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If $X$ is a smooth irreducible projective variety and its Picard group is $0$, can we conclude that $X$ is a point? (For example, when $X=\mathbb P^n$, then Pic($\mathbb P^n$)=$\mathbb Z$ unless $n=0$)

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This seems to be true...

Given a projective variety $X$, consider the invertible sheaf $\varphi^*(\mathcal O(1)),$ where $\varphi: X\to\Bbb P^n$ is the closed immersion given by definition of projective, and $\mathcal O=\mathcal O_{\Bbb P^n}$. By hypothesis, we must have $\varphi^*(\mathcal O(1))\cong\mathcal O_X,$ and so we know that the structure sheaf of $X$ is generated by the global sections $\varphi^*(x_i), i=0,1,\ldots,n.$ However, by $X$ being projective, we have $\Gamma(X,\mathcal O_X)\cong k,$ which implies that all $\varphi^*(x_i)$ are constant. By the equivalent formulation of the morphism $\varphi:X\to\Bbb P^n$ as a line bundle generated by global sections (cf. Hartshorne Theorem II.7.1 for example), we see that $\varphi$ must indeed be a constant map. Hence, $X$ is a point.

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    Proper geometrically reduced and geometrically connected is enough @AlexYoucis.2016-07-06