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I have a, probably very simple, question: My intuition tells me that the following statement should be true, but I couldn't find it anywhere and I wanted to make sure I am not missing something.

Let $\pi:Y\to X$ be a finite, étale morphism of nonsingular varieties over some algebraically closed field $\Bbbk$. Is it true that every point $P\in X$ has an affine neighbourhood $U$ such that $\pi^{-1}(U)$ consists of $\deg(\pi)$ irreducible components, each of which is isomorphic to $U$ via $\pi$?

Of course, if it is true, I would also be happy if you could provide a proof, in literature or otherwise.

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No, it's not true. Consider the map $x\mapsto x^2$ as a map from $X=\mathbb A^1-\{0\}$ to itself. The "problem" is that Zariski neighborhoods are too big. Any open subset of $X$ has exactly one irreducible component (in general, an open subset cannot have more components than the ambient space), so there is no hope to get the preimage of an open set to have two components.

However, if you refine your topology to allow "etale open neighborhoods" (i.e. you allow pullback by etale maps $U\to X$, not just open immersions $U\to X$), then the answer is yes. Perhaps the easiest way to prove that is to pull back by $\pi$ itself, after which you can "peel off" the diagonal component of $Y\times_X Y$. Now you have a finite etale map $(Y\times_X Y -\Delta)\to Y$ which has degree one less. Repeat until the degree is $1$, at which point you have an etale map $U\to \cdots \to Y\to X$ so that $Y\times_X U$ is $deg(\pi)$ copies of $U$.

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    $\begingroup$ Another way to see the last fact is to note that finite etale covers of a strictly henselian ring are split, together with the fact that the strict henselianization at a point is the "limit" of etale neighborhoods (and a "noetherian descent" argument to argue that the splitting must descend to some etale neighborhood after all). $\endgroup$
    – Akhil Mathew
    Commented Aug 8, 2011 at 16:26
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    $\begingroup$ It's worth pointing out that it's a general philosophy that things that you expect to be true because of your experience with the analytic topology are usually true in the etale topology. $\endgroup$
    – Anton Geraschenko
    Commented Aug 8, 2011 at 16:38
  • $\begingroup$ @Anton Geraschenko:How to see an finite etale morphism of degree one is an isomorphism? I think you use this fact in your answer. $\endgroup$
    – Xiaobo Zhuang
    Commented Jan 2, 2013 at 2:30
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    $\begingroup$ @ZhuangXiaobo: Let's work locally, so that the map is $Spec(S)\to Spec(R)$. By definition, "degree $n$" means that the corresponding ring homomorphism $R\to S$ makes $S$ into a free module of rank $n$ over $R$. If $n=1$, the map is bijective, so an isomorphism of rings. $\endgroup$
    – Anton Geraschenko
    Commented Jan 2, 2013 at 14:58
  • $\begingroup$ @Anton Geraschenko Is the finite etale map $(Y\times_X Y -\Delta)\to Y$ surjective (if $deg(\pi)>1$)? I think you need this to see that the $U$ you get in the end turns out to be an etale cover of $X$. I'm unable to show this. $\endgroup$
    – Lao-tzu
    Commented Apr 9, 2020 at 14:36
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This is almost never true. For example if $X,Y$ are irreducible, then for any (affine) Zariski open $U$ of $X$, the inverse image is open in $Y$ and hence irreducible.

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