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In an integral domain, if a|c and b|c then LCM(a,b)|c? What if GCD(a,b)=1 and ac=bd, may I assume that a|d?

I am needing to assure this for a demonstration but I can't prove or find an counterexample. What I'm willing to prove is that GCD exists if and only if LCM exists. I know that there is a similar question in the forum but I want to know what is above.

Bryant
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    All integral domains do not have gcds and lcms. – Bernard Oct 12 '19 at 22:44
  • How do you define lcm and gcd in an integral domain? I only know how to define them for principal ideal domains. – Joppy Oct 12 '19 at 22:45
  • @Bernard I'm assuming that they exist in a certain Integral domain. I should have written it better, thanks for your attention. – Bryant Oct 12 '19 at 23:07
  • @Joppy well, the question only stats that they exist. I've updated the description. – Bryant Oct 12 '19 at 23:08
  • Re: your edit: that's not true. What is true is that $,{\rm lcm}(a,b)$ exists $\iff \gcd(ac,bc)$ exists for all $,c\neq 0\ \ \ $ – Bill Dubuque Oct 13 '19 at 00:05
  • @BillDubuque oh, so there is an error in the textbook we are using and in the proof I wrote hahaha. Thank you again! Now I have to check what mistake I've made so I could "prove" this. – Bryant Oct 13 '19 at 00:16
  • @Bryant What is the textbook and page number? Maybe the assertion is that every pair of elements has a gcd iff every pair of elements has an lcm, which is true. – Bill Dubuque Oct 13 '19 at 00:36
  • @BillDubuque The problem is: Suppose R is an integral domain and a, b ∈ R \ {0}. Show that gcd(a, b) exists if and only if lcm(a, b) exists. It's on page 66, exercise 3.1.18 of this textbook https://alistairsavage.ca/mat3143/notes/MAT3143-Ring_theory.pdf – Bryant Oct 13 '19 at 09:09
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    @Bryant That claim is false, e.g. as here, in $,\Bbb Z[\sqrt{-3}],$ we have $,a,b = 2,,1!-!\sqrt{-3},$ have gcd $= 1$ but $,\gcd(ab',bb') = \gcd(2!+!2\sqrt{-3},,4)$ doesn't exist, so $,\ell := {\rm lcm}(a,b),$ doesn't exist ( by the equivalence in my first comment). More explicitly, if the lcm $\ell$ existed then

    $$2,b\mid 4,2b,\Rightarrow,\ell \mid 4,2b,\Rightarrow, \ell/2\mid 2,b,\Rightarrow, \ell/2 = 1,\Rightarrow, \ell = 2,\Rightarrow, b\mid 2,\Rightarrow,2\mid b',\Rightarrow!\Leftarrow$$

     – Bill Dubuque Oct 13 '19 at 14:47
  • @BillDubuque wow! Now I understand this. Thank you very much! – Bryant Oct 14 '19 at 09:20
  • @Joppy, you can define the LCM of $a,b$ such as the LCM being the element satisfying (1.) $a|\text{lcm}(a,b), b|\text{lcm}(a,b)$ and (2.) $a|c, b|c \Rightarrow \text{lcm}(a,b)|c$, for $a,b,c$ elements of the integral domain. Similarly, the GCD must satisfy (1.) $\gcd(a,b)|a, \gcd(a,b)|b$ and (2.) $c|a,c|b \Rightarrow c|\gcd(a,b)$. – Albert Aug 20 '21 at 18:00

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In an integral domain, if a|c and b|c then LCM(a,b)|c?

Yes, that follows by the definition of LCM (assuming it exists), namely

$$ a,b\mid c\color{#c00}\iff {\rm lcm}(a,b)\mid c$$

What if GCD(a,b)=1 and ac=bd, may I assume that a|d?

Yes if ${\rm lcm}(a,b)$ exists then $\gcd(a,b)$ exists and $\gcd(a,b){\rm lcm}(a,b) = ab$

In particular we have $\gcd(a,b)=1\Rightarrow {\rm lcm}(a,b) = ab\,$ hence

$$a\mid bd\iff a,b\mid bd\color{#c00}\iff \underbrace{{\rm lcm}(a,b)}_{\textstyle ab}\mid bd\iff a\mid d,\ \ {\rm when}\ \ b\neq 0$$

Bill Dubuque
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  • Awesome! Thank you very much!!! – Bryant Oct 12 '19 at 23:21
  • Pehaps you could help me with a problem I've been turning around for hours---and it's driving me crazy---: prove that for a commutative, integral domain $D$, for all $a,b\in D$, if a, b have both a lcm and a gcd, then $\text{lcm}(a,b)\gcd(a,b)$ and $ab$ are associates. The proof should be only based on the definition of integral domain as well as of lcm and gcd. – Albert Aug 20 '21 at 12:15