TITLE: $M$ maximal in a ring $R$, what is $R/M$?
QUESTION [5 upvotes]: I just proved that if $R$ is a commutative ring with unity, then $M$ maximal ideal implies that $R/M$ is a field, and the converse is also true.
I have the following questions: 

If $R$ is a ring with unity, would $R/M$ be a division ring? Is the converse true?
If $R$ is a ring (possibly without unity), then what is $R/M$?
If $R/M$ is a field, what we can say about $R$?

REPLY [2 votes]: If $R$ is not commutative, but with identity, it's not true in general that, for $M$ a maximal (two-sided) ideal, $R/M$ is a division ring.
The classical example is $R=M_2(F)$, the ring of $2\times2$ matrices over a field $F$. The zero ideal is maximal (prove it), but $R$ is not a division ring.
What you can say is that $R/M$ is a simple ring, that is, $R/M$ has no non trivial two sided ideals.
In the case when $R$ has no identity, maximal ideals need not exist. The quotient $R/M$, assuming $M$ is a maximal ideal, need not be a simple ring: if $R$ is a zero ring and $M$ is a maximal ideal, the ring $R/M$ has no non trivial ideals, but $R/M$ is still a zero ring, which is not simple.
Of course, if $R/M$ happens to be a field, then $M$ is a maximal ideal, by the correspondence theorem, because a field has no non trivial ideals.