## Definition and basic properties of Bezout domains

An integral domain is called a Bezout domain if every ideal generated by two elements is principal.

1. Prove that an integral domain $R$ is a Bezout domain if and only if every pair of elements $a,b$ has a greatest common divisor $d$ such that, for some $x,y \in R$, $d = ax + by$.
2. Prove that every finitely generated ideal in a Bezout domain is principal.
3. Let $R$ be a Bezout domain, and let $D^{-1}R$ denote the field of fractions of $R$. Prove that every element of $F$ can be written in the form $\frac{a}{b}$ where $a$ and $b$ are relatively prime. (That is, 1 is a greatest common divisor of $a$ and $b$.)

1. Let $R$ be an integral domain.

Suppose first that $R$ is a Bezout domain, and let $a,b \in R$. Now $(a,b) = (d)$ for some $d$. Note in particular that we have $ax + by = d$ for some $x,y \in R$, and that $d|a$ and $d|b$. Suppose now that $e|a$ and $e|b$. Then $(a),(b) \subseteq (e)$, and hence $(d) = (a,b) \subseteq (e)$. Thus $e|d$, and in fact $d$ is a greatest common divisor of $a$ and $b$.

Conversely, suppose that every pair of elements $a,b \in R$ has a greatest common divisor $d$ such that $d = ax + by$ for some $x,y \in R$. Certainly then $d \in (a,b)$, so that $(d) \subseteq (a,b)$. Now let $z \in (a,b)$, with $z = ar + bs$. we have $a = a^\prime d$ and $b = b^\prime d$, so that $z = d(a^\prime r + b^\prime s)$; then $z \in (d)$, and we have $(a,b) = (d)$. Hence $R$ is a Bezout domain.

2. We proceed by induction on the cardinality of a (finite) generating set $A$. For the base case $|A| = 1$, certainly $(A)$ is principal. Suppose now that for some $k$, for all sets $A$ of cardinality $k$, $(A)$ is principal. Let $\{\alpha\} \cup A$ be a set of cardinality $k+1$. Then $(\{\alpha\} \cup A) = (\alpha) + (A)$ $= (\alpha) + (\beta)$ $= (\alpha,\beta) = (\gamma)$, since $R$ is a Bezout domain. Thus by induction, every finitely generated ideal in a Bezout domain is principal.
3. Let $\frac{a}{b} \in D^{-1}R$. We have $ax + by = d$, where $d$ is the greatest common divisor of $a$ and $b$ . If $d = 0$, then by part 1 above, $a = b = 0$, a contradiction. So we can assume that $d \neq 0$. We can then write $a^\prime dx + b^\prime d y =d$, so that $a^\prime x + b^\prime y = 1$. By this previous exercise, $a^\prime$ and $b^\prime$ are relatively prime. (That is, a greatest common divisor is 1.) Finally, note that $\frac{a}{b} = \frac{a^\prime d}{b^\prime d} = \frac{a^\prime}{b^\prime}$.