Timeline for Residue fields of attached to coefficients of modular forms
Current License: CC BY-SA 3.0
7 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
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| Jun 3, 2013 at 4:07 | comment | added | Orac | The unique (up to conjugation) normalized eigenform in $S_2(\Gamma_0(23))$ has coefficients in $\mathbf{Q}(\sqrt{5})$, and the corresponding mod-$2$ representation lands in $S_3$ (the representation comes from the Hilbert class field of $\mathbf{Q}(\sqrt{-23})$). It follows that all the $a_p$ with $p$ prime lie in $\mathbf{Z}[\sqrt{5}]$ and their mod $2$ reduction lands in $\mathbf{F}_2$. But $a_2 = (-1 + \sqrt{5})/2$, so $a_2 \mod 2$ generates $\mathbf{F}_4$. | |
| May 26, 2013 at 21:13 | comment | added | MF1 | @Dror: I agree -- dropping a single $a_q$ for $q$ an unramified prime for the residual representation won't change the residue field. But I don't know how to deal with ramified primes -- especially the prime $p$. | |
| May 26, 2013 at 15:43 | comment | added | Dror Speiser | Looking at the mod p representation attached by Shimura, Deligne, Deligne-Serre, corresponding to the inclusion $K_f\rightarrow\bar{\mathbb{Q}}_p$, we see that for any $q\not | pN$, we have, from the Eichler-Shimura relation, that $a_q\ \text{mod p}$ depends only on the conjugacy class of Frobenius of $q$. By Chebotarev's theorem, there are infinitely many $q$'s in every such conjugacy class. Hence, if you delete such $q$'s, the residue field doesn't change. I'm not sure what happens at the ramified primes of the finite representation. | |
| May 26, 2013 at 14:33 | comment | added | MF1 | @Dror: I think one has to be careful about which rings (as opposed to fields) are being generated by these Fourier coefficients in the char 0 case. Indeed, I'm worried about the possibility where say all of the $a_q$ generate the full ring of integers, but if one removes a single $a_q$ then you get a smaller order which has a smaller residue field. | |
| May 25, 2013 at 18:35 | comment | added | MF1 | @Dror: could you elaborate? I don't see the argument. | |
| May 25, 2013 at 3:59 | history | asked | MF1 | CC BY-SA 3.0 |