# 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?

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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

$\mathbb Z[\sqrt{d}]$ is a free abelian group of rank 2. Every ideal is in particular a subgroup and so has rank at most 2.

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Ca y est. $\ \$ –  Pete L. Clark Dec 7 '11 at 16:13

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.

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Interesting indeed but unfortunately it does not answer the question because $\mathbb Z [\sqrt{d}]$ is not always equal to the ring of integers of $\mathbb Q (\sqrt{d})$. –  lhf Dec 7 '11 at 22:31