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Let $R$ be a graded ring and $M$ a graded module. Then for sufficently large $n$, we have $$H^0(\operatorname{Proj}(R), \widetilde{M}(n))\cong M_n.$$ Hence if I want to show that $M_{>0}$ is non-trivial, I can use cohomology. However, this fails for the negative part $M_{<0}$. Even if $H^0(\operatorname{Proj}(R),\widetilde{M}(-n))\cong M_{-n}$, then most of my tools to deal with cohomology are useless, as the usally deal with sheaves of the form $\mathcal{F}(n)$ for $n\gg0$. Of course, if I could relate $H^0(\operatorname{Proj}(R),\widetilde{M}(-n))$ with $H^0(\operatorname{Proj}(R),\widetilde{M}^{\vee}(n))$, then I could deal with it. But I'm unaware of such a relation. So I guess I'm here for guidance or any sort of hint.

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  • $\begingroup$ Your whole premise is wrong: there are nonzero graded modules $M$ which have associated sheaf zero. Consider the $k[x_0,x_1]$-module given by $k$ in degree $0$ and $0$ elsewhere: it's associated sheaf on $\Bbb P^1$ is just the zero sheaf. $\endgroup$ – KReiser Jun 2 at 23:01
  • $\begingroup$ True, but I what I meant (apologies for not being clear) is that I'm interested in elements which are not of degree zero. We can test cohomologically if there exist non-trivial elements of degree strictly greater then zero the way I describe above. I'll edit the question accordingly. $\endgroup$ – curious math guy Jun 2 at 23:24
  • $\begingroup$ That's still wrong - just shift the previous module around (for instance, place it in degree $n$ instead of degree $0$). $\endgroup$ – KReiser Jun 2 at 23:27
  • $\begingroup$ Fair, but I belive my premise and thus my question is still valid if we assume that $R_{>0}$ is not acting trivially on $M$, i.e. is not contained in the annihilator of $M$ as then the module can not be concentrated in one single degree, and in fact the if $M$ is non-trival in one degree, then it is non-trival in in all high degrees (which we can detect cohomologically). $\endgroup$ – curious math guy Jun 3 at 0:31
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This is maybe not quite an answer to your question, but let me take the time to let you know about some things to file under "$\operatorname{Proj}$ exhibits interesting behavior with quasicoherent sheaves not found with $\operatorname{Spec}$". See here for some more examples.

Let us explain an important feature of the functor $\widetilde{-}:R\text{-mod}\to \operatorname{QCoh}(\operatorname{Proj} R)$ which takes graded modules $M$ on a graded ring $R$ to their associated sheaf on $\operatorname{Proj R}$. If $M$ is a graded module which is nonzero in only finitely many degrees, then $\widetilde{M}$ is the zero sheaf on $\operatorname{Proj} R$.

Proof: by definition, the sections of $\widetilde{M}$ on $D(f)$ for $f$ homogeneous of positive degree are given by $M_{(f)}$, the degree-zero elements of $M_{f}$. If $s$ is an element of $M_{(f)}$, then $s=\frac{f^ns}{f^n}$ is also such an element for $n$. But by picking $n$ large enough, we get that $f^ns$ lies in a degree where $M$ is zero, so $s=0$ and in fact $M_{(f)}=0$. So our sheaf is the zero sheaf.

This means that there's no invariant of quasi-coherent sheaves on $\operatorname{Proj} R$ which can tell apart the sheaves associated to two graded modules which differ by some module supported in finitely many degrees - we already destroyed that information just by applying the associated sheaf functor.

In particular, if $R$ has no elements of negative degree (a very common assumption) then unless $M$ is nonzero in infinitely many negative degrees, you can't tell it apart from a module which is zero in negative degrees. As a consequence, $M$ has to be infinitely generated as an $R$-module with generators in arbitrarily large negative degrees (and you'll need to say something about the annihilators of these generators not being the whole of $R_+$, etc). None of this is so bad, but you need to know what's up.

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  • $\begingroup$ The last paragraph seems to be an excellent answer to the question. Can you explain or say a few words on how much (or which) information of a quasi-coherent sheaf can be read off from its cohomology? In other words, what do we lose compared to the case of coherent sheaves? $\endgroup$ – Youngsu Jun 5 at 4:57

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