 Solution

A left-primitive ideal is a two-sided ideal of the formfor a nonzero simple module $V$ on the left. Equivalently, the quotient ring $R/P$ has a faithful module that is a simple module, and is therefore a primitive ring.

To prove that $P$ is a prime ideal of a noncommutative ring, let $A,B$ be two-sided ideals with $AB\subseteq P$. Since $B$ is two-sided, $BV$ is a submodule of $V$: for $r\in R$, $b\in B$ and $v\in V$, one has $r(bv)=(rb)v\in BV$. If $B⊈P$, then $BV\ne 0$, so simplicity gives $BV=V$. Consequentlyand $A\subseteq P$. Thus $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$. Also $P\ne R$, since $1$ acts as the identity element on the nonzero module $V$. These are precisely the required prime ideal of a noncommutative ring conditions.