Lemma 28.18.4. Let $X$ be a scheme. Then $X$ is quasi-affine if and only if the canonical morphism

$X \longrightarrow \mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$

from Schemes, Lemma 26.6.4 is a quasi-compact open immersion.

Proof. If the displayed morphism is a quasi-compact open immersion then $X$ is isomorphic to a quasi-compact open subscheme of $\mathop{\mathrm{Spec}}(\Gamma (X, \mathcal{O}_ X))$ and clearly $X$ is quasi-affine.

Assume $X$ is quasi-affine, say $X \subset \mathop{\mathrm{Spec}}(R)$ is quasi-compact open. This in particular implies that $X$ is separated, see Schemes, Lemma 26.23.9. Let $A = \Gamma (X, \mathcal{O}_ X)$. Consider the ring map $R \to A$ coming from $R = \Gamma (\mathop{\mathrm{Spec}}(R), \mathcal{O}_{\mathop{\mathrm{Spec}}(R)})$ and the restriction mapping of the sheaf $\mathcal{O}_{\mathop{\mathrm{Spec}}(R)}$. By Schemes, Lemma 26.6.4 we obtain a factorization:

$X \longrightarrow \mathop{\mathrm{Spec}}(A) \longrightarrow \mathop{\mathrm{Spec}}(R)$

of the inclusion morphism. Let $x \in X$. Choose $r \in R$ such that $x \in D(r)$ and $D(r) \subset X$. Denote $f \in A$ the image of $r$ in $A$. The open $X_ f$ of Lemma 28.17.1 above is equal to $D(r) \subset X$ and hence $A_ f \cong R_ r$ by the conclusion of that lemma. Hence $D(r) \to \mathop{\mathrm{Spec}}(A)$ is an isomorphism onto the standard affine open $D(f)$ of $\mathop{\mathrm{Spec}}(A)$. Since $X$ can be covered by such affine opens $D(f)$ we win. $\square$

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