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The Stacks project

Lemma 55.8.2. Let X be a model of a geometrically normal variety V over K. Then the normalization \nu : X^\nu \to X is finite and the base change of X^\nu to the completion R^\wedge is the normalization of the base change of X. Moreover, for each x \in X^\nu the completion of \mathcal{O}_{X^\nu , x} is normal.

Proof. Observe that R^\wedge is a discrete valuation ring (More on Algebra, Lemma 15.43.5). Set Y = X \times _{\mathop{\mathrm{Spec}}(R)} \mathop{\mathrm{Spec}}(R^\wedge ). Since R^\wedge is a discrete valuation ring, we see that

Y \setminus Y_ k = Y \times _{\mathop{\mathrm{Spec}}(R^\wedge )} \mathop{\mathrm{Spec}}(K^\wedge ) = V \times _{\mathop{\mathrm{Spec}}(K)} \mathop{\mathrm{Spec}}(K^\wedge )

where K^\wedge is the fraction field of R^\wedge . Since V is geometrically normal, we find that this is a normal scheme. Hence the first part of the lemma follows from Resolution of Surfaces, Lemma 54.11.6.

To prove the second part we may assume X and Y are normal (by the first part). If x is in the generic fibre, then \mathcal{O}_{X, x} = \mathcal{O}_{V, x} is a normal local ring essentially of finite type over a field. Such a ring is excellent (More on Algebra, Proposition 15.52.3). If x is a point of the special fibre with image y \in Y, then \mathcal{O}_{X, x}^\wedge = \mathcal{O}_{Y, y}^\wedge by Resolution of Surfaces, Lemma 54.11.1. In this case \mathcal{O}_{Y, y} is a excellent normal local domain by the same reference as before as R^\wedge is excellent. If B is a excellent local normal domain, then the completion B^\wedge is normal (as B \to B^\wedge is regular and More on Algebra, Lemma 15.42.2 applies). This finishes the proof. \square


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