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 ?
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.