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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