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?).
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.