TITLE: What is the structure of the group of rational points of an abelian variety over a Laurent series field?
QUESTION [9 upvotes]: 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.

REPLY [5 votes]: 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.