# Hartshorne Exercise II.6.10 Proving existence of a filtration for coherent sheaves supported on a closed subscheme.

Fix $$X$$ a Noetherian scheme and $$\mathscr{F}$$ a coherent sheaf on $$X$$ whose support is contained inside some closed subscheme $$Y\subseteq X$$. I would like to show that there is a filtration $$0=\mathscr{F}_n\subseteq\dots\subseteq \mathscr{F}_1\subseteq \mathscr{F}_0=\mathscr{F}$$ in which each $$\mathscr{F}_i/\mathscr{F}_{i+1}$$ is an $$\mathcal{O}_Y$$-module. It is easy to see how this implies the main result (exactness of the sequence $$K(Y)\rightarrow K(X)\rightarrow K(X-Y)\rightarrow 0$$ of Grothendieck groups in the middle). Unfortunately, I have been unable to establish existence of this finite filtration.

Following the general principle that I should try to reduce problems down to commutative algebra, I am aware of the following result (taken from Bourbaki's Commutative Algebra text)

[Bourbaki's Algebre Commutative, Chap. IV n$$^\circ$$4 Theoreme 2] Fix $$A$$ a noetherian ring and $$M$$ a finitely generated $$A$$-module. Then there is a filtration of modules $$0=M_n\subseteq \dots\subseteq M_1\subseteq M_0=M$$ with $$M_i/M_{i+1}\cong A/ \mathfrak{p}_i$$ where $$\operatorname{Ass}(M)\subseteq\{\mathfrak{p}_0,\mathfrak{p}_2,\dots,\mathfrak{p}_{n-1}\}\subseteq\operatorname{Supp}(M).$$

This result easily gives the existence of a filtration in the affine case. However, I am unable to "glue" together these filtrations to a filtration of $$\mathscr{F}$$ itself.

I am also unable to locate a reference which proves existence of the filtration outside of this "solutions manual" which I have not read yet. On a quick glance, it appears sound to me, but I am curious to see if the above proof can be made to work.

I will appreciate either a reference or a good explanation why the first approach could not work (or even a proof that it could work!). Thank you.

The problem I think is that the affine filtrarion you are quoting is not "canonically constructed", and so it is not clear how can we glue them. For instance, imagine that $$R=\mathbb{Z}$$ and $$M=\mathbb{Z}^2$$. We can take the filtration $$0 \subseteq 2\mathbb{Z} \times \{0\} \subseteq \mathbb{Z}\times \{0\} \subseteq \mathbb{Z} \times 3\mathbb{Z} \subseteq \mathbb{Z}^2$$, which has $$\mathfrak{p}_0=3\mathbb{Z}$$, $$\mathfrak{p}_1=0$$, $$\mathfrak{p}_2=2\mathbb{Z}$$ and $$\mathfrak{p}_3=(0)$$. Of course, we also have the filtration $$0 \subseteq \mathbb{Z} \times 0 \subseteq \mathbb{Z}^2$$, with $$\mathfrak{p}_0=\mathfrak{p}_1=(0)$$. This is I think the problem with taking an aribtrary filtration, as a module might admit a lot of weird filtrations. (This example is essentially from Vakil, Exercise 5.5.M.)
I don't know an exact reference for the trick you are mentioning, but something similar is as a hint in Exercise III.3.1 in Hartshorne. The basic idea here is that you have a sheaf $$\mathscr{F}$$ on $$X$$ supported on a closed subscheme $$Y \subseteq X$$ with ideal sheaf $$\mathscr{I}$$. The hint tells you to consider the filtration $$\mathscr{F} \supseteq \mathscr{I}\mathscr{F} \supseteq \mathscr{I}^2\mathscr{F} \supseteq \dots$$ As you can easily check, the quotients are supported in $$Y$$. So, it only remains to show that the fibration stops. (This might be tricky! Here is where the solution manual you linked uses that $$\operatorname{Supp}\mathscr{F}$$ is contained in $$Y$$.)