# Are finite type analytic sheaves coherent?

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?

• I think Oka's theorem will say that $\mathcal{O}_X$ is coherent and then the rest will easily follow. Aug 21, 2019 at 16:47
• Hi Mohan, thanks for the reply! I would like to know how to use the coherence of structure sheaf to prove this. Btw is my question well known to be true? Aug 21, 2019 at 16:52
• What definition of coherence are you using? Aug 21, 2019 at 16:53
• This might be stupid, but why does the coherence of $\mathcal O^n_X$ imply that any of its subsheaf is finitely generated? Aug 21, 2019 at 17:10
• I don't think so. Let $n>2$ and let $K\subset \mathbb C^n=X$ be a closed, nowhere dense, non-analytic subset, like $S^1\times 0$. Moreover, let $I\subset \mathcal O_X$ be the ideal of germs vanishing along $K$. Then I think the support of $\mathcal F=\mathcal O_{X}/I$ should be $K$, but if $\mathcal F$ were coherent, the support would be analytic. In other words, $I$ is not locally finitely generated.
– Ben
Aug 21, 2019 at 19:50

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.)