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 a $\mathcal{O}_ X$-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 10.8.12.
$\square$

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).
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

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).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)