The Stacks project

Proof. Let $\mathop{\mathrm{Spec}}(C) \subset Z$ be an affine open and let $\mathop{\mathrm{Spec}}(B) \subset Y$ be an affine open which maps into $\mathop{\mathrm{Spec}}(C)$. The assumption on $X \to Y$ implies we can find a standard affine fppf covering $\{ \mathop{\mathrm{Spec}}(B_ j) \to \mathop{\mathrm{Spec}}(B)\}$ and lifts $x_ j : \mathop{\mathrm{Spec}}(B_ j) \to X$. Since $\mathop{\mathrm{Spec}}(B_ j)$ is quasi-compact we can find finitely many affine opens $\mathop{\mathrm{Spec}}(A_ i) \subset X$ lying over $\mathop{\mathrm{Spec}}(B)$ such that the image of each $x_ j$ is contained in the union $\bigcup \mathop{\mathrm{Spec}}(A_ i)$. Hence after replacing each $\mathop{\mathrm{Spec}}(B_ j)$ by a standard affine Zariski coverings of itself we may assume we have a standard affine fppf covering $\{ \mathop{\mathrm{Spec}}(B_ i) \to \mathop{\mathrm{Spec}}(B)\}$ such that each $\mathop{\mathrm{Spec}}(B_ i) \to Y$ factors through an affine open $\mathop{\mathrm{Spec}}(A_ i) \subset X$ lying over $\mathop{\mathrm{Spec}}(B)$. In other words, we have ring maps $C \to B \to A_ i \to B_ i$ for each $i$. Note that we can also consider

and that the ring map $B \to \prod B_ i$ is faithfully flat and of finite presentation.

The case $P = flat$. In this case we know that $C \to A$ is flat and we have to prove that $C \to B$ is flat. Suppose that $N \to N' \to N''$ is an exact sequence of $C$-modules. We want to show that $N \otimes _ C B \to N' \otimes _ C B \to N'' \otimes _ C B$ is exact. Let $H$ be its cohomology and let $H'$ be the cohomology of $N \otimes _ C B' \to N' \otimes _ C B' \to N'' \otimes _ C B'$. As $B \to B'$ is flat we know that $H' = H \otimes _ B B'$. On the other hand $N \otimes _ C A \to N' \otimes _ C A \to N'' \otimes _ C A$ is exact hence has zero cohomology. Hence the map $H \to H'$ is zero (as it factors through the zero module). Thus $H' = 0$. As $B \to B'$ is faithfully flat we conclude that $H = 0$ as desired.

The case $P = locally\ finite\ type$. In this case we know that $C \to A$ is of finite type and we have to prove that $C \to B$ is of finite type. Because $B \to B'$ is of finite presentation (hence of finite type) we see that $A \to B'$ is of finite type, see Algebra, Lemma 10.6.2. Therefore $C \to B'$ is of finite type and we conclude by Lemma 35.14.2.

The case $P = locally\ finite\ presentation$. In this case we know that $C \to A$ is of finite presentation and we have to prove that $C \to B$ is of finite presentation. Because $B \to B'$ is of finite presentation and $B \to A$ of finite type we see that $A \to B'$ is of finite presentation, see Algebra, Lemma 10.6.2. Therefore $C \to B'$ is of finite presentation and we conclude by Lemma 35.14.1. $\square$