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.
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.