# The Picard group of projective variety

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 $n=0$)

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.
• And the smoothness and irreducibility hypotheses are not needed. If $X$ is quasi-projective, then the same proof says that $Pic=0$ implies that $X$ is affine.
• The proof seems to use that $k$ is algebraically closed. Or don't we need it? Jun 30, 2013 at 23:20
• Dear @MartinBrandenburg, it seems to me that the only place the proof might use $k$ algebraically closed is to conclude $\Gamma(X,\mathcal O_X)\cong k.$ To be honest, I only remember seeing this latter fact proved for $k$ algebraically closed (and $X$ complete), so I'm not sure if it is true more generally... though in the special case $X=\mathbb P^n$ it is true, according to Hartshorne section III.5. Jul 4, 2013 at 19:21
• @Andrew Yes, and proper. Just do the proof over $\overline{k}$, and use the Cech description of $H^0(X,\mathcal{O}_X)$ (on a finite affine open cover) to show that $H^0(X_{\overline{k}},\mathcal{O}_{\overline{k}})=H^0(X,\mathcal{O}_X)\otimes_k \overline{k}$ (this is true, more generally, for any flat extension of base rings--this is a way, way, way simple version of flat base extension). For the proof of the proper case, you only need to note that every morphism $X\to\mathbb{A}^1$ must be constant, which shows that $\mathcal{O}_X(X)$ is integral over $\overline{k}$, and so is $\overline{k}$. Nov 4, 2014 at 2:27