The number of ideals in $\mathcal{O}_K$ whose norm is a prime number $p$ is bounded by $[K:Q]$.

Question

Let $K$ be a number field of degree $n$, $p$ be a prime number. Define the set $$ A_p=\{I \subset \mathcal{O}_K: N(I)=p\}. $$ I want to show that $|A_p| \leq n$.

Answer 1

So $O_K$ is a Dedekind domain, and thus its ideals uniquely factor into prime ideals. If $I$ is an ideal of $O_K$, then $I=P_1^{e_1}\cdot \ldots \cdot P_k^{e_k}$. Now we claim if $P$ is a prime ideal with $N(P)=p^f$, then $P | pO_K$.

It is straightforward to show $P \cap \mathbb{Z}$ is a prime ideal in $\mathbb{Z}$, and since $O_K/P$ has characteristic $p$, this prime ideal must be $p\mathbb{Z}$. In particular, $p \in P$ and thus, $pO_K \subseteq P$ i.e. $P|pO_K$.

Now back to $I$. If $N(I)=p$, then since $f=1$ in this case and the norm is multiplicative, $I$ must be a prime ideal, and thus by the claim $I |pO_K$.

Now let $$pO_K=P_1^{e_1}\ldots P_k^{e_k}$$ Clearly, $I$ must be one of the $P_i$, so $|A_p|= k$. Now if $N(P_i)=p^{f_i}$ for all $i$, then since $$p^n=|N_{K/\mathbb{Q}}(p)|=N(pO_K)=N(P_1)^{e_1}\ldots N(P_k)^{e_k}=p^{\sum_{i=1}^k e_if_i}$$ we have that $$n=\sum_{i=1}^k e_if_i\geq \sum_{i=1}^k 1=k$$

so by transitivity, $|A_p| \leq n$.