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Let $f \colon X \to Y$ be an ├ętale morphism of schemes.
We know:
(1) if $Y$ is normal, then $X$ is normal.
(2) if $Y$ is regular, then $X$ is regular.
(3) if $Y$ is reduced, then $X$ is reduced.

Question: If $Y$ is integral, then is $X$ integral?

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    $\begingroup$ This question would be more suitable on Math.SE. (Hint: the properties (1)-(3) are local, but integral isn't.) $\endgroup$ – user5117 Jun 5 '14 at 11:21
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the answer is no for silly reasons: take X to be the disjoint union of two copies of Y.

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    $\begingroup$ right, I gave a somewhat more difficult example :-) $\endgroup$ – Francesco Polizzi Jun 5 '14 at 11:41
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    $\begingroup$ I think we were writing at the same time. Your answer is only testament to the fact that you're a proper algebraic geometer (where proper is meant in the non-technical sense -- you might not even be quasi-compact for all I know). $\endgroup$ – bananastack Jun 5 '14 at 11:47
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    $\begingroup$ I would even have taken the sum of zero copies of $Y$. $\endgroup$ – ACL Jun 5 '14 at 15:31
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    $\begingroup$ @ACL: that is just superb! But wait, is Spec{0} not irreducible? $\endgroup$ – bananastack Jun 5 '14 at 16:19
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    $\begingroup$ @user125763: No it's not! The empty topological space is not irreducible, but some say it is connected, and the zero ring is not integral. $\endgroup$ – ACL Jun 6 '14 at 19:14
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Here is an example where $X$ is not integral but connected.

Let $Y$ be the plane nodal curve $y^2=x^2(x+1)$. Then $Y$ admits an ├ętale cover $f \colon X \to Y$ of degree $2$, where $X$ is the union of the two copies of the normalization of $Y$ (see [Hartshorne, Algebraic Geometry], Exercise 10.6 page 276). These two components intersect in two points, both mapped to the node by $f$.

In particular $X$ is not an integral scheme, so the answer to your question is no.

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    $\begingroup$ Dear Francesco, this is a nice example, but $Y \sqcup Y \rightarrow Y$ (for any $Y$) seems a little simpler. $\endgroup$ – user5117 Jun 5 '14 at 11:41
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    $\begingroup$ Dear Francesco, I envy your mental processes! $\endgroup$ – user5117 Jun 5 '14 at 11:44
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    $\begingroup$ Dear Francesco and Artie, Thank you for nice examples. $\endgroup$ – user51649 Jun 5 '14 at 12:04
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    $\begingroup$ $X$ is not a disjoint union; it has a single connected component, but two irreducible components. $\endgroup$ – Ben Wieland Jun 6 '14 at 0:58
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    $\begingroup$ @Ben:of course you are right (actually, this is what makes this example interesting). I corrected the answer, thank you. $\endgroup$ – Francesco Polizzi Jun 6 '14 at 5:57

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