Let $G$ be a connected, linear, semisimple algebraic group over an algebraically closed field of characteristic zero and $P$ be the maximal parabolic subgroup. We know that the quotient $Z=G/P$ is a projective scheme. Is it true that the Picard group of $Z$ is isomorphic to $\mathbb{Z}$?

1$\begingroup$ You can't say "the" maximal parabolic; there's usually more than one, even up to conjugacy. But for every such $P$ this is true. $\endgroup$– Allen KnutsonJun 30 '15 at 18:30
Yes. We can assume that $G$ is simply connected, and in this case, any line bundle on $G/P$ is equivariant for $G$ (if $G$ isn't simply connected, you get silly things like $\mathcal{O}(1)$ on $\mathbb{P}^n$ isn't equivariant for $PSL_{n+1}$). For any maximal parabolic subgroup, $P/[P,P]\cong \mathbb{G}_m$, and every line bundle is gotten by taking $(G\times V)/P$ for $V$ a 1dimensional representation of $P$.