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

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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