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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.

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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. –  BCnrd Mar 12 '10 at 8:06
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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. –  BCnrd Mar 12 '10 at 17:14
    
Thanks a lot.!! –  Yuhao Huang Mar 12 '10 at 22:26
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1 Answer 1

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."

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