Characterization of prime ideals in ring which is not necessarily commutative

Question

This is a question in Hungerford's Algebra book (question 14 in Section 2 (ideals) in chapter 3 (rings)).

If $P$ is an ideal in a not necessarily commutative ring $R$, then the following conditions are equivalent.
(a) $P$ is a prime ideal.
(b) If $(r)$ and $(s)$ are principal ideals of $R$ such that $(r)(s) \subseteq P$, then $r \in P$ or $s \in P$

$(a) \Rightarrow (b)$ is obvious form the definition. But I have no idea how to prove the other implicaiton.

Answer 1

Suppose b) holds but a) does not. Then there are ideals $A,B$ such that $AB\subseteq P$ but neither of $A$ or $B$ are contained in $P$. Pick $r\in A\setminus P$ and $s\in B\setminus P$.

Now, $(r)(s)\subseteq AB\subseteq P$... Can you see your way to the contradiction?

Therefore b) implies a).