For integral domain $R$, show that $a \mid c, b \mid c \implies ab \mid c$ when $(a, b)$ unit

Question

This kind of question has been asked before, but for a UFD. I want to show the result, more generally, for an integral domain.

Let $R$ be an integral domain (not necessarily a UFD).

Suppose $a, b \in R$ with $d \mid a,\;d \mid b \implies d$ unit.

Show that

$a \mid c,\; b \mid c \implies ab \mid c$

EDIT: Or find a counterexample. Turns out the result is not true, in general!

Answer 1

This is not true in all domains since it implies atoms (irreducibles) $\,p\,$ are prime, i.e.

$$p\mid ab,\ p\nmid a\,\Rightarrow\, (a,p)=1,\ \ {\rm so}\,\ \ a,p\mid ab\,\Rightarrow\, ap\mid ab\,\Rightarrow\, p\mid b$$

So any non-UFD number ring yields a counterexample, e.g. see here for that and more.

Answer 2

This is not true in general. The ring $\mathbb{Z}[\sqrt{-5}]$ is not an UFD, and for example you have two factorizations of $21=3\times 7=(4+\sqrt{-5})\times(4-\sqrt{-5})$. Now put $a=3$, $b=4+\sqrt{-5}$ and $c=21$.