Lemma 32.12.6. Let $k$ be a field. Let $X$ be a scheme of finite type over $k$. Let $x \in X$. Then $X$ is geometrically regular at $x$ if and only if $X \to \mathop{\mathrm{Spec}}(k)$ is smooth at $x$ (Morphisms, Definition 28.32.1).

**Proof.**
The question is local around $x$, hence we may assume that $X = \mathop{\mathrm{Spec}}(A)$ for some finite type $k$-algebra. Let $x$ correspond to the prime $\mathfrak p$.

If $A$ is smooth over $k$ at $\mathfrak p$, then we may localize $A$ and assume that $A$ is smooth over $k$. In this case $k' \otimes _ k A$ is smooth over $k'$ for all extension fields $k'/k$, and each of these Noetherian rings is regular by Algebra, Lemma 10.138.3.

Assume $X$ is geometrically regular at $x$. Consider the residue field $K := \kappa (x) = \kappa (\mathfrak p)$ of $x$. It is a finitely generated extension of $k$. By Algebra, Lemma 10.44.3 there exists a finite purely inseparable extension $k \subset k'$ such that the compositum $k'K$ is a separable field extension of $k'$. Let $\mathfrak p' \subset A' = k' \otimes _ k A$ be a prime ideal lying over $\mathfrak p$. It is the unique prime lying over $\mathfrak p$, see Algebra, Lemma 10.45.7. Hence the residue field $K' := \kappa (\mathfrak p')$ is the compositum $k'K$. By assumption the local ring $(A')_{\mathfrak p'}$ is regular. Hence by Algebra, Lemma 10.138.5 we see that $k' \to A'$ is smooth at $\mathfrak p'$. This in turn implies that $k \to A$ is smooth at $\mathfrak p$ by Algebra, Lemma 10.135.18. The lemma is proved. $\square$

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## Comments (1)

Comment #4267 by Tim Holzschuh on