Under the condition specified below, is $\mathcal{O}_X(X-V(I))=R$, where $X=\mathrm{Spec}R$? Is it true that given a noetherian normal domain $R$ and an ideal $I$ of height $\geq 2$ we have $\mathcal{O}_X(X-V(I))=R$, where $X=\mathrm{Spec}R$?
 A: See [Hartshorne], Exercise III.3.5 if $I$ is maximal.
A: Yes, it is true. This is a special case of [EGA IV$_2$, 5.10.5] (see also 5.9.9 if needed) combined with Serre's criterion for normality (see http://stacks.math.columbia.edu/tag/031S). The latter tells you that the depth of the local rings of $X$ at the points in $V(I)$ is $\ge 2$, so the EGA reference applies.
A: I think that one could also prove this with weaker tools than Serre's criterion for normality. That is, consider the restriction map
$$ \mathcal{O}(X) \longrightarrow \mathcal{O}(X\setminus V(I)) $$
then it is easy to see that it is injective and it is surjective by the Algebraic Hartogs Lemma (see Vakil's notes at http://math.stanford.edu/~vakil/0708-216/216class20.pdf) which tells you that
$$ R = \bigcap_{P\text{ prime of height } 1} R_{P} $$
where the intersection takes place inside the fraction field $Frac\,\, R$.
More explicitly, suppose that you have an $f\in \mathcal{O}(X\setminus V(I))$ and you want to prove that this is actually defined in $\mathcal{O}(X)$. Thanks to the Hartogs Lemma it is enough to prove that $f$ is defined at every prime $p\in X$ of height 1, but the open subset $X\setminus V(I)$ contains every such prime $p$, otherwise it would be that $p\in V(I)$, so that $p\supseteq I$ and $height(p)\geq height(I)=2$, absurd.
