$\gcd(ac,bc)=\gcd(a,b)c$ in an integral domain

Question

$A$ an integral domain, $a,b,c\in A$. If $d$ is a greatest common divisor of $a$ and $b$, is it true that $cd$ is a greatest common divisor of $ca$ and $cb$? I know it is true if $A$ is a UFD, but can't think of a counterexample in general situation.

Answer 1

In a general integral domain the best we can say is

Lemma $\rm\ \ (a,b) = (ac,bc)/c\ \,$ if $\rm\,\ (ac,bc)\, $ exists.

Proof $\rm\ \ d\ |\ a,b \iff dc\ |\ ac,bc \iff dc\ |\ (ac,bc) \iff d\ |\ (ac,bc)/c$

See here for a few other proofs of this fundamental GCD Disributive Law.

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Generally $\rm\ (ac,bc)\ $ need not exist, as is most insightfully viewed as failure of

Euclid's Lemma $\rm\ \ a\ |\ bc\ $ and $\rm\ (a,b)\!=\!1\ \Rightarrow\ a\ |\ c\ \ $ if $\rm\ (ac,bc)\ $ exists.

Proof $\ \ $ If $\rm\ (ac,bc)\ $ exists then $\rm\ a\ |\ ac,bc\ \Rightarrow\ a\ |\ (ac,bc) = (a,b)\,\!c = c\ $ by the Lemma.

Therefore if $\rm\, a,b,c\, $ fail to satisfy Euclid's Lemma, namely if $\rm\ a\ |\ bc\ $ and $\rm\ (a,b) = 1\ $ but $\rm\ a\nmid c,\,$ then we immediately infer that the gcd $\rm\ (ac,bc)\ $ fails to exist. For the special case $\rm\,a\,$ is an atom (i.e. irreducible), the implication reduces to: atom $\Rightarrow$ prime. So it suffices to find a nonprime atom in order to exhibit a pair of elements whose gcd fails to exist. This task is a bit simpler, e.g. for $\rm\ \omega = 1 + \sqrt{-5}\ \in\ \mathbb Z[\sqrt{-5}]\ $ we have: $ $ atom $\rm\, 2\ |\ \omega'\omega = 6,\,$ but $\rm\ 2\nmid \omega',\omega,\,$ so $\rm\,2\,$ is not prime. Therefore we deduce that the gcd $\rm\, (2\,\!\omega,\, \omega'\omega) = (2+2\sqrt{-5},\,6)\ $ fails to exist in $\rm\, \mathbb Z[\sqrt{-5}]$.

Note that if the gcd $\rm\, (ac,bc)\,$ fails to exist then this implies that the ideal $\rm\, (ac,bc)\,$ is not principal. Therefore we've constructively deduced that the failure of Euclid's lemma immediately yields both a nonexistent gcd and a nonprincipal ideal.

That the $\Rightarrow$ in Euclid's lemma implies that Atoms are Prime $\rm(:= AP)$ is denoted $\rm\ D\Rightarrow AP\, $ in the list of domains closely related to GCD domains in this answer, which has links to further literature on domains closely related to GCD domains. See especially the referenced comprehensive survey by D.D. Anderson: GCD domains, Gauss' lemma, and contents of polynomials, 2000.

See also this answer for the general universal definitions of $\rm GCD,\, LCM$ and for further remarks on how such $\iff$ definitions enable concise proofs (another simple example is here).