipotently on $V$. In fact, a stronger result is true: for any $g \in I$ we have $(g1)^2=0$ on $V$. It follows that if $U$ is any Galois stable subspace of $V$ then $(g1)^2=0$ 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 procyclic. 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. ◾
Remark. In the case $B=J_0(N)$, which is the only case of interest to us, we know that $V$ decomposes into a sum of $V_{f,\lambda}$, which are each two dimensional. Thus the fact that $(g1)^2=0$ for $g \in I$ follows from $g$ being unipotent in this case.
Remark. The above lemma shows that the inertia group at $N$ acts unipotently and nontrivially on $V_{f,\lambda}$. This is an instance of localglobal compatibility in the Langlands program.
Proposition. $\delta(\cA[p^n]_{\fa})=nd+O(1)$.
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 quasifinite 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 $\tfrac{1}{2} \dim(V_{\fa})=\rk(\hat{\bT}_{\fa})=d$. ◾
Finiteness of the Eisenstein piece of the MordellWeil group
Proposition. The group $\hat{\bT}_{\fa} \otimes_{\bT} A(\bQ)$ is finite.
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
$$ \alpha(\cG_n)=\alpha(\cA[p^n]_{\fa})=nd+O(1) $$and
$$ \delta(\cG_n)=\delta(\cA[p^n]_{\fa})+O(1)=nd+O(1) $$(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
$$ h^1(\cG_n)  h^0(\cG_n) \le \delta(\cG_n)  \alpha(\cG_n) = O(1). $$Now, $\cG_n(\bZ) \subset \cA(\bZ)[p^n]=A(\bQ)[p^n]$ has bounded cardinality as $n$ varies, by the MordellWeil theorem. Thus $h^0(\cG_n)$ is bounded, and so $h^1(\cG_n)$ is bounded as well.
From the short exact sequence
$$ 0 \to \cA^{\circ}[p^n] \to \cA^{\circ} \to \cA^{\circ} \to 0 $$we obtain an injection
$$ \cA^{\circ}(\bZ) \otimes \bZ/p^n \bZ \to \rH^1_{\rm fppf}(\bZ, \cA^{\circ}[p^n]). $$Taking inverse limits over $n$, we have an injection
$$ \cA^{\circ}(\bZ) \otimes \bZ_p \to \varprojlim \rH^1_{\rm fppf}(\bZ, \cA^{\circ}[p^n]). $$Applying the idempotent $e$, we obtain an injection
$$ \cA^{\circ}(\bZ) \otimes_{\bT} \hat{\bT}_{\fa} \to \varprojlim \rH^1_{\rm fppf}(\bZ, \cG_n). $$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. ◾
Completion of proof
Lemma. Let $\cO$ be an order in a number field and let $\fa$ be a maximal ideal in $\cO$. Suppose that $M$ is a finitely generated $\cO$ module and $M \otimes_{\cO} \hat{\cO}_{\fa}$ is finite. Then $M$ is finite.
Lemma. Let $M$ be a $\bT/I$module which is finitely generated as a $\bZ$module, and suppose that $M \otimes_{\bT} \hat{\bT}_{\fa}$ is finite. Then $M$ is finite.
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. ◾
Proposition. $A(\bQ)$ has rank 0.
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. ◾