Lemma 35.14.1. Let $R \to A \to B$ be ring maps. Assume $R \to B$ is of finite presentation and $A \to B$ faithfully flat and of finite presentation. Then $R \to A$ is of finite presentation.

**Proof.**
Consider the algebra $C = B \otimes _ A B$ together with the pair of maps $p, q : B \to C$ given by $p(b) = b \otimes 1$ and $q(b) = 1 \otimes b$. Of course the two compositions $A \to B \to C$ are the same. Note that as $p : B \to C$ is flat and of finite presentation (base change of $A \to B$), the ring map $R \to C$ is of finite presentation (as the composite of $R \to B \to C$).

We are going to use the criterion Algebra, Lemma 10.127.3 to show that $R \to A$ is of finite presentation. Let $S$ be any $R$-algebra, and suppose that $S = \mathop{\mathrm{colim}}\nolimits _{\lambda \in \Lambda } S_\lambda $ is written as a directed colimit of $R$-algebras. Let $A \to S$ be an $R$-algebra homomorphism. We have to show that $A \to S$ factors through one of the $S_\lambda $. Consider the rings $B' = S \otimes _ A B$ and $C' = S \otimes _ A C = B' \otimes _ S B'$. As $B$ is faithfully flat of finite presentation over $A$, also $B'$ is faithfully flat of finite presentation over $S$. By Algebra, Lemma 10.168.1 part (2) applied to the pair $(S \to B', B')$ and the system $(S_\lambda )$ there exists a $\lambda _0 \in \Lambda $ and a flat, finitely presented $S_{\lambda _0}$-algebra $B_{\lambda _0}$ such that $B' = S \otimes _{S_{\lambda _0}} B_{\lambda _0}$. For $\lambda \geq \lambda _0$ set $B_\lambda = S_\lambda \otimes _{S_{\lambda _0}} B_{\lambda _0}$ and $C_\lambda = B_\lambda \otimes _{S_\lambda } B_\lambda $.

We interrupt the flow of the argument to show that $S_\lambda \to B_\lambda $ is faithfully flat for $\lambda $ large enough. (This should really be a separate lemma somewhere else, maybe in the chapter on limits.) Since $\mathop{\mathrm{Spec}}(B_{\lambda _0}) \to \mathop{\mathrm{Spec}}(S_{\lambda _0})$ is flat and of finite presentation it is open (see Morphisms, Lemma 29.25.10). Let $I \subset S_{\lambda _0}$ be an ideal such that $V(I) \subset \mathop{\mathrm{Spec}}(S_{\lambda _0})$ is the complement of the image. Note that formation of the image commutes with base change. Hence, since $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(S)$ is surjective, and $B' = B_{\lambda _0} \otimes _{S_{\lambda _0}} S$ we see that $IS = S$. Thus for some $\lambda \geq \lambda _0$ we have $IS_{\lambda } = S_\lambda $. For this and all greater $\lambda $ the morphism $\mathop{\mathrm{Spec}}(B_\lambda ) \to \mathop{\mathrm{Spec}}(S_\lambda )$ is surjective.

By analogy with the notation in the first paragraph of the proof denote $p_\lambda , q_\lambda : B_\lambda \to C_\lambda $ the two canonical maps. Then $B' = \mathop{\mathrm{colim}}\nolimits _{\lambda \geq \lambda _0} B_\lambda $ and $C' = \mathop{\mathrm{colim}}\nolimits _{\lambda \geq \lambda _0} C_\lambda $. Since $B$ and $C$ are finitely presented over $R$ there exist (by Algebra, Lemma 10.127.3 applied several times) a $\lambda \geq \lambda _0$ and an $R$-algebra maps $B \to B_\lambda $, $C \to C_\lambda $ such that the diagram

is commutative. OK, and this means that $A \to B \to B_\lambda $ maps into the equalizer of $p_\lambda $ and $q_\lambda $. By Lemma 35.3.6 we see that $S_\lambda $ is the equalizer of $p_\lambda $ and $q_\lambda $. Thus we get the desired ring map $A \to S_\lambda $ and we win. $\square$

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## Comments (2)

Comment #1662 by Lucas Braune on

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