An integral domain is called a Bezout domain if every ideal generated by two elements is principal.
- Prove that an integral domain is a Bezout domain if and only if every pair of elements has a greatest common divisor such that, for some , .
- Prove that every finitely generated ideal in a Bezout domain is principal.
- Let be a Bezout domain, and let denote the field of fractions of . Prove that every element of can be written in the form where and are relatively prime. (That is, 1 is a greatest common divisor of and .)
- Let be an integral domain.
Suppose first that is a Bezout domain, and let . Now for some . Note in particular that we have for some , and that and . Suppose now that and . Then , and hence . Thus , and in fact is a greatest common divisor of and .
Conversely, suppose that every pair of elements has a greatest common divisor such that for some . Certainly then , so that . Now let , with . we have and , so that ; then , and we have . Hence is a Bezout domain.
- We proceed by induction on the cardinality of a (finite) generating set . For the base case , certainly is principal. Suppose now that for some , for all sets of cardinality , is principal. Let be a set of cardinality . Then , since is a Bezout domain. Thus by induction, every finitely generated ideal in a Bezout domain is principal.
- Let . We have , where is the greatest common divisor of and . If , then by part 1 above, , a contradiction. So we can assume that . We can then write , so that . By this previous exercise, and are relatively prime. (That is, a greatest common divisor is 1.) Finally, note that .