What is the structure of the group of rational points of an abelian variety over a Laurent series field?

Let $K = \mathbb{F}_q((t))$, and let $A_{/K}$ be a nontrivial abelian variety. Then $A(K)$ is a compact $K$-adic Lie group. What can be said about its structure?

By way of comparison, if $K/\mathbb{Q}_p$ is an extension of degree $d$ and $A$ has dimension $g$, then $p$-adic Lie theory shows that $A(K) \cong \mathbb{Z}_p^{dg} \oplus T$, where $T$ is a finite group. I am looking for a similar description in the positive characteristic case.

I've wondered about this off and on for years but all of a sudden I have a good reason to know: in particular, I would like to know the structure of $A(K)/p^aA(K)$, which I suspect is always an infinite group of exponent $p^a$. This is the case for e.g. $p$-adically uniformized abelian varieties, unless I am very much mistaken.

In particular, in the elliptic curve case it would be nice if the height of the formal group told the majority of the story, in the sense that e.g. if $E_1$ and $E_2$ were any two ordinary elliptic curves over $K$, then $E_1(K)$ and $E_2(K)$ would admit isomorphic finite-index subgroups. Is this true?

Added: After seeing Professor Lubin's answer I can be more precise. I would like a proof that $A(K) \cong \left(\prod_{i=1}^{\infty} \mathbb{Z}_p \right) \oplus T$, where $T$ is a finite group.

• Since $A(K)$ is profinite separable and pro-$p$ near 0 with finite $p$-torsion, a small compact open subgroup $U$ in $A(K)$ has $U[p]=1$, so $U$ is a countable direct product of copies of $\mathbf{Z}_p$. It is a countably infinite product if $A(K)/(p)$ is infinite. Suppose $A(K)/(p)$ is finite, so $p\cdot A(K)$ is open in $A(K)$. In the formal group of the Neron model $N$ over $R$ with characteristic $p$, $[p]=V_N \circ F_N$. Thus, $[p]^{\ast}(X_j) \in R[[X_1^p,\dots,X_g^p]]$ for all $j$, so openness is impossible by measure-theoretic reasons. Thus, $U=\prod_{n=1}^{\infty}\mathbf{Z}_p$. Nov 11, 2013 at 6:19
• I should have mentioned that in 4.3 in Expose VII$_{\rm{A}}$ in SGA3 there is a discussion and construction of the relative Verscheibung homomorphism (and associated factorization of $[p]$ as $V \circ F$) for any flat commutative group scheme over an $\mathbf{F}_p$-scheme. This yields what I claimed at the level of formal groups above, when applied to the smooth Neron model over $R$. Nov 11, 2013 at 6:28

I’m feeling fairly fuzzy on this, but it seems to me that everything should be just what you expect, and without regard to the particular shape of the formal group. I’ll just deal with the case of elliptic curves over $R=k[[t]]$, and for the constant field $k=\mathbb F_p$. I’m sure you can do the mutatis mutandis.

We always have the exact sequence $$0\longrightarrow \hat E(R)\longrightarrow E(K)\longrightarrow E(k)\longrightarrow0\,,$$ and I think you’re just asking about the shape of $\hat E(R)$, by which I mean the points of the formal group $\hat E$ of $E$ with values in $R$, and these have to be elements of the maximal ideal $tR$.

Seems to me that in the ordinary case, the story is exactly the same as for the group $1+tR$, the $R$-valued points of the formal multiplicative group. You get one more generator for each exponent prime to $p$, so the group has structure $$\prod_iZ_i\,,$$ where each group $Z_i$ is a free $\mathbb Z_p$ module of rank one, generated by $t^i$. Remember that for $\alpha\in\mathbb Z_p$, $\alpha\cdot t^i=[\alpha](t^i)$, where the right-hand side refers to the $\alpha$-endomorphism of the formal group. And in the case of height one (ordinary), the indices $i$ are not all positive $i$, but just those prime to $p$.

I’m a little nervous about the height-two (supersingular) case, but since in the case $k=\mathbb F_p$, the endomorphisms of the formal group are exactly $\mathbb Z_p[\pi]$, where $\pi$ is the Frobenius endomorphism $x^p$, it’s not too bad there. In this case, the points of the formal group seem to be $$\prod_iY_i\,,$$ where each $Y_i$ is a free $\mathbb Z_p$-module of rank two, with basis elements $t^i$ and $\pi(t^i)=t^{pi}$. Same indices as before.

It seems to me that the moral of the story is that in the category of topological groups, the points of the formal group are always the same.

• Thanks for this quick answer. I agree: the answer should be the direct product of countably infinitely many copies of $\mathbb{Z}_p$ in all cases. I wonder what is a good method to prove that? In general, I may need a bit more help on the proofs of these statements, but I'll think about it for a bit and get back to you if necessary. Nov 11, 2013 at 5:59
• Well, in fact there can be some torsion in the formal group -- e.g. suppose $E$ is ordinary and has supersingular reduction -- so what you write may not literally be true (if I am understanding correctly). However, I know that if you go up high enough in the filtration the torsion subgroup becomes trivial (I believe an old paper of mine, with Xavier Xarles, contains a proof of this), and the problem is to show that as soon as you restrict to a torsionfree finite index subgroup it is isomorphic to a product of $\mathbb{Z}_p$'s. But I'm a bit rusty on the techniques necessary to show this... Nov 11, 2013 at 6:05
• @PeteL.Clark: If $G$ is pro-$p$ commutative with trivial $p$-torsion then its Pontryagin dual is $p$-divisible and consists of $p$-power torsion, so it is a divisible module over $\mathbf{Z}_{(p)}$ and hence is an injective module. Thus, as over any noetherian ring (see 18.5 in Matsumura's "Commutative ring theory"), it is a direct sum of copies of the injective modules associated to prime ideals; i.e., copies of $\mathbf{Q}/\mathbf{Z}_{(p)}$ or $\mathbf{Q}$. (I know, overkill.) None of the latter by the $p$-power torsion condition. Dualizing back turns direct sum into direct product, etc. Nov 11, 2013 at 6:53
• @Marguax: thanks for all three of your comments! Your last one helped me understand your first one. I may have one or two further questions later on, but what you've said is already extremely helpful. Nov 11, 2013 at 7:02
• @PeteL.Clark, “some torsion in the formal group”, yes you’re absolutely right. This becomes clear when you look at the Newton polygon of $[p]$, and its finitenes drops right out from that too. Nov 11, 2013 at 12:21