# The greatest common divisor of $a$ and $b$ is a linear combination of $a$ and $b$. In general, in what kind of ring does this hold?

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

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.

$\qquad\qquad$

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)?