Definition 26.9.1. 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

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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