The Stacks project

Proof. We can find a cartesian diagram

and a finitely presented $\mathcal{O}_{X_0}$-module $\mathcal{F}_0$ which pulls back to $\mathcal{F}$ such that $X_0$ and $S_0$ are of finite type over $\mathbf{Z}$. See Limits, Proposition 32.5.4 and Lemmas 32.10.1 and 32.10.2. Thus we may assume $X$ and $S$ are of finite type over $\mathbf{Z}$ and $\mathcal{F}$ is a coherent $\mathcal{O}_ X$-module.

Assume $X$ and $S$ are of finite type over $\mathbf{Z}$ and $\mathcal{F}$ is a coherent $\mathcal{O}_ X$-module. In this case every quasi-coherent ideal is of finite type, hence we do not have to check the condition that $S_ i$ is cut out by a finite type ideal. Set $S_0 = S_{red}$ equal to the reduction of $S$. By generic flatness as stated in Morphisms, Proposition 29.27.2 there is a dense open $U_0 \subset S_0$ such that $\mathcal{F}$ pulled back to $X \times _ S U_0$ is flat over $U_0$. Let $S_1 \subset S_0$ be the reduced closed subscheme whose underlying closed subset is $S \setminus U_0$. We continue in this way, provided $S_1 \not= \emptyset $, to find $S_0 \supset S_1 \supset \ldots $. Because $S$ is Noetherian any descending chain of closed subsets stabilizes hence we see that $S_ t = \emptyset $ for some $t \geq 0$. $\square$

Proof. Apply Lemma 37.54.1 with $\mathcal{F} = \mathcal{O}_ X$. $\square$

Proof. We omit the proof that (1) implies (2). Assume (2). To prove (1) is a local question, hence we may assume that $S$ is affine. Choose a stratification $S \supset S_0 \supset S_1 \supset \ldots \supset S_ t = \emptyset $ as in Lemma 37.54.2. Set $T_ i = S_ i \setminus S_{i - 1}$ and $Y_ i = X \times _ S T_ i$. Denote $g_ i : T_ i \to S$ and $h_ i : Y_ i \to X$ the inclusion morphisms. Since $f$ is surjective, the morphisms $f_ i : X_ i \to T_ i$ are surjective and flat. Since $Lh_ i^* \circ Lf^* = Lf_ i^* \circ Lg_ i^*$ we conclude from (2) that $Lf_ i^* Lg_ i^*M$ is isomorphic to a flat $\mathcal{O}_{Y_ i}$-module placed in degree $0$. Since $f_ i$ is flat and surjective, we see that $Lg_ i^*M$ is isomorphic to a flat $\mathcal{O}_{T_ i}$-module placed in degree $0$.

We will prove this implies the result by induction on $t$. We will use the following notation: $S = \mathop{\mathrm{Spec}}(A)$ and $M$ is the object of $D_\mathit{QCoh}(\mathcal{O}_ S)$ corresponding to $N$ in $D(A)$, see Derived Categories of Schemes, Lemma 36.3.5.

Base case: $t = 1$. Write $S_0 = \mathop{\mathrm{Spec}}(A/I)$ where $I$ is a finite generated ideal. Then $I$ is nilpotent as $S_0 = S$ set theoretically. The assumption is that $N \otimes _ A^\mathbf {L} A/I$ has tor amplitude in $[0, 0]$. By More on Algebra, Lemma 15.67.20 the same is true for $N$.

Induction step. Assume $t > 1$. Write $S_{t - 1} = \mathop{\mathrm{Spec}}(A/I)$ where $I = (f_1, \ldots , f_ r)$ is a finitely generated ideal. We will argue by induction on $r$. Observe that $N_{f_ r} = N \otimes _ A^\mathbf {L} A_{f_ r}$ in $D(A_{f_ r})$ has tor amplitude in $[0, 0]$ by induction hypothesis (because the stratification on the principal open $D(f_ r)$ has length at most $t - 1$). On the other hand, consider $N' = N \otimes _ A^\mathbf {L} A/f_ rA$ in $D(A/f_ rA)$. Setting $S' = \mathop{\mathrm{Spec}}(A/f_ rA)$ we obtain a stratification

where $S' \cap S_{t - 1}$ is cut out by $r - 1$ equations in $S'$. Similarly to before, the derived pullbacks of $M$ to the parts $S' \cap S_ i \setminus S' \cap S_{i + 1}$ are of tor amplitude in $[0, 0]$. Whence by induction on $r$ we conclude that $N'$ has tor amplitude in $[0, 0]$. Then we finally conclude that $N$ has tor amplitude in $[0, 0]$ by More on Algebra, Lemma 15.103.7. $\square$

Proof. By replacing $M$ by the image of $M \to N$, we may assume $M \subset N$. Choose a filtration $0 = N_0 \subset N_1 \subset \ldots \subset N_ t = N$ such that $N_ i/N_{i - 1} = B/\mathfrak q_ i$ for some prime ideal $\mathfrak q_ i \subset B$, see Algebra, Lemma 10.62.1. Set $M_ i = M \cap N_ i$. Then $Q = N/M$ has a filtration by the submodules $Q_ i = N_ i/M_ i$. It suffices to prove $Q_ i/Q_{i - 1}$ becomes flat after localizing at a nonzero element of $f$ (since extensions of flat modules are flat by Algebra, Lemma 10.39.13). Since $Q_ i/Q_{i - 1}$ is isomorphic to the cokernel of the map $M_ i/M_{i - 1} \to N_ i/N_{i - 1}$, we reduce to the case discussed in the next paragraph.

Assume $B$ is a domain and $M \subset N = B$. After replacing $A$ by the image of $A$ in $B$ we may assume $A \subset B$. By generic flatness, we may assume $A$ and $B$ are flat over $R$ (Algebra, Lemma 10.118.1). It now suffices to show $M \to B$ becomes $R$-universally injective after replacing $R$ by a principal localization (Algebra, Lemma 10.82.7). By generic freeness, we can find a nonzero $g \in A$ such that $B_ g$ is a free $A_ g$-module (Algebra, Lemma 10.118.1). Thus we may choose a direct summand $M' \subset B_ g$ as an $A_ g$-module, which is finite free as an $A_ g$-module, and such that $M \to B \to B_ g$ factors through $M'$. Clearly, it suffices to show that $M \to M'$ becomes $R$-universally injective after replacing $R$ by a principal localization.

Say $M' = A_ g^{\oplus n}$. Since $M \subset M'$ is a finite $A$-module, we see that $M$ is contained in $(1/g^ m)A^{\oplus n}$ for some $m \geq 0$. After changing our basis for $M'$ we may assume $M \subset A^{\oplus n}$. Then it suffices to show that $A^{\oplus n}/M$ and $A_ g/A$ become $R$-flat after replacing $R$ by a principal localization. Namely, then $M \to A^{\oplus n}$ and $A^{\oplus n} \to A_ g^{\oplus n}$ are universally injective by Algebra, Lemma 10.39.12 and consequently so is the composition $M \to M' = A_ g^{\oplus n}$.

By generic flatness (see reference above), we may assume the module $A^{\oplus n}/M$ is $R$-flat. For the quotient $A_ g/A$ we use the fact that

and the module $A/g^ mA$ has a filtration of length $m$ whose successive quotients are isomorphic to $A/gA$. Again by generic flatness we may assume $A/gA$ is $R$-flat and hence each $A/g^ mA$ is $R$-flat, and hence so is $A_ g/A$. $\square$

Proof. We can find a commutative diagram

with cartesian squares such that $U$, $V$, $W$ are of finite type over $\mathbf{Z}$. Namely, first write $S$ as a cofiltered limit of finite type schemes over $\mathbf{Z}$ with affine transition morphisms using Limits, Proposition 32.5.4 and then descend the morphism $X \to Y$ using Limits, Lemma 32.10.1. This reduces us to the case discussed in the next paragraph.

Assume $S$ is Noetherian. In this case every quasi-coherent ideal is of finite type, hence we do not have to check the condition that $S_ i$ is cut out by a finite type ideal. Set $S_0 = S_{red}$ equal to the reduction of $S$. Let $\eta \in S_0$ be a generic point of an irreducible component of $S_0$. By Noetherian induction on the underlying topological space of $S_0$, we may assume the result holds for any closed subscheme of $S_0$ not containing $\eta $. Thus it suffices to show that there exists an open neighbourhood $U_0 \subset S_0$ such that the base change $f_0$ of $f$ to $U_0$ has property (3).

Let $R$ be a Noetherian domain. Let $f : X \to Y$ be a morphism of finite type schemes over $R$. By the discussion in the previous paragraph it suffices to show that after replacing $R$ by $R_ g$ for some $g \in R$ nonzero and $X$, $Y$ by their base changes to $R_ g$, formation of the scheme theoretic image of $f/R$ commutes with arbitrary base change.

Let $Y = V_1 \cup \ldots V_ n$ be an affine open covering. Let $U_ i = f^{-1}(V_ i)$. If the statement is true for each of the morphisms $U_ i \to V_ i$ over $R$, then it holds for $f$. Namely, the scheme theoretic image of $U_ i \to V_ i$ is the intersection of $V_ i$ with the scheme theoretic image of $f : X \to Y$ by Morphisms, Lemma 29.6.3. Thus we may assume $Y$ is affine.

Let $X = U_1 \cup \ldots U_ n$ be an affine open covering. Then the scheme theoretic image of $X \to Y$ is the same as the scheme theoretic imge of $\coprod U_ i \to Y$. Thus we may assume $X$ is affine.

Say $X = \mathop{\mathrm{Spec}}(A)$ and $Y = \mathop{\mathrm{Spec}}(B)$ and $f$ corresponds to the $R$-algebra map $\varphi : A \to B$. Then the scheme theoretic image of $f$ is $\mathop{\mathrm{Spec}}(A/\mathop{\mathrm{Ker}}(\varphi ))$ and similarly after base change (by an affine morphism, but it is enough to check for those). Thus formation of the scheme theoretic image commutes with base change if $\mathop{\mathrm{Ker}}(\varphi \otimes _ R R') = \mathop{\mathrm{Ker}}(\varphi ) \otimes _ R R'$ for all ring maps $R \to R'$.

After replacing $R$, $A$, $B$ by $R_ g$, $A_ g$, $B_ g$ for a suitable nonzero $g$ in $R$, we may assume $A$ and $B$ are flat over $R$. By Lemma 37.54.4 we may also assume $B/A$ is a flat $R$-module. Then $0 \to \mathop{\mathrm{Ker}}(\varphi ) \to A \to B \to B/A \to 0$ is an exact sequence of flat $R$-modules, which implies the desired base change statement. $\square$