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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 $x_0, \dots, x_n$.

Future
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    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.

     – Remy Jan 02 '16 at 02:21
  • 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 $x_0, \dots, x_n$. – Future Jan 02 '16 at 02:25
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    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! – Future Jan 02 '16 at 02:45
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    @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.

     – Luke Oct 31 '17 at 00:38
  • 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. – Future Oct 31 '17 at 01:07
  • I believe my epiphany occurred when I reread the proof of this theorem in Hartshorne, regarding the global sections of the twisted sheaf (I think this is a typo in my original comment, which only mentioned the nontwisted sheaf, yet I'm summing it over $\mathbf{Z}$?), but this time, rather than reading it line by line as I originally did, I thought about what the big picture here is: why did Hartshorne go through the trouble of picking out the distinguished opens so carefully and checking so many technical conditions? – Future Oct 31 '17 at 01:09
  • Sorry, this got a bit lengthy and verbose, partly due to my attempts to retrace my own path. For me, my confusion lied in my shaky foundation in algebraic geometry: I didn't understand the connection between giving a section, in this case a global section, and the sheaf condition of building it up from its local pieces while checking the compatibility of the gluing. This may or may not be your confusion, but once I went back to review the proof again, with that in mind, both the proof of the theorem and my original question became clear. @Remy 's comment may also be helpful to you. – Future Oct 31 '17 at 01:14
  • @Future Thank you, this is actually very helpful. It seems like I am at exactly the position you were in when you asked this question. I've been going over the proof in Hartshorne a few times and am having the same trouble you were identifying exactly where this obstruction is, and how this relates to being generated by the global sections of $\mathcal{O}(1)$. I am thinking I might make a full question about this with a few other details and issues I am having which I might put a link to here when I am done. – Luke Oct 31 '17 at 01:32

1 Answers1

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Let $A$ be a commutative unital ring and let $E:=A\{e_0,..,e_n\}$ be the free $A$-module of rank $n+1$ on the elements $e_i$. Let $E^*:=A\{x_0,..,x_n \}$ be the dual of $E$ and let $S:=Sym_A(E^*)\cong A[x_0,..,x_n]$. Let $\mathbb{P}(\mathcal{E}^*):=Proj(S)$ be the projective space bundle of $\mathcal{E}$ (the sheafification of $E$). There is a canonical exact sequence of graded $S$-modules

$\phi: \oplus_{i=0}^n S x_i \rightarrow S(1) \rightarrow 0$

defined by

$\phi( f_0,f_1,..,f_n):= \sum_{i=0}^n f_i x_i .$

Let $\pi:\mathbb{P}(\mathcal{E}^*) \rightarrow Y$ (here $Y:=Spec(A)$) be the projection map. When sheafifying the map $\phi$ we get the tautological sequence

$\pi^*\mathcal{E}^* \rightarrow^{\phi} \mathcal{O}(1) \rightarrow 0.$

Since $\pi^*\mathcal{E}^*$ is a free $\mathcal{O}$-module it follows this proves the invertible sheaf $\mathcal{O}(1)$ is generated by the global sections $s_i:=\phi(x_i)$.

Lemma: Let $X$ be a scheme. For a surjection $\phi: \mathcal{O}_X \{y_1,..,y_l\} \rightarrow \mathcal{L}$ where $\mathcal{L}$ is an invertible sheaf on $X$ and $\mathcal{O}_X\{y_1,..,y_l\}$ is the free sheaf of rank $l$ on the elements $y_i$, it follows $\phi(y_i)$ are global sections generating $\mathcal{L}$. Conversely given a set of global sections $s_1,..,s_l$ of $\mathcal{L}$ that generated $\mathcal{L}$ it follows there is a canonical surjective map of $\mathcal{O}_X$-modules $\phi: \mathcal{O}_X\{y_1,..,y_l\}\rightarrow \mathcal{L}$ defined on an open set $U$ by $\phi(\sum_i u_iy_i):=\sum_i u_i(s_i)_U$ where $(s_i)_U$ is the restriction of $s_i$ to $U$. The details may be found in Hartshorne's book Chapter II.7.

hm2020
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