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Let $A \to B$ be a finitely generated homomorphism between two commutative noetherian rings.

As far as I understand, in various generalizations of this situation, such a map is called smooth if $B$ is a perfect object in $D(B\otimes_A B)$. (See for example Definition 2.2 of http://arxiv.org/pdf/1006.4721v2.pdf).

Is this true for commutative noetherian rings?

That is, is a finite type map $A\to B$ is smooth if and only if $B$ has a finite projective dimension over $B\otimes_A B$? Note that one direction (that smoothness implies finite projective dimension) is true because of EGA IV Proposition 17.12.4 which says that the kernel of $B\otimes_A B\to B$ is locally generated by a regular sequence. But does the converse holds?

Edit: the answer below shows that surjections, for example, are a counterexample. What if in addition we assume flatness of the map $A\to B$, so that $B\otimes^L_A B \cong B\otimes_A B$?

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Yes to the second question. More generally, if $f:A \to B$ is a flat homomorphism of noetherian commutative rings such that the flat dimension of $B$ over $B\otimes_A B$ is finite, then $f$ is regular. See Rodicio: Smooth algebras and vanishing of Hochschild homology, Comm. math. Helv. 65 (1990) 474-477. When $f$ is of finite type then you get your answer, since in this case regularity and smoothness are the same thing.

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No: if $B$ is a quotient of $A$, then $B \otimes_A B = B$, but $B$ is very rarely smooth over $A$. The correct homological characterization of smooth maps involves André-Quillen cohomology.

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    $\begingroup$ However $B\otimes^L_A B$ is not B in this case, which I think is the relevant question in the derived setting. Moreover the data of Andre-Quillen cohomology is essentially equivalent to the data of the self-Tor of B over this algebra (via the Hochschild-Kostant-Rosenberg theorem). $\endgroup$ – David Ben-Zvi May 20 '13 at 0:06
  • $\begingroup$ will the answer be different if we assume flatness? $\endgroup$ – the L May 20 '13 at 8:21

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