Picard group of the grassmannian

I would like some help in finishing this argument, as sketched in Dolgachev's book "Lectures in Invariant Theory" (page 165). Everything here is done over $\mathbb{C}$.

The claim is that $Pic(Gr_{k,n}) \cong \mathbb{Z}$. The argument goes as follows: Consider $Gr_{k,n}$ embedded into $\mathbb{P}^{n}$ via Plucker coordinates $p_{i_{1}..i_{k}}$. Given a line bundle $L$ on $Gr_{k,n}$, one may trivialize the line bundle over the open set (plucker coordinate) $p_{1...k} \neq 0$ which is isomorphic to an affine space. Thus, one may produce a global section $s$ of $L$ such that the corresponding divisor $Z(s)$ is supported in the hyperplane section $p_{1..k} = 0$. If this hyperplane section were smooth or irreducible, then we may conclude that $Z(s)$ is a multiple of this hyperplane section and then we would be finished.

However, Dolgachev does not address the fact that the hyperplane section may not be irreducible, and I think that without this fact the argument may fall apart. I am aware of Bertini's theorem, which tells me that there will exist smooth hyperplane sections.. however, the complement of an arbitrary hyperplane section may not be isomorphic to affine space.

Could anybody shed some light on this?

Thanks!

• The hyperplane section $\{ p_{1\ldots k}=0 \} \subset Gr_{k,n}$ consists of those $k$-planes in $\mathbf P^n$ which intersect a certain $(n-k-1)$-plane (namely the set of points of $\mathbf P^n$ where the first $k+1$ coordinates are zero). Try to understand this set of $k$-planes geometrically. Google "Schubert variety" if you get stuck. Mar 7, 2016 at 12:57