Let $X$ be a $k$-scheme and $\mathcal{L},\mathcal{N}$ two invertible sheaves on $X$. Assume that there exist a morphism $\mathcal{L} \to \mathcal{N}$ of $\mathcal{O}_X$-modules. Assume that $X$ has for every coherent sheaf $Coh(X)$ finite dimensional cohomology groups. (for example this holds when $X$ is projective or proper)
(*) My question is why does this already imply that $deg(\mathcal{L}) \le deg(\mathcal{N})$.
Remark: The degree of an invertible sheaf $\mathcal{L}$ is defined by $$deg(\mathcal{L}):= \chi(\mathcal{L}) - \chi(\mathcal{O}_X)$$
with Euler-characteristics
$$\chi(\mathcal{L}) := \sum _{i \ge 0} (-1)^i dim_k H^i(X, \mathcal{L})$$
My ideas:
The Euler -cahracteristic is additive in the sense that if
$$0 \to \mathcal{L'} \to \mathcal{L} \to \mathcal{L''} \to 0$$
is a short exact sequence then $\chi(\mathcal{L})= \chi(\mathcal{L'})+\chi(\mathcal{L''})$.
therefore it suffice to show that $\mathcal{L} \to \mathcal{N}$ is injective. is it always true? Since it is a local problem and invertible sheaves are locally trivial we can assume that
-$X= Spec(R)$
-$\mathcal{L},\mathcal{N}=\mathcal{O}_X$
So the problem boils down to show that any $R$-module morphism $\phi: R \to R$ is injective. Everything is told on the image of $1$. But if $R$ isn't an integral domain $\phi$ might be not injective so my considerations fail.
Does anybody see an argument why (*) nevertheless must hold? Or do we need an extra assumption for $X$ like beeing integral or smooth (to garantee that local sections are integral). In the original script $X$ was a ruled surface (as defined in Hartshorne).
Or is there another way to verify (*)?
you have maps$\mathcal{L}\to \phi(\mathcal{L})\to \mathcal{L'}$ the first one is surjective the last one is injective and now you can use short exact sequence($\phi(\mathcal{L})$is an invertible sheaf)