Let $A\rightarrow B$ be a commutative $A$-algebra. If $A$ is a field and $B$ Noetherian and formally smooth over $A$, then it is known that $B$ must be a regular ring. Is there a partial converse of this result, asserting that "$B$ is a regular ring + some further conditions on $A$ and $B$ $\implies$ $B$ is formally smooth over $A$?" (perhaps if $A$ is a perfect field and something else?).
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$\begingroup$ If I'm not mistaken, a regular variety over a perfect field is smooth (ie A \to B, A perfect field, B finite type, reduced, regular A-algebra). But in general it's probably wild, as there many non-smooth morphisms between smooth varieties (even if you impose some flatness). $\endgroup$– bananastackCommented Dec 3, 2014 at 21:17
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$\begingroup$ @user125763: I somehow expected that it is wild in general, but could you (or someone else) provide a reference for what is a fact if you are not mistaken? This would be a good answer. $\endgroup$– Mister Benjamin DoverCommented Dec 3, 2014 at 21:32
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1 Answer
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Here is EGA IV$_1$, Chap. 0, Th. 22.5.8:
Let $k$ be a field with characteristic exponent $p$, and let $A$ be a local Noetherian $k$-algebra. Then the following are equivalent:
- $A$ is a formally smooth $k$-algebra for its preadic topology,
- $A$ is geometrically regular over $k$,
- for all finite extensions $k'/k$ such that $k'^p\subset k$, the ring $A'=A\otimes_k k'$ is regular.
answered Dec 3, 2014 at 21:47
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$\begingroup$ A reference for a simpler proof (due to Faltings) of the equivalence of (1) and (2) is Theorem 28.7 of Matsumura's CRT book (which actually gives (3) implies (1), so equivalence of all three conditions). $\endgroup$ Commented Dec 4, 2014 at 3:46