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. $\square$

Comment #4325 by Thomas Brazelton on

The notation $f$ is overloaded in this proof.

Comment #4326 by Laurent Moret-Bailly on

A shorter way to prove that $C:=B_{red}\otimes_A\kappa(\mathfrak{p})$ is reduced: since $A\to B_{red}$ factors through $A_{red}$, we have $C=B_{red}\otimes_{A_{red}}\kappa(\mathfrak{p})$. But since $\mathfrak{p}$ is a minimal prime of $A$, $\kappa(\mathfrak{p})$ is the localization of $A_{red}$ at $\mathfrak{p}$, so $C$ is a localization of $B_{red}$, hence reduced.

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