Quotients of non-commutative rings.

Question

All of the ring theory I know comes from a commutative perspective. I'm learning some non-commutative algebra right now, and I'm wondering whether the following theorem holds.

Let $R$ be a commutative ring, $I$ an ideal. Let $R/I \cong K$. Then $K$ is a field if and only if $I$ is a maximal ideal.

I'm wondering if, say $A$ is a non-commutative ring, and $I$ a left ideal, does the theorem still hold? Or does the theorem only work in the commutative sense? Is there an analogous way to show an ideal is maximal in the non-commutative sense, if the theorem doesn't hold?

Answer 1

To go another direction with it, it is true that for a left ideal $I$, $R/I$ is a simple left $R$ module if and only if $I$ is a maximal left ideal. This generalizes the commutative case.

But usually when people talk about this they are interested in $I$ being an ideal and $R/I$ being a ring, not just a module. As mentioned already, $I$ will need to be an ideal for $R/I$ to have a ring structure. And then, the question of whether or not $R/I$ has to be a division ring has been well covered already:

Prove $R/M$ is a division ring for a non-commutative ring $R$ with max ideal $M$

What can we say about a non-commutative ring modulo a max ideal?

Answer 2

As Brian says, if $R$ is a noncommutative ring you need $I$ to be a two-sided ideal for $R/I$ to inherit a ring structure. Now the statement is that $I$ is maximal as a two-sided ideal iff $R/I$ is a simple ring.