Convergence of Dirichlet series expression of Dedekind zeta function The Dedekind $\zeta$-function of an algebraic number field $K$ is traditionally defined in the domain $\textrm{Re}(s)>1$ in terms of a Dirichlet series expression:
$$
\zeta_K(s) := \sum_{\mathfrak{a} \subset \mathcal{O}_K}\frac{1}{N(\mathfrak{a})^s},
$$
where the sum is taken over all integral ideals in the ring of integers $\mathcal{O}_K$, and $N$ denotes the ideal norm.
It is clear by the unique factorisation of ideals in a Dedekind domain that when both expressions converge, the Dirichlet series expression of the Dedekind $\zeta$-function coincides with its Euler product expression, i.e.:
$$
\sum_{\mathfrak{a} \subset \mathcal{O}_K}\frac{1}{N(\mathfrak{a})^s} = \prod_{\mathfrak{p} \subset \mathcal{O}_K}\frac{1}{1-N(\mathfrak{p})^{-s}},
$$
where the product is taken over all non-zero prime ideals $\mathfrak{p} \subset \mathcal{O}_K$.
Now most proofs that I have seen that show that $\zeta_K(s)$ is analytic in the domain $\textrm{Re}(s)>1$ look at its Euler product expression, and observe that if $d$ is the degree of $K$, then:
$$
\left\lvert\prod_{\mathfrak{p} \subset \mathcal{O}_K}\frac{1}{1-N(\mathfrak{p})^{-s}} \right\rvert \leq \prod_{p}\frac{1}{\lvert1-p^{-s}\rvert^d} = \lvert \zeta(s) \rvert^d,
$$
where the second product is taken over all rational primes $p \in \mathbb{N}$. This is because every prime in $\mathbb{N}$ has at most $d$ primes in $\mathcal{O}_K$ above it. Then they use the well known fact that the Riemann $\zeta$-function is absolutely and uniformly convergent in the domain to obtain the result.
But this only proves the convergence of the Euler product expression of $\zeta_K(s)$. It does not necessarily follow that $\sum_{\mathfrak{a} \subset \mathcal{O}_K}\frac{1}{N(\mathfrak{a})^s}$ is convergent in the domain.
So does anyone know of a proof of the convergence of the Dirichlet series expression of the Dedekind $\zeta$-function?
My idea was to write
$$
\sum_{\mathfrak{a} \subset \mathcal{O}_K}\frac{1}{N(\mathfrak{a})^s} = \sum_{n=1}^{\infty}\frac{c_n}{n^s}
$$
where the $c_n$ denote the number of integral ideals $\mathfrak{a} \subset \mathcal{O}_K$ with norm $n$. Then define $M:=\textrm{max}(c_n)$ to arrive at:
$$
\left \lvert \sum_{n=1}^{\infty}\frac{c_n}{n^s} \right \rvert \leq M \lvert \zeta(s) \rvert.
$$
But I think the $c_n$ can become arbitrarily large, so this does not work.
Any ideas?
 A: Absolutely convergent sequences converge whatever order you write the terms in. So thinking of $\zeta_K(s)$ as a limit of sums of $N(\mathfrak{a})^{-s}$ over ideals divisible only by primes $N(\mathfrak{p}) \le X$, then you can indeed compare to the Euler product. On the other hand, in your Euler product you have to include $p$ copies of the factor $(1 - p^{-s})$, so thie upper bound should be  $|\zeta(s)|^d$.
If you want to proceed directly with the Dirichlet series, you can, and you get exactly the same bound. Write
$$\zeta(s)^d = \sum \frac{d_n}{n^s}$$
This has an Euler product $\prod (1 - p^{-s})^{-d}$, so $d_{ab} = d_a d_b$ for $(a,b) = 1$, and $d_{p^n}$ is the coefficient of $p^{-ns}$ in $(1 - p^{-s})^{-d}$, or equivalently the coefficient of $X^n$ in $(1 - X)^{-d}$, which happens to be $\binom{d+n-1}{n}$.
On the other hand, by algebraic number theory you certainly have:


*

*$c_{ab} = c_a c_b$ if $(a,b) = 1$, 

*$c_p \le d:=[K:\mathbf{Q}]$, the worst case when $p$ splits completely.

*More generally, $c_{p^n}$ is bounded by the case when $p$ splits completely, and the upper bound is exactly the coefficient of $X^n$ in $(1-X)^{-d}$, so you get


$$|\zeta_K(s)| \le |\zeta(s)|^d.$$
