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.

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