sections of an affine scheme over a non-principal open set

Question

For an affine scheme $A$ is there a purely ring-theoretic interpretation of the sections $\mathcal{O}_{Spec A}(U)$ over an open subset $U$ which is not of the form $D(f)$ ? The textbook definition is that it is the result of sheafification over the basis of principal open sets, so one can express it formally as the finite sum of such sections, but it would be nice to have a more direct interpretation.

Answer 1

To answer my own question (reference is here) one may construct the structure sheaf directly over the whole topology as follows,

$$\Delta(U) := A \setminus \bigcup_{\mathfrak{p} \in U} \mathfrak{p} $$

One may show that this is a saturated multiplicatively closed (m.c.) set (see Atiyah MacDonald Ch V Ex 7). Namely $xy \in \Delta(U) \iff x \in \Delta(U) \wedge y \in \Delta(U)$. Moreover it's evident that $V \subseteq U \iff \Delta(U) \subseteq \Delta(V)$ therefore we may construct the pre-sheaf

$$ \mathcal{O}_X(U) := \Delta(U)^{-1} A $$

with the canonical restriction morphisms. It may be shown that $\Delta(D(f))$ is the minimal saturation of the usual m.c. set $S_f = \{1, f, f^2 \ldots \}$, whence one obtains a canonical isomorphism (Atiyah-Macdonald Ch V Ex 8)

$$ A_f \xrightarrow{\sim} \mathcal{O}_X(D(f)) $$

This may be seen more directly by noticing that

$$x \in \Delta(D(f)) \iff D(f) \subseteq D(x) \iff f \in \sqrt{x}$$

and show that the canonical morphism is bijective. The stalks may be constructed naturally in this setting because

$$ \Delta(\{x\}) = \bigcup_{U \ni x} \Delta(U) $$

implying that naturally one has

$$ A_{\mathfrak{p}_x} = \Delta(\{x\})^{-1} A \xrightarrow{\sim} \varprojlim_{U \ni x} (\Delta(U)^{-1} A) = \mathcal{O}_{X,x} $$

Of course one must still show that $\mathcal{O}_X$ is a sheaf in the usual way, it being enough to show on the basis of open sets $\mathcal{B}$.

Answer 2

Expanding on david r.'s answer with a proof that it gives a sheaf. For every open $U$ and finite principal open cover by $D(f_i)$ we want exactness of $$0\to\Delta(U)^{-1}A\to\prod A_{f_i}\to\prod A_{f_if_j}$$

The right part follows because we can write $1\in\Delta(U)^{-1}A$ as a partition of unity $\sum b_if_i$ with coefficients in $\Delta(U)^{-1}A$, and a preimage of the family $(a_i/f_i)_i$ will be $\sum b_ia_i$ (replacing $f_i$ by a power and $a_i$ by a multiple of $f_i$ we may assume $a_if_j=a_jf_i$). Nothing new here.

For the left part: We expect to use that $U$ is small, i.e. that $U \subset\cup D(f_i)$, i.e. that $I\subset\sqrt{(f_i)}$. If a section $s\in A$ on $U$ goes to zero, its annihilator $J$ contains a power of each $f_i$. We want $\Delta(U)^{-1}J=(1)$. A prime ideal containing $J$ contains all $f_i$ and thus $I$. Yet the prime ideals of $\Delta(U)^{-1}A$ are precisely those that do not contain $I$, and we're done.

For well-definedness, we wanted $\Delta(U)$ to be not too large (contained in the saturation of each $\{1,f_i,f_i^2,\ldots\}$). For injectivity, we wanted $\Delta(U)$ to be large enough so that it does not contain prime ideals that contain all $f_i$.