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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 $X/S_m=\mathbb{C}P^m$?

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    $\begingroup$ The standard hint is to use the elementary symmetric functions on $n$ variables. $\endgroup$ – Ted Shifrin Jan 16 at 21:11
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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.

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    $\begingroup$ Just to make this explicit, take $$ \big([a_1:b_1], [a_2:b_2], \ldots, [a_m: b_m]\big) \mapsto [c_0: c_1: \ldots: c_m] $$ where $c_i$ is the coefficient of $x^iy^{m-i}$ in $\prod_i (a_ix + b_iy)$. $\endgroup$ – hunter Jan 16 at 21:35

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