Let E/Q_p$E/\mathbf{Q}_p$ be an elliptic curve having split multiplicative reduction. Then the Tate uniformization gives a surjective homomorphism of p$p$-adic analytic groups G_m \to E$G_m \to E$, with infinite cyclic kernel. Is there an analogue of this fact for E$E$ having nonsplit multiplicative reduction, perhaps replacing Gm with a nonsplit torus? E.g., can one uniformize E$E$ over the quadratic extension where the reduction splits, and then somehow descend?
(My intuition was as follows. Take E/Q_p$E/\mathbf{Q}_p$ with nonsplit multiplicative reduction, and let K/Q_p$K/\mathbf{Q}_p$ be quadratic so that E$E$ becomes split semistable over K$K$, and let E'$E'$ be the K$K$-twist of E$E$ (which has split multiplicative reduction). Then one has a short exact sequence
0 \to Z \to G_m \to E' \to 0$$0 \to Z \to \mathbf{G}_m \to E' \to 0$$
(where Z$Z$ is the constant analytic group of integers). Extending scalars to K$K$ then applying Weil restriction of scalars, we get
0 \to X \to T \to A \to 0,$$0 \to X \to T \to A \to 0$$
where X$X$ is an etale-locally-constant analytic group, T$T$ is a torus, and A$A$ is an abelian variety, each of rank 2$2$ 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$0 \to Z' \to \mathbf{G}_m' \to E \to 0$,
where ' still denotes twisting by K/Q_p$K/\mathbf{Q}_p$. Since Z'$Z'$ has trivial Q_p$\mathbf{Q}_p$-points, then, one should have something like G_m'(Q_p) = E(Q_p)$\mathbf{G}_m'(\mathbf{Q}_p) = E(\mathbf{Q}_p)$, modulo any descent used in forming the quotient.
Does this sound sensical?
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