Can someone explain the construction of the global $\mathbf{Proj}$ to me? Although this question has been asked here, I still have several questions.

  • For each open affine subset $U = \mathrm{Spec} A$ of $X$, let $\mathcal{S}(U) = \Gamma(U, \mathcal{S}|_{U})$, which is a graded $A$-algebra. Then we have a natural morphism $\pi_{U}: \mathrm{Proj} \mathcal{S}(U) \rightarrow U$.

This map is from $A \rightarrow \mathrm{Proj} \mathcal{S}(U)_{f}: a \rightarrow a/1 $, right?

  • Let $f \in A$ and $U_{f} = \mathrm{Spec} A_{f}$. Since $\mathcal{S}$ is quasi-coherent we have $\mathrm{Proj} \mathcal{S}(U_{f}) \cong \pi_{U}^{-1}(U_{f})$.

I don't really understand the explanation under that question. Why $\mathcal S(U_f)=\mathcal S(U)_f$? And why we need $\mathcal{S}$ to be quasi-coherent?

  • How to glue the invertible sheaves $\mathcal{O}(1)$ on each $\mathrm{Proj} \mathcal{S}(U)$ to get an invertible sheaf $\mathcal{O}(1)$ on $\mathbf{Proj}S$

Can someone explain more explicitly? Thank you very much!

Here is my understanding:

Suppose $\mathcal{S}|_{U}=\tilde{M}$, then $\mathcal{S}|_{U}(U)_f=M_f=\mathcal{S}|_{U}(U_f)$. Then we have this commutative diagram: $\require{AMScd}$ $$\begin{CD} \mathrm{Proj}\mathcal{S}(U_f)=\mathrm{Proj}\mathcal{S}(U)\times_{\mathrm{Spec}A}\mathrm{Spec}\,\mathcal A_f @>>> \mathrm{Spec}\,\mathcal A_f\\ @VVV @VVV \\ \mathrm{Proj}\mathcal{S}(U) @>>> \mathrm{Spec}\,\mathcal A. \end{CD}$$


(1) The "old" $\operatorname{Proj}$ of a graded ring $S$ came with a canonical morphism $\operatorname{Proj} S \to \operatorname{Spec} S_0$. It seems to me that one has to define this on standard opens via $S_0 \to S_{(f)}$ and then glue. Maybe that's what your notation means.

(2) Having $\mathscr{F}(\operatorname{Spec} A_f) = \mathscr{F}(\operatorname{Spec} A)_f$ is one characterization of quasi-coherence. This is a good exercise and in the end the solution is very short.

(3) The gluing is similar to that for $\operatorname{\mathbf{Proj}}$. The key step is to show that $\mathscr{O}(1)$ on $\operatorname{Proj}$ is compatible with base change: in the notation of the referenced answer, the pullback of $\mathscr{O}_{\operatorname{Proj} B}(1)$ to $\operatorname{Proj} (B \otimes_A A')$ is $\mathscr{O}_{\operatorname{Proj} (B \otimes_A A')}(1)$. Some details are given in the Stacks Project.

  • $\begingroup$ Thanks for your help. I add my understanding. I think it makes sense now. $\endgroup$ – WWK Mar 9 '14 at 20:58

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