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.
Yes, that follows by the definition of LCM (assuming it exists), namely
$$ a,b\mid c\color{#c00}\iff {\rm lcm}(a,b)\mid c$$
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
$$a\mid bd\iff a,b\mid bd\color{#c00}\iff \underbrace{{\rm lcm}(a,b)}_{\textstyle ab}\mid bd\iff a\mid d,\ \ {\rm when}\ \ b\neq 0$$