Can a scheme be defined by gluing open affines such that the intersections are affine?

One way to think of a manifold is as a family of of open subsets $U_i \subset \mathbb{R}^n$, together with distinguished subsets $V_{ij} \subset U_i$ and isomorphisms $\psi_{ij}: V_{ij} \to V_{ji}$ that satisfy the cocycle condition. This may not be useful practically, but occasionally it might be an intuitive crutch. Now, just as manifolds are obtained by gluing together subsets of $\mathbb{R}^n$, a scheme is obtained by gluing together affines. In other words, we have open affines $U_i = \mathrm{Spec} A_i$ for suitable rings $A_i$, open subsets $V_{ij} \subset U_i$, and isomorphisms $\psi_{ij}$ (of locally ringed spaces) as before. However, the open subsets $V_{ij}$ need not be themselves affine.

Question: Is it possible to formulate this definition such that the sets $V_{ij}$ are affine? I know this can be done if the scheme is separated (because the intersection of open affines is affine).

One of the nice things about this is that wouldn't have to worry about the isomorphisms $\psi_{ij}$ being isomorphisms of locally ringed spaces, just isomorphisms of the corresponding rings.

This is probably a bad way of thinking of schemes in general; the only reason I was interested in it was because then the fibered product could perhaps be thought of more "explicitly."

Question': Is there an easy way to tell when the complement of $V(\mathfrak{a}) \subset \mathrm{Spec} A$ is affine? Of course, this is true when $\mathfrak{a}$ is principal. (Answered: see the comments of Matthew Emerton and David Speyer.)

• Under fairly general hypotheses (locally Noetherian is okay), one finds that if the complement of a closed subscheme $Y$ of a scheme $X$ is affine, $\mathcal O_{X,\eta}$ has dimension at most 1, for each generic point $\eta$ of $Y$. If $X =$ Spec $A$, say with $A$ Noetherian, and $Y= V(\mathfrak a)$ has affine complement, this says that the local ring $A_{\mathfrak p}$ has dimension at most 1 for each minimal prime $\mathfrak p$ of $\mathfrak a$. This is a fairly restrictive condition on $\mathfrak a$ (it says that each component of $V(\mathfrak a)$ is of codimension at most 1). – Emerton Jul 12 '10 at 14:52
• All schemes you get this way will be "semi-separated," meaning that the intersection of two affines is affine (of course). In particular, you won't get the plane with the doubled origin. This is the problem that Charles Staats solves, but it may not be worth the bother to remove the hypothesis. – Ben Wieland Jul 12 '10 at 16:32
• Akhil, the best "explicit" way to think about fiber products is functorially (coupled with concrete knowledge of how it interacts with open and closed immersions). That is more useful in many cases than mucking around with open affines. To amplify Emerton's comment, the spectrum of the local noetherian ring $A = O_ {X,\eta}$ has the property that the complement of its closed point is affine, which forces "local cohomology" modules ${\rm{H}}^i_ {\mathfrak{m}}(A,M)$ to vanish for all $i > 1$ and all $A$-modules $M$. Theory of local cohomology then leads to the dim. bound Emerton mentions. – BCnrd Jul 12 '10 at 17:14
• Regarding Emerton's comment: Beware that, if $Y$ is codimension $1$ at every minimal prime, this does not imply that $X \setminus Y$ is affine. For example, take $X=\mathrm{Spec} \ k[w,x,y,z](wz−xy)$ and take `$Y=\{w=x=0\}$. Then $X$ is irreducible of dimension 3 , $Y$ is irreducible of dimension 2 , and $X \setminus Y$ is isomorphic to $3$ -space with a line removed. – DES-SupportsMonicaAndTransfolk Jul 12 '10 at 18:45
• Akhil, concerning your parenthetical comment, connectedness of $X \otimes_k \overline{k}$ is not a consequence of mere irreducibility of $X$. Consider $X = {\rm{Spec}}(k')$ for a nontrivial finite separable extension $k'/k$; then $X \otimes_k \overline{k}$ is disconnected but $X$ is irreducible. That sequence of exercises in Hartshorne is super-duper important, so well worth the investment of time to figure out for yourself (I won't tell you where it is all done in EGA). – BCnrd Jul 12 '10 at 20:11

You can, if you use a slightly more general notion of gluing. (The notion of gluing you present is "wrong", or at least simplistic, in roughly the same way that it is "wrong" to require that a basis for a topology be closed under intersections. E.g., if you do this, then the set of open balls in $\mathbb{R}^n$ for $n > 1$ does not form a "basis.")
Let $X$ be a scheme. Consider the diagram whose objects are open affine subschemes of $X$, and whose morphisms are inclusions $U \hookrightarrow V$ such that $U$ is a distinguished open subset of $V$. Whenever $U$ and $V$ are two objects and $x \in U \cap V$, there exists an object $W \subset U \cap V$ such that $x \in W$ and $W \hookrightarrow U$, $W \hookrightarrow V$ are both morphisms: Since the distinguished open subsets of $U$ form a basis for the topology, there is a distinguished open $W'$ in $U$ such that $x \in W' \subset U \cap V$. Similarly, there is a section $f$ over $V$ such that $x \in V_f \subset W'$. But then $V_f = W'_f$ is a distinguished open subset of both $U$ and $V$, so we let $W = W'_f$.
In particular, the fiber product is obtained by gluing together schemes of the form $\mathrm{Spec} A \otimes_C B$, where $\mathrm{Spec} C$ contains the images of both $\mathrm{Spec} A$ and $\mathrm{Spec} B$, with "overlap inclusions" specified by morphisms $A \otimes_C B \to A_f \otimes_C B_g$. An important note here: if $C \to D$ is a ring epimorphism (e.g., corresponds to an open immersion), and $A, B$ are $D$-algebras, then $A \otimes_C B$ is naturally isomorphic to $A \otimes_D B$.
• Thanks! Is there some other condition you need for a ring epimorphism to be an open immersion? (E.g. $\mathbb{Z} \to \mathbb{Q}$) – Akhil Mathew Jul 12 '10 at 18:33