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.

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.

• Every principal ideal domain is a Bézout unique factorization domain.

• The ring of entire holomorphic functions on the complex plane

Non-examples

• For any principal ideal 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