Ideals of $ \mathbb Z [\sqrt{d}] $ 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?
 A: $\mathbb Z[\sqrt{d}]$ is a free abelian group of rank 2. Every ideal is in particular a subgroup and so has rank at most 2.
A: 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 Theorem to show that if $\mathfrak{P}=\prod_{i=1}^k \mathfrak{p}_i^{n_i}$, $i\neq j \Rightarrow \mathfrak{p}_i \neq \mathfrak{p}_j$, is a product of prime ideals and if $b_i \in \mathfrak{p}_i^{n_i}-\mathfrak{p}_i^{n_i+1}$ ($-$ denoting the difference of sets), there is a $x \in \mathfrak{P}$ which fulfills the conditions $x\equiv b_i \mod{\mathfrak{p}_i^{n_i+1}}$ for all $1\leq i \leq k$.
2) If $\mathfrak{a}\subseteq \mathfrak{b}$ are two fractional ideals of $K$, then there is an $x\in\mathfrak{b}$ so that $\mathfrak{b}=\mathfrak{a}+(x)$.
This can be done by assuming $\mathfrak{b}$ to be an integral ideal. Using unique factorisation assume that $\mathfrak{a}$ is a product of the same prime ideals (maybe some of them have exponent 0 in the prime ideal decomposition). Show that any primde ideal of $\mathbb{Z}_K$ divides $\mathfrak{b}$ and $\mathfrak{a}+(x)$ the same number of times each.
3) Now you can easily conclude that any fractional ideal of $K$ is generated by at most 2 elements.
