# Can we characterise affine open subschemes of ${\rm Spec}(A)$?

Let $A$ be any ring, commutative with identity, and let $I\subset A$ be an ideal $\neq A$. Let $U\subset{\rm Spec}(A)$ be the open subscheme obtained by "removing" the closed set $V(I)$ of all the prime ideals of $A$ containing $I$. Is there a criterion on the ideal $I$ that enables one to decide whether $U$ is an affine scheme?

• This was discussed here: mathoverflow.net/questions/20782/… – Martin Brandenburg Sep 25 '10 at 9:04
• Assume $U$ is q-compact; equivalently, choose $I$ finitely generated. Inclusion $j:U \rightarrow X := {\rm{Spec}}(A)$ is a q-compact open immersion, so $j_{\ast}(F)$ is q-coh. on $X$ for all q-coh. $F$ on $U$. Restricting back to $U$ gives $F$, so $F = \widetilde{M}|_U$ for an $A$-module $M$. Then by excision ${\rm{H}}^i(U,F) = {\rm{H}}^i(U,\widetilde{M}) = {\rm{H}}^{i+1}_{X-U}(X,\widetilde{M}) = \injlim {\rm{Ext}}^{i+1}(A/I^n,M)$ (limit over $n \rightarrow \infty$), final equality since $I$ f. gen'td (univ. $\delta$-functor argument). Vanishing for $i > 0$ and all $M$ seems "impractical"... – BCnrd Sep 25 '10 at 9:11
• A necessary condition: $I$ has to be of pure codimension one (see e.g. jstor.org/pss/1970814) – auniket Sep 25 '10 at 10:12
• In fact, BCnrd's argument implies aniket's as well (at least in sufficiently geometric settings). This is because $H^{i+1}_{X\setminus U}(X, \widetilde{M})$ is non-zero at the generic points of $X \setminus U$ for appropriate $i$ (depending on the dimension of those generic points). – Karl Schwede Sep 25 '10 at 21:53

For example, 4.4 of the first paper (see also 5.5 of the second) states that if $A$ is a domain of finite over $\mathbb C$ and $U\subset X = \text{Spec} A$ open such that $\Gamma(U,\mathcal O_X)$ is a finite generated $\mathbb C$-algebra. Then $U$ is affine iff $U^{an}$ is Stein.
Here is a simple necessary condition that is of interest to some algebraists. For simplicity I will assume our rings are Noetherian. Since preimage of affine is affine, one conclude that for any ring homomorphism $A\to B$, the ideal $IB$ also has height one. This gives simple non-examples. Take $A=k[x,y,u,v]/(xu-v)$, $I=(x,y)$, $B=A/(u,v)=k[x,y]$. Then the height of $IB$ is $2$, so $U$ can't be affine.
The condition that the height of $I$ remains at most one in all extensions is called "$I$ has superheight one". Krull's Hauptidealsatz can be viewed as saying the superheight of $I=(T)$ in $\mathbb Z[T]$ is one! This point of view leads to many non-trivial generalizations, and some of the most notorious open questions in the field can be stated as a very simple question involving superheights.