Lemma 15.48.6. Let $A$ be a Noetherian complete local domain. Then $A$ is J-0.
Proof. By Algebra, Lemma 10.160.11 we can find a regular subring $A_0 \subset A$ with $A$ finite over $A_0$. The induced extension $K/K_0$ of fraction fields is finite. If $K/K_0$ is separable, then we are done by Lemma 15.47.5. If not, then $A_0$ and $A$ have characteristic $p > 0$. For any subextension $K/M/K_0$ there exists a finite subextension $A_0 \subset B \subset A$ whose fraction field is $M$. Hence, arguing by induction on $[K : K_0]$ we may assume there exists $A_0 \subset B \subset A$ such that $B$ is J-0 and $K/M$ has no nontrivial subextensions. In this case, if $K/M$ is separable, then we see that $A$ is J-0 by Lemma 15.47.5. If not, then $K = M[z]/(z^ p - b_1/b_2)$ for some $b_1, b_2 \in B$ with $b_2 \not= 0$ and $b_1/b_2$ not a $p$th power in $M$. Choose $a \in A$ nonzero such that $az \in A$. After replacing $z$ by $b_2 a^ p z$ we obtain $K = M[z]/(z^ p - b)$ with $z \in A$ and $b \in B$ not a $p$th power in $M$. By Lemma 15.48.5 we can find a derivation $D : B \to B$ with $D(b) \not= 0$. Applying Lemma 15.48.4 we see that $A_\mathfrak p$ is regular for any prime $\mathfrak p$ of $A$ lying over a regular prime of $B$ and not containing $D(b)$. As $B$ is J-0 we conclude $A$ is too. $\square$
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