In the ring of integers (or the monoid of natural numbers under multiplication), I believe that the following theorem holds:
Lemma Set $m$, $a$, $b$. If $m | ab$ then there exist $u$, $v$ such that
- $m = uv$
- $u | a$
- $v | b$
Note that there isn't a unique choice of $u, v$ (for example, consider $m = a = b = 2$).
A proof for natural numbers might go like this: Set $u = \mathrm{gcd}(m, a)$. Then $u | a$. Also $u | m$, which means we can set $v = m/u$. Then we know that $m = uv$. It remains to show that $v | b$.
This is the point where the proof gets a bit finicky. Since $m | ab$, it follows that $m/u$ divides $(ab)/u$. Write $\bar{a} = a/u$ (defined because $u | a$) so that $v | \bar{a}b$. If you then decompose $v$ into its prime factorisation, each prime must either divide $\bar{a}$ or $b$. Of course, none can divide $\bar{a}$ (because that would contradict the maximality of the GCD) so all must divide $b$.
Firstly: Do we really need to use decomposition into primes to prove this? It seems like rather "big guns" for what sounds like an easy claim.
Secondly: I could write down the claim in the lemma in any semigroup. Is there a nice description for the semigroups where it holds?