# If $\gcd(a,b)=d$, then $\gcd(ac,bc)=cd$?

$A$ 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.

Here is the best that one can say for arbitrary integral domains:

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

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

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

EUCLID'S LEMMA $$\rm\quad a\ |\ bc\$$ and $$\rm\ (a,b)=1\ \Rightarrow\ a\ |\ c\quad$$ if $$\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 the Euclid Lemma $$\Rightarrow\:$$, namely if $$\rm\ a\ |\ bc\$$ and $$\rm\ (a,b) = 1\$$ but $$\rm\ a\nmid c\:$$, then one immediately deduces 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 that the 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 my post here. There you will find 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 my post here for the general universal definitions of $$\rm GCD,\: LCM$$ and for further remarks on how such $$\iff$$ definitions enable slick proofs, and see here for another simple example of such.

• I don't see how this helps. – ashpool Jan 1 '11 at 17:23
• @ashpool: I revised it to be more explicit (and more general). – Bill Dubuque Jan 1 '11 at 17:33
• Thanks. I was aware of your lemma, and my problem is basically showing the existence of (ac,bc). – ashpool Jan 1 '11 at 17:40
• Why do you think that such a gcd should exist in an arbitrary domain? – Bill Dubuque Jan 1 '11 at 17:44
• @athos Now updated to a local link, thanks. – Bill Dubuque Aug 7 '19 at 12:33