Finite, Étale Morphism Of Varieties - MathOverflow

Finite, Étale Morphism Of Varieties

2011-08-08T15:45:47Z

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.

Answer by Anton Geraschenko for Finite, Étale Morphism Of Varieties

2011-08-08T16:00:16Z

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 $U$.

Answer by Mohan for Finite, Étale Morphism Of Varieties

2011-08-08T16:01:21Z

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.