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Is there a simple description of the group Q_p*/(Q_p*)^2 where Q_p denotes the p-adic integers?

I am doing descent calculations for elliptic curves, and so am most interested in the case p = 2. However, I would also like to know the answer for other p.

I am trying to understand the behavior of the Kummer map

k: E(Q)/2E(Q) ---> Q*/(Q*)^2 x Q*/(Q*)^2

and what happens when we pass from k into a local field

k_p: E(Q_p)/2E(Q_p) ---> Q_p*/(Q_p*)^2 x Q_p*/(Q_p*)^2

In particular, I would like to know a way to compute the kernel of the map Q*/(Q*)^2 ---> Q_p*/(Q_p*)^2 for small p. I am hoping to use this kernel to deduce facts about im(K) (namely, the rank) from knowledge of im(K_p).

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I set this sort of question as homework. Here are some hints. $p=2$ is the hardest case. The surjection $v:Q_p^*\to Z$ splits (choose a uniformiser) so $Q_p^*=Z\times Z_p^*$. If $p$ is odd then p-adic log and exp identify $1+pZ_p$ with $pZ_p$ as topological groups. If $p=2$ then you identify $1+4Z_2$ with $4Z_2$ the same way. Now put everything together. For $p$ odd the group has order 4, generated by a non-residue and a uniformiser. For $p=2$ it has order 8. Finally, the kernel you want to compute is gigantic because the LHS is infinite-dimensional as a vector space over Z/2 and the RHSisfini –  Kevin Buzzard Jun 9 '11 at 23:17

1 Answer 1

Kevin has answered your question in comments, but it might help to make some further remarks:

If $p$ is odd and $E$ has good reduction at $p$, then the image of $K_p$ is independent of $E$ (i.e. does not depend on the particular $E$ other then requiring that it has good reduction). To be precise, the image will be $\mathbb Z_p^{\times}/(\mathbb Z_p^{\times})^2 \times \mathbb Z_p^{\times}/(\mathbb Z_p^{\times})^2.$

Thus I don't think that there is much chance that you will be able to extract any information about the global elliptic curve $E$ from knowing the image of $K_p$. (Even if $p = 2$ and/or the reduction is bad, there is very little information specific to $E$ in the image of $K_p$; it will just depend on generalities about the reduction type of $E$.)

Have you looked at the discussion of descent and Selmer groups in (e.g.) Silverman's book? If you do, you'll see that the problem of doing descent involves looking at every prime (the point being that every prime intervenes in the definition of the Selmer group).

Concretely, in the case you are looking at, which I guess is an $E$ all of whose $2$-torsion is defined over $\mathbb Q$, what one sees is that if $P \in E(\mathbb Q)$, and we solve $P = 2Q$, then $Q$ is defined over a biquadratic extension of $\mathbb Q$ which is unramified away from $2$ and the primes of bad reduction. This greatly limits the possibilities, and is the basis for why descent works. In particular, the weak Mordell--Weil theorem --- i.e. the statement that $E(\mathbb Q)/2 E(\mathbb Q)$ is finite --- in this context is essentially the statement that there are only finitely many biquadratic extensions of $\mathbb Q$ with prescribed ramification.

If you focus on just one (or even a finite number) of $p$, you are throwing away this basic fact (i.e. that the field of definition of $Q$ is unramified away from finitely many primes).

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