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I am looking for a reference for the following fact (I hope it is a fact, if not, can someone give a more precise formulation)?

Suppose $V$ and $W$ are two algebraic varieties over an algebraically closed field. Then a morphism $f:V\to W$ is an etale map if and only if, the induced map on each Zariski cotangent space is an isomorphism.

By an etale map I mean a flat unramified map of finite type.

Thank you!

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    As well as the references given below, this in Mumford's Red Book. The general statement is that maps on completions of local rings are isomorphisms. However, if the source and target are smooth, this will follow from the weaker statement about Zariski tangent spaces.2011-05-04

2 Answers 2

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I think you need to assume that $V$ and $W$ are smooth varieties; then you'll find the statement (or, at least, a very similar one about stalks) in SGA 1, II, Corollaire 4.6 or (in English) Bosch, Lütkebohmert, Raynaud, Néron Models, 2.2, Proposition 10.

I've been trying to come up with a good counterexample with non-smooth varieties. Perhaps you can do something like this: take $W = l_1 \cup l_2$ to be the union of two lines in the plane, and V = l'_1 \cup l'_2 \cup l'_3 the union of three lines in the plane meeting in a point. They both have one singular point, at which the tangent space has dimension 2. Map $V \to W$ by sending l'_1 to $l_1$, and both l'_2 and l'_3 to $l_2$, so that the singular point of $V$ maps to the singular point of $W$. Then I believe you get an isomorphism of tangent spaces at each point, but the map isn't étale.

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    Why does the singular point need to be mapped to the singular point?2018-06-12
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I believe the correct version of this is Proposition I.2.9 in Milne's Lectures on Etale Cohomology. (I see that there is already a reference given, but this one is available online freely.)