Let $X$ be a projective integral scheme over $K=\bar{K}$. Suppose that $\Gamma(X,\mathcal L)\neq 0$, $\Gamma(X,\mathcal L^{*})\neq 0$. We want to show that $\mathcal L\cong \mathcal O_X$. There is an answer here on a pretty good website but I don't fully understand it. It defines non zero (why ?) morphisms $\mathcal L\to \mathcal L$ and $\mathcal O_X\to \mathcal O_X$ using composition of maps we get with two non zero global sections. Why would the regular sections defining these morphisms would be constant ? Why is an endomorphism of line bundles would an isomorphism ?

In my case we suppose more than in the answer, we have projective. With these assumptions is there another way ? I wanted to say that $X$ is a closed subscheme of some projective space but I think we would need more assumptions on $S$ if $X=\operatorname{Proj}S$.

Any help ?


1 Answer 1


Take a point $x\in X$ where the section of $\mathcal L$ and the section of $\mathcal L^*$ do not vanish. Since the fiber of a line bundle over $x$ is a 1-dimensional vector space, it is easy to see that compositions $\mathcal L\to\mathcal L$ and $\mathcal O_X\to\mathcal O_X$ induce non-zero endomorphisms of fibers over $x$, hence they are non-zero. Any endomorphism of line bundle is a multiplication with a regular function (one can prove it locally, using local triviality of a line bundle), and any regular function on a proper connected variety is constant.

  • $\begingroup$ Why would there be a point where they both don't vanish ? $\endgroup$
    – raisinsec
    Sep 30 at 10:46
  • 1
    $\begingroup$ Perhaps another way to phrase this argument could be: let $s, s’$ be nonzero global sections of $\mathcal{L}$ and its dual, respectively. Then $f=s’(s)$ is a global function on $X$, hence is constant. If $f=0$, then $X=V(s) \cup V(s’)$ is not irreducible (since $s,s’$ are nonzero). Hence $f$ is a nonzero constant, so $s$ never vanishes. $\endgroup$
    – Aphelli
    Sep 30 at 11:42
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    $\begingroup$ There is such a point, because vanishing locus of a section is closed, and X is irreducible, as Aphelli said. $\endgroup$
    – danneks
    Sep 30 at 12:55
  • $\begingroup$ @Aphelli Indeed this is nicely stated, thank you both. $\endgroup$
    – raisinsec
    Oct 3 at 13:53
  • $\begingroup$ @Aphelli Actually why does it imply $X=V(s)\cup V(s')$ ? If $x\notin V(s)$ I get that $s_x\in \operatorname{Ker}s'_x$, not necessarily that $s'_x=0$ $\endgroup$
    – raisinsec
    Oct 10 at 9:51

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