Lemma 10.107.9. Let R \to S be an epimorphism of rings. Then
\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R) is injective, and
for \mathfrak q \subset S lying over \mathfrak p \subset R we have \kappa (\mathfrak p) = \kappa (\mathfrak q).
Proof. Let \mathfrak p be a prime of R. The fibre of the map is the spectrum of the fibre ring S \otimes _ R \kappa (\mathfrak p). By Lemma 10.107.3 the map \kappa (\mathfrak p) \to S \otimes _ R \kappa (\mathfrak p) is an epimorphism, and hence by Lemma 10.107.8 we have either S \otimes _ R \kappa (\mathfrak p) = 0 or S \otimes _ R \kappa (\mathfrak p) = \kappa (\mathfrak p) which proves (1) and (2). \square
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