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 $f = g$. $\square$


Comments (0)


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01PY. Beware of the difference between the letter 'O' and the digit '0'.