# Finitely many ideals of given norm in rings of integers

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.

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.
• One typo, should be $p_t^{r_t f_t}$ Commented Jul 2 at 22:24