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The Stacks project

Lemma 37.24.6. Let f : X \to Y be a morphism of schemes. Let \eta \in Y be a generic point of an irreducible component of Y. Then (X_\eta )_{red} = (X_{red})_\eta .

Proof. Choose an affine neighbourhood \mathop{\mathrm{Spec}}(A) \subset Y of \eta . Choose an affine open \mathop{\mathrm{Spec}}(B) \subset X mapping into \mathop{\mathrm{Spec}}(A) via the morphism f. Let \mathfrak p \subset A be the minimal prime corresponding to \eta . Let B_{red} be the quotient of B by the nilradical \sqrt{(0)}. The algebraic content of the lemma is that C = B_{red} \otimes _ A \kappa (\mathfrak p) is reduced. Denote I \subset A the nilradical so that A_{red} = A/I. Denote \mathfrak p_{red} = \mathfrak p/I which is a minimal prime of A_{red} with \kappa (\mathfrak p) = \kappa (\mathfrak p_{red}). Since A \to B_{red} and A \to \kappa (\mathfrak p) both factor through A \to A_{red} we have C = B_{red} \otimes _{A_{red}} \kappa (\mathfrak p_{red}). Now \kappa (\mathfrak p_{red}) = (A_{red})_{\mathfrak p_{red}} is a localization by Algebra, Lemma 10.25.1. Hence C is a localization of B_{red} (Algebra, Lemma 10.12.15) and hence reduced. \square


Comments (3)

Comment #4325 by Thomas Brazelton on

The notation is overloaded in this proof.

Comment #4326 by Laurent Moret-Bailly on

A shorter way to prove that is reduced: since factors through , we have . But since is a minimal prime of , is the localization of at , so is a localization of , hence reduced.


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