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The Stacks project

Lemma 33.6.8. Let k be a field. Let X be a scheme over k.

  1. If x' \leadsto x is a specialization and X is geometrically reduced at x, then X is geometrically reduced at x'.

  2. If x \in X such that (a) \mathcal{O}_{X, x} is reduced, and (b) for each specialization x' \leadsto x where x' is a generic point of an irreducible component of X the scheme X is geometrically reduced at x', then X is geometrically reduced at x.

  3. If X is reduced and geometrically reduced at all generic points of irreducible components of X, then X is geometrically reduced.

Proof. Part (1) follows from Lemma 33.6.2 and the fact that if A is a geometrically reduced k-algebra, then S^{-1}A is a geometrically reduced k-algebra for any multiplicative subset S of A, see Algebra, Lemma 10.43.3.

Let A = \mathcal{O}_{X, x}. The assumptions (a) and (b) of (2) imply that A is reduced, and that A_{\mathfrak q} is geometrically reduced over k for every minimal prime \mathfrak q of A. Hence A is geometrically reduced over k, see Algebra, Lemma 10.43.7. Thus X is geometrically reduced at x, see Lemma 33.6.2.

Part (3) follows trivially from part (2). \square


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