1
$\begingroup$

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 (*)?

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Mohan
    Apr 11, 2019 at 0:11
  • $\begingroup$ @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. $\endgroup$
    – user267839
    Apr 11, 2019 at 0:32
  • $\begingroup$ 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. $\endgroup$
    – Mohan
    Apr 11, 2019 at 0:43

1 Answer 1

0
$\begingroup$

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)

$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – user267839
    Apr 10, 2019 at 21:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .