Let consider a curve $C$ over $k$ and the degree map on it's Picard Group $\deg: \operatorname{Pic}(C) \to \mathbb{Z},\; \mathcal{L} \mapsto \chi(\mathcal{L}) - \chi(\mathcal{O}_C)$.
Here $\chi(\mathcal{L}) := \sum _{i \ge 0} (-1)^i \dim_k H^i(X, \mathcal{L})= \dim_k H^0(X, \mathcal{L})- \dim_k H^1(X, \mathcal{L})$ because $C$ has dimension $1$, so $\dim_k H^i(X, \mathcal{L})=0$ for $i >1$ by Grothendieck's vanishing theorem.
The aim is to show that $\deg$ is a group homomorphism, so I have to show that $$\deg(\mathcal{L} \otimes \mathcal{L}') = \deg(\mathcal{L}) + \deg(\mathcal{L}').$$
My question refers to following reduction step:
Why we can wlog replace the curve $C$ by a subcurve $C' \subset C$ which arises from a zero-dimensional ideal sheaf $\rho$?
My ideas:
We have the exact sequence
$$0 \to \rho \to \mathcal{O}_C \to \mathcal{O}_{C'} \to 0.$$
We know that $\chi$ is additive, so $\chi(\mathcal{O}_{C})=\chi(\rho)+ \chi(\mathcal{O}_{C'})$, futhermore $\dim_k H^i(C, \rho)=0$ for $i >0$ by Grothendieck, but this lead me not to a solution...