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If $a$ and $b$ are elements in an integral domain with unity 1$\neq$0. Show that $a$ and $b$ have a least common multiple if $a$ and $b$ have a highest common factor.

More generally there is a problem of showing that if any finite non-empty non-zero subset of the ring has a highest common factor, then any finite non-empty non-zero subset of the ring has a least common multiple. (Actually the converse of the preceding sentence is also true.)

user26857
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    HINT: Think in the integers, if $a$ and $b$ were integers, then $gcd(a,b)lcm(a,b)=ab$. This should serve as an inspiration. – Josué Tonelli-Cueto Apr 14 '12 at 22:04
  • Yes that is fantastic. I would love my candidate for $lcm(a,b)$ to be $hcf(a,b)a'b'$ where $hcf(a,b)a'=a$ and $hcf(a,b)b'=b$. The problem is, when I consider $c$ to be another multiple, I want to show that $c$ actually is a multiple of $a$ and $b$ too. This amounts to showing that $hcf(a,b)a'b' \mid c$, which is unclear how to do, since I can't even break things down into finitely many irreducibles (as the ring may not be Noetherian). –  Apr 14 '12 at 22:11
  • Note that your comment in the prior question has it the wrong way around. – Bill Dubuque Apr 14 '12 at 22:13

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It's true that $\rm\,lcm(a,b)\:$ exists $\Rightarrow$ $\rm\:gcd(a,b)\:$ exists - see the Theorem below. But the converse fails, e.g. as here, in $\Bbb Q[x^2,x^3]$ we have $\,\gcd(x^2,x^3)=1\,$ but ${\rm lcm}(x^2,x^3)$ does not exist. Or, for simple well-known counterexamples in quadratic number fields see the paper linked below.

Theorem $\rm\;\; (a,b) = ab/[a,b] \;\;$ if $\;\rm\ [a,b] \;$ exists.

Proof: $\rm\quad\quad d\:|\:a,b \;\iff\; a,b\:|\:ab/d \;\iff\; [a,b]\:|\:ab/d \;\iff\; d\:|\:ab/[a,b] \quad\;\;$ QED

For further discussion see this post and see also Khurana, On GCD and LCM domains, and for basic properties of gcd and lcm (in cancellative commutative monoids) see also Section 1.6, Factorization in Commutative Monoids, in Robert Gilmer, Commutative Semigroup Rings, 1984.

Bill Dubuque
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  • @Bill Thanks for your help; I'm still unclear though. Why are you allowed to "divide" $ab$ by $[a,b]$ or by $d$. This is the issue I'm facing. –  Apr 14 '12 at 22:24
  • @user710587 $\rm:d:|:b:$ $\Rightarrow$ $\rm:ad:|:ab,:$ $\Rightarrow$ $\rm:d:|:a(b/d).:$ By definition of LCM, $\rm:a,b:|:ab:$ $\Rightarrow$ $\rm:[a,b]:|:ab.:$ And $\rm:d:|:ab/[a,b]:$ $\Rightarrow$ $\rm:d[a,b]:|:ab:$ $(\Rightarrow$ $\rm:d:|:ab):$ $\Rightarrow$ $\rm:[a,b]:|:ab/d,:$ etc. These are inferences one can make mentally with a little practice, analogous to scaling and reducing fractions. See the linked post for the universal definitions of GCD and LCM and further discussion. – Bill Dubuque Apr 14 '12 at 23:06