Lemma 54.11.5. Let $(A, \mathfrak m)$ be a local Noetherian ring. Let $X$ be a scheme over $A$. Assume
$A$ is analytically unramified (Algebra, Definition 10.162.9),
$X$ is locally of finite type over $A$, and
$X \to \mathop{\mathrm{Spec}}(A)$ is étale at the generic points of irreducible components of $X$.
Then the normalization of $X$ is finite over $X$.
Proof. Since $A$ is analytically unramified it is reduced by Algebra, Lemma 10.162.10. Since the normalization of $X$ depends only on the reduction of $X$, we may replace $X$ by its reduction $X_{red}$; note that $X_{red} \to X$ is an isomorphism over the open $U$ where $X \to \mathop{\mathrm{Spec}}(A)$ is étale because $U$ is reduced (Descent, Lemma 35.18.1) hence condition (3) remains true after this replacement. In addition we may and do assume that $X = \mathop{\mathrm{Spec}}(B)$ is affine.
The map
is injective because $A \to A^\wedge $ is faithfully flat (Algebra, Lemma 10.97.3) hence induces a surjective map between sets of minimal primes (by going down for flat ring maps, see Algebra, Section 10.41). Both sides are finite products of fields as our rings are Noetherian. Let $L = \prod _{\mathfrak q \subset B\text{ minimal}} \kappa (\mathfrak q)$. Our assumption (3) implies that $L = B \otimes _ A K$ and that $K \to L$ is a finite étale ring map (this is true because $A \to B$ is generically finite, for example use Algebra, Lemma 10.122.10 or the more detailed results in Morphisms, Section 29.51). Since $B$ is reduced we see that $B \subset L$. This implies that
Then $M$ is the total ring of fractions of $C$ and is a finite product of fields as a finite separable algebra over $K^\wedge $. It follows that $C$ is reduced and that its normalization $C'$ is the integral closure of $C$ in $M$. The normalization $B'$ of $B$ is the integral closure of $B$ in $L$. By flatness of $A \to A^\wedge $ we obtain an injective map $B' \otimes _ A A^\wedge \to M$ whose image is contained in $C'$. Picture
As $A^\wedge $ is Nagata (by Algebra, Lemma 10.162.8), we see that $C'$ is finite over $C = B \otimes _ A A^\wedge $ (see Algebra, Lemmas 10.162.8 and 10.162.2). As $C$ is Noetherian, we conclude that $B' \otimes _ A A^\wedge $ is finite over $C = B \otimes _ A A^\wedge $. Therefore by faithfully flat descent (Algebra, Lemma 10.83.2) we see that $B'$ is finite over $B$ which is what we had to show. $\square$
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