nLab Bézout domain

Definition

A Bézout domain is an integral domain that is also a Bézout ring.

In constructive mathematics

In constructive mathematics, there are different types of integral domains, yielding different types of Bézout domains: the gcd function and Bézout coefficient functions are no longer valued in $R \setminus \{0\}$ in one variable, but in $\{x \in R \vert x \neq 0\}$, $\{x \in R \vert x \# 0\}$, or some other definition, depending on what the base integral domain ends up being (classical, Heyting, discrete, residue, et cetera).

Properties

Every Bézout domain is a GCD domain.

Proof

Let $R$ be a Bézout domain and let $a,b \in R$. There exists $c \in R$ such that $aR+bR=cR$. We see that $c$ divides $a$ and $b$, and that for every $r \in R$, if $r$ divides $a$ and $b$, then $r$ divides $c$. Therefore, $c$ is a gcd of $a$ and $b$.

Every GCD domain of dimension at most 1 is a Bézout domain.

In classical mathematics

In classical mathematics, every Bézout domain $R$ that satisfies one of the following conditions is a principal ideal domain

1. $R$ is a unique factorization domain

2. $R$ is a Noetherian ring

3. $R$ is an atomic domain?

4. $R$ satisfies the ascending chain condition on principal ideals.

In constructive mathematics

In constructive mathematics, Bézout domains are usually better behaved because many important rings may fail to be principal ideal domains.

For instance, the ring of integers is a principal ideal domain if and only if the law of excluded middle holds: In one direction, the usual proofs rely on being able to decide whether any particular integer belongs to the ideal or not.

For the converse, let $\varphi$ be an arbitrary proposition. Consider the ideal $\{ x \in \mathbb{Z} | (x = 0) \vee \varphi \}$. By assumption, it is generated by some number $n$. Since the integers are discrete, it holds that $n = 0$ or $n \neq 0$. In the first case $\neg\varphi$ holds, in the second $\varphi$.

However, this ideal cannot be proved to be finitely generated either. If an ideal is generated by $n_1, \ldots, n_k$, then we may form their gcd one step at a time, which we can do algorithmically. Therefore, $\mathbb{Z}$ remains a Bézout domain.

As a result, in constructive mathematics not every Bézout domain that is a unique factorization domain is a principal ideal domain, as $\mathbb{Z}$ is both a Bézout domain and a unique factorization domain, but is not a principal ideal domain.

Some authors have tried to define a principal ideal domain as a Noetherian Bézout domain, but it is unknown if this still coincides in constructive mathematics with the definition of principal ideal domain as a integral domain whose ideals are all principal ideals.

Examples

• Every field $K$ is a Bézout domain where for all elements $a \in K$ and $b \in K$, $\gamma(a, b) = a \cdot b$, $\beta_1(a, b) = 1$, and $\beta_2(a, b) = 1$

• The ring of integers $\mathbb{Z}$

• For any discrete field $K$, the polynomial ring $K[x]$ on one generator is a Bézout domain.

• The ring of entire holomorphic functions on the complex plane

Non-examples

• For any Bézout unique factorization domain $R$, the polynomial ring $R[x]$ on one generator is a unique factorization domain which is not a Bézout domain. The ideal generated by $2$ and $x$ is not a principal ideal, as $1$ is the greatest common divisor of $2$ and $x$, but $1$ is not in the ideal generated by $2$ and $x$.

• In particular, any integral domain extension $\mathbb{R}[x, y]$ of the rational numbers by a transcendental number $x \in \mathbb{R}$is not a Bézout domain.

• Similarly, for any field $K$, the polynomial ring $R[x, y]$ on two generators is a unique factorization domain which is not a Bézout domain. The ideal generated by $x$ and $y$ is not a principal ideal, as $1$ is the greatest common divisor of $x$ and $y$, but $1$ is not in the ideal generated by $x$ and $y$.

• In particular, any integral domain extension $\mathbb{Q}[x, y]$ of the rational numbers by two transcendental numbers $x \in \mathbb{R}$ and $y \in \mathbb{R}$ which are linearly independent as basis $\mathbb{Q}$-vectors is not a Bézout domain.

See also

• Bézout ring

• GCD domain

• Euclidean domain

• principal ideal domain

• function field analogy

References

• Henri Lombardi, Claude Quitté (2010): Commutative algebra: Constructive methods (Finite projective modules) Translated by Tania K. Roblo, Springer (2015) (doi:10.1007/978-94-017-9944-7, pdf)

See also:

• Wikipedia, Bézout domain