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Let $X$ be an irreducible scheme and $Y \to X$ a finite étale morphism. Is there some finite étale cover $Z \to X$ which trivializes $Y$ (i.e. $Y \times_X Z$ is a union of copies of $Z$) such that $Z$ is also irreducible?

I already know that the corresponding statement for "connected" instead of "irreducible" is true.

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up vote 3 down vote accepted

This is not always possible.

Suppose you have such a trivialization and suppose $Y$ is connected. Composing with the projection $Y\times_X Z\to Y$, you get a morphism of $X$-schemes $Z\to Y$. It is necessarily étale because $Z, Y$ are étale (SGA 1, Exposé I, cor. 4.8). So $Z\to Y$ is finite and open (by flatness). Its image is thus a connected component of $Y$, hence equal to $Y$. Therefore $Y$ itself is irreducible.

Now there are examples of reducible finite étale covers of irreducible schemes. Let $Y$ be the union of two copies of $\mathbb A^1$ meeting transversally at $0, 1$. Let $\sigma$ be the involution on $Y$ sending $t$ of the first component to $1-t$ on the second component and vice-versa. The action is free and we have a finite étale cover $Y\to X:=Y/\langle \sigma \rangle$ = $\mathbb A^1$. Obviously $X$ is irreducible but not $Y$.

Edit More intuitively, $\sigma$ permutes the components of $Y$ and also permutes the two intersection points.

Edit 2 Sorry, I didn't write what I had in mind ! The quotient of $Y$ by the involution is the affine line with $0$ identified with $1$ (hence a nodal rational curve). This example appears as the reduction of some étale double cover of an elliptic curve with multiplicative reduction.

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Thank you! I have some questions: 1) Why not taking $0,0$ as the points of intersection, i.e. $Y=V(x*y)$? This should not change anything. 2) Why is $Y \to Y/\langle \sigma \rangle$ etale? I have checked it by writing down the corresponding algebra, but somehow it turns out to be not separable. Perhaps because of 1). 3) Actually my question is motivated by another question, may I send you an e-mail about this? – Martin Brandenburg Nov 26 '12 at 11:23
@MartinBrandenburg, yes email-me. – user18119 Nov 26 '12 at 12:45
(1): if you take the union of axes, I don't know how to make it an étale cover of an irreducible curve. (2) If a finite group acts freely (all stabilizers are trivial), then the quotient morphism is étale. – user18119 Nov 26 '12 at 13:32
Ah, thanks a lot! And this is even locally trivial w.r.t. to the Zariski topology, i.a. a covering in the topological sense, right? – Martin Brandenburg Nov 26 '12 at 18:55
No it is not Zariski locally trivial as above the (integral) node, the two points are non-integral. – user18119 Nov 26 '12 at 20:46

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