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?

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