The following is an exercise in a number theory book (by Childress):

Let $K/\mathbb{Q}$ be a number field. Show that there series below representing $\zeta_K(s)$ $$\sum\limits_{n=1}^\infty \frac{\gamma_n}{n^s}\quad \text{where. } \gamma_n=\#\{\text{ideals }\mathfrak{a} \subseteq \mathcal{O}_K: N\mathfrak{a}=n\} $$ is absolutely convergent for $\Re(s)>1$. Are the $\gamma_n$ bounded?

Here’s my attempted write up, I can show by example that $\gamma_n$ need not be bounded: Take $K=\mathbb{Q}[i]$, and recall that any prime $p\equiv 1 \bmod 4$ splits $p=\mathfrak{p}_1\mathfrak{p}_2$ in $\mathcal{O}_K$. So $g=2, e=1,f=1$ and $N(\mathfrak{p}_i)=p$. Then $N(\mathfrak{p}_1^i\mathfrak{p}_2^{m-i})=p^m$ for $i=0, \ldots m$, and therefore $\gamma_{p^m} = m+1$. Hence letting $m\to \infty$, we see that $\gamma_{p^m}$ is not bounded.

I’m stuck on the initial problem of absolute convergence and the case of a general field $K$. Let $d=[K:\mathbb{Q}]$. The ``worst case'' scenario (i.e. bound on $\gamma_n$) is where all primes $p$ split completely $p=\mathfrak{p}_1\cdots \mathfrak{p}_d$. Then $\gamma_{p^m}$ is the number of solutions $(n_1, \ldots, n_d)$ in nonnegative integers to $n_1 + \cdots n_d =m$, and this is $\binom{m}{d-1}$. Note that $m=\log_p p^m = \frac{\log p^m}{\log p}$ so for $p\geq 3$ we have $m \leq \log(p^m)$. Also, $\binom{m}{d-1} \leq m^{d-1}$. Therefore $$\gamma_{p^m} \leq (\log p^m)^{d-1}$$

Now let $n=p_1^{m_1} \cdots p_r^{m_r}$ be the prime factorization of an positive integer $n \in \mathbb{Z}$. Then $$ \gamma_n = \prod_{i=1}^r \gamma_{p_i^{m_i}} \leq \prod (\log p_i^{m_i})^{d-1} $$

By the AM-GM inequality we have $$ \prod \log p_i^{m_i} \leq \left( \frac{\sum_i \log p_i^{m_i}}{r} = \right)^r = \left(\frac{\log n}{r} \right) ^r $$ where $r$ is the number of primes dividing $n$.

So our bound on $\gamma_n$ is $$ \gamma_n \leq \left(\frac{\log n}{r} \right)^{r(d-1)}$$

A calculus exercise shows that $f(r)=(C/r)^r$ is maximized at $r=C/e$ (where $C$ is a constant). Hence

$$ \gamma_n \leq e^{(\log n)(d-1)/e}$$ Hmm, it seems our bound has turned into an exponential function, so that is not good.

  • $\begingroup$ The sum is "connected" to an Euler product, and the splitting of primes can be controlled by algebraic means. (Or maybe use this insight to get the corresponding bound for $\gamma_n$ in terms of $\gamma_p$ and work only with the Dirichlet sum.) $\endgroup$ – dan_fulea Aug 29 '18 at 17:25

The right way to think about this is by considering the Euler product. The $p$-part of this is $$\prod_{i}\left(1-\frac1{p^{a_is}}\right)^{-1} =\prod_{i}\left(1+\frac1{p^{a_is}}+\frac1{p^{2a_is}}+\cdots\right)$$ where the $p^{a_i}$ are the norms of the primes in $\mathcal{O}_K$ dividing $p$. As there are at most $d$ of these, then for $s>1$ $$1+\frac1{p^{a_is}}+\frac1{p^{2a_is}}+\cdots \le 1+\frac1{p^{s}}+\frac1{p^{2s}}+\cdots=\left(1-\frac1{p^s}\right)^{-1}$$ and so $$\prod_{i}\left(1+\frac1{p^{a_is}}+\frac1{p^{2a_is}}+\cdots\right) \le\left(1-\frac1{p^s}\right)^{-d}.$$ For $s>1$ then $\zeta_K(s)\le\zeta(s)^d$.

This inequality holds for the coefficients in the Dirichlet series. The coefficient of $1/n^s$ in the Dirichlet series of $\zeta(s)^d$ is $c_d(n)$ where this denotes the number of $d$-tuples $(n_1,\ldots,n_d)$ of positive integers with product $n$. You should be able to get this from your formula for $\gamma_{p^m}$. Once you have the inequality $\gamma_n\le c_d(n)$, then you win, since even if you didn't know the Euler product then it's elementary that $\zeta(s)^d=\sum_n c_d(n)/n^s$.

I suspect your estimates on the $\gamma_n$ are a bit too loose to get a manifestly convergent series.


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