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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