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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...2012-07-07

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