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In $\mathbb{Z}$, the greatest common divisor of $a$ and $b$ is a linear combination of $a$ and $b$.

This generalizes to Euclidean domains since Euclid's algorithm works. Moreover this statement generalizes to PIDs, for if ideals $(c)=(a)+(b)$ then $c$ is a linear combination of $a$ and $b$, and $c$ is the gcd of $a$ and $b$.

My question is: how far can we generalize the statement above? In the conventional classification of commutative rings with unit, what is the best generalization?

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What's a greatest common divisor? Not all Euclidean domains have any sort of order, for one of the divisors to be greatest. – Karolis Juodelė Mar 30 '13 at 13:23
The common divisors of two elements in a ring can always be ordered by divisibility. The greatest common divisor, by definition, is the greatest one under this ordering (if it exists). – Montez Mar 30 '13 at 13:27
I'm not sure, but are you looking for Bézou domain? – Hans Giebenrath Mar 30 '13 at 13:27
@HansGiebenrath. Yes, I think you are right. Thank you. – Montez Mar 30 '13 at 13:32
@Hans: Nicely done. Make that an answer. – Cameron Buie Mar 30 '13 at 13:37
up vote 5 down vote accepted

Rings in which every two-generated ideal is principal $\rm\:(a,b) = (c)\:$ are called Bezout rings, since they are precisely the rings where gcds exist and have linear (Bezout) form. For suppose that $\rm\:(a,b) = (c).\:$ Then $\rm\:(c)\supseteq (a),(b)\:\Rightarrow\: c\mid a,b,\:$ so $\rm\:c\:$ is a common divisor of $\rm\:a,b.\:$ Conversely $\rm\:(a,b)\supseteq (c)\:\Rightarrow\: c = ja + k b\:$ so $\rm\:d\mid a,b\:\Rightarrow\:d\mid c,\:$ so $\rm\:c\:$ is a greatest common divisor (greatest in terms of divisibility order).

Bezout domains lie between PIDs and GCD domains in the following list of domains closely related to GCD domains.

$\qquad\qquad$ enter image description here

PID: $\ \ $ every ideal is principal

Bezout: $\ \ $ every ideal (a,b) is principal

GCD: $\ \ $ (x,y) := gcd(x,y) exists for all x,y

SCH: $\ \ $ Schreier = pre-Schreier & integrally closed

SCH0: $\ \ $ pre-Schreier: a|bc $\, \Rightarrow\, $ a = BC, B|b, C|c

D: $\ \ $ (a,b) = 1 & a|bc $\,\Rightarrow\,$ a|c

PP: $\ \ $ (a,b) = (a,c) = 1 $\,\Rightarrow\,$ (a,bc) = 1

GL: $\ \ $ Gauss Lemma: product of primitive polys is primitive

GL2: $\ \ $ Gauss Lemma holds for all polys of degree 1

AP: $\ \ $ atoms are prime [i.e. PP restricted to atomic a]

Since atomic & AP $\,\Rightarrow\,$ UFD, reversing the above UFD $\,\Rightarrow\,$ AP path shows that in atomic domains all these properties (except PID, Bezout) collapse, becoming all equivalent to UFD.

There are also many properties known equivalent to D, e.g.

[a] $\ \ $ (a,b) = 1 $\,\Rightarrow\,$ a|bc $\,\Rightarrow\,$ a|c

[b] $\ \ $ (a,b) = 1 $\,\Rightarrow\,$ a,b|c $\,\Rightarrow\,$ ab|c

[c] $\ \ $ (a,b) = 1 $\,\Rightarrow\,$ (a)/\(b) = (ab)

[d] $\ \ $ (a,b) exists $\,\Rightarrow\,$ lcm(a,b) exists

[e] $\ \ $ a + b X irreducible $\,\Rightarrow\,$ prime for b $\ne$ 0 (deg = 1)

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I think you are looking for Bézou domain, a well known concept in ring theory, generalizing the notion of principal ideal rings.

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I think the question is: What are the rings $R$ in which d=GCD(a,b) implies dR=(a,b)?

By the definition of the GCD we know that if d = GCD(a,b) then a=rd and b=sd and so d=GCD(rd,sd) which leads to 1=GCD(r,s) if d is a not a zero divisor. Restricting to rings without zero divisors and specifically to integral domains we see that in a domain D, d=GCD(a,b) implies dR=(a,b) if and only if in D, 1=GCD(x,y) implies R=(x,y). The domains in which 1=GCD(x,y) implies R=(x,y) were called pre-Bezout domains by Cohn in [C].

Obviously a Bezout domain is a pre-Bezout domain, but a pre-Bezout domain is far from being Bezout. Here is a simple example: Let V be a discrete rank 1 valuation domain, with maximal ideal M=pV, and let L be a field that contains the quotient field of V properly and consider R=V+XL[[X]] the ring of power series f in L[[X]] such that f(0) is in V. It is easy to see that R is a quasi local domain such that an element f in R is a unit if and only if f(0)=1 if and only if f is not divisible by any positive power of p. We claim that R is pre-Bezout. For if f and g in R are such that GCD(f,g)=1 then either f is not divisible by p or g is. But then f is a unit or g is and in either case (f,g)=R.

Now in some cases pre-Bezout is as potent as Bezout. Recall that a Noetherian Bezout domain is a PID, it was shown in [MZ] that a Noetherian pre-Bezout domain is a PID (Corollary 6.6 of [MZ] deals with more general atomic domains.) . A special case of a pre-Bezout domain is studied as a GCD-Bezout domain in [PT]. (D is GCD-Bezout if d=GCD(a₁,a₂,...,a_{n}) implies that dD=(a₁,a₂,...,a_{n})). It was stated in [PT] that a Noetherian GCD-Bezout domain is a PID.

  • [C] P.M. Cohn, Bezout rings and their subrings, Proc. Cambridge Philos. Soc. 64 (1968) 251-264.
  • [PT] M.H. Park and F. Tartarone, Divisibility properties related to star-operations on integral domains. Int. Electron. J. Algebra 12 (2012), 53--74.
  • [MZ] J.L. Mott and M. Zafrullah, On Prufer v-multiplication domains, manuscripta math. 35(1-2) (1981) 1- 26.
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