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When looking at an imaginary quadratic field, its zeta function is not too difficult to explicitly define (ex: $ \zeta_{\mathbb{Q}(i)}(s) = \frac{1}{4} \sum_{(m,n) \in \mathbb{Z}^2\(0,0)} \frac{1}{(m^2+n^2)^s}$). Unfortunately, I realized that in any other field it's a lot harder to define the zeta function due to the fact that there's an infinitude of elements with equal norm due to the group of units. I attempted to find a closed form of $\zeta_{\mathbb{Q} (\sqrt 2)}(s)$ but couldn't find a way to create an index that omits elements having $(1+\sqrt2)^{\mathbb{Z}}$ (units) as a factor. Is there any way to find an expression for the zeta function of real quadratic fields similar to that of imaginary quadratic fields?

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  • $\begingroup$ In the particular case of an abelian extension $\Bbb{Q}(\sqrt{2})/\Bbb{Q}$ , quadratic reciprocity gives $\zeta_{\Bbb{Q}(\sqrt{2})}(s) = \zeta(s)L(s,(\frac{.}{\Delta})$). But the answer in general is no. For $\zeta_k(s)$ we have only : the Euler product, the sum over ideals, and the integral over $\Bbb{R}^n/L$ of the $\Theta$ function. Thus it is much more complicated than $\zeta(s)$, despite the (conjectured) very similar properties $\endgroup$ – reuns Jul 1 at 2:09

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