Prove that, if they exist, $\text{lcm}(a,b)\gcd(a,b)$ is an associate of $ab$.

Question

Similar questions have been asked but didn't helped me because either they didn't answer my problem, either they answered a specific case of my problem, not helping me prove the most general conclusion. The problem is the following.

Let $D$ be a commutative, integral domain. Prove that for any $a,b\in D$, if both $\text{lcm}(a,b)=:m$ and $\gcd(a,b)=:d$ exist, then $dm$ is an associate of $ab$.

I've been turning around the problem for hours, without success. Noting $a/b$ the element such that $(a/b)b = b (a/b) = a$, the further I've got was that $m = d (b/d) (m/b) = d (a/d) (m/a)$ which in turn implies that $$(b/d) m = (m/d)b$$ $$(a/d) m = (m/d) a$$ $$\Rightarrow (m/d)(b-a) = m(b/d - a/d)$$ where I'm stuck. I was trying to prove that the ideal generated by $ab$ is equal to the one generated by $dm$, or equivalently that $ab|dm,\ dm|ab$. By biggest problem, I guess, is that intuitively I don't see why the conclusion should be true. Thank you in advance for your time.

Answer 1

Following Dietrich Burde's comment to my question, I read the answer he referred me to and I now consider Marc van Leeuwen's answer to be applicable to my question too. Indeed, following his proof$-$that I won't restate here to avoid plagiarism or deteriorate his elegant reasoning$-$we arrive to the equality $$\gcd(a,b)\text{lcm}(a,b) = ab$$ up to a unit factor, i.e. if $m,m'$ and $d,d'$ are respectively two LCMs and two GCDs of $a,b$, then$-$since the ring $D$ under consideration is an integral, commutative domain$-$there are two units $u_m, u_d\in D^*$ such that $d = d' u_d, m = m' u_m$ so $$a b = d m= d' m' u_d u_m$$ whence one must conclude $$(dm) = (ab)$$ which implies any product of a LCM and a GCD of $a$ and $b$ is an associate of $ab$.