Lemma 10.126.1. Let $R \to S$ be a ring map. Let $R \to R'$ be a faithfully flat ring map. Set $S' = R'\otimes _ R S$. Then $R \to S$ is of finite type if and only if $R' \to S'$ is of finite type.
Proof. It is clear that if $R \to S$ is of finite type then $R' \to S'$ is of finite type. Assume that $R' \to S'$ is of finite type. Say $y_1, \ldots , y_ m$ generate $S'$ over $R'$. Write $y_ j = \sum _ i a_{ij} \otimes x_{ji}$ for some $a_{ij} \in R'$ and $x_{ji} \in S$. Let $A \subset S$ be the $R$-subalgebra generated by the $x_{ij}$. By flatness we have $A' := R' \otimes _ R A \subset S'$, and by construction $y_ j \in A'$. Hence $A' = S'$. By faithful flatness $A = S$. $\square$
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