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Suppose $K$ is a quadratic field and $a_K(n)$ denotes the number of ideals in the ring of integers of $K$ whose norm is equal to $n$. Then I need to show that $$\sum_{n\leq x} a_K(n)=O(x).$$ Clearly the above claim will imply that the Dedekind zeta function $\zeta_K(z):=\sum_{n\geq 1}\frac{a_K(n)}{n^z}$ converges for $\mathrm{Re}(z)>1$. Is there an analytic continuation of $\zeta_K$ so that it can be defined in some punctured disc around $1$ and does this analytic continuation have a simple pole at $z=1$ like Riemann zeta function?

If the proofs are technical and long, I only need some easy intuitions and ideas to see why the above statements should hold true, instead of complete proofs.

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1 Answer 1

To answer your questions in a backwards order:

Yes, there is an analytic continuation similar to the regular $\zeta$ function. Further, it has a simple pole at $z=1$, although the residue isn't $1$ anymore (it depends a bit on which field $K$ we use, to no surprise).

A bit of intuition might be gotten from writing it out in the case $\mathbb{Q}(\sqrt{D})$ for $D$ squarefree:

$$\zeta_K(s) = \sum_{\mathfrak{a}} \frac{1}{\mathcal N\mathfrak{a}^s} = \prod_\mathfrak{p} \left(1 - \frac{1}{\mathfrak{\mathcal Np}^s}\right)^{-1},$$

where $\mathcal N$ is the norm operator, and where $\mathfrak{a,p}$ indicate ideals and prime ideals, respectively. Then one can classify how primes split in $\mathbb{Q}(\sqrt D)$ based on the Legendre symbol $\left( \dfrac{d}{p} \right)$, where $d$ is the discriminant of the extension. It turns out that

$$\zeta_K(s) = \prod_\mathfrak{p} \left(1 - \frac{1}{\mathfrak{\mathcal Np}^s}\right)^{-1} = \prod_{(\frac{d}{p}) = 1} \left(1 - \frac{1}{p^{2s}}\right)^{-1}\prod_{(\frac{d}{p}) = -1} \left(1 - \frac{1}{p^{s}}\right)^{-2}\prod_{p \mid d} \left(1 - \frac{1}{p^{s}}\right)^{-1},$$

where $p$ is the rational prime below $\mathfrak{p}$. This can be recognized as the same expansion as

$$ \zeta(s) L(\chi_d, s),$$

where

$$ L(\chi_d, s) = \sum_{n \geq 1} \frac{(\frac{d}{n})}{n^s},$$

which is to say that $\chi_d$ is the Legendre symbol with $d$ in the numerator. This is quite surprising, and should perhaps be gone through - but it's the source of (my, at least) intuition on the subject.

The important thing here is that we know that $L$ functions don't have poles at $1$ because of the oscillatory nature of their numerators. In addition, $L(\chi_d, 1) \neq 0$, so that the simple pole of the Riemann zeta indicates to us the simple pole of the Dedekind zeta function.

Now that we (heuristically) know that the Dedekind zeta function has a simple pole, taking the inverse Mellin transform (or rather applying the Residue theorem with some large contours) would tell you that

$$\sum_{n\leq x} a_K(n)=O(x).$$

In fact, you could do better, and get the exact asymptotic.

For more, you should look at most books on algebraic number theory. I'm pretty sure Neukirch starts with this, but I think that even Rosen's Intro to Modern Number Theory gets to this by the end (depending on your familiarity with algebraic number theory - these two books are at different ends of the spectrum). These books will also include the more complicated cases when $D$ isn't square free, or worse.

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