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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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    Thanks for the comments! The most important examples for me are $A,B$ regular rings over an algebraically closed field, but not necessarily finitely generated. I'm happy to assume they "come from geometry" in some sense: e.g., that they are f.g. over $k$ or else localizations or completions or something like that. Anyway, the question is just about terminology. I edited the question to include smoothness as a hypothesis (though I believe it's not necessary) in my assertion.2012-10-11

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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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    Excellent! I was not aware of that reference. Thanks very much!2012-10-12