Let $k$ be an algebraically closed field. If $S$ is a positively graded $k$-algebra which is finitely generated by $S_1$ over $S_0 = k$ then quasi-coherent sheaves on $\operatorname{Proj}S$ are equivalent to graded $S$-modules modulo the finite $k$-dimensional modules.

I have a positively graded finitely generated $k$-algebra whose generators are not necessarily in degree $1$. Does the statement above still hold? I've been googling around and every reference assumes that the ring is generated by $S_1$ but makes no mention of why. Is it just for convenience or is there a known counterexample?

If it fails I assume the functor that takes a graded module $M$ to it's sheaf $\widetilde M$ is at least still exact. Is it maybe still faithful? Or even full? Any references would be greatly appreciated.

  • 1
    $\begingroup$ The proofs definitely use generation in degree $1$. Even $\widetilde{M \otimes N} \cong \tilde{M} \otimes \tilde{N}$ uses it (EGA II, 2.5.13), and many other basic results. With this assumption it suffices to consider homogeneous localizations at elements of degree $1$, which are easier to handle as those with arbitrary degree. But I don't have any counterexamples. $\endgroup$ – Martin Brandenburg Mar 18 '13 at 0:43

Even though this question is old, I'd like to point out the following counterexample; see my other answer for details.

Let $S = k[x,y]$ where $x$ has degree $1$ and $y$ has degree $2$. Consider the module $M = (S/(x))[1]$, which is not isomorphic to $0$. Then, $\widetilde{M} = 0$ since $M_{(x)} = 0$ and $M_{(y)} = \{f(y)/y^n \mid \deg f(y) = 2n - 1\} = 0$ since $\deg f(y)$ is even. Thus, the functor $N \to \widetilde{N}$ is not fully faithful.

  1. $M\mapsto \widetilde{M}$ is exact: this is because the exactness is a local property and $\widetilde{M}(D_+(f))=M_{(f)}$ for any homogeneous $f$. Taking $M_{(f)}$ is easily seen to be exact.

  2. It is faithfull modulo the part supported by the irrelevent maximal ideal $\mathfrak m$ exactly as in the case of $S$ generated by $S_1$. Indeed, if a graded $S$-linear map $\phi: M\to N$ is zero when passing to $\widetilde{M}\to \widetilde{N}$, then $\phi(M)_{(f)}=0$ for all homogeneous $f \in S_+$. So each element $x$ of $\phi(M)$ is annihilated by a power of $\mathfrak m$: if $\mathfrak m$ is generated by $f_1, \dots, f_n$ and $f_i^Nx=0$, then $\mathfrak m^{nN}x=0$.

  3. It is "asymptotically" full for finitely generated $M$ as in the case of $S$ generated by $S_1$. What is important here is for any $d\ge 1$, $X:=\mathrm{Proj}(S)$ is also $\mathrm{Proj}(S(d))$ (non standard notation) where $S(d)=\oplus_{n\ge 0} S_{nd}$. As $S$ is finitely generated over $S_0$, $S(d)$ is finitely generated by $S(d)_1$ for some $d$. So using the special case where $S$ is generated by $S_1$, you see that $\oplus_{n\ge n_0} M_{dn}$ can be recovered from $\widetilde{M}$ when $n_0$ is big enough. As $M(d):=\oplus_{n\ge 0} M_{nd}$ and $\oplus_{n\ge n_0} M_{nd}$ differ by a finite-dimensional vector space, and $M/M(d)$ is supported in $\{ \mathfrak m\}$ (its $\widetilde{\kern 1pt}$ is trivial), you recover almost $M$ (not less than in the special case anyway).

I didn't check all the details, but I think you can work out them by yourself. Otherwise just tell.

  • $\begingroup$ So you get an exact fully faithful functor. What about essentially surjective? Is that where Serre's theorem possibly fails? $\endgroup$ – Jim Mar 18 '13 at 20:35
  • $\begingroup$ @Jim: I don't know. $\endgroup$ – user18119 Mar 19 '13 at 10:27

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.