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Many classical arithmetic functions can be thought of as functions on the set of (non-zero) ideals of $\mathbb{Z}$ rather than as functions on $\mathbb{N}$.

Example: For $n \in \mathbb{N}$ the divisor function $d(n)$ is defined to equal the number of divisors of $n$. Equivalently, we could define $d(I)$ to the number of ideals dividing the ideal $I \lhd \mathbb{Z}$, where we count once the fact that $(1)|(n)$.

Now, the associated Dirichlet could be written as $$ \sum^\infty_{n=1} d(n) n^{-s} = \sum_{I \lhd \mathbb{Z}} d(I) N(I)^{-s} = \zeta(s)^2. $$

If $K$ is a number field with ring of integers $\mathcal{O}$ we can extend the definition of $d(I)$ to the ideals of $\mathcal{O}$. Furthermore, we can consider the associated Dirichlet series: $$ \sum_{I \lhd \mathcal{O}} d(I) N(I)^{-s}. $$

QUESTIONS: Is this Dirichlet series equal to $\zeta_K(s)^2$, the square of the Dedekind zeta function of the field?

Does this phenomena generalize? ie if $a(n)$ is arithmetic function that can be equivalently defined on the ideals of $\mathbb{Z}$ and the Dirichlet series associated to $a(n)$ is a quotient of Riemann zeta functions (= Dedekind zeta function of $\mathbb{Q}$). Can we simply `replace' the Riemann zeta functions with the appropriate Dedekind zeta functions to obtain the Dirichet series associated to the extended a(n)?

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  • $\begingroup$ What do you mean exactly by a quotient of Riemann Zeta functions ? $\endgroup$
    – François Brunault
    Commented Oct 4, 2012 at 11:27
  • $\begingroup$ Exactly? I am not sure. I mean expressions similar to those for the the Dirichlet series of many classical arithmetic functions. Most of the example here: en.wikipedia.org/wiki/Dirichlet_series#Examples . If there is a precise term for "quotients of shifted Riemann zeta function" I would like very much to know it. $\endgroup$
    – Steve Pandarus
    Commented Oct 4, 2012 at 22:42
  • $\begingroup$ @Steve : Thanks for the clarification. I don't know a precise term for these functions. $\endgroup$
    – François Brunault
    Commented Oct 5, 2012 at 7:32

1 Answer 1

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The answer to your Question 1 is "yes". It's clear that the number of ideals of $\mathcal{O}$ of norm $\le M$ is bounded above by a polynomial in $M$, so one can manipulate Dirichlet series term-by-term for $Re(s) \gg 0$ and argue that

$$ \zeta_K(s)^2 = \sum_{A, B} N(A)^{-s} N(B)^{-s} = \sum_{C} \#\{ (A, B) : AB = C\} N(C)^{-s} = \sum_C d(C) N(C)^{-s}.$$

As for Question 2 it's not entirely clear to me what your precise question is, but philosophically at least the answer is "yes" -- any arithmetical function $a$ definable purely in terms of ideals of $\mathbb{Z}$ will have a natural generalization to ideals of a number field, and if you can express $\sum_n a(n) n^{-s}$ in terms of the Riemann zeta then the same argument should give you an expression for $\sum_{I \triangleleft \mathcal{O}} a(I) N(I)^{-s}$ in terms of $\zeta_K(s)$.

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  • $\begingroup$ Hi David and thanks for your answer. Let me try to give a picture of whats really going on: I am calculating some Dirichlet series, they turn out to be a product/quotient of shifted Dedekind zeta functions e.g. $\zeta_K(S)^2$. A priori, the problem has nothing to do with the number of divisors of ideals in rings of integers. Having done some work, it turns out that the series I am looking at equals $\zeta_K(S)^2$. Given your answer, this means that maybe my problem `really' is about divisors. Now I can return to my original problem with new insight and possibly find something new. $\endgroup$
    – Steve Pandarus
    Commented Oct 4, 2012 at 22:49

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