This articleneeds additional citations forverification. Please helpimprove this article byadding citations to reliable sources. Unsourced material may be challenged and removed. Find sources: "Bézout domain" – news ·newspapers ·books ·scholar ·JSTOR(August 2024) (Learn how and when to remove this message)

Inmathematics, aBézout domain is anintegral domain in which the sum of twoprincipal ideals is also a principal ideal. This means thatBézout's identity holds for every pair of elements, and that everyfinitely generated ideal is principal. Bézout domains are a form ofPrüfer domain.

Anyprincipal ideal domain (PID) is a Bézout domain, but a Bézout domain need not be aNoetherian ring, so it could have non-finitely generated ideals; if so, it is not aunique factorization domain (UFD), but is still aGCD domain. The theory of Bézout domains retains many of the properties of PIDs, without requiring the Noetherian property.

Bézout domains are named after theFrenchmathematicianÉtienne Bézout.

• All PIDs are Bézout domains.
• Examples of Bézout domains that are not PIDs include the ring ofentire functions (functions holomorphic on the whole complex plane) and the ring of allalgebraic integers.^[1] In case of entire functions, the only irreducible elements are functionsassociated 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 (for instance) 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 domainS that is not a UFD from any Bézout domainR that is not a field, for instance from a PID; the caseR =Z is the basic example to have in mind. LetF be thefield of fractions ofR, and putS =R +XF[X], the subring of polynomials inF[X] with constant term inR. This ring is not Noetherian, since an element likeX with zero constant term can be divided indefinitely by noninvertible elements ofR, which are still noninvertible inS, and the ideal generated by all these quotients of is not finitely generated (and soX has no factorization inS). One shows as follows thatS is a Bézout domain.

1. It suffices to prove that for every paira,b inS there exists,t inS such thatas +bt divides botha andb.
2. Ifa andb have a common divisord, it suffices to prove this fora/d andb/d, since the sames,t will do.
3. We may assume the polynomialsa andb nonzero; if both have a zero constant term, then letn be the minimal exponent such that at least one of them has a nonzero coefficient ofXⁿ; one can findf inF such thatfXⁿ is a common divisor ofa andb and divide by it.
4. We may therefore assume at least one ofa,b has a nonzero constant term. Ifa andb viewed as elements ofF[X] are not relatively prime, there is a greatest common divisor ofa andb in this UFD that has constant term 1, and therefore lies inS; we can divide by this factor.
5. We may therefore also assume thata andb are relatively prime inF[X], so that 1 lies inaF[X] +bF[X], and some constant polynomialr inR lies inaS +bS. Also, sinceR is a Bézout domain, the gcdd inR of the constant termsa₀ andb₀ lies ina₀R +b₀R. Since any element without constant term, likea −a₀ orb −b₀, is divisible by any nonzero constant, the constantd is a common divisor inS ofa andb; we shall show it is in fact a greatest common divisor by showing that it lies inaS +bS. Multiplyinga andb respectively by the Bézout coefficients ford with respect toa₀ andb₀ gives a polynomialp inaS +bS with constant termd. Thenp −d has a zero constant term, and so is a multiple inS of the constant polynomialr, and therefore lies inaS +bS. But thend does as well, which completes the proof.

A ring is a Bézout domain if and only if it is an integral domain in which any two elements have agreatest common divisor that is alinear 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 calledBézout'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 aGCD domain and thus Bézout domains are GCD domains. In particular, in a Bézout domain,irreducibles areprime (but as the algebraic integer example shows, they need not exist).

For a Bézout domainR, the following conditions are all equivalent:

1. R is a principal ideal domain.
2. R is Noetherian.
3. R is aunique factorization domain (UFD).
4. R satisfies theascending chain condition on principal ideals (ACCP).
5. Every nonzero nonunit inR factors into a product of irreducibles (R is anatomic domain).

The equivalence of (1) and (2) was noted above. Since a Bézout domain is a GCD domain, it follows immediately that (3), (4) and (5) are equivalent. Finally, ifR 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. (4) and (2) are thus equivalent.

A Bézout domain is aPrüfer domain, i.e., a domain in which each finitely generated ideal is invertible, or said another way, a commutativesemihereditary domain.)

Consequently, one may view the equivalence "Bézout domain iff Prüfer domain and GCD-domain" as analogous to the more familiar "PID iffDedekind domain and UFD".

Prüfer domains can be characterized as integral domains whoselocalizations at allprime (equivalently, at allmaximal) ideals arevaluation domains. So the localization of a Bézout domain at a prime ideal is a valuation domain. Since an invertible ideal in alocal ring is principal, a local ring is a Bézout domain iff it is a valuation domain. Moreover, a valuation domain with noncyclic (equivalently non-discrete) value group is not Noetherian, and everytotally orderedabelian 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 formxR for somex inR. One notable result is that a right Bézout domain is a rightOre domain. This fact is not interesting in the commutative case, sinceevery commutative domain is an Ore domain. Right Bézout domains are also right semihereditary rings.

Some facts about modules over a PID extend to modules over a Bézout domain. LetR be a Bézout domain andM finitely generated module overR. ThenM is flat if and only if it is torsion-free.^[2]

• Semifir (a commutative semifir is precisely a Bézout domain.)
• Bézout ring

• Cohn, P. M. (1968),"Bezout rings and their subrings"(PDF),Mathematical Proceedings of the Cambridge Philosophical Society,64 (2):251–264,doi:10.1017/s0305004100042791,MR 0222065
• Helmer, Olaf (1940), "Divisibility properties of integral functions",Duke Math. J.,6 (2):345–356,doi:10.1215/s0012-7094-40-00626-3,ISSN 0012-7094,MR 0001851
• Kaplansky, Irving (1970),Commutative rings, Boston, Mass.: Allyn and Bacon Inc., pp. x+180,MR 0254021
• Bourbaki, Nicolas (1989),Commutative algebra
• "Bezout ring",Encyclopedia of Mathematics,EMS Press, 2001 [1994]

• Commutative algebra
• Ring theory

• Articles needing additional references from August 2024
• All articles needing additional references