Let $A$ be a local noetherian ring. When (besides when $A$ is excellent) do we have that $\operatorname{Spec}(A[[t]])\rightarrow \operatorname{Spec}(A[t])$ is regular?


As you probably are aware, a sufficient condition is that $A$ is a $G$-ring [Matsumura, Thm. 79]. On the other hand, the following result says that one must put some strong assumptions on $A$ to have $A[t] \to A[[t]]$ be regular.

Theorem [Sharp, Thm. 2.9]. There exist local domains $B$ of dimension two and three such that setting $L = \operatorname{Frac}(B)$, the generic fiber $B[[t_1,\ldots,t_n]] \otimes_B L$ of the homomorphism $B \to B[[t_1,\ldots,t_n]]$ is not Cohen–Macaulay for every integer $n > 0$.

The example is based on Ferrand and Raynaud's construction of a local domain $A$ such that the zero ideal in the completion $\hat{A}$ has an embedded prime [Ferrand–Raynaud, Prop. 3.3]. This ring $A$ therefore has non-Cohen–Macaulay formal fibers, and I believe Sharp's construction gives a ring $B$ that also has non-Cohen–Macaulay formal fibers.

We can now consider the composition $$B \longrightarrow B[t] \longrightarrow B[[t]],$$ where $B$ is as in Sharp's theorem. The first map is a regular homomorphism. If the second map were also regular, then the composition would be regular. But Sharp's theorem says that this cannot be the case.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.