The Stacks project

Lemma 28.26.7. Let $X$ be a scheme and $\mathcal{L}$ 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. We show first that $X$ is quasi-separated. By assumption we can find a covering of $X$ by affine opens of the form $X_ s$. By Lemma 28.26.4, the intersection of any two such sets is affine, so Schemes, Lemma 26.21.6 implies that $X$ is quasi-separated.

To show that $X$ is separated, we can use the valuative criterion, Schemes, Lemma 26.22.2. 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. As $A$ is local, there exists $p, q \ge 1$, $s \in \Gamma (X, \mathcal{L}^{\otimes p})$, and $t \in \Gamma (X, \mathcal{L}^{\otimes q})$ such that $X_ s$ and $X_ t$ are affine, $f(\mathop{\mathrm{Spec}}A) \subseteq X_ s$, and $g(\mathop{\mathrm{Spec}}A) \subseteq X_ t$. We now replace $s$ by $s^ q$, $t$ by $t^ p$, and $\mathcal{L}$ by $\mathcal{L}^{\otimes pq}$. This is harmless as $X_ s = X_{s^ q}$ and $X_ t = X_{t^ p}$, and now $s$ and $t$ are both sections of the same sheaf $\mathcal{L}$.

The quasi-coherent module $f^*\mathcal{L}$ 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 26.7.4). The $A$-modules $M$ and $N$ are locally free of rank $1$ (Lemma 28.20.1) and as $A$ is local they are free (Algebra, Lemma 10.55.8). Therefore we may identify $M$ and $N$ with $A$-submodules of $M \otimes _ A K$ and $N \otimes _ A K$. The equality $f|_{\mathop{\mathrm{Spec}}(K)} = g|_{\mathop{\mathrm{Spec}}(K)}$ determines an isomorphism $\phi \colon M \otimes _ A K \to N \otimes _ A K$.

Let $x \in M$ and $y \in N$ be the elements corresponding to the pullback of $s$ along $f$ and $g$, respectively. These satisfy $\phi (x \otimes 1) = y \otimes 1$. The image of $f$ is contained in $X_ s$, so $x \not\in \mathfrak {m}_ A M$, that is, $x$ generates $M$. Hence $\phi $ determines an isomorphism of $M$ with the submodule of $N$ generated by $y$. Arguing symmetrically using $t$, $\phi ^{-1}$ determines an isomorphism of $N$ with a submodule of $M$. Consequently $\phi $ restricts to an isomorphism of $M$ and $N$. Since $x$ generates $M$, its image $y$ generates $N$, implying $y \not\in \mathfrak {m}_ A N$. Therefore $g(\mathop{\mathrm{Spec}}(A)) \subseteq X_ s$. Because $X_ s$ is affine, it is separated by Schemes, Lemma 26.21.15, and we conclude $f = g$. $\square$

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