Definition 17.11.1. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a sheaf of $\mathcal{O}_ X$-modules. We say that $\mathcal{F}$ is of *finite presentation* if for every point $x \in X$ there exists an open neighbourhood $x\in U \subset X$, and $n, m \in \mathbf{N}$ such that $\mathcal{F}|_ U$ is isomorphic to the cokernel of a map

## 17.11 Modules of finite presentation

This means that $X$ is covered by open sets $U$ such that $\mathcal{F}|_ U$ has a *presentation* of the form

Here presentation signifies that the displayed sequence is exact. In other words

for every point $x$ of $X$ there exists an open neighbourhood such that $\mathcal{F}|_ U$ is generated by finitely many global sections, and

for a suitable choice of these sections the kernel of the associated surjection is also generated by finitely many global sections.

Lemma 17.11.2. Let $(X, \mathcal{O}_ X)$ be a ringed space. Any $\mathcal{O}_ X$-module of finite presentation is quasi-coherent.

**Proof.**
Immediate from definitions.
$\square$

Lemma 17.11.3. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be an $\mathcal{O}_ X$-module of finite presentation.

If $\psi : \mathcal{O}_ X^{\oplus r} \to \mathcal{F}$ is a surjection, then $\mathop{\mathrm{Ker}}(\psi )$ is of finite type.

If $\theta : \mathcal{G} \to \mathcal{F}$ is surjective with $\mathcal{G}$ of finite type, then $\mathop{\mathrm{Ker}}(\theta )$ is of finite type.

**Proof.**
Proof of (1). Let $x \in X$. Choose an open neighbourhood $U \subset X$ of $x$ such that there exists a presentation

Let $e_ k$ be the section generating the $k$th factor of $\mathcal{O}_ X^{\oplus r}$. For every $k = 1, \ldots , r$ we can, after shrinking $U$ to a small neighbourhood of $x$, lift $\psi (e_ k)$ to a section $\tilde e_ k$ of $\mathcal{O}_ U^{\oplus n}$ over $U$. This gives a morphism of sheaves $\alpha : \mathcal{O}_ U^{\oplus r} \to \mathcal{O}_ U^{\oplus n}$ such that $\varphi \circ \alpha = \psi $. Similarly, after shrinking $U$, we can find a morphism $\beta : \mathcal{O}_ U^{\oplus n} \to \mathcal{O}_ U^{\oplus r}$ such that $\psi \circ \beta = \varphi $. Then the map

is a surjection onto the kernel of $\psi $.

To prove (2) we may locally choose a surjection $\eta : \mathcal{O}_ X^{\oplus r} \to \mathcal{G}$. By part (1) we see $\mathop{\mathrm{Ker}}(\theta \circ \eta )$ is of finite type. Since $\mathop{\mathrm{Ker}}(\theta ) = \eta (\mathop{\mathrm{Ker}}(\theta \circ \eta ))$ we win. $\square$

Lemma 17.11.4. Let $f : (X, \mathcal{O}_ X) \to (Y, \mathcal{O}_ Y)$ be a morphism of ringed spaces. The pullback $f^*\mathcal{G}$ of a module of finite presentation is of finite presentation.

**Proof.**
Exactly the same as the proof of Lemma 17.10.4 but with finite index sets.
$\square$

Lemma 17.11.5. Let $(X, \mathcal{O}_ X)$ be a ringed space. Set $R = \Gamma (X, \mathcal{O}_ X)$. Let $M$ be an $R$-module. The $\mathcal{O}_ X$-module $\mathcal{F}_ M$ associated to $M$ is a directed colimit of finitely presented $\mathcal{O}_ X$-modules.

**Proof.**
This follows immediately from Lemma 17.10.5 and the fact that any module is a directed colimit of finitely presented modules, see Algebra, Lemma 10.11.3.
$\square$

Lemma 17.11.6. Let $X$ be a ringed space. Let $I$ be a preordered set and let $(\mathcal{F}_ i, \varphi _{ii'})$ be a system over $I$ consisting of sheaves of $\mathcal{O}_ X$-modules (see Categories, Section 4.21). Assume

$I$ is directed,

$\mathcal{G}$ is an $\mathcal{O}_ X$-module of finite presentation, and

$X$ has a cofinal system of open coverings $\mathcal{U} : X = \bigcup _{j\in J} U_ j$ with $J$ finite and $U_ j \cap U_{j'}$ quasi-compact for all $j, j' \in J$.

Then we have

**Proof.**
An element of the left hand side is given by the equivalence classe of a pair $(i, \alpha _ i)$ where $i \in I$ and $\alpha _ i : \mathcal{G} \to \mathcal{F}_ i$ is a morphism of $\mathcal{O}_ X$-modules, see Categories, Section 4.19. Postcomposing with the coprojection $p_ i : \mathcal{F}_ i \to \mathop{\mathrm{colim}}\nolimits _{i' \in I} \mathcal{F}_{i'}$ we get $\alpha = p_ i \circ \alpha _ i$ in the right hand side. We obtain a map

Let us show this map is injective. Let $\alpha _ i$ be as above such that $\alpha = p_ i \circ \alpha _ i$ is zero. By the assumption that $\mathcal{G}$ is of finite presentation, for every $x \in X$ we can choose an open neighbourhood $U_ x \subset X$ of $x$ and a finite set $s_{x, 1}, \ldots , s_{x, n_ x} \in \mathcal{G}(U_ x)$ generating $\mathcal{G}|_{U_ x}$. These sections map to zero in the stalk $(\mathop{\mathrm{colim}}\nolimits _{i'} \mathcal{F}_ i)_ x = \mathop{\mathrm{colim}}\nolimits _{i'} \mathcal{F}_{i', x}$. Hence for each $x$ we can pick $i(x) \geq i$ such that after replacing $U_ x$ by a smaller open we have that $s_{x, 1}, \ldots , s_{x, n_ x}$ map to zero in $\mathcal{F}_{i(x)}(U_ x)$. Then $X = \bigcup U_ x$. By condition (3) we can refine this open covering by a finite open covering $X = \bigcup _{j \in J} U_ j$. For $j \in J$ pick $x_ j \in X$ with $U_ j \subset U_{x_ j}$. Set $i' = \max (i(x_ j); j \in J)$. Then $\mathcal{G}|_{U_ j}$ is generated by the sections $s_{x_ j, k}$ which are mapped to zero in $\mathcal{F}_{i(x)}$ and hence in $\mathcal{F}_{i'}$. Hence the composition $\mathcal{G} \to \mathcal{F}_ i \to \mathcal{F}_{i'}$ is zero as desired.

Proof of surjectivity. Let $\alpha $ be an element of the right hand side. For every point $x \in X$ we may choose an open neighbourhood $U \subset X$ and finitely many sections $s_ j \in \mathcal{G}(U)$ which generate $\mathcal{G}$ over $U$ and finitely many relations $\sum f_{kj} s_ j = 0$, $k = 1, \ldots , n$ with $f_{kj} \in \mathcal{O}_ X(U)$ which generate the kernel of $\bigoplus _{j = 1, \ldots , m} \mathcal{O}_ U \to \mathcal{G}$. After possibly shrinking $U$ to a smaller open neighbourhood of $x$ we may assume there exists an index $i \in I$ such that the sections $\alpha (s_ j)$ all come from sections $s_ j' \in \mathcal{F}_ i(U)$. After possibly shrinking $U$ to a smaller open neighbourhood of $x$ and increasing $i$ we may assume the relations $\sum f_{kj} s'_ j = 0$ hold in $\mathcal{F}_ i(U)$. Hence we see that $\alpha |_ U$ lifts to a morphism $\mathcal{G}|_ U \to \mathcal{F}_ i|_ U$ for some index $i \in I$.

By condition (3) and the preceding arguments, we may choose a finite open covering $X = \bigcup _{j = 1, \ldots , m} U_ j$ such that (a) $\mathcal{G}|_{U_ j}$ is generated by finitely many sections $s_{jk} \in \mathcal{G}(U_ j)$, (b) the restriction $\alpha |_{U_ j}$ comes from a morphism $\alpha _ j : \mathcal{G} \to \mathcal{F}_{i_ j}$ for some $i_ j \in I$, and (c) the intersections $U_ j \cap U_{j'}$ are all quasi-compact. For every pair $(j, j') \in \{ 1, \ldots , m\} ^2$ and any $k$ we can find we can find an index $i \geq \max (i_ j, i_{j'})$ such that

see Sheaves, Lemma 6.29.1 (2). Since there are finitely many of these pairs $(j, j')$ and finitely many $s_{jk}$ we see that we can find a single $i$ which works for all of them. For this index $i$ all of the maps $\varphi _{i_ ji} \circ \alpha _ j$ agree on the overlaps $U_ j \cap U_{j'}$ as the sections $s_{jk}$ generate $\mathcal{G}$ over this overlap. Hence we get a morphism $\mathcal{G} \to \mathcal{F}_ i$ as desired. $\square$

Remark 17.11.7. In the lemma above some condition beyond the condition that $X$ is quasi-compact is necessary. See Sheaves, Example 6.29.2.

Lemma 17.11.8. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $x \in X$ such that $\mathcal{F}_ x \cong \mathcal{O}_{X, x}^{\oplus r}$. Then there exists an open neighbourhood $U$ of $x$ such that $\mathcal{F}|_ U \cong \mathcal{O}_ U^{\oplus r}$.

**Proof.**
Choose $s_1, \ldots , s_ r \in \mathcal{F}_ x$ mapping to a basis of $\mathcal{O}_{X, x}^{\oplus r}$ by the isomorphism. Choose an open neighbourhood $U$ of $x$ such that $s_ i$ lifts to $s_ i \in \mathcal{F}(U)$. After shrinking $U$ we see that the induced map $\psi : \mathcal{O}_ U^{\oplus r} \to \mathcal{F}|_ U$ is surjective (Lemma 17.9.4). By Lemma 17.11.3 we see that $\mathop{\mathrm{Ker}}(\psi )$ is of finite type. Then $\mathop{\mathrm{Ker}}(\psi )_ x = 0$ implies that $\mathop{\mathrm{Ker}}(\psi )$ becomes zero after shrinking $U$ once more (Lemma 17.9.5).
$\square$

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