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.
Best Answer
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 $.