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Let $B$ be a graded ring an $M$ be a $\mathbb Z$-graded $B$-module. We can associate to $M$ a sheaf of modules on $\operatorname{Proj} B$ by defining the sheaf on the principal open sets $D_+(f)$ to be $(M_{(f)})^\sim$ (where this is the associated module in the affine case) and checking that these agree on overlaps.

However, there is a second construction, which appears on page $165$ of Qing Liu's Algebraic Geometry and Arithmetic Curves. Consider the canonical injection $f:\operatorname{Proj B}\rightarrow \operatorname{Spec} B$. This is continuous. Consider the sheaf $\mathcal F = M^\sim$ on $\operatorname{Spec} B$ (again, this is the associated sheaf in the affine case). We can take the subsheaf of $f^{-1}\mathcal F$ consisting of all degree 0 elements to be the sheaf associated to $M$ on $\operatorname{Proj} B$.

(As @Hanno notes in the comments, there is an issue here. There doesn't seem to be a way to turn this into a sheaf of modules, because $f$ is a merely a continuous map, not a morphism. I would also appreciate comments on this issue.)

I would like to know why these constructions agree. I know that $(f^{-1} \mathcal F)_{\mathfrak p} = M_\mathfrak p$, so taking the homogeneous elements, we see that the two sheafs have the same stalks. But I am rather uncomfortable with the inverse image construction, and do not see how to construct an isomorphism.

Potato
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  • Isn't a morphism of sheaves an isomorphism iff it's an isomorphism on stalks? I seem to recall being told that something like that is true (although I don't know the details). – Qiaochu Yuan Jul 05 '13 at 05:44
  • @QiaochuYuan That's very true. But here we only know the stalks are isomorphic individually. We don't yet have a sheaf morphism. – Potato Jul 05 '13 at 06:01
  • Could you tell in more detail how you want to turn $f^{-1}{\mathcal F}$ into a sheaf of ${\mathcal O}{\text{Proj}(B)}$-modules? It is a priori only a module over $f^{-1}{\mathcal O}{\text{Spec}(B)}$, and $f$ only a map of topological spaces, not equipped with an algebraic component usually taken to induce from a module over $f^{-1}{\mathcal O}_Y$ a module over ${\mathcal O}_X$. – Hanno Jul 05 '13 at 06:58
  • @Hanno I am not sure about that myself! This appears on Liu, Algebraic Geometry and Arithmetic Surfaces, page 165. – Potato Jul 05 '13 at 07:54
  • @Hanno: it is nowhere said that $f^{-1}F$ is a module on Proj$(B)$, but only some specific subgroup. –  Jul 07 '13 at 13:41
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    Dear Potato, As a side comment, have you thought geometrically about what these various constructions mean?

    For example, suppose that $B = \mathbb C[x,y]$ with the usual grading, so that its Proj is $\mathbb P^1$ and its Spec is $\mathbb A^2$, both over $\mathbb C$.

    In this case, it might help you to think about what the map $f: \mathbb P^1 \to \mathbb A^2$ actually is. (Certainly, this example makes it clear that it's not a morphism of schemes.) Regards,

     – Matt E Jul 08 '13 at 06:29
  • @MattE Honestly, I have not. I find it very hard to get geometric intuition for many of the things I am reading about. I do not see what the map is here, even in your restricted case. I know that $\mathbb P^1$ is a sphere and $\mathbb A^2$ a plane, but I do not know what the injection looks like. – Potato Jul 08 '13 at 06:51
  • @Potato: if $b$ is a homogeneous element in $B$, then $f^{-1}(D(b))=D_+(b)$. –  Jul 08 '13 at 08:19
  • @QiL'8 Sorry, I knew that much. I meant I don't know how it looks geometrically. – Potato Jul 08 '13 at 19:16
  • @Potato: the map $f$ is not so intuitive, as it maps closed points to non-closed points. –  Jul 08 '13 at 19:46
  • @QiL'8 Oh, so if $f^{-1}(D(b))=D_+(b)$ for homogeneous $b$, then since the map is an injection, we know that on $D_+(f)$, the inverse image is $M_f$, and then taking the degree zero elements gives $M_{(f)}$? So the sheaf as defined originally and the sheaf we get by taking the degree zero elements of the inverse image sheaf agree on a basis, and hence must agree? Or something like that? – Potato Jul 11 '13 at 21:01
  • @Potato: yes, you just have to check the details. –  Jul 12 '13 at 10:39

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