Rational integer belongs to at most certain number of ideals Suppose that $K$ is an extension of $\mathbb{Q}$ of degree $n$. How can one go about proving that a non-zero rational number $q$ belongs to at most $q^{n}$ ideals of $\mathcal{O}_{K}$?
Would you be so kind as to provide me with a hint for this exercise? How can I link the decomposition of $q$ as product of irreducible elements of $\mathcal{O}_{K}$ with the decomposition of $\langle q \rangle$ as a product of prime ideals? Does it help at all?
Thanks for leaving your suggestions.
 A: Actually it is easier to find the answer with the correct level of generality: take $q\in \mathcal{O}_K$ and replace $q^n$ with the norm $|N_{K/\mathbb{Q}}(q)|$.
The ideals of $\mathcal{O}_K$ that contain $q$ are in bijection with the ideals of $\mathcal{O}_K/q\mathcal{O}_K$. Moreover, every ideal of $R = \mathcal{O}_K/q\mathcal{O}_K$ is principal. Now count how many principal ideals there can be in $R$.
A: To elaborate on Mathmo's comment, decompose $n$ into its prime factors in $\mathbb Z $  $$n=\prod_{i=1}^kp_i^{r_i} $$ and you know that each prime splits in $O_K$ as $$p_iO_K = \prod_{j=1}^{g_i} \mathfrak p_{ij}^{e_{ij}} $$
such all the $\mathfrak p_{ij}$ are distinct and that $\sum_{j=1}^{g_i} e_{ij}f_{ij}\leq n$ where $f_{ij}$ are the respective inertial degrees. We can ignore them to get a coarse bound $\sum_{j=1}^{g_i} e_{ij}\leq n$. So the ideal generated by $n$ in $O_K$ is $$nO_K=\prod_{i,j}\mathfrak p_{ij}^{r_ie_{ij}} $$
Now for $n$ to be contained in an ideal $I$, it needs to be $$I=\prod_{i,j}\mathfrak p_{ij}^{a_{ij}} $$ 
with the non-negative integers $0\leq a_{ij}\leq r_ie_{ij}$. So we are down to showing that $$\prod_{i=1}^k\prod_{j=1}^{g_i} (r_ie_{ij}+1) \leq q^n.$$ I don't feel like combinatorics rn, so can you help me finish?
