## Tate uniformization of nonsplit semistable elliptic curves

Let E/Q_p be an elliptic curve having split multiplicative reduction. Then the Tate uniformization gives a surjective homomorphism of p-adic analytic groups G_m \to E, with infinite cyclic kernel. Is there an analogue of this fact for E having nonsplit multiplicative reduction, perhaps replacing Gm with a nonsplit torus? E.g., can one uniformize E over the quadratic extension where the reduction splits, and then somehow descend?

(My intuition was as follows. Take E/Q_p with nonsplit multiplicative reduction, and let K/Q_p be quadratic so that E becomes split semistable over K, and let E' be the K-twist of E (which has split multiplicative reduction). Then one has a short exact sequence

0 \to Z \to G_m \to E' \to 0

(where Z is the constant analytic group of integers). Extending scalars to K then applying Weil restriction of scalars, we get

0 \to X \to T \to A \to 0,

where X is an etale-locally-constant analytic group, T is a torus, and A is an abelian variety, each of rank 2 in the appropriate sense. The latter short exact sequence contains the former short exact sequence as a sub (direct factor?); the quotient sequence should be something like

0 \to Z' \to Gm' \to E \to 0,

where ' still denotes twisting by K/Q_p. Since Z' has trivial Q_p-points, then, one should have something like G_m'(Q_p) = E(Q_p), modulo any descent used in forming the quotient.

Does this sound sensical?

-
 I have nothing sensical to say, except that I'm not sure if you have the permissions set up correctly for your wave - at least, I do have access, but your link doesn't work for me. – TG Nov 3 2009 at 22:07 I think I fixed it. Try again. – Jay Pottharst Nov 4 2009 at 1:50

A form of this is contained in Silverman, second book, Chapter V, Corollary 5.4. I guess that the image of Gm' in E (at the level of Q_p-points) may have index 2.

-
 Thanks for the reference. I'd love to see Silverman's statement presented as a presentation of analytic groups, though. Any thoughts on that? (BTW, re: the possible index 2, see Scott Carnahan's comment about Galois cohomology.) – Jay Pottharst Nov 3 2009 at 18:55 I'll give this answer the credit because of the explicit citation, but really both this and Scott's were helpful. The better approach than what I wrote above is to take the entire S.E.S. 0 \to Z \to G_m \to E' \to 0 and twist it by K/Q_p (avoiding the extension+restriction of scalars) to get 0 \to Z' \to G_m' \to E \to 0. (Twisting of Z,G_m is done exactly as it is for E.) Taking Galois cohomology and computing, one gets 0 \to G_m'(Q_p) \to E(Q_p) \to Z/2 \to 0, and if one wants to know E(Q_p) one is left with explicitly describing G_m'(Q_p), e.g. via the kernel of the norm – Jay Pottharst Nov 4 2009 at 14:01

Most of it makes sense. Elliptic curves with non-split reduction can be analytically uniformized by the norm torus. There is a "nice" picture of this using the Berkovich spectrum for a non-split torus. I have my doubts about the statement concerning rational points - you should have a Galois cohomology exact sequence.

-
 Re: your doubts, I'll lump those under "modulo descent". – Jay Pottharst Nov 3 2009 at 18:45 The revised version of that sentence is much clearer, thanks. Unfortunately, I still get a headache whenever the word "descent" is used. – S. Carnahan♦ Nov 3 2009 at 19:42