# Formally smooth map from a regular ring

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?

• It depends on the definition of "formal smoothness". The answer will be "yes" if your are considering the definition w.r.t. the discrete topology, or if the homomorphism is local and the topology is the one induced by its maximal ideal, or if some finiteness hypothesis are implicit in your definition, etc. – Vinteuil Jul 10 '14 at 14:34
• do you have any reference for these statements? (I do not have finiteness assumptions implict in my situation) – wonderman Jul 10 '14 at 14:49

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.
• Thanks, this is helpful. Can you just explain why $H_2(A,l,l) = 0$? I know that if $(A,m)$ is regular then $H_n(A,A/m,-) = 0$. – wonderman Jul 10 '14 at 15:18
• Because of a similar exact sequence: $0=H_2(A,k,l) \to H_2(A,l,l) \to H_2(k,l,l)=0$. – Vinteuil Jul 10 '14 at 15:20