 Solution

Let $P=\mathrm{Ann}(m)$ be maximal among annihilators of nonzero elements of $M$. It is a proper ideal, since $1m=m\ne 0$. Suppose $ab\in P$ and $b\notin P$. Then $bm\ne 0$ andThe inclusion holds because every $r\in P$ satisfies $r(bm)=b(rm)=0$. Maximality in the stated set forces $\mathrm{Ann}(bm)=P$, so $a\in P$. This proves the defining implication for a prime ideal:This is a maximal annihilator of a module element is prime argument. Maximality among element annihilators is sufficient; $P$ need not be a maximal ideal of $R$.