Let $S.$ be a graded ring, finitely generated by degree 1 elements as a $S_0$-algebra. Let $M.$ be a finitely generated graded $S.$-module. There exists a natrual map $M_n\to\Gamma(\operatorname{Proj}S.,\widetilde{M(n).})$ . How to prove it is an isomorphism when $n$ is sufficiently large?

  • $\begingroup$ Could it be that you meant $\operatorname{Proj} S.$ rather than $Proj S.$? If so, that can be set in TeX. $\endgroup$ – Michael Hardy Aug 8 '11 at 22:49
  • $\begingroup$ Yes, I have edited that, thanks! $\endgroup$ – Li Zhan Aug 9 '11 at 1:36
  • 1
    $\begingroup$ You changed it to \textrm, but then I changed it to \operatorname. There's a difference: \textrm makes it non-italicized, but \operatorname does that and also provides proper spacing between that and what follows it. (Last I recall, this didn't work on Wikipedia; that's a software bug. I don't know if it's been reported or not.) Generally things like \det, \max, \log, \sin, \gcd, \sup, \Pr etc., that are built-in operatornames in LaTeX will not only prevent italicization and provide spacing, but also assure that other standard convetions are followed. E.g., subscripts on \sup ..... $\endgroup$ – Michael Hardy Aug 10 '11 at 1:22
  • $\begingroup$ ....are directly under "sup", at least when it's "displayed" rather than "inline". In fact, if it works one way when "displayed" and another when "inline", then that's yet another way in which standard conventions are built in to these things. $\endgroup$ – Michael Hardy Aug 10 '11 at 1:23

This follows from some cohomology on projective schemes (I shall assume $S_0$ noetherian).

Namely, we can start by assuming without loss of generality that $S$ is a graded algebra $S_0[x_1, \dots, x_n]$, because $\mathrm{Proj} S$ embeds as a closed subscheme of such. So the question is: given an $S_0[x_1, \dots, x_n]$-module $M$ (graded, and of finite type), there is an associated coherent sheaf $\widetilde{M}$ on the projective space $\mathbb{P}^n_{S_0}$. To what extent can we recover $M$ from $\widetilde{M}$?

The claim is that $\widetilde{M}$ determines $M$ in sufficiently large degrees. More fancily, there is an equivalence of categories between the abelian category of coherent sheaves on $\mathbb{P}^n_{S_0}$ and the quotient of the category of finitely generated $S_0[x_1, \dots, x_n]$-modules by the Serre subcategory of modules with only finitely many nonzero graded terms. The quasi-inverse functor sends $\mathcal{F} \mapsto \bigoplus \Gamma(\mathbb{P}^n_{S_0}, \mathcal{F}(n))$.

To prove this is essentially your claim, since the functor $M \mapsto \widetilde{M}$ is essentially surjective (in fact, we know that starting with $\mathcal{F}$, forming the graded module as above, and applying the tilde gives the same thing).
Let's say now that a module $M$ has the property $P$ if the map $P_n \to \Gamma(\mathbb{P}^n_{S_0}, \widetilde{P}(n))$ is an isomorphism for $n \gg 0$. It is easy to check that this is true for free graded modules, and thus for their twists. The claim now is that if $M', M$ have property $P$ and there is an exact sequence $$M' \to M \to M'' \to 0,$$ then $M''$ has $P$ as well. The reason is that the sequence $$\Gamma(\mathbb{P}^n_{S_0}, \widetilde{M'}(n)) \to \Gamma(\mathbb{P}^n_{S_0}, \widetilde{M}(n) )\to \Gamma(\mathbb{P}^n_{S_0}, \widetilde{M''}(n)) \to 0$$ is exact for $n \gg 0$. (This is essentially the theorem of Serre that twisting a lot kills cohomology.) So now a five-lemma argument shows that it is true for $M''$. Since every finitely generated graded module has a free presentation, the claim is proved.

  • $\begingroup$ Surely there's a more direct approach though? The claim appears in a special case as Exercise 5.9 of Chapter II in Hartshorne. (Not that I know how to solve that either...) $\endgroup$ – Zhen Lin Aug 10 '11 at 7:52
  • $\begingroup$ @Mathew Thanks! I roughly get your idea except some details. This is also an exercise in Vakil's notes, nothing about cohomology has been metioned before that. I expect there should be a down to earth, constructive proof indicating which n it should be. Anyway, I apprecitate your efforts! $\endgroup$ – Li Zhan Aug 10 '11 at 8:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.