# Boundedness of this ugly limit (Maybe about analytic number theory)

Let $K$ be a number field. I'm considering the fact that $i(t)=O(t)$ where $i(t)$ is the number of ideals in $R=\mathcal{O}_K$ such that $||I|| \leq t$. I want to show this without the fact that $i_C(t)=\kappa t + O(t^{1-1/n})$ where $n=[K:\mathbb Q]$.

My try is the following. Let $I=P_1^{e_1} \cdots P_r^{e_r}$ where $P_i$ are prime ideals. Since $P_i$ is lying over some $p_i \in \mathbb Z$, $p_i$ is a prime number, so if $||I|| \leq t$, then $p_i^{ne_i} \leq t$, hence $p_i \leq \sqrt[n] t$. And also we have $e_i \leq \frac{1}{n} \log_{p_i}t$. Therefore there are at most $\prod_{p \leq \sqrt[n]t,\\ p:\mathbf{prime}}(\frac{1}{n}\log_p t+1)$ ideals those norms are less than $t$, so $$\frac{i(t)}{t} \leq \frac{1}{n^{\pi(\sqrt[n]t)}t} \prod_{p \leq \sqrt[n]t,\\ p:\mathbf{prime}}(\log_pt +n).$$ I wonder whether the limit of the right term is bounded but it's very ugly and it looks hard. I don't have any knowledge about analytic number theory so I haven't any idea about this.

## 1 Answer

Let $K(t)$ be the right side. We show that $\log K(t) \rightarrow\infty$. \begin{align} \log\left[\frac1{n^{\pi(\sqrt[n]t)}t}\prod_{p^n\leq t} \frac{\log t + n\log p}{\log p} \right]&= -\pi(\sqrt[n]t) \log n -\log t + \sum_{p^n\leq t} \left[ \log(\log t + n\log p)-\log\log p\right]\\ &=-\pi(\sqrt[n]t) \log n -\log t +\sum_{p^n\leq t}\log\left[ 1+\frac{n\log p}{\log t}\right] \\ & +\pi(\sqrt[n]t)\log\log t-\sum_{p^n\leq t} \log\log p\\ &=\pi(\sqrt[n]t)\log\log\sqrt[n]t-\log t + \sum_{p^n\leq t}\log\left[ 1+\frac{n\log p}{\log t}\right] \\ &-\sum_{p^n\leq t} \log\log p. \end{align} By $\log(1+x) \asymp x$ and Chebyshev estimate, we have $$\sum_{p^n\leq t}\log\left[ 1+\frac{n\log p}{\log t}\right]\asymp \sum_{p^n\leq t}\frac{n\log p}{\log t} \asymp \frac{n\sqrt[n]t}{\log t}.$$ By partial summation, \begin{align} \sum_{p^n\leq t} \log\log p &= \int_{2-}^{\sqrt[n]t} \log\log x \ d\pi(x) \\ &=\pi(x)\log\log x \Bigg\vert_{2-}^{\sqrt[n]t} - \int_2^{\sqrt[n]t} \frac{\pi(x)}{x\log x} dx \\ &=\pi(\sqrt[n]t)\log\log \sqrt[n]t + O\left( \frac{\sqrt[n]t}{\log^2 (\sqrt[n] t)}\right).\end{align} Combining these estimates together, we have $$\log\left[\frac1{n^{\pi(\sqrt[n]t)}t}\prod_{p^n\leq t} \frac{\log t + n\log p}{\log p} \right] \asymp \frac{n\sqrt[n]t}{\log t}.$$ This estimate shows that your terms on the right is going to $\infty$ as $t\rightarrow\infty$. This maybe due to possible over-counting on the right side. Note that you considered each $e_i$ separately to the maximal possible value.

• Thank you so much. – LWW Jul 12 '18 at 3:47