Lemma 15.50.5. Let $k$ be a field of characteristic $p$. Let $A = k[[x_1, \ldots , x_ n]][y_1, \ldots , y_ n]$ and denote $K$ the fraction field of $A$. Let $\mathfrak p \subset A$ be a prime. Then $A_\mathfrak p^\wedge \otimes _ A K$ is geometrically regular over $K$.
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. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.