example of a Bezout domain that is not a PID

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

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

Let $I$ be a finitely generated ideal of $\mathbb{A}$ . Then there exists a positive integer $n$ and $\alpha_1, \dots , \alpha_n \in \mathbb{A}$ with $I=\langle \alpha_1, \dots , \alpha_n \rangle$ . Let $K=\mathbb{Q}(\alpha_1, \dots , \alpha_n)$ , and let $\mathcal{O}_K$ denote the ring of integers of $K$ . Then $\alpha_1, \dots , \alpha_n \in \mathcal{O}_K$ and $I \cap \mathcal{O}_K$ is an ideal of $\mathcal{O}_K$ . Let $h$ denote the class number of $K$ . Then $(I \cap \mathcal{O}_K)^h=\langle \beta \rangle \cap \mathcal{O}_K$ for some $\beta \in \mathcal{O}_K$ . Let $L=K(\sqrt[h]{\beta})$ , and let $\mathcal{O}_L$ denote the ring of integers of $L$ . Then

Since unique factorization of ideals holds in $\mathcal{O}_L$ , $I \cap \mathcal{O}_L=\langle \sqrt[h]{\beta} \rangle \cap \mathcal{O}_L$ . Since $\mathcal{O}_K \subseteq \mathcal{O}_L$ and $\alpha_1, \dots , \alpha_n \in I \cap \mathcal{O}_K \subseteq I \cap \mathcal{O}_L=\langle \sqrt[h]{\beta} \rangle \cap \mathcal{O}_L$ , there exist $\gamma_1, \dots , \gamma_n \in \mathcal{O}_L$ with $\alpha_j = \gamma_j \sqrt[h]{\beta}$ for all positive integers $j$ with $j \le n$ . Thus, $I=\langle \alpha_1, \dots , \alpha_n \rangle = \langle \gamma_1 \sqrt[h]{\beta}, \dots , \gamma_n \sqrt[h]{\beta} \rangle \subseteq \langle \sqrt[h]{\beta} \rangle$ . Since $I \subseteq \langle \sqrt[h]{\beta} \rangle$ and $I \cap \mathcal{O}_L =\langle \sqrt[h]{\beta} \rangle \cap \mathcal{O}_L$ , $I=\langle \sqrt[h]{\beta} \rangle$ . Hence, $I$ is principal. It follows that $\mathbb{A}$ is a Bezout domain.

On the other hand, $\mathbb{A}$ is not a principal ideal domain (PID). For example, the ideal generated by all of the $n$ th roots of $2$ , $J=\langle 2, \sqrt{2}, \sqrt[3]{2}, \dots \rangle$ , is an ideal of $\mathbb{A}$ that is not principal.