GCD domain is LCM domain

Question

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?

Answer 1

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, $$a=\frac{ab}{m}\frac{m}{b}\implies \frac{ab}{m}\mid a,$$ $$b=\frac{ab}{m}\frac{m}{a}\implies \frac{ab}{m}\mid b.$$

Thus, $ab/m$ is a common divisor of $a$ and $b$.

Let $e\in D$ such that $e\mid a$ and $e\mid b$. Then, $e \mid ab$, so that $ab/e$ is an integer. Moreover, $a \mid ab/e$ (since $e \mid b$) and $b \mid ab/e$ (likewise), whence $m \mid ab/e$. That is, $em\mid ab$. Hence $e\mid ab/m$, and this shows 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 $s\in D$ such that $s\mid ra$ and $s\mid rb$. Then, $rab/s$ is an integer and satisfies $a \mid rab/s$ (since $s \mid rb$) and $b \mid rab/s$ (likewise), so that $m \mid rab/s$. Hence, $sm\mid rab$. 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 $n\in D$ such that $a\mid n$ 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 (Dinesh Khurana, On GCD and LCM in domains -- A conjecture of Gauss, Resonance, 8(6), 72–79) 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 $n\ge 3$.

Answer 2

I want to make a few supplements to the above good answer by Xam. If $R$ is an integral domain and $gcd(a,b)$ exists, then $lcm(a,b)$may not exist.

Consider $Z[\sqrt{-3}]$. Let $a = 2$ and $b = 1+\sqrt{-3}$. If $lcm(2,1+\sqrt{-3})$ exists, then $lcm(2,1+\sqrt{-3})$ $\mid$ $2(1+\sqrt{-3})$ and $lcm(2,1+\sqrt{-3})$ $\mid$ $2(1-\sqrt{-3})$(because $2(1-\sqrt{-3}) = -(1+\sqrt{-3})(1+\sqrt{-3})$). However, the factors of $2(1+\sqrt{-3})$ are $2, (1+\sqrt{-3})$ and $(1-\sqrt{-3})$.

Thus $lcm(2,1+\sqrt{-3})$ must be equal to $2(1+\sqrt{-3})$. In the same way, according to $lcm(2,1+\sqrt{-3})$ $\mid$ $2(1-\sqrt{-3})$, we have $lcm(2,1+\sqrt{-3})$ must be equal to $2(1-\sqrt{-3})$. This is a contradiction.

Actually, $gcd(4,2(1+\sqrt{-3}))$ doesn't exist，but$gcd(2,1+\sqrt{-3}) =1$