MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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 '10 at 1:15
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 '10 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 '10 at 2:06

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 $f \in \sqrt{I}$.

share|cite|improve this answer

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 $\mathbb{C}^2\setminus\{(0,0)\}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.