Q_p*/(Q_p*)^2 and descent for elliptic curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T12:47:36Z http://mathoverflow.net/feeds/question/67393 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/67393/q-p-q-p2-and-descent-for-elliptic-curves Q_p*/(Q_p*)^2 and descent for elliptic curves unknown (yahoo) 2011-06-09T23:10:57Z 2011-06-10T01:04:11Z <p>Is there a simple description of the group Q_p*/(Q_p*)^2 where Q_p denotes the p-adic integers?</p> <p>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.</p> <p>I am trying to understand the behavior of the Kummer map</p> <p>k: E(Q)/2E(Q) ---> Q*/(Q*)^2 x Q*/(Q*)^2</p> <p>and what happens when we pass from k into a local field</p> <p>k_p: E(Q_p)/2E(Q_p) ---> Q_p*/(Q_p*)^2 x Q_p*/(Q_p*)^2</p> <p>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).</p> http://mathoverflow.net/questions/67393/q-p-q-p2-and-descent-for-elliptic-curves/67397#67397 Answer by Emerton for Q_p*/(Q_p*)^2 and descent for elliptic curves Emerton 2011-06-10T00:05:08Z 2011-06-10T01:04:11Z <p>Kevin has answered your question in comments, but it might help to make some further remarks:</p> <p>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 $\mathbb Z_p^{\times}/(\mathbb Z_p^{\times})2 \times \mathbb Z_p^{\times}/(\mathbb Z_p^{\times})^2.$ </p> <p>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 $E$.)</p> <p>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 <em>every</em> prime (the point being that every prime intervenes in the definition of the Selmer group). </p> <p>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 $Q = 2P$, 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. </p> <p>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). </p>