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All varieties are assumed to be projective over $\mathbb{C}$. Let $f_1: Y \to X$ and $f_2: Y' \to X$ be étale morphisms with same finite Galois groups (to be honest, I don't know what does Galois group really mean in this context), then can we conclude $Y,Y'$ are birational?

Even if the general situation might have negative answer, are there any results concerning problemss along this line?

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    $\begingroup$ The answer for the general question is no: take an elliptic curve $E$ and a subgroup $G$ of $E$ isomorphic to $\mathbb{Z}/n\mathbb{Z}$, let $E'=E/G$. Then you have cyclic degree $n$ covers $E\to E'$ but also $E'\to E$ (by killing the remaining part of the $n$-torsion), but $E'$ will almost never be birational (=isomorphic) to $E$. $\endgroup$ Sep 23, 2013 at 14:45
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    $\begingroup$ @PiotrAchinger You are not actually answering the question as the OP fixes the target. $\endgroup$ Sep 23, 2013 at 15:54
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    $\begingroup$ Suppose for simplicity that $X$ is normal. Hence $Y, Y'$ are normal. Then $Y$ birational to $Y'$ implies $Y$ isomorphic to $Y'$ because both are then the normalization of $X$ in the function field of $Y$. $\endgroup$
    – Cantlog
    Sep 23, 2013 at 20:30

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I think, for an elliptic curve $E/\mathbf{C}$, there are étale coverings $E_i \to E$, $i = 1,2$ of degree a prime number (hence with the same Galois group; the étale fundamental group of an elliptic curve is Abelian), with $E_1 \not\cong E_2$.

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    $\begingroup$ For most elliptic curves $E$, all degree $p$ etale covers of $E$ (as in your example) are distinct. This follows (is equivalent to in fact) to the irreducibility of the modular curve $X_0(p)$. $\endgroup$ Sep 23, 2013 at 15:55
  • $\begingroup$ What am i missing ? On a birational level, doesn't the question amount to- given a field $K$ and two Galois extensions, $L$ and $M$ with the same Galois group asking if $L$ and $M$ are isomorphic. In particular aren't all elliptic curves double covers of $P^1$ and hence counter examples ? $\endgroup$
    – meh
    Sep 23, 2013 at 18:47
  • $\begingroup$ @aginensky The coverings of $P^1$ you are speaking about are not étale. $\endgroup$
    – Sasha
    Sep 23, 2013 at 20:02
  • $\begingroup$ @Timo Keller Thank you! How to construct an etale morphism between elliptic curves? I usually think of etale cover as a covering space, but this does not seem to be a correct intuition. $\endgroup$
    – Li Yutong
    Sep 24, 2013 at 0:08

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