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 $ST$).


2 Answers 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...)

  • $\begingroup$ That proof is actually identical to @Bill Dubuque's one. Now I see another proof based on $(ct,st)\mid(c,s)t \land (c,s)t\mid(ct,st)$… $\endgroup$
    – beroal
    May 11, 2011 at 17:04
  • $\begingroup$ @beroal Probably your "other" proof of the GCD distributive law is equivalent to the proof I gave above, except that you have torn it apart into two separate cases, one for each arrow direction. But that goes against the whole point of the proof - to unify both directions using the universal $\iff$ definition of the GCD, $\rm\ a\ |\ b,c\ \iff\ a\ |\ (b,c)\:.\:$ If you master this technique, all these proofs become quite trivial. See the links I gave. $\endgroup$ May 11, 2011 at 17:20
  • 1
    $\begingroup$ Note The proof in Pete's notes, excerpted from Haible: Gauss's Lemma without primes, 1990 is essentially the same as the OP's linked Wikipedia proof (attributed to Richman 1988). It's hinted exercise 1-6-8 in Kaplansky's Commutative Rings 1970, but is surely much older than that. It deserves to be better known that it can be proved more simply and more conceptually in the same way as the UFD case. I'll post this later when I have some spare time. $\endgroup$ May 11, 2011 at 18:17

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..

  • 1
    $\begingroup$ Can you please prove $(ct,st)=(c,s)t$? I see why $(c,s)t$ is a common divisor of $ct,st$, but why it is the greatest common divisor? $\endgroup$
    – beroal
    May 10, 2011 at 13:09
  • $\begingroup$ @beroal: this follows directly from the definition of 'ideal generated by', as $xct+yst=(xc+ys)t$. $\endgroup$ May 10, 2011 at 16:51
  • 1
    $\begingroup$ @wil No, GCD domains generally are not Bezout, i,e. finitely generated ideals needn't be principal. $\endgroup$ May 10, 2011 at 18:00
  • $\begingroup$ It turned out quite simple. Thanks. $\endgroup$
    – beroal
    May 11, 2011 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.