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

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

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. $\square$

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