# Quasi-coherent sheaf on $Proj\ S$

Given a graded ring $$S$$ and a quasi-coherent sheaf $$\mathcal{F}$$ on $$Proj\ S$$, does there exist a graded $$S$$-module $$M$$ such that $$\mathcal{F}\cong \widetilde{M}$$?

I know the result is true when $$S$$ is finitely generated by $$S_1$$ as an $$S_0$$-algebra (and so I am guessing the result to be false if we remove this hypothesis), but two cases I would like to know is when $$S$$ is generated by $$S_1$$, but not finitely and the second is when $$S$$ is not generated by $$S_1$$ at all.

This may not be a full definitive answer to your question, but I hope it can shed some light on how these questions you ask can be a little more complicated than they might initially seem. If someone else has a good resolution to the non-quasicompact case discussed at the end, I would welcome reading about it.

The first thing to note is that the construction of $$M$$, if it exists, should be $$M:= \bigoplus_{n\geq 0} \Gamma(\operatorname{Proj} S, \mathcal{F}(n))$$. There's a canonical map $$\widetilde{M}\to \mathcal{F} \qquad (*)$$ regardless of whatever's going on with $$\operatorname{Proj} S$$. (See Stacks for the proof.) We're interested in when this is an isomorphism - as you state in your post, the condition that $$S$$ is finitely generated as an $$S_0$$ algebra by $$S_1$$ is sufficient. We'll see that this is far from a necessary condition, though, by demonstrating cases where $$S$$ is not necessarily finitely generated nor necessarily generated in degree one where this map is an isomorphism.

One funny thing about the Proj construction is that it does a really bad job at distinguishing between graded rings. That is, if $$\operatorname{Proj} S\cong \operatorname{Proj} S'$$, the graded rings $$S$$ and $$S'$$ can be very different. Here are two constructions that change $$S$$ significantly, but do not alter $$\operatorname{Proj} S$$:

• Replacing $$S_0$$ with another ring $$A_0$$ with a homomorphism $$f:A_0\to S_0$$ and defining the action of $$A_0$$ on an element $$s\in S$$ by $$a\cdot s = f(a)S$$.

• Replacing $$S$$ with $$S^{(d)}:=\bigoplus_{n\geq 0} S_{dn}$$ for $$d>0$$, a thinned out version of $$S$$.

Consider applying the first construction to $$\Bbb C[x,y]$$, replacing the copy of $$\Bbb C$$ in degree zero by $$\Bbb Z$$ and calling this new ring $$S'$$. Then $$\operatorname{Proj} \Bbb C[x,y] \cong \operatorname{Proj} S' \cong \Bbb P^1_{\Bbb C}$$, but the first is Proj of an algebra finitely generated in degree one and the second is not ($$S'$$ is uncountable, but any finitely-generated $$\Bbb Z$$-algebra is countable). On the other hand, the procedure which determines $$M$$ from the data of a sheaf $$\mathcal{F}$$ on this scheme doesn't see the difference between $$\Bbb C[x,y]$$ and $$S'$$. This means that asking about whether $$S$$ is finitely generated by $$S_1$$ as a $$S_0$$-algebra is not always the right question.

We may also apply the second construction to reduce any situation in which there's a $$n>0$$ so that all the generators of $$S$$ as an $$S_0$$-algebra are of degree $$\leq n$$ to a situation where $$S$$ is generated in degree 1 without changing Proj: replacing $$S$$ with $$S^{(m)}$$ for an appropriately-chosen $$m$$ (this should be the LCM of all degrees in which there's a generator), we may replace $$S$$ with a ring which is in fact generated in degree one without changing the projective space. For an example involving this construction, consider ring $$S=\Bbb C[x,y,z]$$ with $$x$$ in degree 1, $$y$$ in degree 2, and $$z$$ in degree 3. Let $$S^{(6)}=k[x^6,x^4y,x^3z,x^2y^2,xyz,y^6,z^6]$$ where each of the generating monomials is in degree one. Then $$\operatorname{Proj} S=\operatorname{Proj} S'$$, but one is generated in degree one and one is not. Since the construction of $$M$$ from $$\mathcal{F}$$ doesn't see this change, this means that asking about whether $$S$$ is generated in degree one is not always the right question either.

As mentioned in the comments by Ben, Stacks 0AG5 shows that this map $$(*)$$ from $$\widetilde{M}\to \mathcal{F}$$ is an isomorphism when $$\operatorname{Proj} S$$ is quasicompact. In particular, asking about quasicompactness of $$\operatorname{Proj} S$$ is a good way to eliminate the "tricks" from the previous section - rephrasing our conditions to depend on the scheme instead of the ring means we won't get fooled by the examples that violated the "$$S$$ is finitely generated as an $$S_0$$-algebra by $$S_1$$" condition if there's another choice of $$S'$$ which gives the same scheme after taking Proj and satisfies the quoted condition.

Restricting to quasicompact projective schemes should not be such a big deal if you're just starting out learning these things (I see this is your first question or answer in the algebraic-geometry tag, welcome!). Strange things can happen with non-quasicompact schemes - there are examples of such with no closed points, for instance. In general, once one leaves the confines of nice finiteness assumptions (noetherian/locally noetherian/quasicompact/quasiseparated are common such assumptions), a lot of machinery can no longer be guaranteed to work. Until you decide to work on a problem which needs the removal of these assumptions, it can be a lot easier to let sleeping dogs lie.

In an attempt to point out a meaningful example when $$(*)$$ is not an isomorphism, I would guess that a ring $$S$$ which does not satisfy "$$S$$ is generated by elements of degree $$\leq n$$ for some fixed integer $$n$$" cannot have $$\widetilde{M}\to \mathcal{F}$$ be an isomorphism. A key ingredient of the proof that $$(*)$$ is an isomorphism in the quasi-compact case is picking a $$d$$ so that $$\mathcal{O}(d)$$ is invertible, so that tensoring with $$\mathcal{O}(d)$$ is an isomorphism. By setting it up so that no such sheaf is invertible, we should be guaranteeing that we destroy information about $$\mathcal{F}$$ every time we tensor by $$\mathcal{O}(d)$$ which will cause $$(*)$$ to fail to be an isomorphism.

The reason having generators of unbounded degree should cause this to fail is given by Stacks 01MU which states that for $$X=\operatorname{Proj} S$$, the largest open set $$W_d\subset X$$ so that $$\mathcal{O}_X(dn)|_{W_d}$$ is invertible is exactly the union of all open sets $$D_+(fg)$$ for $$f,g$$ homogeneous with $$\deg(f)=\deg(g)+d$$. If we can't cover $$X$$ by such subsets for any $$d>0$$, this shows that no $$\mathcal{O}_X(d)$$ is invertible.

The case of "generated by $$S_1$$ but not finitely" and $$\operatorname{Proj} S$$ not quasicompact seems trickier. For instance, $$\Bbb P^\infty = \operatorname{Proj} k[x_0,x_1,\cdots]$$ with each $$x_i$$ in degree one doesn't seem so bad - there's a number of sub-results used in the proof that $$(*)$$ is an isomorphism which depend on quasicompactness which I wouldn't be surprised either way if they just went through because of the relatively nice situation we had, or instead broke in interesting ways. If anyone has treated this specific case, I would be interested in hearing about it - please leave a link in the comments.

• I was thinking the same thing. Perhaps it would be easier to come up with an extraordinary module on $\mathrm{Proj}(k[x_1,x_2,...])$ with $x_i$ of degree $i$, but I haven't had any time to think about it any further. – Ben Aug 29 '19 at 19:32
• Thank you. I believe this largely clears up my doubts. – Abhijit A J Aug 30 '19 at 5:15
• My only problem (which I don't really believe affects your answer at all), is that you have written in the second paragraph that $M:= \Gamma(Proj\ S, \mathcal{F}(n))$. Did you mean $M:= \bigoplus\Gamma(Proj\ S, \mathcal{F}(n))$? Even so, we could have some other $M'$ which is not isomorphic to the mentioned module and still generate the same $\mathcal{O}_X$-module as shown by you later in the answer and also shown in (Hartshorne, Chapter 2, Excercise 5.9). It could be that I have misunderstood what you wrote, but it would be great if you could clarify. – Abhijit A J Aug 30 '19 at 5:26
• Yes, that first one was a typo and it's been corrected. We definitely could have other $M'$ - any graded module $N$ which is nonzero in only finitely many degrees will give $\widetilde{N}=0$, so we can take a direct sum with one of those and that will change $M$ but not $\widetilde{M}$. I didn't really get in to this since the question was more focused on properties of $S$ (and my answer was already pretty long!), but I can definitely add this if you would like. – KReiser Aug 30 '19 at 5:30
• Thanks a lot for the prompt reply – Abhijit A J Aug 30 '19 at 5:31