Timeline for What is the structure of the group of rational points of an abelian variety over a Laurent series field?
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| Nov 11, 2013 at 12:21 | comment | added | Lubin | @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 7:02 | comment | added | Pete L. Clark | @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 6:53 | comment | added | Marguax | @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 6:05 | comment | added | Pete L. Clark | 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 5:59 | comment | added | Pete L. Clark | 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 5:10 | history | answered | Lubin | CC BY-SA 3.0 |