Let $K$ be an algebraic number field and let $I$ be a non-zero ideal of $O_K$ with $N(I)$ (norm of ideal) a prime number, prove that $I$ is a prime ideal.
$\textbf{My Try-}$ Let $a,b \in O_K$ such that $ab \in I$ which implies $\langle ab \rangle = \langle a\rangle \langle b \rangle\subseteq I \implies I \ |\ \langle ab \rangle \implies \exists\ \text{an integral ideal}\ J \subseteq O_K$ such that $\langle ab \rangle=I.J$.
Now ${N(\langle ab \rangle)=N(\langle a \rangle)N (\langle b \rangle)=N(I).N(J)} \hspace{8cm} *$ ,
but $N(I)$ is prime so as $N(I)$ divides R.H.S of $*$ it also divides $N(\langle a \rangle)N (\langle b \rangle) \implies N(I)\ |\ N(\langle a \rangle)\ \text {or}\ N(I)\ |\ N (\langle b \rangle)$.
Now W.L.G let $N(I)\ |\ N(\langle a \rangle)$.
Now I am STRUCK here. From here I cannot say that $I\ |\ <a>$, what I missed, is this approach OK?