If $P$ is an ideal in a ring $R$ such that $P \neq R$ and (1) for all $a,b \in R$ with $ab \in P$ implies $a \in P$ or $b \in P$, then $P$ is prime. Conversely if $P$ is prime and R is commutative, then P satisfies condition (1)
The forward implication was easy to prove. However, I am having trouble with the converse, and Hungerford's explanation doesn't seem to be helping. Here is a quote:
"If $R$ is commutative, this implies that $(a)(b) \subseteq (ab)$, whence $(a)(b) \subseteq P$. If $P$ is prime, then either $(a) \subseteq P$ or $(b) \subseteq P$, whence $a \in P$ or $b \in P$.
Note that $(a) = \{ra + na \mid r \in R, n \in \Bbb{Z} \}$ is the ideal generated by $a$. So I agree that $(a)(b) \subseteq (ab)$, and therefore $(a)(b) \subseteq P$. I also agree that this implies $(a) \subseteq P$ or $(b) \subseteq P$, since $P$ is a prime ideal and $(a)$ and $(b)$ are ideals. But I don't see how we can conclude from this that either $a \in P$ or $b \in P$. Specifically, I don't see how $a \in (a)$ and $b \in (b)$ can be true if $R$ isn't assumed to be unital.
Since (a) is the ideal generated by a, then a∈(a) by definition, because (a) is the intersection of all ideals containing a.