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For simplicity let's say $A$ is a noetherian ring, $S = A[x_0, \ldots, x_r]$, and $X = \operatorname{Proj} S = \mathbb{P}^r_A$. I want to understand what, if anything, the sheaf $$\mathscr{F} := \bigoplus_{n \in \mathbb{Z}} \mathcal{O}_X(n)$$ means geometrically.

First, I think that if I put $$Y := \operatorname{Spec}(S) \setminus V(S_+) = \mathbb{A}^{n+1}_A \setminus V(S_+)$$ then I have a morphism $q : Y \to X$ (which takes prime ideals corresponding to closed points to their homogenizations) and that $\mathscr{F} = q_\ast \mathcal{O}_Y$. Is this correct? I'm not confident enough with these things yet to trust myself on something like that yet. [EDIT: To be more specific, I'm having difficulty constructing the underlying function $q : Y \to X$ in a way that makes it tractable to show that $q$ is Zariski-continuous, to the point that I'm wondering whether such a map exists in general.]

Second, whether the above is correct or not, does $\mathscr{F}$ have a nice geometric meaning of some sort?

Daniel McLaury
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  • The answer is yes for the first question. – Cantlog Sep 23 '13 at 20:23
  • For an arbitrary noetherian ring $A$? I had a bit of trouble checking that that actually gave a continuous map if it wasn't just over an algebraically closed field. Do you have a source? – Daniel McLaury Sep 23 '13 at 22:31
  • This holds for any commutative ring $A$. I will write a proof if you like. – Cantlog Sep 24 '13 at 13:41
  • Could you? I was having trouble defining the map $q$ in a way that made it easy to check that the map was even continuous... – Daniel McLaury Sep 24 '13 at 22:26
  • Dear Daniel, it seems that I misundertood your real question. This is not really transparent in your original post. I will delete my answer and probably you should edit your question or post another one. – Cantlog Sep 26 '13 at 19:39

1 Answers1

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The map $q$ is well defined for any positive graded ring $S$.

For any prime ideal $\mathfrak p$ of $S$, denote by $\mathfrak p^h$ the homogeneous ideal associated to $\mathfrak p$. It is also prime. For any homogeneous ideal $I$, by construction we have:

$I\subseteq \mathfrak p^h $ if and only if $I\subseteq \mathfrak p$.

Taking $I=S_+$, we see that $q(\mathfrak p):=\mathfrak p^h\in \mathrm{Proj}(S)$ if $\mathfrak p\notin V(S_+)$. This shows that $q$ is actually a map
$$q: \mathrm{Spec}(S) \setminus V(S_+)\to \mathrm{Proj}(S).$$ Moreover the above equivalence shows that $$ q^{-1}(V_+(I))=V(I) \setminus V(S_+). $$ Hence $q$ is continuous. Finally, for any $f\in S_+$ homogeneous, we have $$q^{-1}(D_+(f))=D(f).$$ This helps to show the equality of sheaves you are after.

Cantlog
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  • Can you expand on what's meant by $\mathfrak{p}^h$? I took it to mean the ideal generated by the homogeneous parts of each element of $\mathfrak{p}$, but then it appears for instance that in $k[x, y]$ we have $(y-x^2)$ prime but $(y-x^2)^h = (x^2, y)$ not even radical. Or am I making a mistake or working with the wrong definition? – Daniel McLaury Sep 26 '13 at 21:28
  • Ah, I think I see -- I want to work with something larger, not smaller, so I should take the maximal homogeneous ideal contained inside of $\mathfrak{p}$. – Daniel McLaury Sep 26 '13 at 21:42
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    Sorry, I forgot to mention. Your second comment is correct: $\mathfrak p^h$ is the ideal generated by the homogeneous elements of $\mathfrak p$ – Cantlog Sep 26 '13 at 22:04