Finitely many ideals of given norm in rings of integers

Question

I'm trying to prove that there are only finitely many ideals of any given norm in the ring of integers $\mathcal{O}_k$ over a numberfield $K$.

I know there are "standard proofs" (e.g. How many elements in a number field of a given norm?), but I'm just wondering would a induction proof on the norm be valid:

(i) True for $n=1$.

(ii) For $n=k+1$, if $n$ is prime then it's either rammified, inert or split in $K$ which would mean there are 1, 0 or 2 ideals of norm $n$ respectively. If $n$ is composite then it can be factored into prime ideals of smaller norms (since $\mathcal{O}_k$ is a Dedekind domain), where there are only finite ideals of smaller norm, hence it follows by induction.

Thanks.

Answer 1

Your argument is more or less correct. Be careful, however - there could be a lot more than two prime ideals above a given prime (but there is a finite number). Also, induction is really awkward and not necessary here.

A much cleaner proof is just to use unique factorization directly: an arbitrary ideal of $\mathcal O_K$ can be written as $\wp_1^{r_1}\cdots\wp_t^{r_t}$, and the norm of this is $p_1^{r_1f_1}\cdots p_t^{r_1f_t}$. If you require that this equals $n$ for some $n$, then there are finitely many possibilities for $t$, for the $\wp$'s, and for the $r$'s.