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?).



Regular rings and formally smooth algebras
Here is EGA IV$\_1$, Chap. 0, Th. 22.5.8:



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> Let $k$ be a field with characteristic exponent $p$, and let $A$ be a local Noetherian $k$-algebra. Then the following are equivalent:
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> 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.
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