# The Stacks Project

## Tag: 0367

This tag has label descent-lemma-finite-type-local-source-fppf-algebra and it points to

The corresponding content:

Lemma 31.10.2. 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 9.154.2 there exists a commtutative 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. $\square$

\begin{lemma}
\label{lemma-finite-type-local-source-fppf-algebra}
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.
\end{lemma}

\begin{proof}
By
Algebra, Lemma \ref{algebra-lemma-descend-faithfully-flat-finite-presentation}
there exists a commtutative 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.
\end{proof}


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