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

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

and

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

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

$$ax+by=\gcd(a,b).$$

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?

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  • $\begingroup$ 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). $\endgroup$ – Joel Cohen Mar 18 '12 at 13:27
  • $\begingroup$ how do you deduce that gcd($x,y$) = $1$? $\endgroup$ – David Wheeler Mar 18 '12 at 13:40
  • $\begingroup$ 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). $\endgroup$ – Joel Cohen Mar 18 '12 at 13:51
  • $\begingroup$ 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 $\langle X \rangle + \langle Y\rangle \ne R$. $\endgroup$ – Joel Cohen Mar 18 '12 at 13:54
  • $\begingroup$ @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? $\endgroup$ – user23211 Mar 18 '12 at 14:17
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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).

Remark $\ $ For a comprehensive survey of integral domains closely related to GCD domains see D.D. Anderson: GCD domains, Gauss' lemma, and contents of polynomials, 2000. See also

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

$(1)\ \ $ prime ideals are maximal if nonzero, $ $ i.e. $\rm\ dim\,\ D \le 1$
$(2)\ \ $ prime ideals are principal
$(3)\ \ $ maximal ideals are principal
$(4)\ \ \rm\ gcd(a,b) = 1\, \Rightarrow\, (a,b) = 1, $ i.e. $ $ coprime $\Rightarrow$ comaximal
$(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)\ $ where $\rm\,p_i,\,P\,$ denote primes $\neq 0$

$(1\Rightarrow 2)$ $\rm\ \ p_1^{e_1}\cdots p_n^{e_n}\in P\,\Rightarrow\,$ some $\rm\,p_j\in P\,$ so $\rm\,P\supseteq (p_j)\, \Rightarrow\, P = (p_j)\:$ by dim $\le1$
$(2\Rightarrow 3)^{\phantom{|^i}}\!\!\!$ $ \ $ max ideals are prime, so principal by $(2)$
$(3\Rightarrow 4)^{\phantom{|^i}}\!\!\!$ $\ \rm \gcd(a,b)=1\,\Rightarrow\,(a,b) \subsetneq (p) $ for all max $\rm\,(p),\,$ so $\rm\ (a,b) = 1$
$(4\Rightarrow 5)^{\phantom{|^|}}\!\!\!$ $\ \ \rm c = \gcd(a,b)\, \Rightarrow\, (a,b) = c\ (a/c,b/c) = (c)$
$(5\Rightarrow 6)^{\phantom{|^|}}\!\!\!$ $\ $ Ideals $\neq 0$ in Bezout UFDs are generated by an elt with least #prime factors
$(6\Rightarrow 1)^{\phantom{|^|}}\!\!\!$ $\ \ \rm (d) \supsetneq (p)$ properly $\rm\Rightarrow\,d\mid p\,$ properly $\rm\,\Rightarrow\,d\,$ unit $\,\rm\Rightarrow\,(d)=(1),\,$ so $\rm\,(p)\,$ is max

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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).

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  • $\begingroup$ Thank you. I think we need to consider $d=0$ separately. But then $a=0=b$ and (1) holds. $\endgroup$ – user23211 Mar 18 '12 at 14:15

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