Is the first chern class the same as the degree of the Divisor?
Say, $C$ is some divisor on $M$, is $c_1(\mathcal O (C)) = \text{deg }C$?
And say I have some Divisor $D$ with first chern class $c_1(\mathcal{O}(D)) = k[S]$ where $[S]$ is some class in $H^2(M)$. Is it true that the integral first chern class is just $k\cdot \text{deg }S$?