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}.
Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: