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In constructing the relative Proj for a graded ring $S$, one inevitably has that the distinguished opens, $D_+ (f)$, where $f \in S_+$ gives rise to a scheme that is isomorphic to $\textrm{Spec } S_{(f)}$, the latter denoting the degree zero elements in the localization $S_f$.

In all constructions I have seen of this, including Hartshorne and Liu, they usually leave it at that with no mention of global sections. I was tempted to handle this by saying $f = 1$ then $D_+ (f) = \textrm{Proj } S$ is isomorphic to $\textrm{Spec } S_{(1)} = \textrm{Spec } S_0$ (degree 0 elements of $S$). The Wikipedia article mentions this without proof or justification under the "Twisting Sheaf of Serre" section (https://en.wikipedia.org/wiki/Proj_construction#The_twisting_sheaf_of_Serre) lending credence to this fact.

However, the condition that $f \in S_+$, where $1$ typically does not reside (take the natural grading on the polynomial ring for example) seems to invalidate this proof. What justification should I use instead?

Background: I am interested in justifying that for $X = \textrm{Proj } A[x_0,\dots,x_n]$, that $\mathcal{O}_X (1)$ can be generated by the global sections $x_0, \dots, x_n$.

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    $\begingroup$ Globally, $\operatorname{Proj} S$ is almost never an affine scheme (otherwise this construction wouldn't provide any additional value over the $\operatorname{Spec}$ construction). Thus, you cannot hope to write down a ring for which $\operatorname{Proj} S \cong \operatorname{Spec} R$. If $\operatorname{Proj} S$ is a variety (a finite type integral scheme over an algebraically closed field, or more generally a finite type, geometrically integral scheme over a field), then the only global sections of $\mathcal O_X$ on $X = \operatorname{Proj} S$ are the constants. $\endgroup$ – Remy Jan 2 '16 at 2:21
  • $\begingroup$ I asked myself this question when considering that for $S = A[x_0,\dots,x_n]$ and $X = \textrm{Proj } S$. Why can $\mathcal{O}_X (1)$ be generated by global sections $x_0, \dots, x_n$. $\endgroup$ – Future Jan 2 '16 at 2:25
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    $\begingroup$ Aha! I finally understand that theorem in Hartshorne now which states that $\bigoplus_{n \in \mathbf{Z}} \Gamma (X, \mathcal{O}_X) = S$. I always thought it was trivial because I thought it held true for all global sections, but I see now that it is very specific for polynomial things. Thank you f or clearing that up! $\endgroup$ – Future Jan 2 '16 at 2:45
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    $\begingroup$ @Future This may be long shot since this question is so old, but I was wondering if you could share the insight you gained about Hartshorne's claim that $ \oplus_{n \in \mathbb{Z}} \Gamma(X, \mathcal{O}_{X}) = S $? It seems like a comment that you're referring to was deleted. I'm having the same problem understanding the claim that you seemed to be. Why is this so special to polynomial rings? He later shows that the equality holds for sufficiently highly graded components, but only in the case that $A$ is finitely-generated over a field, but no mention of that hypothesis is made here. $\endgroup$ – Luke Oct 31 '17 at 0:38
  • $\begingroup$ It's been almost a year, so I can't remember exactly what I was thinking. But rereading what I have, I seem to have been initially confused on what the global sections of arbitrary Proj might be described as, thinking that it would always be analogous to polynomial rings. In fact, if I recall correctly (I don't have Hartshorne with me at the moment and it's been awhile since I studied the proof, so this needs double checking!), there is some obstruction to gluing the distinguished opens, which is able to be overcome in the very special polynomial case. $\endgroup$ – Future Oct 31 '17 at 1:07

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