$L$-function absolutely convergent for $\text{Re}(s) > 1$, condition for $L(s, \chi)$ converging for $\text{Re}(s) > 0$?

Question

I have two questions related to here.

Let $K$ be a number field, $Cl(K)$ the ideal class group, $\chi: Cl(K) \to \mathbb{C}^\times$ a homomorphism. If $\mathfrak{a} \subset \mathcal{O}_K$ is any ideal, let $[\mathfrak{a}]$ denote its ideal class in $Cl(K)$, and define $\chi(\mathfrak{a}) = \chi([\mathfrak{a}])$.

Let$$L(s, \chi) = \sum_{\mathfrak{a} \subset \mathcal{O}_K} {{\chi(\mathfrak{a})}\over{N\mathfrak{a}^s}}.$$

1. How do I see that $L(s, \chi)$ is absolutely convergent for $\text{Re}(s) > 1$?
2. What would we need to know in order to show that $L(s, \chi)$ converges for $\text{Re}(s) > 0$?

Answer 1

For (1), note that $|\chi(\mathfrak{a})| = 1$ (since $\operatorname{Cl}(K)$ is finite so $\chi$ takes values in the unit circle), so it suffices to prove that the Dedekind zeta function $\zeta_K(s) = \sum_{\mathfrak{a} \subset \mathcal{O}_K} \frac{1}{\left(N\mathfrak{a}\right)^s}$ of $K$ converges absolutely for $\operatorname{Re}(s) > 1$.

To see this convergence, note that by the unique factorization of ideals in $\mathcal{O}_K$, we have $$ \zeta_K(s) = \sum_{\mathfrak{a} \subset \mathcal{O}_K} \frac{1}{\left(N\mathfrak{a}\right)^s} = \prod_{\mathfrak{p}} \left(1 + \frac{1}{\left(N\mathfrak{p}\right)^2} + \frac{1}{\left(N \mathfrak{p}\right)^{2 s}} + \cdots\right) = \prod_{\mathfrak{p}} \frac{1}{1 - \left(N\mathfrak{p}\right)^{-s}}. $$ Here all the products are taken over nonzero prime ideals of $\mathcal{O}_K$. Thus, it suffices to prove that the product $\prod_{\mathfrak{p}} \big(1 - \left(N\mathfrak{p}\right)^{-s}\big)$ converges absolutely for $\operatorname{Re}(s) > 1$, and, by the theory of infinite products, this is equivalent to the absolute convergence of the series $\sum_{\mathfrak{p}} \left(N\mathfrak{p}\right)^{-s}$ for $\operatorname{Re}(s) > 1$. Next, for each prime number $p$, there are at most $[K : \mathbb{Q}]$ prime ideals in $K$ above $p$ and for each such prime ideal $\mathfrak{p}$, we have $N\mathfrak{p}=p$. Therefore, $$ \sum_{\mathfrak{p}} \left|\left(N\mathfrak{p}\right)^{-s}\right| = \sum_{\mathfrak{p}} \frac{1}{\left(N\mathfrak{p}\right)^{\operatorname{Re}(s)}} \leq \sum_{p} \frac{[K : \mathbb{Q}]}{p^{\operatorname{Re}(s)}} \leq [K : \mathbb{Q}] \sum_{n=1}^\infty \frac{1}{n^{\operatorname{Re}(s)}} = [K : \mathbb{Q}] \zeta(\operatorname{Re}(s)). $$ So it suffices to prove that the Riemann zeta function $\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}$ converges absolutely for $\operatorname{Re}(s) > 1$. This follows easily from, e.g., the integral test.