Lemma 33.27.6. Let k be a field. Let X be a locally algebraic k-scheme. Let \nu : X^\nu \to X be the normalization of X. Let x \in X be a point such that (a) \mathcal{O}_{X, x} is reduced, (b) \dim (\mathcal{O}_{X, x}) = 1, and (c) for every x' \in X^\nu with \nu (x') = x the extension \kappa (x')/k is separable. Then X is geometrically reduced at x and X^\nu is geometrically regular at x' with \nu (x') = x.
Proof. We will use the results of Lemma 33.27.1 without further mention. Let x' \in X^\nu be a point over x. By dimension theory (Section 33.20) we have \dim (\mathcal{O}_{X^\nu , x'}) = 1. Since X^\nu is normal, we see that \mathcal{O}_{X^\nu , x'} is a discrete valuation ring (Properties, Lemma 28.12.5). Thus \mathcal{O}_{X^\nu , x'} is a regular local k-algebra whose residue field is separable over k. Hence k \to \mathcal{O}_{X^\nu , x'} is formally smooth in the \mathfrak m_{x'}-adic topology, see More on Algebra, Lemma 15.38.5. Then \mathcal{O}_{X^\nu , x'} is geometrically regular over k by More on Algebra, Theorem 15.40.1. Thus X^\nu is geometrically regular at x' by Lemma 33.12.2.
Since \mathcal{O}_{X, x} is reduced, the family of maps \mathcal{O}_{X, x} \to \mathcal{O}_{X^\nu , x'} is injective. Since \mathcal{O}_{X^\nu , x'} is a geometrically reduced k-algebra, it follows immediately that \mathcal{O}_{X, x} is a geometrically reduced k-algebra. Hence X is geometrically reduced at x by Lemma 33.6.2. \square
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