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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 $k\cdot \text{deg }S$?

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The Chern class cannot be the degree because the latter is a number and the former is a cohomology class... – Mariano Suárez-Alvarez Jul 7 '12 at 7:27

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.

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Nice answer. May I ask: why don't you write $\text{deg}(\mathcal{O}(D))=\text{deg}(D)$? – wildildildlife Jul 7 '12 at 11:56
Dear Georges, I was under the impression that you (in agreement with Mario's comment) are making a careful distinction between $c_1(L)$, which lives in $H^2(M,\mathbb{Z})$, and the integer $\text{deg}(L)$ obtained from it by pairing with the fundamental class (as in your second to last equation). But then your last equation seems to ignore that distinction (with $L=\mathcal{O}(D)$). – wildildildlife Jul 7 '12 at 12:50
Dear @wildildildlife, you are absolutely right! I misunderstood your first comment, but now that I understand what you mean, I completely agree with you and have edited my answer accordingly (and have deleted my irrelevant comment!) . Many thanks for your much appreciated contribution! – Georges Elencwajg Jul 7 '12 at 13:59

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