Timeline for Rationality of moduli spaces of rational curves

6 events

when toggle format	what		by	license	comment
Dec 5, 2014 at 20:11	comment	added	Piotr Achinger		@AbdelmalekAbdesselam good point! :D
Dec 5, 2014 at 14:16	comment	added	Abdelmalek Abdesselam		@Piotr: where did the SL2 action go?
Dec 4, 2014 at 22:16	comment	added	Piotr Achinger		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).
Dec 4, 2014 at 15:14	answer	added	Abdelmalek Abdesselam		timeline score: 4
Dec 4, 2014 at 15:13	comment	added	Will Sawin		$M_{0,n} / S_{n-3}$ 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 .
Dec 4, 2014 at 14:57	history	asked	user58018	CC BY-SA 3.0