# Commutative Ring Theory/Bézout domains

Let ${\displaystyle R}$ be a Bézout domain. Then ${\displaystyle R}$ is a GCD domain.
Proof: Given any two elements ${\displaystyle a,b\in R}$, we may consider the ideal ${\displaystyle I=\langle a,b\rangle }$ generated by ${\displaystyle a}$ and ${\displaystyle b}$. By the definition of Bézout domains, ${\displaystyle I=\langle c\rangle }$ for at least one ${\displaystyle c\in R}$ (which is moreover unique up to similarity). Then ${\displaystyle \langle c\rangle \leq \langle a\rangle }$ and ${\displaystyle \langle c\rangle \leq \langle b\rangle }$, so that by the characterisation of divisibility by principal ideals, ${\displaystyle c}$ is a common divisor of ${\displaystyle a}$ and ${\displaystyle b}$. Moreover, if ${\displaystyle d\in R}$ is another common divisor of ${\displaystyle a}$ and ${\displaystyle b}$, then ${\displaystyle a,b\in \langle d\rangle }$, so that ${\displaystyle \langle a,b\rangle =\langle c\rangle \leq \langle d\rangle }$, so that ${\displaystyle d|c}$. Hence, ${\displaystyle c}$ is a greatest common divisor of ${\displaystyle a}$ and ${\displaystyle b}$. ${\displaystyle \Box }$