Prime and Maximal Ideals in Rings Q. We know that if $R$ is commutative ring with unity, then 
(*)  $M$ is maximal ideal $\Longleftrightarrow$ $R/M$ is field.
(**)  $M$ is prime ideal $\Longleftrightarrow$ $R/M$ is integral domain.
Suppose we drop one of these two conditions - "$R$ is commutative" or "$R$ contains unity", or both the conditions, does any implication in statement (*) and (**) remains true? 
 A: If $R$ is a not necessarily commutative ring with unity then:
$M$ is a maximal ideal $\iff R/M$ is a simple ring, i.e. it has no non-zero two-sided ideals.
There are many examples of noncommutative rings that are simple but not division rings (a division ring is a noncommutative ring in which every element has a two-sided inverse, so the noncommutative analogue of a field) - $M_2(\mathbb{C})$, as pointed out above, is one example.
For prime ideals, the definition of "prime" is generally slightly different for noncommutative rings: $P$ is prime iff for any two (left / right / two-sided) ideals $A$ and $B$ then $AB \subset P$ implies $A \subset P$ or $B \subset P$. This contrasts with the usual commutative algebra definition that $P$ is prime iff for any two elements $a$ and $b$ then $ab \in P$ implies $a \in P$ or $b \in P$. The latter definition is usually called "completely prime" (and does imply prime, but not vice versa; however if $R$ is commutative then the two definitions of "prime" do coincide).
With that in mind, $M$ is completely prime iff $R/M$ has no zero-divisors, i.e. is a domain (I would say the term "integral domain" implied commutativity, and that "domain" is the more general term, but that is debatable). But there are prime ideals $M$ such that $R/M$ does have zero-divisors. As Konstantin points out below, the $0$ ideal in $M_2(\mathbb{C})$ is again an example.
For commutative rings without unity, $4\mathbb{Z}$ is a maximal ideal in $2\mathbb{Z}$, but $2\mathbb{Z}/4\mathbb{Z}$ is not a field (as it cannot be, since it doesn't have a $1$, so how can elements have multiplicative inverses?).
However "$M$ completely prime $\iff R/M$ has no zero divisors" does still hold.
