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I'm just learning the language of schemes, so I'm sorry if this seems a little elementary. Consider an affine scheme $\text{Spec}(R)$. For an ideal $I$ of $R$, denote by $U(I)$ the open subset of $\text{Spec}(R)$ consisting of prime ideals $p$ that do not contain $I$.

Is the ring of regular functions on $U(I)$ simply $R_I$ (the localization of $R$ with respect to $I$)? If $I$ is a principal ideal, then this is one of the earliest results in Hartschorne. Also, it is easy to see that $R_I$ injects into the ring of regular functions on $U(I)$. My guess is that this injection is not surjective, but I can't seem to come up with any examples. Thanks!

flag	asked Mar 29 2010 at 0:58 Albert N 49●1

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What do you mean by "the localization of $R$ with respect to $I$?" The result Hartshorne proves is that the ring of sections of the structure sheaf over a principal open set $D(f)$, $f\in R$, is $A_f$, which is $A$ localized at $f$, not $A$ localized at the ideal $(f)$. If $f$ happens to be a prime element, then $A_{(f)}$ makes sense, but for an arbitrary ideal $I$ (namely a non-prime ideal), $A_I$ doesn't make sense. – Keenan Kidwell Mar 29 2010 at 1:15

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Elaborating on Keenan's comment, if you use $R_S$ to denote the localization other people write as $S^{-1}R$ then $R_I$ is not a reasonable thing to consider, because while $I$ is certainly a multiplicatively closed subset of $R$, it also contains zero, so inverting it gives the zero ring. If you use $R_I$ by analogy with the notation $R_{\mathfrak p}=(R-\mathfrak p)^{-1}R$ for a prime ideal $\mathfrak p$ of $R$, note that $(R-I)$ is only a multiplicatively closed subset when $I$ is prime. When $I$ is prime, this localization is actually sort of opposite to reg. fn's on $U(I)$. – Sam Lichtenstein Mar 29 2010 at 2:02

Geometrically, $R_{\mathfrak p}$ is regular functions defined near the closed subscheme $V(\mathfrak p)$, rather than away from it. – Sam Lichtenstein Mar 29 2010 at 2:06

auniket · Answer 1 · 2010-03-29 01:43:24Z UTC

Well, even if $I$ is a prime ideal, elements of $R_I$ are NOT in general regular functions on $U_I$. For example if $I = \langle x, y \rangle$ in $\mathbb{C}[x,y]$ then $f := 1/(1+x) \in R_I$, but clearly $f$ is not regular everywhere on $\mathbb{C}^2\setminus{(0,0)}$.

babubba · Answer 2 · 2010-03-29 07:04:04Z UTC

I think a good reference might be Eisenbud-Harris, Geometry of Schemes. They construct the structure sheaf $\mathcal{O}$ by specifying it on principal open subsets (viz. the 'important' ones) and extending it uniquely to other open subsets.

On a given ring $R$, you have a basis of open sets of Spec $R$ consisting of the $\text{D}(f)$'s.

($D(f) = Spec R - V(R\cdot f)$, where $R\cdot f$ stands for the ideal generated by $f$).

With each $D(f)$ we associate the localization $R_f$.

With a general open subset $U$ we associate the inverse limit of the $R_f$, for $D(f) \subseteq U$.

More concretely, if $U = Spec R - V(I)$, then $D(f) \subseteq U$ if and only if $V(I) \subseteq V(R\cdot f)$ if and only if $f \in \sqrt{I}$. So $\mathcal{O}(U)$ is the inverse limit of the rings $R_f$, for $f \in \sqrt{I}$.

http://en.wikipedia.org/wiki/Inverse_limit

Regular functions on affine schemes

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