The Stacks project

Lemma 17.12.4. Let $(X, \mathcal{O}_ X)$ be a ringed space.

  1. Any finite type subsheaf of a coherent sheaf is coherent.

  2. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism from a finite type sheaf $\mathcal{F}$ to a coherent sheaf $\mathcal{G}$. Then $\mathop{\mathrm{Ker}}(\varphi )$ is finite type.

  3. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of coherent $\mathcal{O}_ X$-modules. Then $\mathop{\mathrm{Ker}}(\varphi )$ and $\mathop{\mathrm{Coker}}(\varphi )$ are coherent.

  4. Given a short exact sequence of $\mathcal{O}_ X$-modules $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ if two out of three are coherent so is the third.

  5. The category $\textit{Coh}(\mathcal{O}_ X)$ is a weak Serre subcategory of $\textit{Mod}(\mathcal{O}_ X)$. In particular, the category of coherent modules is abelian and the inclusion functor $\textit{Coh}(\mathcal{O}_ X) \to \textit{Mod}(\mathcal{O}_ X)$ is exact.

Proof. Condition (2) of Definition 17.12.1 holds for any subsheaf of a coherent sheaf. Thus we get (1).

Assume the hypotheses of (2). Let us show that $\mathop{\mathrm{Ker}}(\varphi )$ is of finite type. Pick $x \in X$. Choose an open neighbourhood $U$ of $x$ in $X$ such that $\mathcal{F}|_ U$ is generated by $s_1, \ldots , s_ n$. By Definition 17.12.1 the kernel $\mathcal{K}$ of the induced map $\bigoplus _{i = 1}^ n \mathcal{O}_ U \to \mathcal{G}$, $e_ i \mapsto \varphi (s_ i)$ is of finite type. Hence $\mathop{\mathrm{Ker}}(\varphi )$ which is the image of the composition $\mathcal{K} \to \bigoplus _{i = 1}^ n \mathcal{O}_ U \to \mathcal{F}$ is of finite type.

Assume the hypotheses of (3). By (2) the kernel of $\varphi $ is of finite type and hence by (1) it is coherent.

With the same hypotheses let us show that $\mathop{\mathrm{Coker}}(\varphi )$ is coherent. Since $\mathcal{G}$ is of finite type so is $\mathop{\mathrm{Coker}}(\varphi )$. Let $U \subset X$ be open and let $\overline{s}_ i \in \mathop{\mathrm{Coker}}(\varphi )(U)$, $i = 1, \ldots , n$ be sections. We have to show that the kernel of the associated morphism $\overline{\Psi } : \bigoplus _{i = 1}^ n \mathcal{O}_ U \to \mathop{\mathrm{Coker}}(\varphi )$ has finite type. There exists an open covering of $U$ such that on each open all the sections $\overline{s}_ i$ lift to sections $s_ i$ of $\mathcal{G}$. Hence we may assume this is the case over $U$. We may in addition assume there are sections $t_ j$, $j = 1, \ldots , m$ of $\mathop{\mathrm{Im}}(\varphi )$ over $U$ which generate $\mathop{\mathrm{Im}}(\varphi )$ over $U$. Let $\Phi : \bigoplus _{j = 1}^ m \mathcal{O}_ U \to \mathop{\mathrm{Im}}(\varphi )$ be defined using $t_ j$ and $\Psi : \bigoplus _{j = 1}^ m \mathcal{O}_ U \oplus \bigoplus _{i = 1}^ n \mathcal{O}_ U \to \mathcal{G}$ using $t_ j$ and $s_ i$. Consider the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & \bigoplus _{j = 1}^ m \mathcal{O}_ U \ar[d]_\Phi \ar[r] & \bigoplus _{j = 1}^ m \mathcal{O}_ U \oplus \bigoplus _{i = 1}^ n \mathcal{O}_ U \ar[d]_\Psi \ar[r] & \bigoplus _{i = 1}^ n \mathcal{O}_ U \ar[d]_{\overline{\Psi }} \ar[r] & 0 \\ 0 \ar[r] & \mathop{\mathrm{Im}}(\varphi ) \ar[r] & \mathcal{G} \ar[r] & \mathop{\mathrm{Coker}}(\varphi ) \ar[r] & 0 } \]

By the snake lemma we get an exact sequence $\mathop{\mathrm{Ker}}(\Psi ) \to \mathop{\mathrm{Ker}}(\overline{\Psi }) \to 0$. Since $\mathop{\mathrm{Ker}}(\Psi )$ is a finite type module, we see that $\mathop{\mathrm{Ker}}(\overline{\Psi })$ has finite type.

Proof of part (4). Let $0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0$ be a short exact sequence of $\mathcal{O}_ X$-modules. By part (3) it suffices to prove that if $\mathcal{F}_1$ and $\mathcal{F}_3$ are coherent so is $\mathcal{F}_2$. By Lemma 17.9.3 we see that $\mathcal{F}_2$ has finite type. Let $s_1, \ldots , s_ n$ be finitely many local sections of $\mathcal{F}_2$ defined over a common open $U$ of $X$. We have to show that the module of relations $\mathcal{K}$ between them is of finite type. Consider the following commutative diagram

\[ \xymatrix{ 0 \ar[r] & 0 \ar[r] \ar[d] & \bigoplus _{i = 1}^{n} \mathcal{O}_ U \ar[r] \ar[d] & \bigoplus _{i = 1}^{n} \mathcal{O}_ U \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \mathcal{F}_1 \ar[r] & \mathcal{F}_2 \ar[r] & \mathcal{F}_3 \ar[r] & 0 } \]

with obvious notation. By the snake lemma we get a short exact sequence $0 \to \mathcal{K} \to \mathcal{K}_3 \to \mathcal{F}_1$ where $\mathcal{K}_3$ is the module of relations among the images of the sections $s_ i$ in $\mathcal{F}_3$. Since $\mathcal{F}_3$ is coherent we see that $\mathcal{K}_3$ is finite type. Since $\mathcal{F}_1$ is coherent we see that the image $\mathcal{I}$ of $\mathcal{K}_3 \to \mathcal{F}_1$ is coherent. Hence $\mathcal{K}$ is the kernel of the map $\mathcal{K}_3 \to \mathcal{I}$ between a finite type sheaf and a coherent sheaves and hence finite type by (2).

Proof of (5). This follows because (3) and (4) show that Homology, Lemma 12.9.3 applies. $\square$


Comments (8)

Comment #975 by JuanPablo on

To prove that the category of coherent modules is abelian we need that the sum of two coherent modules is coherent, I did not find that fact immediate so I put here the proof I have:

Suppose and coherent, then is of finite type. Now take , which corresponds to a pair of morphisms and , with kernels and . The kernel of the original map is . Now we have the exact sequence: where the arrow in the right is the difference. Then is of finite type because is of finite type and is of finite presentation (tag 01BP (2)).

Comment #976 by JuanPablo on

No, wait;

is not surjective so it does not work. Now I don't know how to prove that the sum of two coherent modules is coherent.

Comment #977 by JuanPablo on

Ok. I think is as follows:

The kernel of is which is of finite type because of (2) in this lemma.

Comment #1008 by on

Hi! In stead of directly arguing this in this case, I have put in a reference to Lemma 12.9.3 which says that in the situation where you have a full subcat preserved under kernels and cokernels and extensions, you always have an abelian subcategory. Hope this clarifies things. The change is here.

Comment #2366 by Katharina on

In the proof of part (3) it is implicitly assumed that is surjective, which is not the case in general.

Comment #2429 by on

Sorry, I do not understand your question. Can you clarify? Thanks.

Comment #3045 by Dongryul Kim on

I agree with Katharina here. In the diagram of the proof of (3), there is a on the right of the top row. The map is not necessarily surjective, and neither is because we picked arbitrary sections. Here is what Serre does in FAC instead:

Because is finite type, around each there is a surjective map after possibly making smaller. Composing it with gives a surjective morphism . On the other hand, because is surjective, we can lift sections to . Then we can define the map mapping the first components by and the last components by . By the snake lemma we get an exact sequence . Because is coherent, is of finite type. Thus is also of finite type.

I think the same wrong argument is used in 05CW as well.

Comment #3150 by on

@#3045: OK, I found a very simple proof of this step which I think is the same as what you suggested. See here and here for changes.


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01BY. Beware of the difference between the letter 'O' and the digit '0'.