Are finite type analytic sheaves coherent?

Question

Let $X$ be an open subset of $\mathbb C^n$, and let $\mathcal O_X$ be the structure sheaf of germs of holomorphic functions. Suppose that $\mathcal F$ is a finitely generated $\mathcal O_X$-modules, i.e., there is an exact sequence $\mathcal O_X^m\rightarrow\mathcal F\rightarrow 0$ (where $m\in\mathbb N$). Is it true that $\mathcal F$ is coherent? What about the algebraic case?

Answer 1

Summing up the comments: Let $X$ be any complex space, e. g., an open subset of $\mathbb C^n$ with its sheaf of germs of holomorphic functions. The support of a coherent sheaf on $X$ is an analytic subset. Thus, given $A\subset X$ closed but not analytic, it suffices to construct a finitely generated sheaf supported exactly on $A$.

Let $i\colon A\to X$ be the inclusion map; I claim that $i_*i^{-1}\mathcal O_X$ is such a sheaf. Consider the natural morphism $\mathcal O_X\to i_*i^{-1}\mathcal O_X$. Since $A$ is closed in $X$, $(i_*i^{-1}\mathcal O_X)_x = \mathcal O_{X,x}$ whenever $x\in A$ and otherwise $(i_*i^{-1}\mathcal O_X)_x = 0$. Therefore, $\mathcal O_X\to i_*i^{-1}\mathcal O_X$ is surjective, hence, $i^{-1}\mathcal O_X$ is finitely generated and indeed $A = \mathrm{supp}(i_*i^{-1}\mathcal O_X)$.

Alternatively, as the OP has proposed, we may simply define the ideal $I\subset\mathcal O_X$ to be $I(U) = \mathcal{O}_X(U)$ if $U\cap A = \emptyset$ and $I(U) = 0$ otherwise. It is then easy to see that this is not finitely generated near $A$; hence, the quotient $\mathcal O_X/I$ is not coherent. (This works even is $A$ is analytic.)

In the "algebraic case", the analogous construction produces non-quasi-coherent ideals, hence, non-quasi-coherent finitely generated sheaves.