The Stacks project

Proof. To see the equivalence of (1) and (2) it suffices to check conditions (1)(a), (b), (c) of Morphisms, Definition 29.14.1 for the property of being a pseudo-coherent ring map. These properties follow (using localization is flat) from More on Algebra, Lemmas 15.81.12, 15.81.11, and 15.81.16.

If (1) holds, then $f$ is locally of finite type as a pseudo-coherent ring map is of finite type by definition. Moreover, (1) implies via Lemma 37.59.7 and the definitions that $\mathcal{O}_ X$ is pseudo-coherent relative to $S$. Conversely, if (3) holds, then we see that for every $U$ and $V$ as in (2) the ring $\mathcal{O}_ X(U)$ is of finite type over $\mathcal{O}_ S(V)$ and $\mathcal{O}_ X(U)$ is as a module pseudo-coherent relative to $\mathcal{O}_ S(V)$, see Lemmas 37.59.6 and 37.59.7. This is the definition of a pseudo-coherent ring map, hence (2) and (1) hold. $\square$

Proof. This translates into the following algebra result: Let $A \to B$ be a pseudo-coherent ring map. Let $A \to A'$ be flat. Then $A' \to B \otimes _ A A'$ is pseudo-coherent. This follows from the more general More on Algebra, Lemma 15.81.12. $\square$

Proof. This translates into the following algebra result: If $A \to B \to C$ are composable pseudo-coherent ring maps then $A \to C$ is pseudo-coherent. This follows from either More on Algebra, Lemma 15.81.13 or More on Algebra, Lemma 15.81.15. $\square$

Proof. Immediate from the definitions. $\square$

Proof. This follows from the fact that a flat ring map of finite presentation is pseudo-coherent (and even perfect), see More on Algebra, Lemma 15.82.4. $\square$

Proof. This translates into the following algebra result: If $R \to A \to B$ are composable ring maps and $R \to A$, $R \to B$ pseudo-coherent, then $R \to B$ is pseudo-coherent. This follows from More on Algebra, Lemma 15.81.15. $\square$

Proof. Translated into algebra this lemma says the following: If $R \to A$ is a finite ring map, then $R \to A$ is pseudo-coherent as a ring map (which means by definition that $A$ as an $A$-module is pseudo-coherent relative to $R$) if and only if $A$ is pseudo-coherent as an $R$-module. This follows from the more general More on Algebra, Lemma 15.81.5. $\square$

Proof. This translates into the following algebra result: If $R \to A$ is a finite type ring map with $R$ Noetherian, then $R \to A$ is pseudo-coherent if and only if $R \to A$ is of finite type. To see this, note that a pseudo-coherent ring map is of finite type by definition. Conversely, if $R \to A$ is of finite type, then we can write $A = R[x_1, \ldots , x_ n]/I$ and it follows from More on Algebra, Lemma 15.64.17 that $A$ is pseudo-coherent as an $R[x_1, \ldots , x_ n]$-module, i.e., $R \to A$ is a pseudo-coherent ring map. $\square$

Proof. We will use the criterion of Descent, Lemma 35.22.4 to prove this. By Definition 37.60.2 being pseudo-coherent is Zariski local on the base. By Lemma 37.60.3 being pseudo-coherent is preserved under flat base change. The final hypothesis (3) of Descent, Lemma 35.22.4 translates into the following algebra statement: Let $A \to B$ be a faithfully flat ring map. Let $C = A[x_1, \ldots , x_ n]/I$ be an $A$-algebra. If $C \otimes _ A B$ is pseudo-coherent as an $B[x_1, \ldots , x_ n]$-module, then $C$ is pseudo-coherent as a $A[x_1, \ldots , x_ n]$-module. This is More on Algebra, Lemma 15.64.15. $\square$

Proof. Choose a presentation $B = A[x_1, \ldots , x_ n]/J$. Note that $B$ is pseudo-coherent as an $A[x_1, \ldots , x_ n]$-module because $A \to B$ is a pseudo-coherent ring map by Lemma 37.60.6. Note that $A \to B/I$ is pseudo-coherent if and only if $B/I$ is pseudo-coherent as an $A[x_1, \ldots , x_ n]$-module. By More on Algebra, Lemma 15.64.11 we see this is equivalent to the condition that $B/I$ is pseudo-coherent as an $B$-module. This proves the lemma as the short exact sequence $0 \to I \to B \to B/I \to 0$ shows that $I$ is pseudo-coherent if and only if $B/I$ is (see More on Algebra, Lemma 15.64.6). $\square$

Proof. We will use the criterion of Descent, Lemma 35.26.4 to prove this. It follows from Lemmas 37.60.6 and 37.60.4 that being pseudo-coherent is preserved under precomposing with flat morphisms locally of finite presentation, in particular under precomposing with syntomic morphisms (see Morphisms, Lemmas 29.30.7 and 29.30.6). It is clear from Definition 37.60.2 that being pseudo-coherent is Zariski local on the source and target. Hence, according to the aforementioned Descent, Lemma 35.26.4 it suffices to prove the following: Suppose $X' \to X \to Y$ are morphisms of affine schemes with $X' \to X$ syntomic and $X' \to Y$ pseudo-coherent. Then $X \to Y$ is pseudo-coherent. To see this, note that in any case $X \to Y$ is of finite presentation by Descent, Lemma 35.14.1. Choose a closed immersion $X \to \mathbf{A}^ n_ Y$. By Algebra, Lemma 10.136.18 we can find an affine open covering $X' = \bigcup _{i = 1, \ldots , n} X'_ i$ and syntomic morphisms $W_ i \to \mathbf{A}^ n_ Y$ lifting the morphisms $X'_ i \to X$, i.e., such that there are fibre product diagrams

After replacing $X'$ by $\coprod X'_ i$ and setting $W = \coprod W_ i$ we obtain a fibre product diagram

with $W \to \mathbf{A}^ n_ Y$ flat and of finite presentation and $X' \to Y$ still pseudo-coherent. Since $W \to \mathbf{A}^ n_ Y$ is open (see Morphisms, Lemma 29.25.10) and $X' \to X$ is surjective we can find $f \in \Gamma (\mathbf{A}^ n_ Y, \mathcal{O})$ such that $X \subset D(f) \subset \mathop{\mathrm{Im}}(h)$. Write $Y = \mathop{\mathrm{Spec}}(R)$, $X = \mathop{\mathrm{Spec}}(A)$, $X' = \mathop{\mathrm{Spec}}(A')$ and $W = \mathop{\mathrm{Spec}}(B)$, $A = R[x_1, \ldots , x_ n]/I$ and $A' = B/IB$. Then $R \to A'$ is pseudo-coherent. Picture

By Lemma 37.60.11 we see that $IB$ is pseudo-coherent as a $B$-module. The ring map $R[x_1, \ldots , x_ n]_ f \to B_ f$ is faithfully flat by our choice of $f$ above. This implies that $I_ f \subset R[x_1, \ldots , x_ n]_ f$ is pseudo-coherent, see More on Algebra, Lemma 15.64.15. Applying Lemma 37.60.11 one more time we see that $R \to A$ is pseudo-coherent. $\square$

Proof. Let $f : X \to S$ be a morphism of schemes. Let $\{ g_ i : X_ i \to X\} $ be an fppf covering such that each composition $f \circ g_ i$ is pseudo-coherent. According to Lemma 37.48.2 there exist

1. a Zariski open covering $X = \bigcup U_ j$,

2. surjective finite locally free morphisms $W_ j \to U_ j$,

3. Zariski open coverings $W_ j = \bigcup _ k W_{j, k}$,

4. surjective finite locally free morphisms $T_{j, k} \to W_{j, k}$

such that the fppf covering $\{ h_{j, k} : T_{j, k} \to X\} $ refines the given covering $\{ X_ i \to X\} $. Denote $\psi _{j, k} : T_{j, k} \to X_{\alpha (j, k)}$ the morphisms that witness the fact that $\{ T_{j, k} \to X\} $ refines the given covering $\{ X_ i \to X\} $. Note that $T_{j, k} \to X$ is a flat, locally finitely presented morphism, so both $X_ i$ and $T_{j, k}$ are pseudo-coherent over $X$ by Lemma 37.60.6. Hence $\psi _{j, k} : T_{j, k} \to X_ i$ is pseudo-coherent, see Lemma 37.60.7. Hence $T_{j, k} \to S$ is pseudo coherent as the composition of $\psi _{j, k}$ and $f \circ g_{\alpha (j, k)}$, see Lemma 37.60.4. Thus we see we have reduced the lemma to the case of a Zariski open covering (which is OK) and the case of a covering given by a single surjective finite locally free morphism which we deal with in the following paragraph.

Assume that $X' \to X \to S$ is a sequence of morphisms of schemes with $X' \to X$ surjective finite locally free and $X' \to Y$ pseudo-coherent. Our goal is to show that $X \to S$ is pseudo-coherent. Note that by Descent, Lemma 35.14.3 the morphism $X \to S$ is locally of finite presentation. It is clear that the problem reduces to the case that $X'$, $X$ and $S$ are affine and $X' \to X$ is free of some rank $r > 0$. The corresponding algebra problem is the following: Suppose $R \to A \to A'$ are ring maps such that $R \to A'$ is pseudo-coherent, $R \to A$ is of finite presentation, and $A' \cong A^{\oplus r}$ as an $A$-module. Goal: Show $R \to A$ is pseudo-coherent. The assumption that $R \to A'$ is pseudo-coherent means that $A'$ as an $A'$-module is pseudo-coherent relative to $R$. By More on Algebra, Lemma 15.81.5 this implies that $A'$ as an $A$-module is pseudo-coherent relative to $R$. Since $A' \cong A^{\oplus r}$ as an $A$-module we see that $A$ as an $A$-module is pseudo-coherent relative to $R$, see More on Algebra, Lemma 15.81.8. This by definition means that $R \to A$ is pseudo-coherent and we win. $\square$