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This kind of question has been asked before, but for a UFD. I want to show the result, more generally, for an integral domain.

Let $R$ be an integral domain (not necessarily a UFD).

Suppose $a, b \in R$ with $d \mid a,\;d \mid b \implies d$ unit.

Show that

$a \mid c,\; b \mid c \implies ab \mid c$

EDIT: Or find a counterexample. Turns out the result is not true, in general!

Oscar
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    Hint: show that the assumption on $a$ and $b$ implies that the ideal $\langle a,b\rangle$ is all of $R$, so that you can write $1=ar+bs$ for some $r,s\in R$. – Greg Martin Feb 01 '20 at 21:16
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    @GregMartin: This is false in general. Counterexample: in $K[X,Y]$ ($K$ being a field), the indeterminates have no common divisor, yet they generate a maximal ideal. – Bernard Feb 01 '20 at 21:23
  • What do you denote $(a,b)$ exactly, in this context? – Bernard Feb 01 '20 at 21:29
  • @Bernard $(a, b)$ denotes (i) $d \mid a, d \mid b$ and (ii) $d' \mid a, d' \mid b \implies d' \mid d$ – Oscar Feb 01 '20 at 21:36
  • So you do not have general integral domain, but a gcd domain? – Bernard Feb 01 '20 at 21:37
  • @Bernard I was thinking for when there are such $a$ and $b$, not any gcd, if that makes sense. – Oscar Feb 01 '20 at 21:41
  • @Bernard $(a,b)$ unit can be interpreted to mean they are coprime, i.e. have only unit common divisors, so the question makes sense as stated, i.e. the domain need not be a gcd domain. – Bill Dubuque Feb 01 '20 at 21:46
  • @BillDubuque I think it is a very ambiguous notation, as it implicitly lets the reader there's a Bézout's relation between $a$ and $b$ – Bernard Feb 02 '20 at 10:18
  • @Bernard Only if you insist on reading $,(a,b),$ as an ideal vs. gcd. But it is quite common notation for gcds also, esp, when studing divisibility theory. Usually any ambiguity is easily resolved from the context (and sometimes it is inbtended when a result holds for both ideals and gcds). – Bill Dubuque Feb 05 '20 at 16:41

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This is not true in all domains since it implies atoms (irreducibles) $\,p\,$ are prime, i.e.

$$p\mid ab,\ p\nmid a\,\Rightarrow\, (a,p)=1,\ \ {\rm so}\,\ \ a,p\mid ab\,\Rightarrow\, ap\mid ab\,\Rightarrow\, p\mid b$$

So any non-UFD number ring yields a counterexample, e.g. see here for that and more.

Bill Dubuque
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This is not true in general. The ring $\mathbb{Z}[\sqrt{-5}]$ is not an UFD, and for example you have two factorizations of $21=3\times 7=(4+\sqrt{-5})\times(4-\sqrt{-5})$. Now put $a=3$, $b=4+\sqrt{-5}$ and $c=21$.

  • Thanks Stinking Bishop and @Bill Dubuque. Now I know why I couldn't prove it! I wanted to use this result for proving that in an integral domain $R$, every nonempty set of nonzero elements of $R$ has a highest common factor if and only if every nonempty set of nonzero elements of $R$ has a least common multiple. – Oscar Feb 01 '20 at 21:56
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    @Oscar If you say more about your context I may be able to be more specific. I discuss related results in many answers, e.g. here and here and here and here. Note: you can't ping people under answers they weren't commenting in (so the above didn't work) – Bill Dubuque Feb 02 '20 at 03:07
  • @BillDubuque Basically, I'm working through the exercises in the Galois theory textbook by Garling, and a particular exercise asks to show that in an integral domain $R$, every nonempty set of nonzero elements of $R$ has a highest common factor if and only if every nonempty set of nonzero elements of $R$ has a least common multiple. From the discussion here, I realize it is asking to show GCD domain $\iff$ LCM domain, and so I'm interested in a GCD domain after all, for the question I've asked. Then I found this link: https://math.stackexchange.com/questions/1449521/gcd-domain-is-lcm-domain – Oscar Feb 02 '20 at 03:25
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    @Oscar The results in Khurana's expository paper are very old and well-known, e.g. see Section 1.6, Factorization in Commutative Monoids, in Robert Gilmer's book: Commutative Semigroup Rings, 1984. – Bill Dubuque Feb 02 '20 at 03:37
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    @Oscar But Gilmer's book was not online, so I often mentioned Khurana's paper instead, e.g way back in 2012 here, and earlier on sci.math. Nice to see that Xam and others have added expositions here (strangely no one ever sparked me to yet - it's more general than most questions here). – Bill Dubuque Feb 02 '20 at 03:41
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    @Oscar An excellent place to start learning more about this and closely related topics is the survey article by D. D. Anderson "GCD domains..." that I cite here. – Bill Dubuque Feb 02 '20 at 03:58