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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.)

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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
    
Your questions will be regarded more kindly if you give an indication of what you’ve tried. –  Brian M. Scott Apr 14 '12 at 22:06
    
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). –  user710587 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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1 Answer

It's true that $\rm\:lcm(a,b)\:$ exists $\Rightarrow$ $\rm\:gcd(a,b)\:$ exists, but the converse fails.

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 discusson see this post and see also Khurana, On GCD and LCM domains

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+1 for linking an article from Resonance. –  user21436 Apr 14 '12 at 22:13
    
@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. –  user710587 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
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