How do I show that the ideals of $\mathbb Z[\sqrt{d}]$ are generated by at most two elements? $d$ is square free integer.
My approach is the following Let $A=\{ a\in \mathbb Z : a + b\sqrt{d} \in I\}$ and $B =\{ b \in \mathbb Z : a + b\sqrt{d} \in I]\}$. $I$ is an ideal of $\mathbb Z[\sqrt{d}]$
Both,A and B are ideals in $\mathbb Z$.
So, let $A=(m)$ and $B = (n)$.
I want to show that $I= (m, n\sqrt{d})$.
Is this approach corect? Can it modified?