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 $x \in U = \text{Spec}(R)$ 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...
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.)