Lemma 54.13.2. Let (A, \mathfrak m) be a local Noetherian ring. Let B \subset C be finite A-algebras. Assume that (a) B is a normal ring, and (b) the \mathfrak m-adic completion C^\wedge is a normal ring. Then B^\wedge is a normal ring.
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
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