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[parent] example of a Bezout domain that is not a PID (Example)

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

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

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 $j \le n$ 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 $n$ roots of $2$ $J=\langle 2, \sqrt{2}, \sqrt[3]{2}, \dots \rangle$ is an ideal of $\mathbb{A}$ that is not principal.




"example of a Bezout domain that is not a PID" is owned by Wkbj79.
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Cross-references: principal ideal domain, Bezout domain, unique factorization, class number, ring of integers, integer, positive, subring, intersection, ideals, algebraic integer, minimal polynomials, algebraic numbers, ring

This is version 10 of example of a Bezout domain that is not a PID, born on 2007-04-19, modified 2007-04-21.
Object id is 9220, canonical name is ExampleOfABezoutDomainThatIsNotAPID.
Accessed 1166 times total.

Classification:
AMS MSC13G05 (Commutative rings and algebras :: Integral domains)
 11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers)
 11R29 (Number theory :: Algebraic number theory: global fields :: Class numbers, class groups, discriminants)

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