Lemma 35.14.2. Let $R \to A \to B$ be ring maps. Assume $R \to B$ is of finite type and $A \to B$ faithfully flat and of finite presentation. Then $R \to A$ is of finite type.

Proof. By Algebra, Lemma 10.168.2 there exists a commutative diagram

$\xymatrix{ R \ar[r] \ar@{=}[d] & A_0 \ar[d] \ar[r] & B_0 \ar[d] \\ R \ar[r] & A \ar[r] & B }$

with $R \to A_0$ of finite presentation, $A_0 \to B_0$ faithfully flat of finite presentation and $B = A \otimes _{A_0} B_0$. Since $R \to B$ is of finite type by assumption, we may add some elements to $A_0$ and assume that the map $B_0 \to B$ is surjective! In this case, since $A_0 \to B_0$ is faithfully flat, we see that as

$(A_0 \to A) \otimes _{A_0} B_0 \cong (B_0 \to B)$

is surjective, also $A_0 \to A$ is surjective. Hence we win. $\square$

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