Lemma 31.9.1. Let $f : T \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ S$-module. Then $f^{-1}\text{Fit}_ i(\mathcal{F}) \cdot \mathcal{O}_ T = \text{Fit}_ i(f^*\mathcal{F})$.

## 31.9 Fitting ideals

This section is the continuation of the discussion in More on Algebra, Section 15.8. Let $S$ be a scheme. Let $\mathcal{F}$ be a finite type quasi-coherent $\mathcal{O}_ S$-module. In this situation we can construct the Fitting ideals

as the sequence of quasi-coherent ideals characterized by the following property: for every affine open $U = \mathop{\mathrm{Spec}}(A)$ of $S$ if $\mathcal{F}|_ U$ corresponds to the $A$-module $M$, then $\text{Fit}_ i(\mathcal{F})|_ U$ corresponds to the ideal $\text{Fit}_ i(M) \subset A$. This is well defined and a quasi-coherent sheaf of ideals because if $f \in A$, then the $i$th Fitting ideal of $M_ f$ over $A_ f$ is equal to $\text{Fit}_ i(M) A_ f$ by More on Algebra, Lemma 15.8.4.

Alternatively, we can construct the Fitting ideals in terms of local presentations of $\mathcal{F}$. Namely, if $U \subset X$ is open, and

is a presentation of $\mathcal{F}$ over $U$, then $\text{Fit}_ r(\mathcal{F})|_ U$ is generated by the $(n - r) \times (n - r)$-minors of the matrix defining the first arrow of the presentation. This is compatible with the construction above because this is how the Fitting ideal of a module over a ring is actually defined. Some details omitted.

**Proof.**
Follows immediately from More on Algebra, Lemma 15.8.4 part (3).
$\square$

Lemma 31.9.2. Let $S$ be a scheme. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ S$-module. Then $\text{Fit}_ r(\mathcal{F})$ is a quasi-coherent ideal of finite type.

**Proof.**
Follows immediately from More on Algebra, Lemma 15.8.4 part (4).
$\square$

Lemma 31.9.3. Let $S$ be a scheme. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ S$-module. Let $Z_0 \subset S$ be the closed subscheme cut out by $\text{Fit}_0(\mathcal{F})$. Let $Z \subset S$ be the scheme theoretic support of $\mathcal{F}$. Then

$Z \subset Z_0 \subset S$ as closed subschemes,

$Z = Z_0 = \text{Supp}(\mathcal{F})$ as closed subsets,

there exists a finite type, quasi-coherent $\mathcal{O}_{Z_0}$-module $\mathcal{G}_0$ with

\[ (Z_0 \to X)_*\mathcal{G}_0 = \mathcal{F}. \]

**Proof.**
Recall that $Z$ is locally cut out by the annihilator of $\mathcal{F}$, see Morphisms, Definition 29.5.5 (which uses Morphisms, Lemma 29.5.4 to define $Z$). Hence we see that $Z \subset Z_0$ scheme theoretically by More on Algebra, Lemma 15.8.4 part (6). On the other hand we have $Z = \text{Supp}(\mathcal{F})$ set theoretically by Morphisms, Lemma 29.5.4 and we have $Z_0 = Z$ set theoretically by More on Algebra, Lemma 15.8.4 part (7). Finally, to get $\mathcal{G}_0$ as in part (3) we can either use that we have $\mathcal{G}$ on $Z$ as in Morphisms, Lemma 29.5.4 and set $\mathcal{G}_0 = (Z \to Z_0)_*\mathcal{G}$ or we can use Morphisms, Lemma 29.4.1 and the fact that $\text{Fit}_0(\mathcal{F})$ annihilates $\mathcal{F}$ by More on Algebra, Lemma 15.8.4 part (6).
$\square$

Lemma 31.9.4. Let $S$ be a scheme. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ S$-module. Let $s \in S$. Then $\mathcal{F}$ can be generated by $r$ elements in a neighbourhood of $s$ if and only if $\text{Fit}_ r(\mathcal{F})_ s = \mathcal{O}_{S, s}$.

**Proof.**
Follows immediately from More on Algebra, Lemma 15.8.6.
$\square$

Lemma 31.9.5. Let $S$ be a scheme. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ S$-module. Let $r \geq 0$. The following are equivalent

$\mathcal{F}$ is finite locally free of rank $r$

$\text{Fit}_{r - 1}(\mathcal{F}) = 0$ and $\text{Fit}_ r(\mathcal{F}) = \mathcal{O}_ S$, and

$\text{Fit}_ k(\mathcal{F}) = 0$ for $k < r$ and $\text{Fit}_ k(\mathcal{F}) = \mathcal{O}_ S$ for $k \geq r$.

**Proof.**
Follows immediately from More on Algebra, Lemma 15.8.7.
$\square$

Lemma 31.9.6. Let $S$ be a scheme. Let $\mathcal{F}$ be a finite type, quasi-coherent $\mathcal{O}_ S$-module. The closed subschemes

defined by the Fitting ideals of $\mathcal{F}$ have the following properties

The intersection $\bigcap Z_ r$ is empty.

The functor $(\mathit{Sch}/S)^{opp} \to \textit{Sets}$ defined by the rule

\[ T \longmapsto \left\{ \begin{matrix} \{ *\} & \text{if }\mathcal{F}_ T\text{ is locally generated by } \leq r\text{ sections} \\ \emptyset & \text{otherwise} \end{matrix} \right. \]is representable by the open subscheme $S \setminus Z_ r$.

The functor $F_ r : (\mathit{Sch}/S)^{opp} \to \textit{Sets}$ defined by the rule

\[ T \longmapsto \left\{ \begin{matrix} \{ *\} & \text{if }\mathcal{F}_ T\text{ locally free rank }r \\ \emptyset & \text{otherwise} \end{matrix} \right. \]is representable by the locally closed subscheme $Z_{r - 1} \setminus Z_ r$ of $S$.

If $\mathcal{F}$ is of finite presentation, then $Z_ r \to S$, $S \setminus Z_ r \to S$, and $Z_{r - 1} \setminus Z_ r \to S$ are of finite presentation.

**Proof.**
Part (1) is true because over every affine open $U$ there is an integer $n$ such that $\text{Fit}_ n(\mathcal{F})|_ U = \mathcal{O}_ U$. Namely, we can take $n$ to be the number of generators of $\mathcal{F}$ over $U$, see More on Algebra, Section 15.8.

For any morphism $g : T \to S$ we see from Lemmas 31.9.1 and 31.9.4 that $\mathcal{F}_ T$ is locally generated by $\leq r$ sections if and only if $\text{Fit}_ r(\mathcal{F}) \cdot \mathcal{O}_ T = \mathcal{O}_ T$. This proves (2).

For any morphism $g : T \to S$ we see from Lemmas 31.9.1 and 31.9.5 that $\mathcal{F}_ T$ is free of rank $r$ if and only if $\text{Fit}_ r(\mathcal{F}) \cdot \mathcal{O}_ T = \mathcal{O}_ T$ and $\text{Fit}_{r - 1}(\mathcal{F}) \cdot \mathcal{O}_ T = 0$. This proves (3).

Assume $\mathcal{F}$ is of finite presentation. Then each of the morphisms $Z_ r \to S$ is of finite presentation as $\text{Fit}_ r(\mathcal{F})$ is of finite type (Lemma 31.9.2 and Morphisms, Lemma 29.21.7). This implies that $Z_{r - 1} \setminus Z_ r$ is a retrocompact open in $Z_ r$ (Properties, Lemma 28.24.1) and hence the morphism $Z_{r - 1} \setminus Z_ r \to Z_ r$ is of finite presentation as well. $\square$

Lemma 31.9.6 notwithstanding the following lemma does not hold if $\mathcal{F}$ is a finite type quasi-coherent module. Namely, the stratification still exists but it isn't true that it represents the functor $F_{flat}$ in general.

Lemma 31.9.7. Let $S$ be a scheme. Let $\mathcal{F}$ be an $\mathcal{O}_ S$-module of finite presentation. Let $S = Z_{-1} \supset Z_0 \supset Z_1 \supset \ldots $ be as in Lemma 31.9.6. Set $S_ r = Z_{r - 1} \setminus Z_ r$. Then $S' = \coprod _{r \geq 0} S_ r$ represents the functor

Moreover, $\mathcal{F}|_{S_ r}$ is locally free of rank $r$ and the morphisms $S_ r \to S$ and $S' \to S$ are of finite presentation.

**Proof.**
Suppose that $g : T \to S$ is a morphism of schemes such that the pullback $\mathcal{F}_ T = g^*\mathcal{F}$ is flat. Then $\mathcal{F}_ T$ is a flat $\mathcal{O}_ T$-module of finite presentation. Hence $\mathcal{F}_ T$ is finite locally free, see Properties, Lemma 28.20.2. Thus $T = \coprod _{r \geq 0} T_ r$, where $\mathcal{F}_ T|_{T_ r}$ is locally free of rank $r$. This implies that

in the category of Zariski sheaves on $\mathit{Sch}/S$ where $F_ r$ is as in Lemma 31.9.6. It follows that $F_{flat}$ is represented by $\coprod _{r \geq 0} (Z_{r - 1} \setminus Z_ r)$ where $Z_ r$ is as in Lemma 31.9.6. The other statements also follow from the lemma. $\square$

Example 31.9.8. Let $R = \prod _{n \in \mathbf{N}} \mathbf{F}_2$. Let $I \subset R$ be the ideal of elements $a = (a_ n)_{n \in \mathbf{N}}$ almost all of whose components are zero. Let $\mathfrak m$ be a maximal ideal containing $I$. Then $M = R/\mathfrak m$ is a finite flat $R$-module, because $R$ is absolutely flat (More on Algebra, Lemma 15.104.6). Set $S = \mathop{\mathrm{Spec}}(R)$ and $\mathcal{F} = \widetilde{M}$. The closed subschemes of Lemma 31.9.6 are $S = Z_{-1}$, $Z_0 = \mathop{\mathrm{Spec}}(R/\mathfrak m)$, and $Z_ i = \emptyset $ for $i > 0$. But $\text{id} : S \to S$ does not factor through $(S \setminus Z_0) \amalg Z_0$ because $\mathfrak m$ is a nonisolated point of $S$. Thus Lemma 31.9.7 does not hold for finite type modules.

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