Norm of Prime Ideal

Question

Show that the norm of a prime ideal in a number field $K$ is a power of some prime number, i.e., if $P$ is a prime ideal in $O_K$ for some number field $K$, then $N_\mathbb{Q}^K(P)=p^n$ for some prime number $p$ and some positive integer $n$.

Here is my approach:

Any prime ideal lies over some prime number $p$. If we consider the ideal decomposition of $pO_K$, and apply the norm operator, we get the following:

$pO_K=p_1^{e_1} \cdots p_r^{e_r}$ for some $r$ since $O_K$ is a Dedekind domain. Applying the norm operator to this, we get

$N(pO_K)=N(p_1^{e_1} \cdots p_r^{e_r}) = N(p_1^{e_1})\cdots N(p_r^{e_r})$ since the norm has the multiplicative property.

This is where I am unsure if I have completely answered the question because I found a list of primes as opposed to the suggested $p^n$ in the problem statement.

Thanks in advance, any help is greatly appreciated.

Answer 1

By definition, the norm $N(P)$ is the cardinality of the field $\mathcal{O}_K/P$. Since this is a finite field (the ideal norm is always finite in the ring of integers $\mathcal{O}_K$), it has characteristic $p$ with a prime $p$. It follows that $$ N(P)=p^{[(\mathcal{O}_K/P):\mathbb{F}_p]}. $$ Here $n=[(\mathcal{O}_K/P):\mathbb{F}_p]$ is the degree of the field extension. For the element norm $N_{\mathbb{Q}}^{K}(p)$ we have, with $P=(p)$, that $N(P)=\mid N_{\mathbb{Q}}^{K}(p)\mid $.

Answer 2

For a prime ideal $\mathfrak{P}$ in $\mathcal{O}_K$, $\mathfrak{P} \cap \mathbb{Z} = p\mathbb{Z}$, where $p$ is a prime in $Z$. So, naturally, there exists a rational prime $p \in \mathfrak{P}$. Then, $(p) \subset \mathfrak{P}$. Since $\mathcal{O}_K$ is a Dedekind domain, this yields $(p) = \mathfrak{P} I$ for some other ideal $I$ in $\mathcal{O}_K$. Considering the norms on both sides, we could see that $N(\mathfrak{P})$ divides $N((p))$, which is $p^n$ where $n = [K:\mathbb{Q}]$. Therefore, the possible values of $N(\mathfrak{P})$ are $p, p^2,...,p^n$ .