# Tate uniformization of nonsplit semistable elliptic curves

Let $E/\mathbf{Q}_p$ be an elliptic curve having split multiplicative reduction. Then 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 $E/\mathbf{Q}_p$ with nonsplit multiplicative reduction, and let $K/\mathbf{Q}_p$ 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

$$0 \to Z \to \mathbf{G}_m \to E' \to 0$$

(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

$0 \to Z' \to \mathbf{G}_m' \to E \to 0$,

where ' still denotes twisting by $K/\mathbf{Q}_p$. Since $Z'$ has trivial $\mathbf{Q}_p$-points, then, one should have something like $\mathbf{G}_m'(\mathbf{Q}_p) = E(\mathbf{Q}_p)$, 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. – TSG Nov 3 '09 at 22:07
• I think I fixed it. Try again. – Jay Nov 4 '09 at 1:50

• 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 Nov 4 '09 at 14:01