# GCD domain is LCM domain

On this Wiki page it is written:

A GCD domain is an integral domain $R$ with the property that any two non-zero elements have a greatest common divisor (GCD). Equivalently, any two non-zero elements of $R$ have a least common multiple (LCM).

How to prove last statement that is equivalence of GCD and LCM for all elements? (I am able to prove in Bezout ring but I am not able to prove in general GCD ring.)

Does the existence of gcd of two elements implies existence of lcm and conversely in any integral domain?

• See this answer (and its comments): math.stackexchange.com/a/81580/1242 – Hans Lundmark Sep 26 '15 at 8:35
• If $a$ and $b$ have an LCM $m$, it's not too difficult to check that $d=ab/m$ is a GCD of $a$ and $b$. But the argument doesn't work the other way around. A classical counterexample: let $R$ be the subring of $\mathbf{Z}[X]$ consisting of the polynomials $\sum c_k X^k$ such that $c_1$ is an even number. Then $d=1$ is a GCD of $a=2$ and $b=2X$ in $R$ (note that $b$ is not divisible by $2$ in $R$), but $a$ and $b$ have no LCM (both $4X$ and $2X^3$ are multiples of $a$ and $b$, but they have no factor in common that is a multiple of both $a$ and $b$). – Hans Lundmark Sep 26 '15 at 8:47
• @HansLundmark How to prove that $ab$ is a multiple of $m$? – user150248 Sep 3 '18 at 1:25
• @user150248, isn't it just the definition of lcm? I.e., $ab$ is a common multiple of $a$ and $b$, so it's divisible by their least common multiple $m$ (provided the lcm exists). See en.wikipedia.org/wiki/… – Barry Cipra Sep 3 '18 at 2:45

The result that is given in Wikipedia without proof is an immediate corollary of the following:

Theorem: Let $D$ be a domain and $a,b\in D$. TFAE $\colon$

i) $\text{lcm}(a,b)$ exists.

ii) For all $r\in D\setminus\{0\}$, $\gcd(ra,rb)$ exists.

Proof: i)$\implies$ ii) Let's call $\text{lcm}(a,b)=m$. We have $a\mid ab$ and $b\mid ab$, then $m\mid ab$. We claim that $\gcd(a,b)=ab/m$. Indeed, $$a=\frac{ab}{m}\frac{m}{b}\implies \frac{ab}{m}\mid a,$$ $$b=\frac{ab}{m}\frac{m}{a}\implies \frac{ab}{m}\mid b.$$

Let $e\in D$ s. t. $e\mid a$ and $e\mid b$, then $eb\mid ab$ and $ea\mid ab$; and thus $\text{lcm}(ea, eb)=e\text{lcm}(a,b)\mid ab$, i.e, $em\mid ab$. Hence $e\mid ab/m$, which means that $\gcd(a,b)=ab/m$.

Set $\gcd(a,b)=ab/m=d$. Given $r\neq 0$, we claim that $\gcd(ra,rb)=rd$. Indeed, $d\mid a$ and $d\mid b$ implies that $rd\mid ra$ and $rd\mid rb$. Let $s\in D$ s. t. $s\mid ra$ and $s\mid rb$, then $sb\mid rab$ and $sa\mid rab$. It follows that $\text{lcm}(sa,sb)=s\text{lcm}(a,b)\mid rab$, i.e., $sm\mid rab$. Therefore $s\mid rab/m=rd$. This proves that $\gcd(ra,rb)=rd$.

ii)$\implies$ i) We claim that $$\text{lcm}(a,b)=\frac{ab}{\gcd(a,b)}.$$

Indeed, set $d=\gcd(a,b)$, then $$\frac{ab}{d}=a\frac{b}{d},$$ $$\frac{ab}{d}=b\frac{a}{d}.$$

So $a\mid ab/d$ and $b\mid ab/d$. Let $n\in D$ s. t. $a\mid n$ and $b\mid n$, thus $ab\mid nb$ and $ab\mid na$. It follows that $ab\mid \gcd(na,nb)=n\gcd(a,b)$, i.e., $ab\mid nd$ and then $\frac{ab}{d}\mid n$. Hence, $\text{lcm}(a,b)=ab/d$.

The above theorem is used in this paper by D. Khurana where he proves that for every $n\ge 3$ non-square $\Bbb{Z}[\sqrt{-n}]$ is not a GCD-domain, and hence not a UFD. More exactly, he shows that if $n+1=pk$ for some prime $p$ and $k\ge 2$, then $\text{lcm}(p,1+\sqrt{-n})$ doesn't exist; and if $n+1$ is prime then $\text{lcm}(2,2+\sqrt{-n})$ doesn't exist. So we have an alternative proof to the fact that $\Bbb{Z}[\sqrt{-n}]$ is not a UFD for $n\ge 3$.

• When you are proxing that ii) follows from i) you have used that $\text{lcm}(ea,eb)=e\text{lcm}(a,b)$. Could you clarify it, please? – ZFR May 25 '18 at 9:19