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Vignéras, in her Arithmetics of quaternion algebras, begin chapter II.4 recalling that we know the number of integer ideals of fixed norm of a quaternion algebra $H$ over a local field $K$, ramified or not (lemma 4.1). What is of interest here is that there are $1+q+\cdots +q^n$ such ideal of fixed norm $q^n$ if $H \cong M(2,K)$. That make possible an expression of $\zeta_H$ in terms of $\zeta_K$.

Then, it seems that Vignéras assume $n$ is always even, writing :

$$\zeta_H(s) = \sum_n \sum_{d \leqslant n} q^{d-2ns} = \sum_{d,d'} q^{d-2(d+d')s} = \zeta_K(2s)\zeta_K(2s-1)$$

And here I am totally lost : why do not write

$$\zeta_H(s) = \sum_n \sum_{d \leqslant n} q^{d-ns} = \sum_{d,d'} q^{d-(d+d')s} = \zeta_K(s)\zeta_K(s-1) \quad ?$$

Hence, is $n$ even ? Even if it is, what about the range of $d$, going to $n$ and not to $2n$ ?

Thanks for your help !

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By "fixed normal" you mean "fixed norm", right? The issue is just that the norm of a (right) ideal is the square of the reduced norm $N(I)=[\mathcal{O}:I]=\mathrm{nrd}(I)^2$.

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  • $\begingroup$ 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 ? $\endgroup$
    – Desiderius Severus
    Commented Oct 9, 2014 at 13:56
  • $\begingroup$ What is $q$? It better be $q^n=\mathrm{nrd}(I)$: the reduced norm is not always a square. $\endgroup$
    – John Voight
    Commented Oct 10, 2014 at 18:36

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