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Let $k$ be a commutative ring (if you like, an algebraically closed field) and let $A$ and $B$ be commutative $k$-algebras. Suppose we are given a homomorphism of $k$-algebras $f:A \rightarrow B$. Then $f$ induces a map on modules of differentials $f: \Omega_{A/k} \rightarrow \Omega_{B/k}$, and hence a map $$B \otimes_A \Omega_{A/k} \rightarrow B \otimes_A \Omega_{B/k} \rightarrow \Omega_{B/k}.$$ Is there an established name (written down somewhere permanent and publicly available) for the class of maps $f$ such that this composite map is an isomorphism?

Evidently, if $A$ and $B$ are the coordinate rings of smooth affine varieties $X$ and $Y$ over an algebraically closed field, and $f$ corresponds to an etale map, then this property holds---I need a slight generalization for technical reasons.

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When $X, Y$ are smooth over $k$, one can conclude that $f$ is étale. How do you prove the general case for algebraic varieties ? –  user18119 Oct 11 '12 at 19:56
    
@QiL, I do not understand the question. What is the assertion you are asking me to prove? –  S123 Oct 11 '12 at 20:28
    
Ah sorry. You said that the isomorphism implies that $f$ is étale when $A, B$ are finitely generated over $k$. I was wondering how to prove this (if Spec($A$) and Spec($B)$ are not necessarily smooth). –  user18119 Oct 11 '12 at 21:26
    
OK, I see. Actually, what I wrote in the question was the converse: if $B$ is an etale $A$ algebra then the map is an iso. –  S123 Oct 11 '12 at 21:44
    
Oops. If $A, B$ are f.g. over a field $k$, then the isomorphism implies that the corresponding morphism of varieties is unramified because then $\Omega_{B/A}=0$. If the corresponding varieties are smooth, then $f$ is étale. Otherwise I don't know. –  user18119 Oct 11 '12 at 21:50

1 Answer 1

up vote 1 down vote accepted

This is more a comment/question.

I don't know whether there is a standard terminology for this situation. I would say no.

But have you an example with $A, B$ regular, $B$ finite type over $A$, $k$ algebraically closed, but $A\to B$ is not étale ? I would be interested.

A trivial counterexample without regularity assumption: let $k$ be of characteristic $p$, $A=k[t]$ and $B=k[t]/(t^p)$. Then $B\otimes_A \Omega_{A/k}\to \Omega_{B/k}$ is an isomorphism, but $A\to B$ is of course not étale.

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I have no such example. –  S123 Oct 12 '12 at 16:26
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@Steve: note that when $A, B$ are local and regular, $B$ essentially f.g. over $A$, the isomorphism implies that $A\to B$ étale provided $\dim A=\dim B$ (see EGA IV, 6.1.5). –  user18119 Oct 12 '12 at 16:52
    
Excellent! I was not aware of that reference. Thanks very much! –  S123 Oct 12 '12 at 21:35

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