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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.

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    $\begingroup$ 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
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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.

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  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.

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  • $\begingroup$ 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

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