# Formally smooth maps between adic rings and regular immersions

Suppose $(A,\mathfrak{a})$ and $(B,\mathfrak{b})$ are two adically complete (commutative) noetherian rings. Let $f:A \to B$ be a continuous formally smooth formally of finite type map (that is, $B/\mathfrak{b}$ is a finitely generated $A$-algebra). In this case it is proved in EGA IV that $\Lambda_{\mathfrak{b}} \Omega^1_{B/A}$ is a projective module of finite rank, where $\Lambda_{\mathfrak{b}}$ denotes the $\mathfrak{b}$-adic completion functor.

In a (not nessecerily) adic case, if $f:A \to B$ is formally smooth and of finite type, it is proved that there that $f$ is actually smooth, so that letting $I = \ker (B\otimes_A B \to B)$, we have that $I$ is locally generated by a regular sequence (from which it follows that $\Omega^1_{B/A} \cong I/I^2$ is locally free).

Now, to my question: Again, assume we are in the adic case as above. As seen above, the module of Kähler differentials, $\Omega^1_{B/A}$ is no longer the "correct" object, but its completion is.

Set $\mathfrak{b}^e = \mathfrak{b} \otimes_A B + B \otimes_A \mathfrak{b} \subseteq B\otimes_A B$, and let $J = \ker( \Lambda_{\mathfrak{b}^e} B\otimes_A B \to B)$. Then we have that $J/J^2 \cong \Lambda_{\mathfrak{b}} \Omega^1_{B/A}$. Is it true that $J$ is locally generated by a regular sequence? that is, is the map $\Lambda_{\mathfrak{b}^e} B\otimes_A B \to B$ a regular immersion?

Thanks!

• Have you checked arxiv.org/abs/math/0604241v1? (It is published in Comm. Alg.) – Leo Alonso Feb 17 '11 at 15:37
• Thank you for your comment. Unfortunately, I could not locate this result in this reference. – the L Feb 17 '11 at 15:52