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Is it possible to find an example of an integral domain $D$ and a pair of non-zero elements $x$ and $y$ in $D$ such that $\text{lcm}(zx, zy)$ exists for some non-zero element $z$ in $D$ but $\text{lcm}(x, y)$ does not?

Definition of least common divisor (lcm): Let $a$ and $b$ be elements of a commutative ring $R$. A common multiple of $a$ and $b$ is an element $m$ of $R$ such that there exist elements $x$ and $y$ of $R$ such that $ax = by = m$. A least common multiple of $a$ and $b$ is a common multiple $m$ of $a$ and $b$ that is minimal in the sense that for any other common multiple $n$ of $a$ and $b$, $n=zm$ for some $z$ in $R$.

user150248
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  • @mvw I have added that to the question. – user150248 Sep 04 '18 at 20:09
  • By now you asked three questions on lcm/gcd. That's not a problem, but it'd be nice if you provided some context to this. What are you trying to achieve over all? – quid Sep 04 '18 at 21:29
  • @quid Nothing. I just don’t have anything better to do now. – user150248 Sep 05 '18 at 03:59
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    @quid A few days ago I accidentally spotted a mistake in my old lecture notes of the abstract algebra course which I took as a undergrad student. It wrote: in an integral domain, $\gcd(x, y)=\gcd(x, z)=1$ implies $\gcd(x, yz)=1$. The lecture got this wrong because he used the identity $\gcd(zx, zy)=z\gcd(x, y)$ without checking the existence of $\gcd(zx, zy)$. Since I have nothing to do recentlly, I decided to explore a little bit on this topic. – user150248 Sep 05 '18 at 12:53
  • This type of comment was exactly the type of information that I was looking for. Thank you, it helps me contextualize the questions. – quid Sep 05 '18 at 15:22

1 Answers1

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Let $M$ be a lowest common multiple of $zx$ and $zy$. It's clear that $z$ divides $M$, say $M=zm$, so we can try and see whether $m$ satisfies the properties for being a lowest common multiple of $x$ and $y$.

First, $M=azx=bzy$, so $m=ax=by$.

Now suppose $r$ is a common multiple of $x$ and $y$. Then $zr$ is a common multiple of $zx$ and $zy$; therefore $M=zm$ divides $zr$. Hence $m$ divides $r$.

egreg
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