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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?

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  • $\begingroup$ See this answer (and its comments): math.stackexchange.com/a/81580/1242 $\endgroup$ – Hans Lundmark Sep 26 '15 at 8:35
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    $\begingroup$ 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 $b$). $\endgroup$ – Hans Lundmark Sep 26 '15 at 8:47
  • $\begingroup$ @HansLundmark How to prove that $ab$ is a multiple of $m$? $\endgroup$ – user150248 Sep 3 '18 at 1:25
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    $\begingroup$ @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/… $\endgroup$ – Barry Cipra Sep 3 '18 at 2:45
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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 $n\ge 3$.

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    $\begingroup$ 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? $\endgroup$ – ZFR May 25 '18 at 9:19
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    $\begingroup$ @ZFR and Xam: I've fixed the uses of lcm's other than of $a$ and $b$ (it's been mostly a notational issue). $\endgroup$ – darij grinberg Aug 15 at 21:25
  • $\begingroup$ 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. $\endgroup$ – darij grinberg Aug 15 at 22:23
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    $\begingroup$ 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 ... $\endgroup$ – darij grinberg Aug 15 at 22:25
  • $\begingroup$ ... obtain $\gcd\left(na,nb\right) = n\gcd\left(a,b\right)$ (up to unit factors). $\endgroup$ – darij grinberg Aug 15 at 22:25

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