Definition 26.9.1.historical remark A scheme is a locally ringed space with the property that every point has an open neighbourhood which is an affine scheme. A morphism of schemes is a morphism of locally ringed spaces. The category of schemes will be denoted \mathit{Sch}.
26.9 Schemes
Let X be a scheme. We will use the following (very slight) abuse of language. We will say U \subset X is an affine open, or an open affine if the open subspace U is an affine scheme. We will often write U = \mathop{\mathrm{Spec}}(R) to indicate that U is isomorphic to \mathop{\mathrm{Spec}}(R) and moreover that we will identify (temporarily) U and \mathop{\mathrm{Spec}}(R).
Lemma 26.9.2. Let X be a scheme. Let j : U \to X be an open immersion of locally ringed spaces. Then U is a scheme. In particular, any open subspace of X is a scheme.
Proof. Let U \subset X. Let u \in U. Pick an affine open neighbourhood u \in V \subset X. Because standard opens of V form a basis of the topology on V we see that there exists a f\in \mathcal{O}_ V(V) such that u \in D(f) \subset U. And D(f) is an affine scheme by Lemma 26.6.6. This proves that every point of U has an open neighbourhood which is affine. \square
Clearly the lemma (or its proof) shows that any scheme X has a basis (see Topology, Section 5.5) for the topology consisting of affine opens.
Example 26.9.3. Let k be a field. An example of a scheme which is not affine is given by the open subspace U = \mathop{\mathrm{Spec}}(k[x, y]) \setminus \{ (x, y)\} of the affine scheme X =\mathop{\mathrm{Spec}}(k[x, y]). It is covered by two affines, namely D(x) = \mathop{\mathrm{Spec}}(k[x, y, 1/x]) and D(y) = \mathop{\mathrm{Spec}}(k[x, y, 1/y]) whose intersection is D(xy) = \mathop{\mathrm{Spec}}(k[x, y, 1/xy]). By the sheaf property for \mathcal{O}_ U there is an exact sequence
0 \to \Gamma (U, \mathcal{O}_ U) \to k[x, y, 1/x] \times k[x, y, 1/y] \to k[x, y, 1/xy]
We conclude that the map k[x, y] \to \Gamma (U, \mathcal{O}_ U) (coming from the morphism U \to X) is an isomorphism. Therefore U cannot be affine since if it was then by Lemma 26.6.5 we would have U \cong X.
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