Prime ideal in a noncommutative ring

Remarks

Equivalently, $P \ne R$ and, for all $a, b \in R$, $aRb \subseteq P$ implies $a \in P$ or $b \in P$; and $P$ is prime if and only if $R/P$ is a prime ring. Example: the zero ideal of $M_2(\Q)$ is prime, since the only two-sided ideals are $0$ and $M_2(\Q)$, and $M_2(\Q) \cdot M_2(\Q) \ne 0$. Non-example: the zero ideal of $\Q \times \Q$ fails clause 2 with $A = \Q \times 0$ and $B = 0 \times \Q$, since $AB = 0$. A completely prime ideal is one with $ab \in P \Rightarrow a \in P$ or $b \in P$; it is prime, but the zero ideal of $M_2(\Q)$ shows the converse fails. For a commutative ring both notions agree with the usual prime ideal.