# Morphism between Invertible Sheaves injective?

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 (*)?

• For arbitrary $k$-schemes (for example when $X$ is affine), your definition of degree is meaningless, since $\chi(L),\chi(\mathcal{O}_X)$ are infinite dimensional. – Mohan Apr 11 '19 at 0:11
• @Mohan: yes that's a good point. I assume that all cohomology groups are finite dimensional. This holds for example when $X$ projective or proper. – KarlPeter Apr 11 '19 at 0:32
• Try $X$ to be $2E\subset X'\to\mathbb{P}^2$, the blowing up of a point and $E$ the exceptional divisor. Then we have $L=O_{2E}(-E)\to O_{E}(-E)\to O_{2E}=L'$ a non-zero map, and you can check that the degree inequality does not hold. – Mohan Apr 11 '19 at 0:43

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)
• in this case we obtain $deg(\phi(\mathcal{L}) ) \le deg(\mathcal{L})$ and $deg(\phi(\mathcal{L}) ) \le deg(\mathcal{L'})$ (with obvious exact seqences). But this doesen't imply $deg(\mathcal{L} ) \le deg(\mathcal{L'})$. Or do I oversee something? – KarlPeter Apr 10 '19 at 21:34