Is Gauss's lemma valid for polynomials with coefficients in a GCD domain?

Question

Wikipedia's proof of Gauss's lemma requires this theorem:

If $(C \mid S\cdot T) \land \lnot \operatorname{invertible}(C)$, $C$ has a non-invertible divisor in common with at least one of $S$ and $T$.

I can prove it for a Bézout domain, but not a GCD domain. Can anybody help me?

Original text:

If the contents $c = c(ST)$ is not invertible, it has a non-trivial divisor in common with the leading coefficient of at least one of $S$ and $T$ (since it divides their product, which is the leading coefficient of $ST$).

Answer 1

HINT $\ $ Simply apply Euclid's Lemma, i.e. $\rm\ (c,s)= 1,\ c\ |\ st\ \Rightarrow\ c\ |\ (ct,st) = (c,s)\:t = t\:.$

Alternatively if $\rm\ (c,s) = 1 = (c,t)\:$ then $\rm\ c\:|\:st\ \Rightarrow\ c\ |\ (cc,cs,ct,st) = (c,s)(c,t) = 1\:.$

Per request, here's a simple proof of the GCD distributive law $\rm\ (a,b)\:c = (ac,bc)\:.$

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

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

The above proof uses the universal definitions of GCD, LCM, which often serve to simplify proofs, e.g. see a similar proof of the GCD * LCM law..

Answer 2

FYI: a proof of Gauss's Lemma for GCD domains is given in Section 15.5 of these notes. (I don't claim any superiority to any other proofs you may have seen or discovered...)