Abstract Algebra – How to Prove the Ring of All Algebraic Integers is a Bézout Domain

Question

I was told that the ring of all algebraic integers (that is, the complex numbers which are roots of a monic polynomials with integral coefficients) is a Bézout domain. But I have no idea how to prove it. Can anyone help me with this? Thanks in advance.

Answer 1

First, an example:

Consider the algebraic number field $\mathbb{Q}(\sqrt{-5})$. Its class number is two. That means there exist ideals of the ring of integers $\mathbb{Z}[\sqrt{-5}]$ of this field that are not principal, but the square of any ideal is principal. For instance, we have the decomposition

$$6 = 2 \cdot 3 = ( 1 + \sqrt{-5})\cdot (1-\sqrt{-5})$$

The ideals $\mathfrak{p}\colon =(3, 1+\sqrt{-5})$ is prime and of order $2$. That means that its square $\mathfrak{p}^2$ is principal. Indeed, we have from first principles \begin{eqnarray} \mathfrak{p}^2 = (\, 9,\, 3( 1 +\sqrt{-5}),\, (1+\sqrt{-5})^2)=\\ = (\, (2+\sqrt{-5})(2-\sqrt{-5}),\, (-1 + \sqrt{-5})(2 -\sqrt{-5}),\, -2(2 - \sqrt{-5})\, )= (2 - \sqrt{-5}) \end{eqnarray}

Let's consider the $\mathbb{Q}(\sqrt{2 - \sqrt{-5}}) $ of the field $\mathbb{Q}(\sqrt{5})$. In the ring of the integers of this number field we have the equality of ideals (extend the ideals) $$(3, 1+\sqrt{-5})^2 =(2 - \sqrt{-5}) = ( \sqrt{2 - \sqrt{-5}})^2$$ Because of the unique decomposition of ideals of any ring of integers of an algebraic number field, the above equality implies $$(3, 1+\sqrt{-5})= ( \sqrt{2 - \sqrt{-5}})$$

Let's check this equality. For comfort, fix an embedding of $\mathbb{Q}(\sqrt{2 - \sqrt{-5}}) $ into $\mathbb{C}$, $\sqrt{2 - \sqrt{-5}} \mapsto \frac{ \sqrt{5} - i }{\sqrt{2}}$ ( so $\sqrt{-5} \mapsto \sqrt{5}i$)

We have $3^2 = 9 = (2+ \sqrt{5}i) \cdot (2- \sqrt{5}i)$ so

$$ 3= \frac{ \sqrt{5} + i }{\sqrt{2}}\cdot \frac{ \sqrt{5} - i }{\sqrt{2}} $$

Note that $\frac{ \sqrt{5} + i }{\sqrt{2}}$ is in the ring of integers of $\mathbb{Q}(\frac{ \sqrt{5} - i }{\sqrt{2}})$. Indeed, the minimal polynomial of $\frac{ \sqrt{5} + i }{\sqrt{2}}$ is $x^4 - 4 x^2 + 9$ and moreover $$\frac{ \sqrt{5} + i }{\sqrt{2}}= 4/3 \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}-1/3 \cdot \left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)^3 $$

Also, from $(1+ \sqrt{5}i)^2 = -2 ( 2 - \sqrt{5}i)$ we get $$1+ \sqrt{5}i = \sqrt{2} i \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}$$ Again, $\sqrt{2}i$ is an algebraic integers and $\mathbb{Q}(\frac{ \sqrt{5} - i }{\sqrt{2}})$, since $$\sqrt{2} i =1 /3 \cdot \frac{ \sqrt{5} - i }{\sqrt{2}}-1/3 \cdot \left(\frac{ \sqrt{5} - i }{\sqrt{2}}\right)^3 $$

So far, both $3$ and $1+ \sqrt{5}i$ are divisible by $\frac{ \sqrt{5} - i }{\sqrt{2}}$. Now we'll see that $\frac{ \sqrt{5} - i }{\sqrt{2}}$ is an (algebraic) integer combination of $3$ and $1+ \sqrt{5}i$. Indeed, we have $$\frac{ \sqrt{5} - 3 i}{\sqrt{2}} \cdot (1 + \sqrt{5} i) + \frac{ \sqrt{5} - i }{\sqrt{2}}\cdot 3 = \frac{ \sqrt{5} - i }{\sqrt{2}}$$

And so it goes in general. Consider finitely many algebraic integers $(\alpha_s)$ . Let $K$ the algebraic number field generated by them, $\mathcal{O_K}$ its ring of integers. The ideal class of $K$ is finite. Therefore, the $m$ power of ideal of $\mathcal{O}_K$ generated by the $\alpha_s$ is principal for some integer $m$ ( for instance, if $m$ is the class number of $K$). There exists $\beta$ in $\mathcal{O}_K$ so that $(\alpha_s)^m_s = (\beta)$ ($\beta$ uniquely determined by this $m$ up to a unit element ( in $\mathcal{O}_K ^{\times}$). Extend $K$ by adding $\beta^{1/m}$ to $L = K(\beta^{1/m})$. We have the equality of $\mathcal{O}_L$ ideals $$(\alpha_s)_s^m = (\beta)$$ and so $$(\alpha_s)_s = (\beta^{1/m})$$