The Stacks project

Proof. Let $L/K$ be a finite purely inseparable field extension. We will show by induction on $[L : K]$ that $A_\mathfrak p^\wedge \otimes L$ is regular. The base case is $L = K$: as $A$ is regular, $A_\mathfrak p^\wedge $ is regular (Lemma 15.43.4), hence the localization $A_\mathfrak p^\wedge \otimes K$ is regular. Let $K \subset M \subset L$ be a subfield such that $L$ is a degree $p$ extension of $M$ obtained by adjoining a $p$th root of an element $f \in M$. Let $B$ be a finite $A$-subalgebra of $M$ with fraction field $M$. Clearing denominators, we may and do assume $f \in B$. Set $C = B[z]/(z^ p -f)$ and note that $B \subset C$ is finite and that the fraction field of $C$ is $L$. Since $A \subset B \subset C$ are finite and $L/M/K$ are purely inseparable we see that for every element of $B$ or $C$ some power of it lies in $A$. Hence there is a unique prime $\mathfrak r \subset B$, resp. $\mathfrak q \subset C$ lying over $\mathfrak p$. Note that

see Algebra, Lemma 10.97.8. By induction we know that this ring is regular. In the same manner we have

the last equality because the completion of $C = B[z]/(z^ p - f)$ equals $B_\mathfrak r^\wedge [z]/(z^ p -f)$. By Lemma 15.48.5 we know there exists a derivation $D : B \to B$ such that $D(f) \not= 0$. In other words, $g = D(f)$ is a unit in $M$! By Lemma 15.48.1 $D$ extends to a derivation of $B_\mathfrak r$, $B_\mathfrak r^\wedge $ and $B_\mathfrak r^\wedge \otimes _ B M$ (successively extending through a localization, a completion, and a localization). Since it is an extension we end up with a derivation of $B_\mathfrak r^\wedge \otimes _ B M$ which maps $f$ to $g$ and $g$ is a unit of the ring $B_\mathfrak r^\wedge \otimes _ B M$. Hence $A_\mathfrak p^\wedge \otimes _ A L$ is regular by Lemma 15.48.4 and we win. $\square$