Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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)?

share|improve this question
    
What do you mean exactly by a quotient of Riemann Zeta functions ? –  François Brunault Oct 4 '12 at 11:27
    
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. –  Steve Pandarus Oct 4 '12 at 22:42
    
@Steve : Thanks for the clarification. I don't know a precise term for these functions. –  François Brunault Oct 5 '12 at 7:32
add comment

1 Answer 1

up vote 7 down vote accepted

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 $\zeta_K(s)$.

share|improve this answer
    
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. –  Steve Pandarus Oct 4 '12 at 22:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.