let $X$ be a compact Riemann surface and $D$ a divisor on $X$. I'm looking for a argument for the statement $c_1(\mathcal{O}_X(D)) = \deg(D)$, where $\mathcal{O}_X(D)$ is the associated line bundle to $D$.
Of course, one can consider the exponential sequence, i.e. $H^1(X, \mathcal{O}_X^*) = Pic(X) \stackrel{c_1}{\to} \underbrace{H^2(X,\mathbb{Z})}_{\cong \mathbb{Z}} \to 0$ and examine the connecting homomorphism $c_1$.
I'm looking for another, nicer way to see the above statement. Is there one?
It is quite immediate if you know the definition of Chern classes via smooth sections: You start with your meromorphic section $\sigma$ and near the poles you replace it with a smooth section $s$ transverse to the zero section ${\bf 0}$, using the fact that on the unit circle $z^{-1}=\bar{z}$. Now, just compute the oriented intersection number of the image of $s$ and of the zero section. The result, by the construction, equals the degree of your line bundle.