Let $X$ be a scheme. Let $Q$ be an $O_X$-ideal (no additional assumptions). Let $F$ be the quotient sheaf $O_X/Q$.
I would like to prove that there exists an affine open cover $\{ U_i \}$ of $X$, such that for each $U_i \subset X$, we have an isomorphism
$$
F|_{U_i} \cong \widetilde{M_i}
$$
of $O_X|_{U_i}$– modules for some $M_i$ a finitely generated $O_X(U_i)$-module.
Any hints and comments are appreciated! Thank you!
Best Answer
You are expecting quit a lot right here. Sadly the quotient even of such a nice sheaf as $\mathcal{O}_X$ in general will not be something nice. As a matter of fact things can go wrong even in very nice situations:
Consider $X=\operatorname{Spec}k[x]_{(x)} = \{(0),(x)\}$. Apart from the trivial ones it only has one further open set, $D(x) = \{(0)\}$. Now lets define a (pre-)sheaf of ideals $$\mathcal{I}(X) = (0) \subset k[x]_{(x)} = \mathcal{O}(X),\hspace{0.5cm} \mathcal{I}(D(x)) = k(x) = \mathcal{O}(D(x)).$$ That it is a sheaf basically follows from the fact, that the topology on $X$ is really simple: There are no open covers except for the trivial ones. Hence the quotient sheaf actually also is just the quotient presheaf: $$ \mathcal{O}_X/\mathcal{I}\hspace{0.1cm}(X) = k[x]_{(x)} \hspace{0.5cm} \mathcal{O}_X/\mathcal{I}\hspace{0.1cm}(D(x)) = 0 $$ Obviously this sheaf isn't even quasi-coherent. This example is taken from the wonderful lecture notes of Vakil located here.
All in all you really want to assume that $\mathcal{I}$ is quasi-coherent.