## 37.54 Generic flatness stratification

We can use generic flatness to construct a stratification of the base such that a given module becomes flat over the strata.

Lemma 37.54.1 (Generic flatness stratification). Let $f : X \to S$ be a morphism of finite presentation between quasi-compact and quasi-separated schemes. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation. Then there exists a $t \geq 0$ and closed subschemes

$S \supset S_0 \supset S_1 \supset \ldots \supset S_ t = \emptyset$

such that $S_ i \to S$ is defined by a finite type ideal sheaf, $S_0 \subset S$ is a thickening, and $\mathcal{F}$ pulled back to $X \times _ S (S_ i \setminus S_{i + 1})$ is flat over $S_ i \setminus S_{i + 1}$.

Proof. We can find a cartesian diagram

$\xymatrix{ X \ar[d] \ar[r] & X_0 \ar[d] \\ S \ar[r] & S_0 }$

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 $t \geq 0$. $\square$

Lemma 37.54.2. Let $f : X \to S$ be a morphism of finite presentation between quasi-compact and quasi-separated schemes. Then there exists a $t \geq 0$ and closed subschemes

$S \supset S_0 \supset S_1 \supset \ldots \supset S_ t = \emptyset$

such that $S_ i \to S$ is defined by a finite type ideal sheaf, $S_0 \subset S$ is a thickening, and $X \times _ S (S_ i \setminus S_{i + 1})$ is flat over $S_ i \setminus S_{i + 1}$.

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

Lemma 37.54.3. Let $R$ be a Noetherian domain. Let $R \to A \to B$ be finite type ring maps. Let $M$ be a finite $A$-module and let $N$ a finite $B$-module. Let $M \to N$ be an $A$-linear map. There exists an nonzero $f \in R$ such that the cokernel of $M_ f \to N_ f$ is a flat $R_ f$-module.

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

$A_ g/A = \mathop{\mathrm{colim}}\nolimits (1/g^ m)A/A \cong \mathop{\mathrm{colim}}\nolimits A/g^ mA$

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 $A_ g/A$. $\square$

Let $f : X \to Y$ be a morphism of schemes over a base scheme $S$. Let $Z \subset Y$ be the scheme theoretic image of $f$, see Morphisms, Section 29.6. Let $g : S' \to S$ be a morphism of schemes and let $f' : X \times _ S S' \to Y \times _ S S'$ be the base change of $f$ by $g$. It is not always true that $Z \times _ S S' \subset Y \times _ S S'$ is the scheme theoretic image of $f'$. Let us say that formation of the scheme theoretic image of $f/S$ commutes with arbitrary base change if for every $g$ as above the scheme theoretic image of $f'$ is equal to $Z \times _ S S'$.

Lemma 37.54.4. Let $S$ be a quasi-compact and quasi-separated scheme. Let $f : X \to Y$ be a morphism of schemes over $S$ with both $X$ and $Y$ of finite presentation over $S$. Then there exists a $t \geq 0$ and closed subschemes

$S \supset S_0 \supset S_1 \supset \ldots \supset S_ t = \emptyset$

with the following properties:

1. $S_ i \to S$ is defined by a finite type ideal sheaf,

2. $S_0 \subset S$ is a thickening, and

3. with $T_ i = S_ i \setminus S_{i + 1}$ and $f_ i$ the base change of $f$ to $T_ i$ we have: formation of the scheme theoretic image of $f_ i/T_ i$ commutes with arbitrary base change (see discussion above the lemma).

Proof. We can find a commutative diagram

$\xymatrix{ X \ar[d] \ar[r] & Y \ar[d] \ar[r] & S \ar[d] \\ U \ar[r] & V \ar[r] & W }$

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 37.54.3 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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).