Geometric reducedness can be checked on local rings.

Lemma 33.6.2. Let $k$ be a field. Let $X$ be a scheme over $k$. Let $x \in X$. The following are equivalent

1. $X$ is geometrically reduced at $x$, and

2. the ring $\mathcal{O}_{X, x}$ is geometrically reduced over $k$ (see Algebra, Definition 10.43.1).

Proof. Assume (1). This in particular implies that $\mathcal{O}_{X, x}$ is reduced. Let $k'/k$ be a finite purely inseparable field extension. Consider the ring $\mathcal{O}_{X, x} \otimes _ k k'$. By Algebra, Lemma 10.46.7 its spectrum is the same as the spectrum of $\mathcal{O}_{X, x}$. Hence it is a local ring also (Algebra, Lemma 10.18.3). Therefore there is a unique point $x' \in X_{k'}$ lying over $x$ and $\mathcal{O}_{X_{k'}, x'} \cong \mathcal{O}_{X, x} \otimes _ k k'$. By assumption this is a reduced ring. Hence we deduce (2) by Algebra, Lemma 10.44.4.

Assume (2). Let $k'/k$ be a field extension. Since $\mathop{\mathrm{Spec}}(k') \to \mathop{\mathrm{Spec}}(k)$ is surjective, also $X_{k'} \to X$ is surjective (Morphisms, Lemma 29.9.4). Let $x' \in X_{k'}$ be any point lying over $x$. The local ring $\mathcal{O}_{X_{k'}, x'}$ is a localization of the ring $\mathcal{O}_{X, x} \otimes _ k k'$. Hence it is reduced by assumption and (1) is proved. $\square$

Comment #3018 by Brian Lawrence on

Suggested slogan: A $k$-scheme is geometrically reduced at a point if and only if the local ring is geometrically reduced.

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• 3 comment(s) on Section 33.6: Geometrically reduced schemes

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