Timeline for Finite generation of module of modular forms
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 24, 2017 at 7:59 | comment | added | David Loeffler | I recall Nils Skoruppa once mentioning some results on this problem, for characteristic 0 field coefficients -- in particular, the assertion that $M_*(\Gamma)$ is free over $M_*(SL2Z)$ and an explicit bound on the generator weights. I don't know if this was ever published though. Maybe you might want to email Skoruppa? | |
| Oct 24, 2017 at 6:34 | comment | added | Lennart Meier | @FrançoisBrunault Thanks! For general $R$, we can just use $GL_2(\mathbb{Z}/n)$ instead. Rustom states finite generation as an algebra only for $\Gamma_1(n)$, but same proof should work for $\Gamma(n)$. As $M(\Gamma_1(n);R)$ is contained in $M(\Gamma(n); R)$ the result for $\Gamma_1(n)$ follows as well. The degrees of the generators seem to be far from optimal though. | |
| Oct 24, 2017 at 4:18 | comment | added | François Brunault | @LennartMeier For any $f \in M_k(\Gamma(n),R)$ we can form the polynomial $\prod_{\gamma \in SL_2(\mathbb{Z}/n\mathbb{Z})} X-f | \gamma$ which has coefficients in $M_*(SL_2(\mathbb{Z}),R)$. This shows that $M_*(\Gamma(n),R)$ is integral over $M_*(SL_2(\mathbb{Z}),R)$. I don't know whether this integrality property is true in general. | |
| Oct 23, 2017 at 20:18 | comment | added | Lennart Meier | @FrançoisBrunault: How does this last implication work? | |
| Oct 23, 2017 at 20:17 | comment | added | Lennart Meier | @WillSawin Yes, indeed as $M_*(\Gamma_1(n); R)$ can differ from $M_*(\Gamma_1(n);\mathbb{Z}[\frac1n]) \otimes R$ only in degree $1$. | |
| Oct 23, 2017 at 17:30 | comment | added | François Brunault | I think that in the case $R$ contains $\mathbb{Z}[\zeta_n,\frac{1}{n}]$, the space $M_*(\Gamma(n),R)$ is stable under the action of $SL_2(\mathbb{Z}/n\mathbb{Z})$. Therefore finite generation as an algebra should imply finite generation as a module in this case. | |
| Oct 23, 2017 at 15:44 | comment | added | Will Sawin | Doesn't finite generation for all $R$ follow immediately from the $\mathbb Z[1/N]$ case? | |
| Oct 23, 2017 at 15:12 | history | edited | Lennart Meier | CC BY-SA 3.0 |
added 38 characters in body
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| Oct 23, 2017 at 13:25 | answer | added | Jeremy Rouse | timeline score: 3 | |
| Oct 23, 2017 at 9:07 | history | asked | Lennart Meier | CC BY-SA 3.0 |