# If coprime elements generate coprime ideals, does it imply for any $a,b\in R$ that $\langle a\rangle+\langle b\rangle=\langle \gcd (a,b)\rangle$?

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?

• 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

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

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).

• 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