# 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. (http://planetmath.org/Ie), 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 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

$\begin{array}[]{rl}(I\cap\mathcal{O}_{L})^{h}&=[(I\cap\mathcal{O}_{K})\mathcal% {O}_{L}]^{h}\\ &=(I\cap\mathcal{O}_{K})^{h}(\mathcal{O}_{L})^{h}\\ &=(\langle\beta\rangle\cap\mathcal{O}_{K})\mathcal{O}_{L}\\ &=\langle\beta\rangle\cap\mathcal{O}_{L}\\ &=(\langle\sqrt[h]{\beta}\rangle\cap\mathcal{O}_{L})^{h}\end{array}$

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 $j\leq n$. 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 all of the $n$th roots (http://planetmath.org/NthRoot) of $2$, $J=\langle 2,\sqrt{2},\sqrt[3]{2},\dots\rangle$, is an ideal of $\mathbb{A}$ that is not principal.

Title example of a Bezout domain that is not a PID ExampleOfABezoutDomainThatIsNotAPID 2013-03-22 16:57:04 2013-03-22 16:57:04 Wkbj79 (1863) Wkbj79 (1863) 13 Wkbj79 (1863) Example msc 11R29 msc 11R04 msc 13G05