Let $R$ be a ring with unity, not necessarily commutative. Show that if $M$ is a maximal two-sided ideal in $R$, then $M$ is prime.
I know several proofs in the commutative case, but I can not come up with one in the general case.
Let $I,J$ be two ideals such that $IJ\subset M$. Assume $I\not\subset M$. Then, there is $a\in I$ such that $R=M+(a)$ as two-sided ideals because $M$ is maximal. In the commutative case it is easy to proceed from here, but here I just get that $1=m+ras$ for some $m\in M, r,s\in R$ and multiplying with an element $b\in J$ doesn't give anything?