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Let $S$ be a graded ring with $S_0=A$ a finitely generated $\mathbb{K}$-algebra and $S_1$ a finitely generated $A$-module. Let $M$ be a graded $S$-module and $\tilde{M}$ the corresponding sheaf on $Proj(S)$. For a sheaf $\mathcal{F}$ on the affine space $X=\mathbb{A}^N_{\mathbb{K}}$ we define $$ \Gamma_*(\mathcal{F})=\bigoplus_{n \in \mathbb{Z}}\Gamma(Proj(S),\mathcal{F}(n)) .$$ We define an equivalence relation $\sim$ on graded $S$-modules: we say that $M \sim M'$ if there exists an integer $k$ such that $M_{\ge k} \simeq M'_{\ge k}$, where $M_{\ge k}:=\bigoplus_{n \ge k}M_n$. If $M$ is a graded $S$-module, we say that $M$ is quasifinitely generated if it is equivalent to a finitely generated module. I have to prove that the functors $\tilde{ }$ and $\Gamma_*$ induce an equivalence of categories between the category of quasifinitely generated graded $S$-modules (modulo the equivalence $\sim$) and the category of coherent $\mathcal{O}_{Proj(S)}$-modules.

  • What have you done so far? One outline for the proof is given in Hartshorne, Exercise II.5.9. There are at least two questions here already about the main parts of the Exercise: http://math.stackexchange.com/q/496162/116766 http://math.stackexchange.com/q/1147772/116766 – Takumi Murayama Mar 11 '15 at 05:41
  • @TakumiMurayama Thank you for informations... but in Hartshorne we have an additional hypotesis: here $S$ is not necessarely generated by $S_1$ as $S_0$- algebra. – Emiliano Inguscio Mar 11 '15 at 11:09

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I think if $S_1$ doesn't generate $S$ then your statement is false. The following is Example 4.6 in "Maps between non-commutative spaces" by S. Paul Smith.

Consider the weighted polynomial ring $S = k[x,y]$, where $x$ has degree 1 and $y$ has degree 2. Let $M = S/(x)$, and consider the graded module $M(1)$, which is $M$ with the grading shifted by $1$. Note that $M(1)$ is not isomorphic to $0$ in the category of quasi-finitely generated graded $S$-modules modulo your equivalence relation.

We claim $\widetilde{M(1)} = 0$, hence the functor $N \mapsto \widetilde{N}$ is not fullly faithful. It suffices to show $$(M(1)_x)_0 = (M(1)_y)_0 = 0,$$ since $$\widetilde{M(1)}\bigr\rvert_{D_+(x)} \simeq \bigl((M(1)_x)_0\bigr)^\sim \qquad \text{and} \qquad \widetilde{M(1)}\bigr\rvert_{D_+(y)} \simeq \bigl((M(1)_y)_0\bigr)^\sim.$$ and the affine charts $D_+(x)$ and $D_+(y)$ cover $\operatorname{Proj} S$. First, $(M(1)_x)_0 = 0$ since $x \in \operatorname{Ann} M$. Next, to show $(M(1)_y)_0 = 0$, we note that $$M(1)_y = \biggl\{ \frac{f(y)}{y^n} \biggm\vert f(y) \in M(1),\ n \in \mathbf{Z}_{\ge 0} \biggr\}$$ has no non-zero degree $0$ elements, since the numerators $f(y)$ have odd degree, while the denominators $y^n$ have even degree. Thus, $(M(1)_y)_0 = 0$.

I also want to point out that similar statements relying on generation by $S_1$ (e.g. Prop. 5.12 in Hartshorne) are also false for similarly defined weighted projective spaces; see §1.5 "Pathologies" in Dolgachev's "Weighted projective spaces".

Finally, if you instead define weighted projective spaces using the stack $\operatorname{\mathbb{P}roj}S$ and look at quasi-coherent sheaves there instead, you do get an equivalence; see, e.g., Prop. 2.3 in "Mirror symmetry for weighted projective planes and their noncommutative deformations" by Auroux, Katzarkov, and Orlov.