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Let $X,Y$ be two elliptic curves over $K:=K(R)$ which have good models over $R$, Char($K$)=$0$, where $R$ complete DVR with algebraically residue field $k$.

If $L$ is a finite extension of $K$, such that $X \otimes L \cong Y \otimes L$.

My question is:

Is there $X \cong Y$ ?

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up vote 9 down vote accepted

I assume that by "good models" you mean models $\mathscr{X}$, $\mathscr{Y}$ which are elliptic $R$-curves. These are pointed curves of genus 1, and in particular stable curves. This implies that the sheaf $$I:=\underline{\rm Isom}_{\text{$R$-ell. curves}}(\mathscr{X},\mathscr{Y})$$ is a scheme, finite and unramified over $R$. With your assumptions on $R$ (strictly henselian is enough), $I$ is then a finite disjoint sum of copies of closed subschemes of $R$. In particular, for every extension $L$ of $K$ (finite or not), we have $I(R)=I(L)$, which gives the result.

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Is there some obvious Galois cohomology way to see this? Since twists are parametrized by $H^1(G_{\overline{K}/K}, Aut(X))$ doesn't this say that this group is trivial? – Matt May 1 '13 at 18:04
It is not trivial since there are always quadratic twists (besides, is not a group!). The point here is that any nontrivial twist of $X$ must have bad reduction. – Laurent Moret-Bailly May 2 '13 at 9:26
Thank you!! That was exactly the point I was missing. – Matt May 2 '13 at 16:27

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