The Stacks project

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. $\square$