Tell me more ×  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?  share|improve this
question No. For example, taking $d=-1$, $(1+\sqrt{-1}) \neq
(1,\sqrt{-1})=\mathbb{Z}[\sqrt{-1}]$. –  Chris Eagle Dec 7 '11
at 12:01 add comment  2 Answers  $\mathbb Z[\sqrt{d}]$ is a free
abelian group of rank 2. Every ideal is in particular a subgroup
and so has rank at most 2.  share|improve this answer Ca y est.
$ \ \ $ –  Pete L. Clark Dec 7 '11 at 16:13 add comment
Interestingly this is true for the fractional ideals of the ring
of integers $\mathbb{Z}_K$ of an algebraic number field $K$:  1)
Using the Chinese Remainder