Suppose X is a curve.


Under sufficiently nice conditions we have that every line bundle on X corresponds to an equivalence class of divisors modulo principal divisors, with tensor product of bundles corresponding to addition of divisors.


Given a line bundle L on X, I will call the Euler Characteristic of L minus the Euler characteristic of the trivial line bundle the cohomological degree of L.
By the first half of the Riemann Roch theorem, we have that this is equal to the degree of the equivalence class of divisors corresponding to L. Thus cohomological degree is a homomorphism from the Picard group of X to Z.


Is there a more direct proof of the fact that cohomological degree is a homomorphism, that does not go through divisors and the Riemann-Roch theorem? Hopefully this is just a matter of homological algebra.


Is there a similar, cohomological definition of degree that works in higher dimensions?


Thank you.



Effect of tensor product on euler characteristic of line bundles
I think there is a complete answer to your question in 
SGA 6 Exp X $\S$ 5 :
for a quasi-compact scheme there is a natural isomorphism (first Chern class) ${\rm Pic} X \simeq Gr^1(X)$ where $Gr^1(X)$ is defined by the filtration of the Grothendieck group $K\_.(X)$ of coherent sheaves on $X$ given by the dimension of supports. It is defined by sending the class of an invertible sheaf $\mathcal L$ to the class of $\mathcal O\_X$ minus the class of $\mathcal L^{-1}$. For curves $Gr^1(X)=Filt^1(X)$, so that you have an homomorphism ${\rm Pic} X \simeq Gr^1(X)=Filt^1(X)\subset K\_.(X)$, that you can compose with the Euler characteristic $K\_.(X)\rightarrow \mathbb Z$. In fact the situation is even nicer for curves : you have a natural isomorphism $K\_.(X)\simeq \mathbb Z \oplus {\rm Pic} X$ given by rank and determinant.
Another classical reference is Yu. I. Manin  

Lectures on the K-functor in algebraic geometry.


