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The Dedekind zeta function of a number field $K$, denoted by $\zeta_K(s)$, is defined for all complex numbers $s$ with $\Re(s) > 1$ by the Dirichlet series \begin{equation*} \zeta_K(s) = \sum_{\mathfrak{a}} \frac{1}{N(\mathfrak{a})^s}, \end{equation*} where the sum is over all non-zero ideals $\mathfrak{a}$ in $\mathcal{O}_K$.

I want to show that the Euler product exists for the Dedekind zeta function $\zeta_K(s)$: \begin{equation*} \zeta_K(s) = \prod_{\mathfrak{p}} \left(1 - \frac{1}{N(\mathfrak{p})^{s}}\right)^{-1}, \end{equation*} where the product is over all prime ideals $\mathfrak{p}$ in $\mathcal{O}_K$.

For this I need to show that \begin{equation*} \left|\sum_{\mathfrak{a}} \frac{1}{N(\mathfrak{a})^s} - \prod_{N(\mathfrak{p}) \leq x} \left(1 - \frac{1}{N(\mathfrak{p})^{s}}\right)^{-1}\right| \rightarrow 0, \end{equation*} as $x \rightarrow \infty$.

Note that the norm is a completely multiplicative function and $\mathcal{O}_K$ is a Dedekind domain.

Since we have \begin{equation*} \left(1 - \frac{1}{N(\mathfrak{p})^{s}}\right)^{-1}= 1 + \frac{1}{N(\mathfrak{p})^s} + \frac{1}{N(\mathfrak{p})^{2s}} + \frac{1}{N(\mathfrak{p})^{3s}} + \cdots \end{equation*} the difference \begin{equation*} \left|\sum_{\mathfrak{a}} \frac{1}{N(\mathfrak{a})^s} - \prod_{N(\mathfrak{p}) \leq x} \left(1 - \frac{1}{N(\mathfrak{p})^{s}}\right)^{-1}\right| \leq \left|\sum_{N(\mathfrak{a}) > x} \frac{1}{N(\mathfrak{a})^s}\right|. \end{equation*}

Here, we need a fact that $\zeta_K(s)$ is absolutely convergent for all complex numbers $s$ with $\Re(s) > 1$. How can I prove this fact?

bozcan
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1 Answers1

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$$\zeta_K(s)=\prod_p \prod_{P\text{ prime of } O_K,p\in P}\frac1{1-N(P)^{-s}}=\prod_p \prod_{j=1}^{g_p} \frac1{1-p^{-sf_{p,j}}}$$ $$ pO_K=\prod_{j=1}^{g_p} (P_{p,j})^{e_{p,j}},\qquad N(P_{p,j})=p^{f_{p,j}}$$ It converges for $\Re(s) > 1$ because $g_p\le [K:\Bbb{Q}]$. Thus there is no fear to expand this absolutely convergent product, using the unique factorization of ideals $$\zeta_K(s)=\prod_{P\text{ prime of } O_K} (1+\sum_{k\ge 1} N(P^k)^{-s})=\sum_{I\text{ ideal of } O_K}N(I)^{-s}, \qquad \Re(s)>1$$

reuns
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