# Averaging the Mobius function on arithmetic progressions

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 $$\mu_{K}$$?

• 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 $1$. – Rodrigo Jan 26 at 1:38
• 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 $O_K$. – KConrad Jan 26 at 2:49
• @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 $\mathbb{Q}(\sqrt{2},\sqrt{3})$? – Wenbo Sun Jan 26 at 3:58
• @KConrad Correct. But is there a way to define arithmetic progressions on ideals? – Wenbo Sun Jan 26 at 4:00
• One way of generalizing arithmetic progressions using ideals is with generalized ideal classes. These really generalize arithmetic professions of the form $an+b$ where $\gcd(a,b)=1$. – KConrad Jan 26 at 4:30