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Let $X$ be a smooth projective variety over $\mathbb{C}$, then the Neron-Severi group $NS(X)$ of $X$ is defined to be the Picard group of $X$ modulo algebraically equivalent relations.

On the other hand, by the exponential sequence, there is a first Chern class map

$$c_1: {\rm Pic}(X) \to H^2(X, \mathbb{Z}).$$ It is claimed that the image of $c_1$ coincides with $NS(X)$.

I want to know why this is true. Any suggestion or reference is greatly welcome!

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  • $\begingroup$ another observation is that ,in fact, $NS(X)\cong H^2(X,\mathbb{Z})\cap H^{1,1}(X)$,this follow from the long exact sequence of cohomology groups induced by the exponential exact sequence of sheaves as Brenin's answer and Hodge decomposition theorem. $\endgroup$
    – Jiabin Du
    May 16 '18 at 3:14
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The kernel of $c_1$ consists exactly of line bundles that are algebraically equivalent to $0$.

Let me expand a little bit. The Picard group $\textrm{Pic }X$ has a subgroup $\textrm{Pic}^0\,X\subset \textrm{Pic }X$ consisting of line bundles algebraically equivalent to zero. Equivalently, $\textrm{Pic}^0\,X$ is the connected component containing the identity element of $\textrm{Pic }X$. (With your assumptions on $X$, $\textrm{Pic}^0\,X$ is an abelian variety; it is called the Picard variety of $X$). The Neron-Severi group is, by your definition, the quotient of the Picard group by the subgroup $\textrm{Pic}^0\,X$. But $\textrm{Pic}^0\,X$ is also the kernel of $c_1:\textrm{Pic }X\to H^2(X,\mathbb Z)$, so that $$NS(X)=\textrm{Pic }X/\textrm{Pic}^0\,X=\textrm{Pic }X/\ker(c_1)=\textrm{Im}(c_1).$$ Remark. The exponential exact sequence $0\to \mathbb Z\to \mathscr O_X\to\mathscr O_X^\times\to 1$ induces an exact piece $$H^1(X,\mathscr O_X)\to \textrm{Pic }X\overset{c_1}{\to}H^2(X,\mathbb Z)$$ which tells us that when $\textrm{Pic}^0\,X$ is a point, and hence $H^1(X,\mathscr O_X)=0$ (this $H^1$ is the tangent space to the Picard variety at any point $[L]\in \textrm{Pic}^0\,X$), we have $NS(X)\cong \textrm{Pic }X$.

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    $\begingroup$ Thank you so much for your answer! However, I expect an argument (or a reference) that explicitly shows that $c_1(L) = 0 \iff L$ is algebraically equivalent to zero. For example, the argument should answer question like $L$ is numerically equivalent to zero, but $c_1(L)$ may not be zero... I really want to see the subtlety of $algebraically$ equivalent appeared here. $\endgroup$
    – Li Yutong
    Dec 21 '15 at 14:04
  • $\begingroup$ As a good reference, I would bet on Griffiths-Harris' Principles of Algebraic Geometry, but unfortunately I do not have it available right now. I might add something to my answer later. $\endgroup$
    – Brenin
    Dec 22 '15 at 19:03
  • $\begingroup$ I don't see this in Griffiths and Harris and I really want an answer. Any additions? $\endgroup$ Aug 21 '19 at 23:34

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