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Let $S$ be a graded ring. I was wondering if someone could please explain me how I can interpret structure sheaf of $Proj \ S$ in terms of compatible stalks? Thank you!

Edit: This is Exercise 4.5.M. on Ravi Vakil's notes on Algebraic Geometry. I have been trying to work it out for a while with some of the comments I received. I still could not figure it out and I would appreciate an explanation/answer. Thank you!

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2 Answers 2

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Yes, see Hartshorne's book, Section II.5. It's like for affine schemes, but localizations are replaced by homogeneous localizations.

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  • $\begingroup$ Thank you for the answer. I was wondering if it would be possible to give me a more explicit explanation by any chance. I haven't been able to figure this out out... $\endgroup$
    – user192077
    Commented Nov 27, 2014 at 2:08
  • $\begingroup$ Have you read the description in Hartshorne's book? Why is not explicit enough? What about the structure sheaf on affine schemes, is it explicit for you? In any case, what one really uses is the local description, i.e. $\mathcal{O}_{\mathrm{Proj}(A)}(D_+(f))=A_{(f)}$. $\endgroup$
    – Martin Brandenburg
    Commented Nov 27, 2014 at 9:25
  • $\begingroup$ Hi, I looked into Section II.5 of Hartshorne's book, but I just couldn't figure out where what you mentioned is located... $\endgroup$
    – user192077
    Commented Nov 27, 2014 at 20:50
  • $\begingroup$ Read more carefully. $\endgroup$
    – Martin Brandenburg
    Commented Nov 27, 2014 at 20:54
  • $\begingroup$ I found it and I will look into it. Thank you! $\endgroup$
    – user192077
    Commented Nov 28, 2014 at 1:54
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I just want to say that you already know how to do this, in the sense that if you specify a sheaf on a base (or something even smaller) for the topology then you know the stalks. Then the sheaf is isomorphic to the one whose sections over $U$ are elements of $\prod_{p \in U} \mathscr{F}_p$ that are locally compatible, using the base.

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  • $\begingroup$ Thank you for the answer. I was wondering if it would be possible to give me a more explicit explanation by any chance. I haven't been able to figure this out out... $\endgroup$
    – user192077
    Commented Nov 27, 2014 at 2:09

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