Showing $\prod_{i=0}^{3} L (s, \chi _i) = \prod _{p \equiv 1 (8)} (1-p^{-s})^{-4} \prod_{p \equiv 3, 5, \text{ or } 7 (8)} (1-p^{-2s})^{-2}.$

Question

Suppose $\chi (a)$ is a complete multiplicative function defined on the residue classes modulo $k$ such that $\chi (a) = 0$ if and only if $\gcd (a,k) > 1$. Define the Dirichlet $L$-functions (for $k=8$) for $s>1$ by $$L(s,\chi _i) = \sum_{n=1}^{\infty} \frac{\chi_i (n)}{n^s} \text{ for } i = 0, 1, 2, 3.$$

Show that for $s>1$, $$\prod_{i=0}^{3} L (s, \chi _i) = \prod _{p \equiv 1 (8)} (1-p^{-s})^{-4} \prod_{p \equiv 3, 5, \text{ or } 7 (8)} (1-p^{-2s})^{-2}.$$

I first showed that $L(s,\chi _i)$ is continuous and nonzero at $s=1$, is there a way to proceed and conclude using this property, or should we use another approach? (In either case, how would we proceed and conclude?) To clarify, (8) in the indices means modulo 8.

Answer 1

For Dirichlet $L$-functions attached to a Dirichlet character $\chi$ mod $N$,

$$ \prod_{\chi} L(s,\chi)=\prod_{\chi}\prod_p \left(1-\chi(p)p^{-s}\right)^{-1}=\prod_p\prod_{\chi}\left(1-\chi(p)p^{-s}\right)^{-1} $$

so the Euler $p$-factors are $\prod_\chi (1-\chi(p)p^{-s})^{-1}$, which depends only on $p$ mod $N$.

In the case of $N=8$,

• $p=2$ yields $\chi(p)=0$ so the Euler factor is $1$
• $p\equiv 1~(8)$ yields $\chi(p)=1$, so the Euler factor is $(1-p^{-s})^{-4}$
• $p\equiv 3,5,7$ yields $\chi(p)=\pm1$, twice for each value (from the four $\chi$s), so the Euler factor is given by the product $(1-p^{-s})^{-2}(1+p^{-s})^{-2}=(1-p^{-2s})^{-2}$.