The Stacks project

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 10.18.2). 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 10.44.3.

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. $\square$

Comments (1)

Comment #3018 by Brian Lawrence on

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

Post a comment




All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 035W. Beware of the difference between the letter 'O' and the digit '0'.