Timeline for Factorisation of local quaternionic zeta functions
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2014 at 10:06 | vote | accept | Desiderius Severus | ||
| Oct 10, 2014 at 18:36 | comment | added | John Voight | What is $q$? It better be $q^n=\mathrm{nrd}(I)$: the reduced norm is not always a square. | |
| Oct 9, 2014 at 14:01 | vote | accept | Desiderius Severus | ||
| Oct 12, 2014 at 10:06 | |||||
| Oct 9, 2014 at 13:56 | comment | added | Desiderius Severus | Thanks pour your answer, it works indeed for this calculation, but seems to break the other one, for the case H being a field. The already mentioned lemma 4.1 states that the number of left ideal of fixed norm (understand : reduced norm) $q^n$ is : if $H$ field, 1 if $n$ even, 0 otherwise ; if $H \cong M_2$, $1 + \cdots +q^n$. If it is right that it is the reduced norm, the calculation for the zeta function in the field-case would give $\zeta_H(s) = \zeta_K(4s)$ because $N(I)=n(I)^2=q^{4n}$ for $n(I)=q^{2n}$. And not $\zeta_K(2s)$ as said. Is there so a confusion between the two cases ? | |
| Oct 9, 2014 at 12:38 | history | answered | John Voight | CC BY-SA 3.0 |