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Let $\overline{M}_{0,n}$ be the moduli space of Deligne-Mumford stable pointed rational curves, and let us consider the quotient $\widetilde{M}_{0,n} = \overline{M}_{0,n}/S_n$.

Clearly, there is a dominant rational map $\overline{M}_{0,n}\dashrightarrow \widetilde{M}_{0,n}$, and $\widetilde{M}_{0,n}$ is unirational. If $n=4,5$ then $\widetilde{M}_{0,n}$ is a curve and a surface respectively. Thus, it is rational.

Is it known that $\widetilde{M}_{0,n}$ is rational for any $n\geq 3$? If so, do you know a reference for this?

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    $\begingroup$ $M_{0,n} / S_{n-3}$ is rational, because $M_{0,n}$ is an open subset of $(\mathbb P^1-\{0,1,\infty\} )^{n-3}$, and the quotient of that by $S_{n-3}$ is an open subset of $\mathbb A^{n-3}$ by taking coefficients of the polynomial. For the same reason, the space you want can also be expressed as a quotient of $\mathbb P^n$ by $SL_2$. This also just gives unirationality, but it might be a more fruitful vantage point for proving rationality . $\endgroup$ – Will Sawin Dec 4 '14 at 15:13
  • $\begingroup$ Maybe I'm wrong, but doesn't the following work? For $a_0, \ldots, a_{n-1}\in k$, consider the subscheme of $\mathbb{A}^1_k$ given by $x^n + a_{n-1}x^{n-1} + \ldots + a_0 = 0$. For generic $a_i$, this is a collection of $n$ distinct points, so we get a rational map from $\mathbb{A}^n$ to $\tilde M_{0, n}$ which is actually birational (as it's dominant and injective). $\endgroup$ – Piotr Achinger Dec 4 '14 at 22:16
  • $\begingroup$ @Piotr: where did the SL2 action go? $\endgroup$ – Abdelmalek Abdesselam Dec 5 '14 at 14:16
  • $\begingroup$ @AbdelmalekAbdesselam good point! :D $\endgroup$ – Piotr Achinger Dec 5 '14 at 20:11
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I don't know if this is exactly the answer you are looking for but for even $n$ this is clearly related to the result by Katsylo for moduli spaces classical binary forms or hyperelliptic curves, see this paper. Another link for accessing the article by Katsylo is this one. Theorem 0.2 therein seems to give a positive answer to your rationality question for all $n$ (even or not) except possibly for $n=10$ (or genus four hyperelliptic curves). I think this exception has been covered later by Bogomolov. Note that for $n=10$ now one has an explicit description for the ring (rather than just field) of invariants, see this article by Brouwer and Popoviciu.

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