The Stacks project

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. $\square$

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

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

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

Proof. The problem is local on $S$ and $Y$. Hence we may assume that $S$ and $Y$ are affine. Since $f$ is flat and locally of finite presentation, we see that $f$ is open (Morphisms, Lemma 29.25.10). Hence, since $Y$ is quasi-compact, there exist finitely many affine opens $X_ i \subset X$ such that $Y = \bigcup f(X_ i)$. Clearly we may replace $X$ by $\coprod X_ i$, and hence we may assume $X$ is affine as well. In this case the lemma is equivalent to Lemma 35.14.1 (resp. Lemma 35.14.2) above. $\square$

Proof. Combine Morphisms, Lemmas 29.30.16, 29.34.19, and 29.36.19 with Lemma 35.14.3 above. $\square$

Proof. Assume (1) and that $p$ is smooth. By Lemma 35.14.3 we see that $q$ is locally of finite presentation. By Morphisms, Lemma 29.25.13 we see that $q$ is flat. Hence now it suffices to show that the fibres of $q$ are smooth, see Morphisms, Lemma 29.34.3. Apply Varieties, Lemma 33.25.9 to the flat surjective morphisms $X_ s \to Y_ s$ for $s \in S$ to conclude. We omit the proof of the étale case. $\square$

Proof. By Lemma 35.14.3 we see that $q$ is of finite presentation. By Morphisms, Lemma 29.25.13 we see that $q$ is flat. By Morphisms, Lemma 29.30.10 it now suffices to show that the local rings of the fibres of $Y \to S$ and the fibres of $X \to Y$ are local complete intersection rings. To do this we may take the fibre of $X \to Y \to S$ at a point $s \in S$, i.e., we may assume $S$ is the spectrum of a field. Pick a point $x \in X$ with image $y \in Y$ and consider the ring map

This is a flat local homomorphism of local Noetherian rings. The local ring $\mathcal{O}_{X, x}$ is a complete intersection. Thus may use Avramov's result, see Divided Power Algebra, Lemma 23.8.9, to conclude that both $\mathcal{O}_{Y, y}$ and $\mathcal{O}_{X, x}/\mathfrak m_ y\mathcal{O}_{X, x}$ are complete intersection rings. $\square$

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. $\square$