3
$\begingroup$

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: $\require{AMScd}$ $$\begin{CD} \mathrm{Proj}\mathcal{S}(U_f)=\mathrm{Proj}\mathcal{S}(U)\times_{\mathrm{Spec}A}\mathrm{Spec}\,\mathcal A_f @>>> \mathrm{Spec}\,\mathcal A_f\\ @VVV @VVV \\ \mathrm{Proj}\mathcal{S}(U) @>>> \mathrm{Spec}\,\mathcal A. \end{CD}$$

$\endgroup$
1
$\begingroup$

(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.

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

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.