Lemma 10.41.9. Let $k$ be a field, and let $R$, $S$ be $k$-algebras. Let $S' \subset S$ be a sub $k$-algebra, and let $f \in S' \otimes _ k R$. In the commutative diagram
the images of the diagonal arrows are the same.
Proof. Let $\mathfrak p \subset R$ be in the image of the south-west arrow. This means (Lemma 10.18.6) that
is not the zero ring, i.e., $S' \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. The ring map $S' \otimes _ k \kappa (\mathfrak p) \to S \otimes _ k \kappa (\mathfrak p)$ is injective. Hence also $S \otimes _ k \kappa (\mathfrak p)$ is not the zero ring and the image of $f$ in it is not nilpotent. Hence $(S \otimes _ k R)_ f \otimes _ R \kappa (\mathfrak p)$ is not the zero ring. Thus (Lemma 10.18.6) we see that $\mathfrak p$ is in the image of the south-east arrow as desired. $\square$
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