= Solution
{c}
A <left-primitive ideal> is a two-sided <ideal> of the form
$$
P=\operatorname{Ann}_R(V)=\{r\in R:rV=0\}
$$
for 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 $B\not\subseteq P$, then $BV\ne0$, so simplicity gives $BV=V$. Consequently
$$
AV=A(BV)=(AB)V=0,
$$
and $A\subseteq P$. 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.
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