# Quasicoherent Subsheaf of a Coherent Sheaf is Coherent

Let $X$ be a Noetherian scheme and $\mathcal{F}$ a coherent sheaf, therefore quasicoherent and locally from finite type, so for every $x \in X$ there exist an affine open neighbourhood $x \in U = Spec(R)$ such that there exist a $R$-module M with $\mathcal{F} |_U = \widetilde{M}$ and $M$ has a finite representation

$R^{\oplus m} \to R^{\oplus n} \to M \to 0$ as exact sequence.

My question is why is every quasicoherent subsheaf $\mathcal{F}' \subset \mathcal{F}$ also coherent?

My attempts:

By shrinking $U$ small enough I can reduce the proof to the case $\mathcal{F}' |_U = \widetilde{M}' \subset \widetilde{M} = \mathcal{F} |_U$, but don't see how to conclude futher, because in general you can't expect that every submodule of a finite presented module is also finite presented...

• Since $\mathcal{F}'$ is quasicoherent it is locally of the form $\tilde{N}$ for some $R$-module $N\subset M$. Since $R$ is Noetherian and $M$ is finitely generated, $M$ is Noetherian. Hence $N$ is finitely generated and all of its submodules are as well. (The point is, for a Noetherian ring -- so locally for a Noetherian scheme -- finitely presented is equivalent with finitely generated). – Eoin Dec 27 '17 at 21:53
• The point is that if $M$ is a Noetherian module, then all of it's submodules are also indeed finitely generated. – Ravi Dec 27 '17 at 21:56
• @Eoin: Do you know how this important theorem that a finite generated module over a Noetherian ring is also Noetherian called? – KarlPeter Dec 27 '17 at 22:05
• @KarlPeter I don't think it has a name. It is Theorem 4.1 of these notes by Keith Conrad; see Theorems 2.1 - 2.4, as well. It follows from the facts that (1) if $R$ is noetherian, then so is $R \oplus R$, and (2) if $M$ is an $R$-module and $N \subseteq M$ is a submodule, $M$ is noetherian iff $N$ and $M/N$ are both noetherian. – André 3000 Dec 27 '17 at 22:56

In general, not every submodule of a finitely presented module is finitely presented. However, in this case $X$ is Noetherian, so the ring $R$ which we are considering modules over is a Noetherian ring. Over a Noetherian ring, any submodule of a finitely presented module is finitely presented. (Over a Noetherian ring, finitely presented is the same as finitely generated, since given any finitely generated module $M$ with a surjection $R^n\to M$, the kernel is automatically finitely generated as well.)