Timeline for Is $M_{1,n}$ affine?
Current License: CC BY-SA 3.0
8 events
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| Jul 17, 2012 at 20:54 | comment | added | Dan Petersen | (and crucially, $Y(N)$ is affine for any $N$, being a curve minus a finite set of closed points.) | |
| Jul 17, 2012 at 20:46 | comment | added | Dan Petersen | More elementary is to use that $M_{1,1} = [Y(N)/\mathrm{SL}(2,\mathbf Z/N\mathbf Z)]$ for any $N$, where $Y(N)$ is the (open) modular curve parametrizing elliptic curves with a full level $N$ structure. Now for $N\geq 3$ this is a fine moduli scheme with a universal family; iterating the argument in my previous comment shows that the moduli space of $n$-pointed genus one curves with level $N$ structure is an affine scheme. Dividing by the action of $\mathrm{SL}(2,\mathbf Z/N\mathbf Z)$ recovers $M_{1,n}$. But the quotient of an affine scheme by a finite group is again affine (elementary). | |
| Jul 17, 2012 at 20:01 | comment | added | Jason Starr |
@OMHF -- Although it is true that there is no universal curve over $M_{g,n}$ for every choice of $g$ and $n$, you can use Chevalley's theorem to fix this. There is a finite flat morphism $B\to M_{1,1}$ such that $B$ is an affine scheme and there is a family over $B$, e.g., the $\lambda$-line. By Dan's argument, the associated (2-)fibered product $B\times_{M_{1,1}} M_{1,n}$ is an affine scheme. Since $B \to M_{1,1}$ is finite and surjective, also `$B\times_{M_{1,1}}M_{1,n} \to M_{1,n}$ is finite and surjective. Thus $M_{1,n}$ (as a coarse space) is an affine scheme by Chevalley's thm.
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| Jul 17, 2012 at 18:56 | comment | added | OldMacdonaldHadaForm | Perhaps there is still something to say about this. Indeed, there is no universal curve over the moduli scheme $M_{g,n}$, which is what we are considering. Do you have a reference? | |
| Jul 17, 2012 at 13:36 | vote | accept | OldMacdonaldHadaForm | ||
| Jul 17, 2012 at 13:02 | comment | added | Dan Petersen | Let $C \to M_{g,n}$ be the universal curve. The union of the $n$ markings on $C$ is a relative Cartier divisor which is relatively ample (this can be checked on closed points). So its complement is affine. | |
| Jul 17, 2012 at 11:51 | comment | added | OldMacdonaldHadaForm | Thank you very much! How do you see that the morphism is affine? | |
| Jul 17, 2012 at 11:30 | history | answered | Dan Petersen | CC BY-SA 3.0 |