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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.2012-04-14
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    Your questions will be regarded more kindly if you give an indication of what you’ve tried.2012-04-14
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    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).2012-04-14
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    Note that your comment in the prior question has it the wrong way around.2012-04-14

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