The Stacks project

Lemma 86.14.12. Let $A$ be an adic Noetherian topological ring. Let $\mathfrak p \subset A$ be a rig-closed prime. For any $n \geq 1$ the ring map

\[ A/\mathfrak p \longrightarrow A\{ x_1, \ldots , x_ n\} \otimes _ A A/\mathfrak p = A/\mathfrak p\{ x_1, \ldots , x_ n\} \]

is regular. In particular, the algebra $A\{ x_1, \ldots , x_ n\} \otimes _ A \kappa (\mathfrak p)$ is geometrically regular over $\kappa (\mathfrak p)$.

Proof. We will use some fact on regular ring maps the reader can find in More on Algebra, Section 15.41. Since $A/\mathfrak p$ is a complete local Noetherian ring it is excellent (More on Algebra, Proposition 15.52.3). Hence $A/\mathfrak p[x_1, \ldots , x_ n]$ is excellent (by the same reference). Hence $A/\mathfrak p[x_1, \ldots , x_ n] \to A/\mathfrak p\{ x_1, \ldots , x_ n\} $ is a regular ring homomorphism by More on Algebra, Lemma 15.50.14. Of course $A/\mathfrak p \to A/\mathfrak p[x_1, \ldots , x_ n]$ is smooth and hence regular. Since the composition of regular ring maps is regular the proof is complete. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0GGJ. Beware of the difference between the letter 'O' and the digit '0'.