|
In mathematics, a Bézout domain is a form of a Prüfer domain. It is an integral domain in which the sum of two principal ideals is again a principal ideal. This means that for every pair of elements a Bézout identity holds, and that every finitely generated ideal is principal. Any principal ideal domain (PID) is a Bézout domain, but a Bézout domain need not be a Noetherian ring, so it could have non-finitely generated ideals (which obviously excludes being a PID); if so, it is not a unique factorization domain (UFD), but still is a GCD domain. The theory of Bézout domains retains many of the properties of PIDs, without requiring the Noetherian property. Bézout domains are named after the French mathematician Étienne Bézout. == Examples == * All PIDs are Bézout domains. * Examples of Bézout domains that are not PIDs include the ring of entire functions (functions holomorphic on the whole complex plane) and the ring of all algebraic integers.〔Cohn〕 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 (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 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 (and so ''X'' has no factorization in ''S''). One shows as follows that ''S'' is a Bézout domain. :# It suffices to prove that for every pair ''a'', ''b'' in ''S'' there exist ''s'',''t'' in ''S'' such that divides both ''a'' and ''b''. :# If ''a'' and ''b'' have a common divisor ''d'', it suffices to prove this for ''a''/''d'' and ''b''/''d'', since the same ''s'',''t'' will do. :# We may assume the polynomials ''a'' and ''b'' nonzero; if both have a zero constant term, then let ''n'' be the minimal exponent such that at least one of them has a nonzero coefficient of ''X''''n''; one can find ''f'' in ''F'' such that ''fX''''n'' is a common divisor of ''a'' and ''b'' and divide by it. :# We may therefore assume at least one of ''a'', ''b'' has a nonzero constant term. If ''a'' and ''b'' viewed as elements of ''F''() are not relatively prime, there is a greatest common divisor of ''a'' and ''b'' in this UFD that has constant term 1, and therefore lies in ''S''; we can divide by this factor. :# We may therefore also assume that ''a'' and ''b'' are relatively prime in ''F''(), so that 1 lies in , and some constant polynomial ''r'' in ''R'' lies in . Also, since ''R'' is a Bézout domain, the gcd ''d'' in ''R'' of the constant terms ''a''0 and ''b''0 lies in . Since any element without constant term, like or , is divisible by any nonzero constant, the constant ''d'' is a common divisor in ''S'' of ''a'' and ''b''; we shall show it is in fact a greatest common divisor by showing that it lies in . Multiplying ''a'' and ''b'' respectively by the Bézout coefficients for ''d'' with respect to ''a''0 and ''b''0 gives a polynomial ''p'' in with constant term ''d''. Then has a zero constant term, and so is a multiple in ''S'' of the constant polynomial ''r'', and therefore lies in . But then ''d'' does as well, which completes the proof. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Bézout domain」の詳細全文を読む スポンサード リンク
|