Lemma 27.26.7. Let $X$ be a scheme. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ X$-module. Assume for every point $x$ of $X$ there exists $n \geq 1$ and $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $x \in X_ s$ and $X_ s$ is affine. Then $X$ is separated.

Proof. By assumption we can find a covering of $X$ by affine opens of the form $X_ s$. To show that $X$ is quasi-separated, by Schemes, Lemma 25.21.6 it suffices to show that $X_ s \cap X_{s'}$ is quasi-compact whenever $X_ s$ is affine. This is true by Lemma 27.26.4. Finally, to show that $X$ is separated, we can use the valuative criterion, see Schemes, Lemma 25.22.2.

Thus, let $A$ be a valuation ring with fraction field $K$ and consider two morphisms $f, g : \mathop{\mathrm{Spec}}(A) \to X$ such that the two compositions $\mathop{\mathrm{Spec}}(K) \to \mathop{\mathrm{Spec}}(A) \to X$ agree. Then $f^*\mathcal{L}$ corresponds to an $A$-module $M$ and $g^*\mathcal{L}$ corresponds to an $A$-module $N$ by our classification of quasi-coherent modules over affine schemes (Schemes, Lemma 25.7.4). The $A$-modules $M$ and $N$ are locally free of rank $1$ (Lemma 27.20.1) and as $A$ is local they are free of rank $1$. We are given an isomorphism $N \otimes _ A K \cong M \otimes _ A K$ because $f|_{\mathop{\mathrm{Spec}}(K)} = g|_{\mathop{\mathrm{Spec}}(K)}$. We fix an isomorphism $M \otimes _ A K \cong K \cong N \otimes _ A K$ compatible with the given isomorphism above, so that we may think of $M$ and $N$ as $A$-submodules of $K$ (free of rank $1$ over $A$). Next, choose $s \in \Gamma (X, \mathcal{L}^{\otimes n})$ such that $\mathop{\mathrm{Im}}(f) \subset X_ s$ and such that $X_ s$ is affine. This is possible by assumption and the fact that $A$ is local, so it suffices to look at the image of the closed point of $\mathop{\mathrm{Spec}}(A)$. Then $s$ corresponds to an element $x \in M^{\otimes n}$ and $y \in N^{\otimes n}$ mapping to the same element of $K^{\otimes n}$ and moreover $x \not\in \mathfrak m_ A M^{\otimes n}$ because $f(\mathop{\mathrm{Spec}}(A)) \subset X_ s$. We conclude that $N^{\otimes n} = Ax = Ay \subset M^{\otimes n}$ inside of $K^{\otimes n}$. Thus $N \subset M$. By symmetry we get $M = N$. This in turn implies that $g(\mathop{\mathrm{Spec}}(A)) \subset X_ s$. Then $f = g$ because $X_ s$ is affine and hence separated, thereby finishing the proof. $\square$

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