$\mathbb{C}P^1\times...\times \mathbb{C}P^1/S_m=\mathbb{C}P^m$

Question

Let $X:=\mathbb{C}P^1\times...\times \mathbb{C}P^1$ be the product of $m$ copies of $\mathbb{C}P^1$ and $S_m$ acts on $X$ by permuting the factors. Then why is $X/S_m=\mathbb{C}P^m$?

Answer 1

Consider the line bundle $\mathcal{O}(n)$ on $\mathbb{CP}^{1}$ whose sections are homogeneous polynomials of degree n in 2 variables. Then if we projectivise the space of sections (i.e. consider $\mathbb{P}(H^{0}(\mathbb{P}^{1},\mathcal{O}(n)))$ we get a projective space of dimension $n$. Furthermore, each element of this projective space gives you $n$ unordered points on $\mathbb{P}^{1}$ since we can factorize a homogeneous polynomial in $2$ variables as a product of degree one polynomials.