1
$\begingroup$

Suppose that $f : X \rightarrow Y$ is a smooth (or even étale) surjective morphism over a field $k$ to a scheme $Y$ of finite type over $k$. I want to show that $X$ is locally integral, i.e. (in the Noetherian case) disjoint union of integral schemes.

I made some progress: certainly $X$ is reduced, so it is enough to show that $X$ is locally irreducible. If $Y$ is normal, by a well known result, $X$ is normal, hence the result holds. More generally, if $Y$ is unibranch, then $X$ is unibranch (https://stacks.math.columbia.edu/tag/0DQ1). This implies that $X$ is locally irreducible.

What about the remaining case? For instance, if $Y$ is the node. Any help is appreciated.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ The node is not etale locally irreducible. So the answer to your question is negative outside the geometrically unibranch case. $\endgroup$
    – Piotr Achinger
    Commented Nov 11, 2022 at 15:30
  • $\begingroup$ Are you saying that there is an étale morphism $f : X \rightarrow Y$, where $Y$ is the node, such that $X$ connected, yet reducible? If so, is it trivial to construct it? Thank you $\endgroup$
    – ofiz
    Commented Nov 11, 2022 at 16:01
  • 3
    $\begingroup$ See Hartshorne "Algebraic Geometry" Chapter III, Exercise 10.6 $\endgroup$
    – Piotr Achinger
    Commented Nov 11, 2022 at 16:05
  • 3
    $\begingroup$ In fact Stacks Project, Lemma 06DM implies that "geometrically unibranch" is equivalent to "etale locally irreducible". $\endgroup$
    – Piotr Achinger
    Commented Nov 11, 2022 at 16:09

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .