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I remember hearing about this statement once, but cannot remember where or when. If it is true i could make good use of it.

Let $\pi: X \rightarrow Y$ be an étale map of (irreducible) algebraic varieties and let $Z \subset Y$ be an irreducible subvariety.

Does it follow that $\pi^{-1}(Z)$ is irreducible? If so, why? If not, do you know a counterexample?

If necessary $X$ and $Y$ can be surfaces over $\mathbb{C}$, the map $\pi$ of degree two, and $Z$ a hyperplane section (i.e. it defines a very ample line bundle).

Thanks!

Edit: I assume the varieties $X$ and $Y$ to be projective.

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I just realize that it will suffice for my purposes if we could show that $\pi^{-1}(Z)$ is connected! So any answer to this question would also be great. –  Joachim Jan 23 '13 at 15:47
    
If you take the analogy that etale maps act like covering spaces, shouldn't you expect the opposite? For example $Spec(\mathbb{C}\times\mathbb{C})=Spec(\mathbb{C})\coprod Spec(\mathbb{C})\to Spec(\mathbb{C})$ is etale...ah, I just read the hypothesis that you want $X$ to be irreducible... –  Matt Jan 23 '13 at 17:34
    
For smooth projective surfaces, as $Z$ is ample, $\pi^*Z$ is ample so is connected. It will be irreducible if $Z$ is smooth. –  user18119 Jan 25 '13 at 8:27
    
Hi @QiL, that is great, exactly what i needed, thanks! Could you maybe give a reference? Or is it possible to explain quickly why an ample divisor needs to be connected on a surface (and not on a curve)? –  Joachim Jan 25 '13 at 18:23
    
@Joachim: See Harthsorne, Corollary III.7.9 for a general statement for projective normal variety in dimension $\ge 2$. For smooth projective surface, I think one can Hodge index theorem (I did'nt check). –  user18119 Jan 25 '13 at 22:40
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1 Answer

up vote 3 down vote accepted

Hmm...what about $\mathbb{A}^1 - 0 \rightarrow \mathbb{A}^1$ - 0, with $z \mapsto z^2$? Then the preimage of 1 is $\pm 1$, which is not irreducible.

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Thanks. But i forgot to mention that $X$ and $Y$ should be projective. Taking the projective closure of your example, this shows that ramification points, hence unfortunately it is not étale. Thanks for pointing this out though, i should of course have said this straight away. –  Joachim Jan 24 '13 at 0:42
    
But, it does lead to the right answer, since there are unramified 2:1 coverings between curves of different genus, and then a point has two points in the preimage... Sigh, i should ask my advisor again what she meant.. Thanks anyway! =) –  Joachim Jan 24 '13 at 0:44
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