Showing that if $(a,b)=1$ and if $a\mid c$ and $b\mid c$ then $ab \mid c$, in GCD domains

Question

Is there a proof for the problem below?

$R$ is a commutative, integral domain with unity in which for each pair $a,b\in R$, g.c.d. $(a,b)$ exists. I want to show that if $(a,b)=1$ and if $a\mid c$ and $b\mid c$ then $ab \mid c$.

In a Bézout domain we can apply Bézout's identity as they do here.

Is there a way to proceed without using Bézout's identity?

Answer 1

As user26857's comment on Kishor J's answer makes clear, the important point is showing that for any $a,b,c$ we have $(ca,cb)=c(a,b)$.

Since we are in an integral domain, we have the following:

If $xy\mid xz$, then $y\mid z$

Now observe that $c(a,b)$ divides both $ca$ and $cb$. So $c(a,b)\mid (ca,cb)$. We want to show that the opposite holds as well.

First observe that we have $c\mid c(a,b)\mid (ca,cb)$. So we may write $(ca,cb)$ as $cg$ for some $g$. Now we have:

• $c(a,b)\mid cg$ - so $(a,b)\mid g$.

It remains to show that $g\mid (a,b)$. We note that:

• $gc=(ca,cb)\mid ca$ - so $g\mid a$
• $gc=(ca,cb)\mid cb$ - so $g\mid b$

Therefore, $g\mid (a,b)$, and so $g=(a,b)$ (up to multiplication by units). We get that $(ca,cb)=c(a,b)$.

Now we need only observe that if $(a,b)=1$ then $(ca,cb)=c$, and the result is clear, since $ab\mid ca,cb$ if $a,b\mid c$.

Answer 2

$ab \mid ca$ and $ab \mid cb$. Hence, $ab \mid (ca, cb)$. But $(ca, cb)$ is $c$ if $(a, b) = 1$. Therefore, $ab \mid c$.