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

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 $$n\ge 3$$.

• 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
• @ZFR and Xam: I've fixed the uses of lcm's other than of $a$ and $b$ (it's been mostly a notational issue). – darij grinberg Aug 15 at 21:25
• While I do understand the proof of i) $\Longrightarrow$ ii) now, here's a question on the other direction: Why is $\gcd\left(na,nb\right) = n\gcd\left(a,b\right)$ ? I am used to proving this using Bezout's identity. – darij grinberg Aug 15 at 22:23
• Ah, I see. From $n \mid na$ and $n \mid nb$, we obtain $n \mid \gcd\left(na,nb\right)$, and thus $\gcd\left(na,nb\right) = nu$ for some $u \in D$. Now, $nu = \gcd\left(na,nb\right) \mid na$, so that $u \mid a$ and similarly $u \mid b$. Hence, $u \mid \gcd\left(a,b\right)$. Thus, $nu \mid n\gcd\left(a,b\right)$. Hence, $\gcd\left(na,nb\right) = nu \mid n\gcd\left(a,b\right)$. Combining this with $n\gcd\left(a,b\right) \mid \gcd\left(na,nb\right)$ (which follows from the obvious divisibilities $n\gcd\left(a,b\right) \mid na$ and $n\gcd\left(a,b\right) \mid nb$), we ... – darij grinberg Aug 15 at 22:25
• ... obtain $\gcd\left(na,nb\right) = n\gcd\left(a,b\right)$ (up to unit factors). – darij grinberg Aug 15 at 22:25