number of divisors of an ideal $I$ of $\mathbb{Z}[\sqrt[3]{2}]$

Question

Let $I$ be an ideal of $\mathbb{Z}[\sqrt[3]{2}]$. Let $\tau$ be the divisor function.

1. I am trying to understand why there are at most $\tau(n)^2$ ideals with $N(I) = n$. How is this the case?

2. How can I prove $\tau(I) \leq \tau(n)^3$?

Thank you!

Answer 1

$$\zeta_K(s)= \sum_{n\ge 1}n^{-s}\sum_{I\subset O_K, N(I)=n}1 =\prod_P \frac1{1-N(P)^{-s}}\le\prod_p \frac1{(1-p^{-s})^3}\\ = \prod_p (1+\sum_{k\ge 1} \frac{(k+1)(k+2)}{2} p^{-sk})=\sum_{n\ge 1} n^{-s}\prod_{p^k\| n} \frac{(k+1)(k+2)}{2}$$ with $\le$ I mean comparing each coefficient not the function of $s$.