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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?

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    No. For example, taking $d=-1$, $(1+\sqrt{-1}) \neq (1,\sqrt{-1})=\mathbb{Z}[\sqrt{-1}]$.2011-12-07

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