example of a Bezout domain that is not a PID

Let $𝔸$ be the ring of all algebraic numbers whose minimal polynomials are in $\mathbb{Z}[x]$; i.e. (http://planetmath.org/Ie), every element of $𝔸$ is an algebraic integer.

In the following example, ideals are considered to be of $𝔸$ unless indicated otherwise via intersection with a subring of $𝔸$.

Let $I$ be a ideal of $𝔸$. Then there exists a positive integer $n$ and ${\alpha }_1,\mathrm{\dots },{\alpha }_n\in 𝔸$ with $I=⟨{\alpha }_1,\mathrm{\dots },{\alpha }_n⟩$. Let $K=\mathbb{Q}({\alpha }_1,\mathrm{\dots },{\alpha }_n)$, and let ${𝒪}_K$ denote the ring of integers of $K$. Then ${\alpha }_1,\mathrm{\dots },{\alpha }_n\in {𝒪}_K$ and $I\cap {𝒪}_K$ is an ideal of ${𝒪}_K$. Let $h$ denote the class number of $K$. Then ${(I\cap {𝒪}_K)}^h=⟨\beta ⟩\cap {𝒪}_K$ for some $\beta \in {𝒪}_K$. Let $L=K(\sqrt[h]{\beta })$, and let ${𝒪}_L$ denote the ring of integers of $L$. Then

Since unique factorization of ideals holds in ${𝒪}_L$, $I\cap {𝒪}_L=⟨\sqrt[h]{\beta }⟩\cap {𝒪}_L$. Since ${𝒪}_K\subseteq {𝒪}_L$ and ${\alpha }_1,\mathrm{\dots },{\alpha }_n\in I\cap {𝒪}_K\subseteq I\cap {𝒪}_L=⟨\sqrt[h]{\beta }⟩\cap {𝒪}_L$, there exist ${\gamma }_1,\mathrm{\dots },{\gamma }_n\in {𝒪}_L$ with ${\alpha }_j={\gamma }_j\sqrt[h]{\beta }$ for all positive integers $j$ with $j\le n$. Thus, $I=⟨{\alpha }_1,\mathrm{\dots },{\alpha }_n⟩=⟨{\gamma }_1\sqrt[h]{\beta },\mathrm{\dots },{\gamma }_n\sqrt[h]{\beta }⟩\subseteq ⟨\sqrt[h]{\beta }⟩$. Since $I\subseteq ⟨\sqrt[h]{\beta }⟩$ and $I\cap {𝒪}_L=⟨\sqrt[h]{\beta }⟩\cap {𝒪}_L$, $I=⟨\sqrt[h]{\beta }⟩$. Hence, $I$ is principal. It follows that $𝔸$ is a Bezout domain.

On the other hand, $𝔸$ is not a principal ideal domain (PID). For example, the ideal all of the $n$th roots (http://planetmath.org/NthRoot) of $2$, $J=⟨2,\sqrt{2},\sqrt[3]{2},\mathrm{\dots }⟩$, is an ideal of $𝔸$ that is not principal.