The Stacks project

Lemma 15.50.9. Let $p$ be a prime number. Let $A$ be a Noetherian complete local domain with fraction field $K$ of characteristic $p$. Let $\mathfrak q \subset A[x]$ be a maximal ideal lying over the maximal ideal of $A$ and let $(0) \not= \mathfrak r \subset \mathfrak q$ be a prime lying over $(0) \subset A$. Then $A[x]_\mathfrak q^\wedge \otimes _{A[x]} \kappa (\mathfrak r)$ is geometrically regular over $\kappa (\mathfrak r)$.

Proof. Note that $K \subset \kappa (\mathfrak r)$ is finite. Hence, given a finite purely inseparable extension $L/\kappa (\mathfrak r)$ there exists a finite extension of Noetherian complete local domains $A \subset B$ such that $\kappa (\mathfrak r) \otimes _ A B$ surjects onto $L$. Namely, you take $B \subset L$ a finite $A$-subalgebra whose field of fractions is $L$. Denote $\mathfrak r' \subset B[x]$ the kernel of the map $B[x] = A[x] \otimes _ A B \to \kappa (\mathfrak r) \otimes _ A B \to L$ so that $\kappa (\mathfrak r') = L$. Then

\[ A[x]_\mathfrak q^\wedge \otimes _{A[x]} L = A[x]_\mathfrak q^\wedge \otimes _{A[x]} B[x] \otimes _{B[x]} \kappa (\mathfrak r') = \prod B[x]_{\mathfrak q_ i}^\wedge \otimes _{B[x]} \kappa (\mathfrak r') \]

where $\mathfrak q_1, \ldots , \mathfrak q_ t$ are the primes of $B[x]$ lying over $\mathfrak q$, see Algebra, Lemma 10.97.8. Thus we see that it suffices to prove the rings $B[x]_{\mathfrak q_ i}^\wedge \otimes _{B[x]} \kappa (\mathfrak r')$ are regular. This reduces us to showing that $A[x]_\mathfrak q^\wedge \otimes _{A[x]} \kappa (\mathfrak r)$ is regular in the special case that $K = \kappa (\mathfrak r)$.

Assume $K = \kappa (\mathfrak r)$. In this case we see that $\mathfrak r K[x]$ is generated by $x - f$ for some $f \in K$ and

\[ A[x]_\mathfrak q^\wedge \otimes _{A[x]} \kappa (\mathfrak r) = (A[x]_\mathfrak q^\wedge \otimes _ A K)/(x - f) \]

The derivation $D = \text{d}/\text{d}x$ of $A[x]$ extends to $K[x]$ and maps $x - f$ to a unit of $K[x]$. Moreover $D$ extends to $A[x]_\mathfrak q^\wedge \otimes _ A K$ by Lemma 15.48.1. As $A \to A[x]_\mathfrak q^\wedge $ is formally smooth (see Lemmas 15.37.2 and 15.37.4) the ring $A[x]_\mathfrak q^\wedge \otimes _ A K$ is regular by Proposition 15.49.2 (the arguments of the proof of that proposition simplify significantly in this particular case). We conclude by Lemma 15.48.2. $\square$

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Your email address will not be published.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 07PU. Beware of the difference between the letter 'O' and the digit '0'.