5
$\begingroup$

We all know that smooth morphisms have sections etale locally. However, the following similar statement is not obvious for me:

If X->Y->Z, X is etale over Y, Y is finite and surjective over Z, then a section of X->Y exists etale locally on Z, i.e. there exists an etale cover U of Z such that X_U->Y_U has a section. Where _U means pullback on U.

I think it is supposed to be easy.

Can anyone explain this to me? Thanks.

$\endgroup$
2
  • 7
    $\begingroup$ First reduce to the case when $Y$ is also finitely presented over $Z$. Then you can use limit arguments, so it becomes an easy application of basic facts about strictly henselian local rings and finite algebras over them. It will be instructive for you to think about it some more for yourself in view of these hints. $\endgroup$
    – BCnrd
    Mar 12, 2010 at 8:06
  • 4
    $\begingroup$ I should have also added that you must be assuming $X \rightarrow Y$ is surjective, or else it clearly isn't true. That condition enters into the argument when doing analysis of the situation with strictly henselian local rings. $\endgroup$
    – BCnrd
    Mar 12, 2010 at 17:14

1 Answer 1

3
$\begingroup$

Brian Conrad says: "First reduce to the case when Y is also finitely presented over Z. Then you can use limit arguments, so it becomes an easy application of basic facts about strictly henselian local rings and finite algebras over them. It will be instructive for you to think about it some more for yourself in view of these hints.

I should have also added that you must be assuming $X\to Y$ is surjective, or else it clearly isn't true. That condition enters into the argument when doing analysis of the situation with strictly henselian local rings."

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.