The Stacks project

Proof. It suffices to prove this when $Y = \mathop{\mathrm{Spec}}(A)$ where $A$ is a Noetherian domain. Let $B$ be the integral closure of $A$ in its fraction field. Set $C = \Gamma (X, \mathcal{O}_ X)$. By Cohomology of Schemes, Lemma 30.19.2 we see that $C$ is a finite $A$-module. As $X$ is normal (Properties, Lemma 28.9.4) we see that $C$ is normal domain (Properties, Lemma 28.7.9). Thus $B \subset C$ and we conclude that $B$ is finite over $A$ as $A$ is Noetherian.

There exists a nonempty open $V \subset Y$ such that $f^{-1}V \to V$ is finite, see Morphisms, Definition 29.51.12. After shrinking $V$ we may assume that $f^{-1}V \to V$ is flat (Morphisms, Proposition 29.27.1). Thus $f^{-1}V \to V$ is faithfully flat. Then $V$ is regular by Algebra, Lemma 10.164.4. $\square$

Proof. Consider the commutative diagram

Recall that $\mathfrak m$-adic completion on the category of finite $A$-modules is exact because it is given by tensoring with the flat $A$-algebra $A^\wedge $ (Algebra, Lemma 10.97.2). We will use Serre's criterion (Algebra, Lemma 10.157.4) to prove that the Noetherian ring $B^\wedge $ is normal. Let $\mathfrak q \subset B^\wedge $ be a prime lying over $\mathfrak p \subset B$. If $\dim (B_\mathfrak p) \geq 2$, then $\text{depth}(B_\mathfrak p) \geq 2$ and since $B_\mathfrak p \to B^\wedge _\mathfrak q$ is flat we find that $\text{depth}(B^\wedge _\mathfrak q) \geq 2$ (Algebra, Lemma 10.163.2). If $\dim (B_\mathfrak p) \leq 1$, then $B_\mathfrak p$ is either a discrete valuation ring or a field. In that case $C_\mathfrak p$ is faithfully flat over $B_\mathfrak p$ (because it is finite and torsion free). Hence $B^\wedge _\mathfrak p \to C^\wedge _\mathfrak p$ is faithfully flat and the same holds after localizing at $\mathfrak q$. As $C^\wedge $ and hence any localization is $(S_2)$ we conclude that $B^\wedge _\mathfrak p$ is $(S_2)$ by Algebra, Lemma 10.164.5. All in all we find that $(S_2)$ holds for $B^\wedge $. To prove that $B^\wedge $ is $(R_1)$ we only have to consider primes $\mathfrak q \subset B^\wedge $ with $\dim (B^\wedge _\mathfrak q) \leq 1$. Since $\dim (B^\wedge _\mathfrak q) = \dim (B_\mathfrak p) + \dim (B^\wedge _\mathfrak q/\mathfrak p B^\wedge _\mathfrak q)$ by Algebra, Lemma 10.112.6 we find that $\dim (B_\mathfrak p) \leq 1$ and we see that $B^\wedge _\mathfrak q \to C^\wedge _\mathfrak q$ is faithfully flat as before. We conclude using Algebra, Lemma 10.164.6. $\square$

Proof. Parts (1) and (2) follow from Lemma 54.13.1. We have to redo part of the proof of that lemma in order to set up notation for the proof of (3). Set $C = \Gamma (X, \mathcal{O}_ X)$. By Cohomology of Schemes, Lemma 30.19.2 we see that $C$ is a finite $A$-module. As $X$ is normal (Properties, Lemma 28.9.4) we see that $C$ is normal domain (Properties, Lemma 28.7.9). Thus $B \subset C$ and we conclude that $B$ is finite over $A$ as $A$ is Noetherian. By Lemma 54.13.2 in order to prove (3) it suffices to show that the $\mathfrak m$-adic completion $C^\wedge $ is normal.

By Algebra, Lemma 10.97.8 the completion $C^\wedge $ is the product of the completions of $C$ at the prime ideals of $C$ lying over $\mathfrak m$. There are finitely many of these and these are the maximal ideals $\mathfrak m_1, \ldots , \mathfrak m_ r$ of $C$. (The corresponding result for $B$ explains the final statement of the lemma.) Thus replacing $A$ by $C_{\mathfrak m_ i}$ and $X$ by $X_ i = X \times _{\mathop{\mathrm{Spec}}(C)} \mathop{\mathrm{Spec}}(C_{\mathfrak m_ i})$ we reduce to the case discussed in the next paragraph. (Note that $\Gamma (X_ i, \mathcal{O}) = C_{\mathfrak m_ i}$ by Cohomology of Schemes, Lemma 30.5.2.)

Here $A$ is a Noetherian local normal domain and $f : X \to \mathop{\mathrm{Spec}}(A)$ is a regular alteration with $\Gamma (X, \mathcal{O}_ X) = A$. We have to show that the completion $A^\wedge $ of $A$ is a normal domain. By Lemma 54.11.2 $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$ is regular. Since $\Gamma (Y, \mathcal{O}_ Y) = A^\wedge $ by Cohomology of Schemes, Lemma 30.5.2, we conclude that $A^\wedge $ is normal as before. Namely, $Y$ is normal by Properties, Lemma 28.9.4. It is connected because $\Gamma (Y, \mathcal{O}_ Y) = A^\wedge $ is local. Hence $Y$ is normal and integral (as connected and normal implies integral for Noetherian schemes). Thus $\Gamma (Y, \mathcal{O}_ Y) = A^\wedge $ is a normal domain by Properties, Lemma 28.7.9. This proves (3).

Proof of (4). Let $\eta \in \mathop{\mathrm{Spec}}(A)$ denote the generic point and denote by a subscript $\eta $ the base change to $\eta $. Since $f$ is an alteration, the scheme $X_\eta $ is finite and faithfully flat over $\eta $. Since $Y = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A^\wedge )$ is regular by Lemma 54.11.2 we see that $Y_\eta $ is regular (as a limit of opens in $Y$). Then $Y_\eta \to \mathop{\mathrm{Spec}}(A^\wedge \otimes _ A \kappa (\eta ))$ is finite faithfully flat onto the generic formal fibre. We conclude by Algebra, Lemma 10.164.4. $\square$