Chern Class = Degree of Divisor? 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 $k\cdot \text{deg }S$?
 A: If $M$ is a complex manifold of dimension $n$, the exact sequence of sheaves 
$$0\to\mathbb Z\to \mathcal O_M\to \mathcal O^*_M\to 0$$
yields a morphism of groups in cohomology (part of a long exact sequence) $$c_1:H^1(M,\mathcal O^*_M)\to H^2(M,\mathbb Z)$$
Since $H^1(M,\mathcal O^*_M)$ can be identified to $Pic(M)$, the group of line bundles on $M$, we get the morphism $$c_1:Pic(M)\to H^2(M,\mathbb Z)$$
This morphism coincides with the first Chern class  defined (for $C^\infty$line bundles) in  differential geometry  in terms of curvature of a connection.  
If now $M$ is a compact Riemann surface (= of dimension $1$), we may evaluate  a cohomology class $c\in H^2(M,\mathbb Z) $ on the fundamental class$[M]$  of the Riemann surface $M$,  thereby  obtaining an isomorphism $I: H^2(M,\mathbb Z) \xrightarrow {\cong} \mathbb Z: c\mapsto \langle c,[M] \rangle$, which composed with the first Chern class yields the degree for line bundles:   $$deg=I\circ c_1:Pic(M)\to \mathbb Z: L \mapsto deg(L)=\langle c_1(L),[M] \rangle$$   
Finally, if $L$ is associated to the divisor$D$  i.e.  $ L=\mathcal O(D)$, we have the pleasantly  down to earth  (but highly non tautological!) formula $$deg(\mathcal O(D))=deg (D)$$
which says that  the degree on the left,  obtained by sophisticated differential geometry, can be computed in a childishly simply way by computing the degree of the divisor $D\in \mathbb Z^{(X)}$ seen purely formally as an element of the free abelian group on $X$.
All this is explained in Griffiths-Harris, Principles of algebraic geometry, Chapter 1.
