# Regular rings and formally smooth algebras

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?).

• 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). – bananastack Dec 3 '14 at 21:17
• @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. – Mister Benjamin Dover Dec 3 '14 at 21:32

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:
1. $$A$$ is a formally smooth $$k$$-algebra for its preadic topology,
2. $$A$$ is geometrically regular over $$k$$,
3. for all finite extensions $$k'/k$$ such that $$k'^p\subset k$$, the ring $$A'=A\otimes_k k'$$ is regular.