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Divisor function

Let $K$ be a number field. For a non-zero ideal $\mathfrak a$, write $t(\mathfrak a)$ for the number of its divisors (in the multiplicative monoid of non-zero ideals). It's called the divisor function. I wanted to type tau but as always have had a TeXnical problem which likely comes from a conflict of mathjax and Hatena Blog.

For positive numbers $n>0$, it is classical that the number of ideals of norm $\le n$ is: \[ \kappa n + a(n), a(n)=O(n^{1-1/d}) \] where $\kappa >0 $ is the residue of the zeta function $\zeta _K (s) $ at $s=1$ and $d=[K:\mathbf Q] $.

It implies the following estimate of the sum: \[ \sum _{N(\mathfrak a) \le N} \frac 1{N(\mathfrak a)} = \sum _{n \le N } \frac 1n \cdot (\kappa +a(n)-a(n-1) ) ) \sim \log (N) +\sum _{n\le N } \frac{a(n)}{n(n+1)} .  \]

Since $\frac{a(n)}{n(n+1)} =O (n^{-1-1/d }) $ we conlude that the second term is bounded as $N\to \infty $. As a result we get \[  \sum _{N(\mathfrak a) \le N} \frac 1{N(\mathfrak a)} \sim \kappa \log N . \]

For (always non-zero) ideals $\mathfrak a_1$ and $\mathfrak a_2$, denote by $[\mathfrak a_1,\mathfrak a_2] $ their least common multiple (in the multiplicative monoid of ideals). There is a natural bijection of the set of ideals $\mathfrak a $ of norm $\le N$ which is a multiple of both $\mathfrak a_1$ and $\mathfrak a_2$ and the set of ideals $\mathfrak b=\frac{\mathfrak a}{[\mathfrak a_1,\mathfrak a_2] } $ of norm $\le \frac{N}{N([\mathfrak a_1,\mathfrak a_2] ) } $. By the first paragraph we know there are $\sim \kappa \frac{N}{N([\mathfrak a_1,\mathfrak a_2] ) } $ such ideals.

This implies the next bound: \[ \sum _{N(\mathfrak a)\le N } {t(\mathfrak a)^2} \lesssim \sum _{N(\mathfrak a_1),N(\mathfrak a_2) \le N } \kappa \frac{N}{N([\mathfrak a_1,\mathfrak a_2 ] ) } . \]

Consider $\mathfrak b_0:=(\mathfrak a_1,\mathfrak a_2)$ (the greatest common divisor), $\mathfrak b_1:=\frac{\mathfrak a_1}{\mathfrak b_0}$ and $\mathfrak b_2:= \frac{\mathfrak a_2}{\mathfrak b_0}$. The pair of ideals $\mathfrak a_1,\mathfrak a_2$ of norms $\le N$ can be recovered from $\mathfrak b_0,\mathfrak b_1,\mathfrak b_2$ of norms $\le N$ by $\mathfrak a_i=\mathfrak b_0\mathfrak b_i$. We also have $[\mathfrak a_1,\mathfrak a_2]=\mathfrak b_0\mathfrak b_1\mathfrak b_2$. It follows that the previous sum is bounded as: \[\begin{array}{rl} &\le \kappa N \displaystyle\sum _{\mathfrak b_0,\mathfrak b_1,\mathfrak b_2, norms\le N} \frac 1{N(\mathfrak b_0\mathfrak b_1\mathfrak b_2)} \\ & = \kappa N \left(  \displaystyle\sum _{\mathfrak b} \frac 1{N(\mathfrak b)} \right)^3 \\ & \sim \kappa ^2 N (\log N )^3, \end{array}\] where the last estimate was obtained in the previous paragraph.

Using this we derive an estimate of the sum $\sum _{N(\mathfrak a)\le N} t(\mathfrak a)^2 / N(\mathfrak a) $. By the dyadic decomposition $[1,N]=[1,2]\cup (2,4]\cup \dots $ we have \[ \sum _{N(\mathfrak a)\le N} t(\mathfrak a)^2 / N(\mathfrak a) \le \sum _{i< \log _2 N} \kappa ^2 2^{i +1}(\log 2^{i+1} )^3 / 2^{i} \ll \kappa ^2 \sum _{i} (\log N)^3  \] which is $\ll (\log N)^4$.