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Let $X \subseteq \mathbb{P}^n$ be an embedded projective variety over some field $k$ with its corresponding homogeneous coordinate ring $R = k[X_0,\dots,X_n]/I(X)$. Let further $X = \bigcup_{i=0}^n U_i$ be the standard covering of $X$ by affine varieties, and $p \in X$ some point. When we want to consider the stalk of $X$ at $p$ we can reduce to the affine case since $\mathcal{O}_{X,p} \cong \mathcal{O}_{U_i,p}$ if $p \in U_i$. Because of this the stalk at $p$ can be described in terms of the coordinate ring by dehomogenizing : Let for example $p = (1 : p_1 : \dots : p_n) \in U_0$ then $\mathcal{O}_{X,p} \cong (R/(X_0-1))_{(X_1-p_1, \dots, X_n-p_n)}$.

This description however depends on the choice of the particular affine open sub-variety $U_i$ which contains $p$. Is there another (unifying) way to give $\mathcal{O}_{X,p}$ in terms of $R$ which does not depend on the location of $p$ in $X$?

Thank you in advance!

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  • $\begingroup$ I would rather write $\mathcal{O}_{U_0,p}$ using the dehomogenized coordinates $x_i=X_i/X_0$. I'm not even sure that your way of writing it gives the same end result. It may, but "assigning a constant value to a homogeneous coordinate" gives me a bad vibe - confusion is way too likely to happen. Anyway, if the point $p$ is not on the most natural affine piece, then what are you to do? This is more or less the same thing you may have seen in complex analysis. If you want to study the local behavior of an object near the point of infinity of the Riemann sphere, you can't use the same coordinate. $\endgroup$ Sep 20, 2014 at 13:57
  • $\begingroup$ Here is a more general question: Let $A$ be a commutative $\mathbb{N}$-graded algebra. Consider the corresponding projective scheme $\mathrm{Proj}(A)$. Its underlying space consists of homoegeneous prime ideals of $A$ not containing the irrelevant ideal $A_+$. If $\mathfrak{p} \in \mathrm{Proj}(A)$, what is the stalk $\mathcal{O}_{\mathrm{Proj}(A),\mathfrak{p}}$? Of course, if $f \in A^+$ is homogeneous such that $\mathfrak{p} \in D_+(f) \cong \mathrm{Spec}(A_{(f)})$, the result will be $(A_{(f)})_{(\mathfrak{p})}$, but is there a formula which doesn't depend on $f$? $\endgroup$ Sep 20, 2014 at 14:00
  • $\begingroup$ @JyrkiLahtonen: I think it is the same. By assigning $(1 : p_1 : \dots : p_n)$ to $(p_1,\dots,p_n)$ we get an embedding $U_0 \to \mathbb{A}^n$ and from this it is easy to see that the vanishing ideal of $U_0$ arises from $I(X)$ by substituting $X_0$ by $1$. So $R/(X_0-1)$ is the affine coordinate ring of $U_0$. But I don't know how to get around this, that is why I am asking. I don't even know if it is possible. $\endgroup$
    – Dune
    Sep 20, 2014 at 14:38
  • $\begingroup$ @MartinBrandenburg: Thank you for pointing that out! This is probably the better question. $\endgroup$
    – Dune
    Sep 20, 2014 at 14:40
  • $\begingroup$ @Martin: Since $A_{(f)}$ is not a graded ring (except in an uninterestingly trivial sense) the notation $(A_{(f)})_{(\mathfrak p)}$ is not really appropriate. Anyway the answer you require that makes no reference to $f$ is simply $A_{(\mathfrak p)}$ . $\endgroup$ Sep 21, 2014 at 9:47

2 Answers 2

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Given a completely arbitrary graded ring $R$ (in particular not supposed to be a domain) and its associated projective scheme $X=\text{Proj}R$, the stalk of the structure sheaf $\mathcal O_X$ at $\mathfrak p\in X$ is $$\mathcal O_{X,\mathfrak p}=R_{(\mathfrak p)}$$ Here $R_{(\mathfrak p)}\subset R_{\mathfrak p}$ is the subring consisting of fractions $\frac r\pi$ where $r\in R, \pi\in R\setminus \frak p$ are homogeneous elements of the same positive degree.

References:
Hartshorne Proposition 2.5 (a), page 76 or EGA II Proposition (2.5.5), page 31 or Miyanishi Lemma 5.3(3), page 106.
Non-references:
After a superficial check, I noted that the result seems to be missing from such textbooks as Bosch, Eisenbud-Harris, Görtz-Wedhorn, Qing Liu.
This is no criticism of these fine references: obviously, writing mathematical books implies tough choices on what to omit.

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Often the structure sheaf of $\mathrm{Proj}(A)$ (where $A$ is a commutative graded ring) is constructed via gluing of the affine pieces $D_+(f) \cong \mathrm{Spec}(A_{(f)})$. But actually, we can write it down explicitly. More generally, if $M$ is some graded $A$-module, we can write down the sheaf $\widetilde{M}$ explicitly. The idea is to make the same as in the affine case, but replace localizations by homogeneous localizations.

Recall that $\mathrm{Proj}(A)$ is the space of homogeneous prime ideals of $A$ not containing $A_+$. Let $M$ be a graded $A$-module. Let $U \subseteq \mathrm{Proj}(A)$ be an open subset. We define $\widetilde{M}(U)$ as a subset of $\prod_{\mathfrak{p} \in U} M_{(\mathfrak{p})}$ (here $M_{(\mathfrak{p})}$ denotes the homogeneous localization of $M$ at $\mathfrak{p}$) consisting of those tuples $(s_\mathfrak{p})$ such that for all $\mathfrak{p} \in U$ there is some $\mathfrak{p} \in V \subseteq U$ open and some homogeneous elements $m \in M$, $r \in A$ of the same degree such that for all $\mathfrak{q} \in V$ we have $r \notin \mathfrak{q}$ and $\frac{m}{r} = s_\mathfrak{q}$ in $M_{(\mathfrak{q})}$.

From this description it is quite clear that $\widetilde{M}_\mathfrak{p} = M_{(\mathfrak{p})}$.

After writing this answer, I realize that this is also done in Hartshorne's book.

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