Timeline for Is $M_g$ finitely covered by a scheme over the integers?
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Jun 13, 2012 at 11:53 | comment | added | André Henriques | @Dan: But if you replace "finite group" by "finite group scheme", then Question (a) still stands... doesn't it? | |
| Jun 12, 2012 at 22:50 | answer | added | Ian Agol | timeline score: 3 | |
| Jun 12, 2012 at 20:36 | comment | added | Dan Petersen | André's comment nicely answers Question (a), though! | |
| Jun 12, 2012 at 18:38 | comment | added | André Henriques | @Angelo: you are right. Computing $\pi_1^{et}$ isn't enough. Is there some version of $\pi_1$ than controls this problem? | |
| Jun 12, 2012 at 16:08 | comment | added | Angelo | Dear André, the question does not require $X$ to be étale over the stack. | |
| Jun 12, 2012 at 15:34 | comment | added | André Henriques | Erratum: the groups whose order I listed above are the kernels of $\pi_1^{et}(M_1[1/2])\to \pi_1^{et}(Spec(\mathbb Z[1/2]))$ and $\pi_1^{et}(M_1[1/3])\to \pi_1^{et}(Spec(\mathbb Z[1/3]))$. | |
| Jun 12, 2012 at 15:16 | comment | added | André Henriques | For g=1 the answer is negative. The moduli stack of elliptic curves over $\mathbb Z$ is simply connected: no etale covers. It becomes non-simply connected if you invert 2 ($\pi_1^{et}$ has order 24), or if you invert 3 ($\pi_1^{et}$ has order 12). | |
| Jun 12, 2012 at 15:10 | comment | added | David E Speyer | Nitpick: In the first part of (b), you should also require that $X \to M_g$ be surjective. Otherwise, there is a degree $(2g+2)!$ finite map from $M_{0,2g+2}$ to the hyperelliptic locus. | |
| Jun 12, 2012 at 13:35 | history | edited | Dan Petersen | CC BY-SA 3.0 |
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| Jun 12, 2012 at 12:56 | comment | added | Jason Starr | In the second part of Question (b), I recommend you change "generically finite" to "proper and generically finite". | |
| Jun 12, 2012 at 12:31 | history | edited | Dan Petersen | CC BY-SA 3.0 |
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| Jun 12, 2012 at 12:25 | history | asked | Dan Petersen | CC BY-SA 3.0 |