Let $A$ and $B$ be ideals in a principal ideal domain $R$, where $A \ne R$ and $B \ne R$. Show that the ideal $AB$ is prime iff $A = \{0_R\}$ or $B = \{0_R\}$.
I know that since $R$ is PID then $A$ and $B$ are generated by one element. But, how to start the proof?
Any ideas ?
If $A$ and $B$ are ideals in a PID $R$, then let $A = (a), B = (b)$. Then $AB = (ab)$ (this needs to be proven, but is not so hard in a PID). Clearly if $A = (0)$, then $a = 0$, so $AB = (0b) = (0)$ is prime and the symmetric argument holds for $B$.
On the other hand, suppose that $AB$ is prime. We know that that product $a\cdot b \in (ab)$, so by the primeness of $AB$ either $a \in AB$ or $b \in AB$. WLOG, $a \in AB$. This means there exists $r \in R$ such that $a = rab$. If $a = 0$ then we're done. If not, since $R$ is a PID, and hence a domain, this implies $rb = 1$, so $b$ is a unit, but this contradicts our assumption that $B \neq R$.