Lecture 11: Criterion for rank 0

$$ \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Pic}{Pic} \newcommand{\bQ}{\mathbf{Q}} \let\ol\overline \newcommand{\fN}{\mathfrak{N}} \newcommand{\rH}{\mathrm{H}} \newcommand{\cA}{\mathcal{A}} \newcommand{\bZ}{\mathbf{Z}} \DeclareMathOperator{\Spec}{Spec} \newcommand{\bF}{\mathbf{F}} \DeclareMathOperator{\Hom}{Hom} \newcommand{\bG}{\mathbf{G}} \DeclareMathOperator{\Pic}{Pic} $$

In this lecture, we establish Theorem B from Lecture 1, which is a criterion for an abelian variety to have rank 0. The idea of the proof is similar to that of the weak Mordell--Weil theorem, but here we control the ramification of the cohomology classes much more carefully. Most of the work goes into understanding a certain class of group schemes (the admissible ones) very precisely.


Statement of criterion

The purpose of today's lecture is to establish the following criterion for an abelian variety to have rank 0 (Theorem B from Lecture 1):

Theorem. Let $A/\bQ$ be an abelian variety, and let $N$ and $p$ be distinct prime numbers, with $N$ odd. Suppose the following conditions hold:

  • $A$ has good reduction away from $N$.

  • $A$ has completely toric reduction at $N$.

  • The Jordan--Hölder constituents of $A[p](\ol{\bQ})$ are 1-dimensional and either trivial or cyclotomic.

Then $A(\bQ)$ has rank 0.

Remark. The proof uses many special properties of $\bQ$, but can be generalized slightly, as follows. Let $K$ be an imaginary quadratic number field, let $p$ be a rational prime, and let $\fN$ be a prime of $K$. Assume that $p$ does not divide the class number of $K$ and if $p \le 3$ then $p$ is unramified in $K$. Then the obvious generalization holds: if $A/K$ be an abelian variety with good reduction away from $\fN$, completely toric reduction at $\mathfrak{N}$, and such that the Jordan--Hölder constituents of $A[p](\ol{K})$ are either trivial or cyclotomic then $A(K)$ has rank 0.

Remark. This theorem, and the proof presented here, comes from III.3 of Mazur's paper "Modular curves and the Eisenstein ideal" (MR488287). It is not stated there explicitly, however.

Idea of proof

Recall the proof of the weak Mordell--Weil theorem. Kummer theory gives an injection of $A(\bQ)/nA(\bQ)$ into $\rH^1(G_{\bQ}, A[n])$, so it suffices to prove the $\rH^1$ is finite. However, it's not, because we have not restricted ramificiation. One can show that there is a finite set of places $S$ such that the image of $A(\bQ)/nA(\bQ)$ lands in $\rH^1(G_{\bQ,S}, A[n])$. This $\rH^1$ is finite, and this proves the weak Mordell--Weil theorem.

As one shrinks $S$, the $\rH^1$ gets smaller and smaller, so it makes sense for us to take $S$ as small as possible. In general, one can take $S$ to be the set of places of bad reduction together with the divisors of $n$. So if work with $p$-power torsion, we can take $S=\{N,p\}$. However, this is still too big for us!

We can improve the situation using the following idea. Let $\cA$ be the Néron model of $A$ over $\bZ$, and let $G_n=\cA[p^n]$. Then $\rH^1(G_{\bQ, S}, A[p^n])$ is the étale cohomology group $\rH^1_{\rm et}(\Spec(\bZ[1/Np]), G_n)$ -- restricting ramification to $S$ corresponds to taking étale cohomology over the ring of integers with the primes in $S$ removed. It is not true that $A(\bQ)/p^n A(\bQ)$ injects into $\rH^1_{\rm et}(\Spec(\bZ), G_n)$: this group is often zero. However, $A(\bQ)/p^n A(\bQ)$ does inject into $\rH^1_{\rm fppf}(\Spec(\bZ), G_n)$, and this is the group we will use. (This is a slight lie that we will correct below.)

The idea is to show that this $\rH^1$ is bounded independent of $n$, which will establish that $A(\bQ)$ has rank 0. To do this, we need to understand the flat cohomology of $G_n$ very well, so we begin by studying groups like $G_n$ and their flat cohomology.

Admissible groups


Example. Let $A$ be an abelian variety with good reduction away from $N$ and let $\cA$ be its Néron model over $\bZ$. Then $\cA[p^n]$ is a pre-admissible group scheme over $\bZ$.

Let $G$ be an pre-admissible group over $\bZ[1/N]$. An admissible filtration on $G$ is a filtration

$$ 0=F^0G \subset F^1G \subset \cdots \subset F^nG=G $$

by closed subgroups such that $F^{n+1}G/F^nG$ is isomorphic to $\bZ/p\bZ$ or $\mu_p$ for each $n$. We say that $G$ is admissible if it has an admissible filtration. We say that a pre-admissible group over $\bZ$ is admissible if its restriction to $\bZ[1/N]$ is.

We make a similar definition for Galois representations. Precisely, let $V$ be a $\Gamma_{\bQ}$-module. We say that $V$ is admissible if it possess a filtration $F^{\bullet} V$ by $\Gamma_{\bQ}$-submodules such that $F^{n+1}V/F^nV$ is a one-dimensional $\bF_p$ vector space on which $\Gamma_{\bQ}$ acts either trivially or through the cyclotomic character.

Detecting admissible filtrations

Proposition. Let $G$ be a pre-admissible group over $\bZ[1/N]$. Then $G$ is admissible if and only if $G(\ol{\bQ})$ is.

Proof. Let $V \subset G(\ol{\bQ})$ be the first piece of an admissible filtration, let $H_0 \subset G_{\bQ}$ be the subgroup it corresponds to, and let $H$ be the closure of $H_0$ in $G$. Over $\bZ[1/Np]$, the group $H$ is finite étale, and therefore isomorphic to either $\mu_p$ or $\bZ/p\bZ$, depending on the Galois action on $H(\ol{\bQ})$. Over $\bZ_p$, the group $H$ is a finite flat commutative group which is generically isomorphic to $\mu_p$ or $\bZ/p\bZ$. If $p \ne 2$, Raynaud's theorem implies that the isomorphism extends over $\bZ_p$; the same is true for $p=2$ by a theorem of Fontaine. (Note: it is important here that we're over $\bQ_p$, so that there is no ramification.) Thus $H$ is isomorphic, over $\bZ[1/N]$, to $\mu_p$ or $\bZ/p\bZ$. Applying the same reasoning to $G/H$, the result follows by induction.


Let $G$ be an admissible group over $\bZ$. Following Mazur, we define several invariants:

Note that everything we're applying $\log_p$ to is a $p$th power.

Remark. Let $G$ be a group scheme over a base scheme $S$. The group $\rH^1_{\rm fppf}(S, G)$ admits a fairly concrete description, as follows. A torsor for $G$ is a scheme $T/S$ equipped with an action of $G$ that is simply transitive, in the following sense: for any scheme $S'/S$ and any section $x \in T(S')$, the map $G(S') \to T(S')$ given by $g \mapsto gx$ is a bijection. An fppf (or étale) torsor is a torsor $T/S$ for which there exists an fppf (or étale) cover $S' \to S$ such that $T(S')$ is non-empty. Then $\rH^1_{\rm fppf}(S, G)$ is naturally in bijection with the set of isomorphism classes of fppf torsors; similarly, $\rH^1_{\rm et}(S, G)$ is in bijection with the set of isomorphism classes of étale torsors. Of course, $\rH^0_{\rm fppf}(S, G)=\rH^0_{\rm et}(S, G)=G(S)$ is even easier to describe.

Elementary admissible groups

We say that an admissible group $G$ is elementary if $\ell(G)=1$. Over $\bZ[1/N]$, there are two elementary admissible groups: $\bZ/p\bZ$ and $\mu_p$. Recall the following result:

Proposition. Let $H$ be a pre-admissible group over $\bQ_N$. Then extensions of $H$ to a pre-admissible group over $\bZ_N$ correspond to unramified Galois submodules of $H(\ol{\bQ}_N)$. In particular, if $H(\ol{\bQ}_N)$ is unramified and one-dimensional, it admits exactly two such extensions: a finite one (corresponding to the full module) and the extension by zero $H^{\flat}$ (correspond to the zero submodule).

This applies to both $\bZ/p\bZ$ and $\mu_p$ over $\bZ[1/N]$ (even though we only stated the above result locally). We thus see that there are four elementary admissible groups over $\bZ$, namely: $\bZ/p\bZ$, $(\bZ/p\bZ)^{\flat}$, $\mu_p$, and $\mu_p^{\flat}$.

Proposition. The invariants of the elementary admissible groups are given as follows:

$$ \begin{array}{ccccc} & \bZ/p\bZ & (\bZ/p\bZ)^{\flat} & \mu_p & \mu_p^{\flat} \\ \hline \delta & 0 & 1 & 0 & 1 \\ \alpha & 1 & 1 & 0 & 0 \\ h^0 & 1 & 0 & \begin{array}{c} \scriptstyle \textrm{0 if $p$ odd} \\ \scriptstyle \textrm{1 if $p=2$} \end{array} & 0 \\ h^1 & 0 & 0 & \begin{array}{c} \scriptstyle \textrm{0 if $p$ odd} \\ \scriptstyle \textrm{1 if $p=2$} \end{array} & \epsilon \end{array} $$

Here $\epsilon$ is 0 if $p$ is odd and $N \not\equiv 1$ mod $p$, or if $p$ is even and $N=3 \pmod{4}$, and 1 otherwise.

Proof. The first three lines are obvious. We explain the fourth. For this proof, let $S=\Spec(Z)$. Since $\bZ/p\bZ$ is étale, $\rH^1_{\rm fppf}(S, \bZ/p\bZ) = \rH^1_{\rm et}(S, \bZ/p\bZ)$: if $T$ is an fppf torsor for $\bZ/p\bZ$ over $S$ then there is an fppf cover $S' \to S$ such that $T_{S'}$ is isomorphic to $(\bZ/p\bZ)_{S'}$, and thus étale; this implies that $T$ is étale since this is an fppf local property. Since $\bZ/p\bZ$ is a constant scheme, we have $\rH^1_{\rm et}(S, \bZ/p\bZ) = \Hom(\pi_1^{\rm et}(S), \bZ/p\bZ)$. However, $\pi_1^{\rm et}(S)$ is trivial: it is the Galois group of the maximal everywhere unramified extension of $\bQ$, which is just $\bQ$. Thus $h^1(\bZ/p\bZ)=0$.

Let $G$ be the quotient of $\bZ/p\bZ$ by $(\bZ/p\bZ)^{\flat}$; this is the push-forward of $\bZ/p\bZ$ along the inclusion $\Spec(\bF_N) \to \Spec(\bZ)$. By what we already have shown, there is a short exact sequence

$$ 0 \to \rH^0_{\rm fppf}(S, \bZ/p\bZ) \to \rH^0_{\rm fppf}(S, G) \to \rH^1_{\rm fppf}(S, (\bZ/p\bZ)^{\flat}) \to 0 $$

Both the $\rH^0$'s are $\bZ/p\bZ$, which shows $h^1((\bZ/p\bZ)^{\flat})=0$. We have the following short exact sequence on the fppf site of $S$:

$$ 0 \to \mu_p \to \bG_m \stackrel{p}{\to} \bG_m \to 0. $$

Taking cohomology, we obtain an exact sequence

$$ 0 \to \bZ^{\times}/(\bZ^{\times})^p \to \rH^1_{\rm fppf}(S, \mu_p) \to \rH^1_{\rm fppf}(S, \bG_m)[p] \to 0. $$

The group on the left is 0 if $p$ is odd, and $\bZ/p\bZ$ if $p=2$. By the theory of fppf descent, $\rH^1_{\rm fppf}(S, \bG_m)=\rH^1_{\rm Zar}(S, \bG_m)=\Pic(S)$, which is just the ideal class group of $\bQ$, which is trivial. Thus $h^1(\mu_p)$ is 1 if $p=2$ and 0 otherwise.

Let $G$ be the quotient of $\mu_p$ by $\mu_p^{\flat}$: this is the push-forward of $\mu_p$ along the inclusion $\Spec(\bF_N) \to \Spec(\bZ)$. We have an exact sequence

$$ 0 \to \rH^0_{\rm fppf}(S, \mu_p) \to \rH^0_{\rm fppf}(S, G) \to \rH^1_{\rm fppf}(S, \mu_p^{\flat}) \to \rH^1_{\rm fppf}(S, \mu_p) \to \rH^1_{\rm fppf}(S, G) $$

If $p \ne 2$, then $\rH^i_{\rm fppf}(S, \mu_p)=0$ for $i=0,1$, and so the map $\rH^0_{\rm fppf}(S, G) \to \rH^1_{\rm fppf}(S, \mu_p^{\flat})$ is an isomorphism. The source is just $\mu_p(\bF_N)$, which has order $p$ if $p \mid N-1$, and vanishes otherwise. Now suppose that $p$ is even. Then the map $\rH^0_{\rm fppf}(S, \mu_p) \to \rH^0_{\rm fppf}(S, G)$ is an isomorphism. Kummer theory shows that the unique non-trivial element of $\rH^1_{\rm fppf}(S, \mu_2)$ is represented by the $\mu_2$ torsor $\Spec(\bZ[\sqrt{-1}])$. This torsor splits over $\bF_N$ if and only if $-1$ is a square mod $N$. Thus the kernel of $\rH^1_{\rm fppf}(S, \mu_p) \to \rH^1_{\rm fppf}(S, G)$ is $\bZ/2\bZ$ if $N$ is 1 mod 4, and 0 if $N$ is 3 mod 4. This completes the proof.

Proposition. Let $G$ be an admissible group over $\bZ$. Then $h^1(G)-h^0(G) \le \delta(G)-\alpha(G)$.

Proof. Let

$$ 0 \to G_1 \to G_2 \to G_3 \to 0 $$

be a short exact sequence of admissible groups. From the first few terms of the long exact sequence in cohomology, we find

$$ h^1(G_2)-h^0(G_2) \le (h^1(G_1)-h^0(G_1)) + (h^1(G_3)-h^0(G_3)), $$

that is, $h^1-h^0$ is sub-additive. It is clear that

$$ \delta(G_2)-\alpha(G_2) = (\delta(G_1)-\alpha(G_1)) + (\delta(G_3)-\alpha(G_3)), $$

i.e., $\delta-\alpha$ is additive (in fact, both $\delta$ and $\alpha$ are additive separately). Thus if the result is true for $G_1$ and $G_3$ then it is true for $G_2$. It thus suffices to prove the result for elementary admissible groups, which follows easily from the computation of the invariants.

Proof of criterion

We now prove the main theorem. Let $\cA$ be the Néron model of $A$ over $\bZ$ and let $\cA^{\circ}$ be its identity component (i.e., throw out the non-identity components in each fiber). Let $G_n=\cA^{\circ}[p^n]$. Since $A$ has good reduction away from $N$, it is clear that $G_n$ is pre-admissible. The condition on $A[p](\ol{\bQ})$ exactly says that it is admissible, and so $A[p^n](\ol{\bQ})$ is as well, since it is an iterated self-extension of $A[p](\ol{\bQ})$. It follows that $G_n$ is admissible.

We now compute the invariants $\alpha$ and $\delta$ for $G_n$. We begin with $\delta$. We have $\ell(G_n)=2gn$, where $g=\dim(A)$. Now, $(G_n)_{\bF_N} = \cA^{\circ}_{\bF_N}[p^n]$. By hypothesis, $\cA_{\bF_N}^{\circ}$ is a torus of dimension $g$, and so its $p^n$ torsion has cardinality $p^{ng}$. Thus $(G_n)_{\bF_N}$ has cardinality $p^{gn}$. We thus find $\delta(G_n)=gn$.

We now compute $\alpha$. Since $\alpha$ is additive and only depends on the group over $\bZ[1/N]$, we see that $\alpha(G_n)=n\alpha(G_1)$, so it suffices to treat the $n=1$ case. Note that $\alpha(G_1)$ is the number of $\bZ/p\bZ$'s appearing in $(G_1)_{\bF_p}$; thus it is $\log_p$ of the order of the étale part of $\cA_{\bF_p}[p]$. Since $G_1$ is admissible, $\cA_{\bF_p}[p]$ has only $\bZ/p\bZ$'s and $\mu_p$'s in it, and so $\cA_{\bF_p}$ is ordinary. This implies that its étale part has order $p^g$, and so $\alpha(G_1)=g$. Thus $\alpha(G_n)=gn$.

We thus have $\delta(G_n)=gn$ and $\alpha(G_n)=gn$. It follows that $h^1(G_n)-h^0(G_n) \le 0$. However, $h^0(G_n)$ is the $p^n$ torsion in $\cA^{\circ}(\bZ) \subset \cA(\bZ)=A(\bQ)$, which is bounded independent of $n$ by the Mordell--Weil theorem. It follows that $h^1(G_n)$ is bounded independent of $n$.

Consider now the short exact sequence of sheaves on the fppf site of $\Spec(\bZ)$:

$$ 0 \to G_n \to \cA^{\circ} \stackrel{p^n}{\to} \cA^{\circ} \to 0. $$

A few remarks:

Taking cohomology, we obtain an injection

$$ \rH^0_{\rm fppf}(\Spec(\bZ), \cA^{\circ}) \otimes \bZ/p^n \bZ \to \rH^1_{\rm fppf}(\Spec(\bZ), G_n). $$

It follows that the cardinality of $\cA^{\circ}(\bZ) \otimes \bZ/p^n \bZ$ is bounded as $n \to \infty$.

Let $C$ be the $\bF_N$-points of the component group of $\cA_{\bF_N}$. Then there is an exact sequence

$$ 0 \to \cA^{\circ}(\bZ) \to \cA(\bZ) \to C $$

It follows that $\cA^{\circ}(\bZ)$ is a finite index subgroup of $\cA(\bZ)=A(\bQ)$. In particular, $\cA^{\circ}(\bZ)$ is finitely generated. Since the cardinality of $\cA^{\circ}(\bZ) \otimes \bZ/p^n \bZ$ is bounded, it follows that $\cA^{\circ}(\bZ)$ has rank 0. Thus $A(\bQ)$ has rank zero, as it contains $\cA^{\circ}(\bZ)$ with finite index.