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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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    $\begingroup$ This is a perfectly reasonable question, but not for this site. The best place for this question is math.stackexchange.com $\endgroup$ Nov 14, 2012 at 17:00
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    $\begingroup$ Also if you post the question on Math stackexchange, you should state what definition of Chern class you are using. $\endgroup$ Nov 14, 2012 at 17:01
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    $\begingroup$ 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. $\endgroup$ Nov 14, 2012 at 19:34
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    $\begingroup$ We have the following short exact sequence $0\rightarrow \Omega ^1_X\rightarrow \Omega ^1_X(\log D)\rightarrow \mathcal{O}_D\rightarrow 0$ and since for the total Chern class we have from pervious short exact sequence $\ c(\Omega ^1_X(\log D))=c(\Omega ^1_X).c(\mathcal{O}_D)$ and since $c(\mathcal{O}_D)=(1-[D])^{-1}$ hence $c_p(\Omega ^1_X(\log D))=c_p(\Omega ^1_X)+c_{p-1}(\Omega ^1_X).[D]+\ldots +[D]^p$ hence $c_1(\Omega_X^1(\log D))= c_1(\Omega_X^1)+[D]$ $\endgroup$
    – user21574
    Jun 5, 2017 at 21:09

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