0

Let $R$ be a UFD. Then for any two elements $a,b \in R \setminus \{0\}$ $ab \sim \gcd (a,b){\operatorname{lcm} (a,b)}$.

My attempt :

Let $d = \gcd (a,b)$.Then $d|a$ and $d|b$.Then $\exists x,y \in R$ such that $a = dx$ and $b = dy$.Now it is to be shown that $c = dxy$ is the lcm of $a,b$.Which I find difficulty to show.It is clear that $a|c$ and $b|c$. Now if $u$ is any common multiple of $a$ and $b$.Then $dx|u$ and also since $y|b$ we have $y|u$ i.e. $dxy|u^2$ $\implies c|u^2$.But I have to show $c|u$ to complete the proof which I fail to prove.Please help me to complete the proof.

Thank you in advance.

user26857
  • 52,094

1 Answers1

3

Hint $\,d\mapsto ab/d\,$ bijects the common divisors of $\,a,b\,$ with the common multiples dividing $ab.\,$ Being order-$\rm\color{#c00}{reversing}$, it maps a $\rm\color{#c00}{Greatest}$ common divisor to a $\rm\color{#c00}{Least}$ common multiple, i.e. $\,{\rm\color{#c00}{G}CD}(a,b)\mapsto ab/{\rm GCD}(a,b) \sim {\rm \color{#c00}{L }CM}(a,b).\,$

See here for more on this involution (reflection) symmetry at the heart of gcd, lcm duality.

Remark $\ $ Your approach seems headed toward the dual of the proof in the linked post, i.e. $$a,b\mid u\iff ab\mid ub,ua\iff ab\mid(ub,ua)\!=\!u(b,a)\iff ab/(b,a)\mid u$$

Community
  • 1
Bill Dubuque
  • 272,048
  • would you please tell me what was wrong in my concept? Because I want to complete it by the process mentioned in my post. –  Jan 12 '17 at 20:14
  • @A.Chattopadhyay I added a remark on that to the answer. – Bill Dubuque Jan 12 '17 at 21:40
  • @A.Chattopadhyay for better understanding see also http://math.stackexchange.com/q/2095325/386602 –  Jan 13 '17 at 07:49
  • @Robmath Is there something that you don't understand about the above proofs? They are simpler and more general than the one you link to.Moreover, they have the advantage of clarifying the fundamental innate duality between gcd and lcm - which is essential to understand in order to master these topics. – Bill Dubuque Jan 13 '17 at 16:13