Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Some users, including me, were thinking in chat about the following conditions for a commutative ring $R$ with $1$:

$$(\forall a,b\in R)\;\;\langle a\rangle+\langle b\rangle=\langle\gcd(a,b)\rangle \tag1$$


$$(\forall a,b\in R)\;\;\gcd(a,b)=1\implies \langle a\rangle+\langle b\rangle=R.\tag2$$

The second condition says that coprime elements generate coprime ideals. Suppose $R$ is a gcd domain for the conditions to make sense. It's clear that $(1)$ implies $(2)$. Does $(2)$ imply $(1)$?

Suppose $(2)$ holds. Let $a,b\in R$ and $d=\gcd(a,b)$. ($d$ is one of the associate elements which satisfy the definition of $\gcd(a,b)$). Let $$(ar+bs)\in \langle a\rangle+\langle b\rangle.$$

We have $a=dx$ and $b=dy$ for some $x,y\in R.$ Therefore

$$ar+bs=dxr+dys=d(xr+ys)\in\langle d\rangle.$$

So we surely have $\subset$ in $(1).$ We need $\supset.$

This would follow if we had $x,y\in R$ such that


Does their existence follow from $(2)$ for general gcd domains? If not, what is the (possibly simple) counter-example? And what stronger condition do we need for the implication to hold? Does unique factorization suffice?

share|improve this question
I think (2) implies (1). If $a, b \in R$, pick $d = \gcd(a,b)$ and write $a = dx$ and $b = dy$. Then from (2), we have $\langle x\rangle + \langle y\rangle = R$. Multiply by $d$ to get (1). –  Joel Cohen Mar 18 '12 at 13:27
how do you deduce that gcd($x,y$) = $1$? –  David Wheeler Mar 18 '12 at 13:40
If $d' | \gcd(x,y)$, then $d'd$ is a common divisor of $a$ and $b$, so we get $d'd |d$. And since $d | dd'$ then $d'$ is a unit (I'm assuming the ring is a domain, but this true if we have unique factorization). –  Joel Cohen Mar 18 '12 at 13:51
Of course, (1) and (2) are true in a Principal Ideal Domain, but unique factorization is not sufficient. For example, set $R = k[X,Y,Z]$, $a = X$ and $b = Y$. Then $a$ and $b$ are coprime but $\langle X \rangle + \langle Y\rangle \ne R$. –  Joel Cohen Mar 18 '12 at 13:54
@JoelCohen Thanks, I think this is correct. There could be $d=0$ but then $a=0=b$ and $(1)$ holds. Do you know perhaps what the rings in which these equivalent conditions hold are called? –  user23211 Mar 18 '12 at 14:17

2 Answers 2

up vote 4 down vote accepted

Yes, $(2)\Rightarrow (1)$ holds. If $\rm\:d = gcd(a,b)\:$ then $\rm\:(a,b)\:=\: d\:\!(a/d,b/d) = d\!\:(1) = (d)\:$ by $(2)$. Hence a GCD domain is Bezout iff coprime elements are comaximal (a domain is called Bezout if two-generated (so finitely generated) ideals are principal).

You may find the following theorem of interest. For a comprehensive survey of domains closely related to GCD domains see D.D. Anderson: GCD domains, Gauss' lemma, and contents of polynomials, 2000.

THEOREM $\rm\ \ \ TFAE\ $ for a $\rm UFD\ D$

$(1)\ \ $ prime ideals are maximal if nonzero
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\ \Rightarrow\ (a,b) = 1$
$(5)\ \ $ $\rm D$ is Bezout
$(6)\ \ $ $\rm D$ is a $\rm PID$

Proof $\ $ (sketch of $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4 \Rightarrow 5 \Rightarrow 6 \Rightarrow 1$)

$(1\Rightarrow 2)$ $\rm\ \ P\supset (p)\ \Rightarrow\ P = (p)$
$(2\Rightarrow 3)$ $\ \: $ Clear.
$(3\Rightarrow 4)$ $\ \ \rm (a,b) \subsetneq P = (p)\ $ so $\rm\ (a,b) = 1$
$(4\Rightarrow 5)$ $\ \ \rm c = \gcd(a,b)\ \Rightarrow\ (a,b) = c\ (a/c,b/c) = (c)$
$(5\Rightarrow 6)$ $\ \ \rm 0 \ne I \subset D\:$ Bezout is generated by an elt with the least number of prime factors
$(6\Rightarrow 1)$ $\ \ \rm P \supset (p),\ a \not\in (p)\ \Rightarrow\ (a,p) = (1)\ \Rightarrow\ P = (p)$

share|improve this answer
Thank you very much! –  user23211 Mar 18 '12 at 14:32

A domain in which every sum of two (or finitely many) principal ideals is again a principal ideal is called a Bézout domain (curiously the English write an accent in that name, whereas the French, aware of the fact that writing accents was very haphazard at the time, don't). A Bézout domain is always a GCD domain, but the converse is not true. It is easy to see that the generator of $\langle a\rangle+\langle b\rangle$ is necessarily a gcd of $a$ and $b$. Your question is therefore whether a GCD domain in which (2) holds is necessarily a Bézout domain. The answer is yes, as indicated in the comment by Joel Cohen: when $d=\gcd(a,b)$, the elements $a/d$ and $b/d$ cannot have a common factor, so $1$ is a gcd of $a/d$ and $b/d$, and by (2) there exist $s,t$ with $s(a/d)+t(b/d)=1$, which implies $as+bt=d$ and therefore (1).

share|improve this answer
Thank you. I think we need to consider $d=0$ separately. But then $a=0=b$ and (1) holds. –  user23211 Mar 18 '12 at 14:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.