Q_p*/(Q_p*)^2 and descent for elliptic curves

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).

• 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

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 $E$.)
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).