Let $K$ be a number field and let $x \in O_{K}$ such that Nm$(x)=ab$ where $a$, $b$ are coprime integers.
Show that $<a,x><b,x>=<x>.$
I recently got asked this question and cannot for the life of me work out a solution. I am trying an inclusion/exclusion argument.
I have shown that $<a,x><b,x> \subseteq <x>$ using the fact that $ab \in <x>$ as Nm$(<x>)=|$Nm$(x)|$, and Nm$(<x>) \in <x>$. Hence all generators, $ab, ax, bx$ and $x^{2}$ lie in the set $<x>$, since $x, a, b \in O_{K}$.
The other direction I am totally stuck on, having tried a method via norms of the individual factors, and as $<a,x><b,x> \subseteq <x>$ (by the first part), if Nm$(<a,x><b,x>)$ = Nm$(<x>)$ then $<a,x><b,x>=<x>$ but no avail! I'm sure that the fact that $a$ and $b$ are coprime is important but I don't know how to incorporate into the argument.
Thanks in advance!
Hint: $\,\ x\mid ab\,\Rightarrow\,(a,x)(b,x) = x(ab/x,\color{#c00}{a,b},x) = (x)\ $ by $\ \color{#c00}{(a,b)=(1)}$