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Led $D$ be a very ample divisor in $X$ projective variety.

I can't understand why the first Chern class $c_1(\mathscr{O}_X(D))$ equals the Poincaré dual of $D$, $\mathscr{P}(D)$

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This is a perfectly reasonable question, but not for this site. The best place for this question is math.stackexchange.com –  Michael Joyce Nov 14 '12 at 17:00
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Also if you post the question on Math stackexchange, you should state what definition of Chern class you are using. –  Michael Joyce Nov 14 '12 at 17:01
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Since this is a graduate level question, it does seem appropriate here. In a nutshell, since $D$ is very ample, you can reduce the problem to $X=\mathbb{P}^n$. Since $H^2(X)=\mathbb{Z}$ in this case, $D$ and $c_1(O(1))$ are multiples of each other. By restricting to a line, you can check the factor is $1$. I may flesh this out later, if I have time. –  Donu Arapura Nov 14 '12 at 19:34

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