The Mobius function $\mu\colon \mathbb{N}\to\{-1,0,1\}$ is given by $\mu(n)=(-1)^{k}$ if $n$ is the product of $k$ distinct prime numbers, and $\mu(n)=0$ otherwise. It is classical that for all $a,b\in\mathbb{N}$, we have that $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\mu(an+b)=0.$$

The Mobius function can also be defined on any integer ring. Let $K$ be an algebraic number field and $O_{K}$ be its ring of integers. We can define the Mobius function $\mu_{K}\colon O_{K}\to\{-1,0,1\}$ by $\mu_{K}(n)=(-1)^{k}$ if the ideal $(n)$ is the product of $k$ distinct prime ideals of $O_{K}$, and $\mu_{K}(n)=0$ otherwise. My question is, is there an generalization of the equality $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^{N}\mu(an+b)=0$$ for $\mu_{K}$?

  • $\begingroup$ This is not true because in some cases $\mu_K$ doesn't seem to be the most interesting function - for instance when $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$ any rational prime splits as the product of either $2$ or $4$ prime ideals, so $\mu_K(n)$ is always $1$. $\endgroup$ – Rodrigo Jan 26 at 1:38
  • $\begingroup$ More specifically, the Moebius function "on $O_K$" should really be a function on nonzero ideals (based on prime ideal factorization) rather than on nonzero elements or principal ideals of $O_K$. $\endgroup$ – KConrad Jan 26 at 2:49
  • $\begingroup$ @Rodrigo Thanks! Is there any reference where I can find the claim that rational prime splits as the product of either 2 or 4 prime ideals in $\mathbb{Q}(\sqrt{2},\sqrt{3})$? $\endgroup$ – Wenbo Sun Jan 26 at 3:58
  • $\begingroup$ @KConrad Correct. But is there a way to define arithmetic progressions on ideals? $\endgroup$ – Wenbo Sun Jan 26 at 4:00
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    $\begingroup$ One way of generalizing arithmetic progressions using ideals is with generalized ideal classes. These really generalize arithmetic professions of the form $an+b$ where $\gcd(a,b)=1$. $\endgroup$ – KConrad Jan 26 at 4:30

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