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.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Ben
    Commented Aug 29, 2019 at 19:32
  • $\begingroup$ Thank you. I believe this largely clears up my doubts. $\endgroup$ Commented Aug 30, 2019 at 5:15
  • $\begingroup$ 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. $\endgroup$ Commented Aug 30, 2019 at 5:26
  • $\begingroup$ 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. $\endgroup$
    – KReiser
    Commented Aug 30, 2019 at 5:30
  • $\begingroup$ Thanks a lot for the prompt reply $\endgroup$ Commented Aug 30, 2019 at 5:31

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