Lemma 33.6.2. Let k be a field. Let X be a scheme over k. Let x \in X. The following are equivalent
X is geometrically reduced at x, and
the ring \mathcal{O}_{X, x} is geometrically reduced over k (see Algebra, Definition 10.43.1).
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
X is geometrically reduced at x, and
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
Comments (1)
Comment #3018 by Brian Lawrence on
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