Examples of Bézout domains that are not PIDs include the ring of entire functions and the ring of all algebraic integers. In case of entire functions, the only irreducible elements are functions associated to a polynomial function of degree 1, so an element has a factorization only if it has finitely many zeroes. In the case of the algebraic integers there are no irreducible elements at all, since for any algebraic integer its square root is also an algebraic integer. This shows in both cases that the ring is not a UFD, and so certainly not a PID.
Valuation rings are Bézout domains. Any non-Noetherian valuation ring is an example of a non-noetherian Bézout domain.
The following general construction produces a Bézout domain S that is not a UFD from any Bézout domain R that is not a field, for instance from a PID; the case is the basic example to have in mind. Let F be the field of fractions of R, and put, the subring of polynomials in F with constant term in R. This ring is not Noetherian, since an element like X with zero constant term can be divided indefinitely by noninvertible elements of R, which are still noninvertible in S, and the ideal generated by all these quotients of is not finitely generated. One shows as follows that S is a Bézout domain.
Properties
A ring is a Bézout domain if and only if it is an integral domain in which any two elements have a greatest common divisor that is a linear combination of them: this is equivalent to the statement that an ideal which is generated by two elements is also generated by a single element, and induction demonstrates that all finitely generated ideals are principal. The expression of the greatest common divisor of two elements of a PID as a linear combination is often called Bézout's identity, whence the terminology. Note that the above gcd condition is stronger than the mere existence of a gcd. An integral domain where a gcd exists for any two elements is called a GCD domain and thus Bézout domains are GCD domains. In particular, in a Bézout domain, irreducibles are prime. For a Bézout domain R, the following conditions are all equivalent:
Every nonzero nonunit in R factors into a product of irreducibles.
The equivalence of and was noted above. Since a Bézout domain is a GCD domain, it follows immediately that, and are equivalent. Finally, if R is not Noetherian, then there exists an infinite ascending chain of finitely generated ideals, so in a Bézout domain an infinite ascending chain of principal ideals. and are thus equivalent. A Bézout domain is a Prüfer domain, i.e., a domain in which each finitely generated ideal is invertible, or said another way, a commutative semihereditary domain.) Consequently, one may view the equivalence "Bézout domain iff Prüfer domain and GCD-domain" as analogous to the more familiar "PID iff Dedekind domain and UFD". Prüfer domains can be characterized as integral domains whose localizations at all prime ideals are valuation domains. So the localization of a Bézout domain at a prime ideal is a valuation domain. Since an invertible ideal in a local ring is principal, a local ring is a Bézout domain iff it is a valuation domain. Moreover, a valuation domain with noncyclic value group is not Noetherian, and every totally ordered abelian group is the value group of some valuation domain. This gives many examples of non-Noetherian Bézout domains. In noncommutative algebra, right Bézout domains are domains whose finitely generated right ideals are principal right ideals, that is, of the form xR for some x in R. One notable result is that a right Bézout domain is a right Ore domain. This fact is not interesting in the commutative case, since every commutative domain is an Ore domain. Right Bézout domains are also right semihereditary rings.
Modules over a Bézout domain
Some facts about modules over a PID extend to modules over a Bézout domain. Let R be a Bézout domain and Mfinitely generated module over R. Then M is flat if and only if it is torsion-free.
