Let $A,B$ be two commutative noetherian rings. Let $f:A\to B$ be a formally smooth homomorphism. If $A$ is a regular ring (in the sense that all its localizations are regular local rings), does this imply that $B$ is a regular ring?



Formally smooth map from a regular ring
As I wrote in the comments, the answer is usually yes, depending on the topology you are considering. The two more common cases are (it should be direct references for them, maybe in EGA $0\_{IV}$):


Local case: assume $f$ is local. Let $l$ be the residue field of $B$. You have a Jacobi-Zariski exact sequence in André-Quillen homology
$$0=H\_2(A,l,l) \to H\_2(B,l,l) \to H\_1(A,B,l)=0$$
(the left module is zero since $A$ is regular and the right one since $f$ is formally smooth for the topology of the maximal ideal). Therefore the middle term is zero and then $B$ is regular.


Discrete case: assume that $f$ is formally smooth w.r.t. the discrete topology. Then $H\_1(A\_{\mathfrak p},B\_{\mathfrak q},k({\mathfrak q}))=H\_1(A,B,k({\mathfrak q}))=0$ for any prime ideal $\mathfrak q$ of $B$ (where $ \mathfrak p = f^{-1} \mathfrak q$), and by a similar exact sequence you deduce that $H\_2(B\_{\mathfrak q},k({\mathfrak q}),k({\mathfrak q}))=0$, that is $B\_{\mathfrak q}$ is regular.


