# Noncommutative failure of IBN: $\,R^n\cong R^m \Rightarrow n - m,$ ring $R$ with maximal ideal $I$ such that $R/I$ is not a division ring?

We're all aware of the theorem that says for a commutative ring $R$ with $1$, $I$ is maximal iff $R/I$ is a field. I'm currently working on a problem where I was asked to show commutative rings have IBN, and I used a lemma saying if $\psi:R\to T$ was a ring homomorphism and $R$ did not have IBN, then $T$ does not have IBN. I used this by sending $R$ to $R/I$, where $I$ is maximal, so $R/I$ is a field and must have IBN, so $R$ has IBN.

This made me wonder why this wouldn't work for any ring $R$. If I could just choose any maximal left ideal $I$ then send it to $R/I$, if $R/I$ is a division ring then I must have $R$ having IBN for the same reason. Therefore there must be a case where $I$ is a maximal ideal but $R/I$ is not a division ring. Does anybody know any examples? Is a "maximal ideal" even a well defined notion if $R$ is not commutative?

• When you choose a maximal left ideal $I$, $R/I$ does not have the structure of a ring in general (it has the structure of a simple module over $R$ though). If you choose a maximal two-sided ideal $I$, then you'll get a simple ring, which need not necessarily be a division ring. Nov 3 '15 at 4:55
• @DustanLevenstein thanks; apparently I need to read into this more if I didn't realize this. Nov 3 '15 at 5:03

If I could just choose any maximal left ideal $$I$$ then send it to $$R/I$$, if $$R/I$$ is a division ring then I must have $$R$$ having IBN for the same reason. Therefore there must be a case where $$I$$ is a maximal ideal but $$R/I$$ is not a division ring.

Well yes, sort of. Part of the problem is that your idea mutated toward the end of your reasoning (or maybe just a momentary lapse while you were writing.)

You probably should have said this instead:

there must be a case where $$I$$ is a maximal left ideal but $$R/I$$ is not a division ring.

As already commented, $$R/I$$ cannot have a ring structure unless $$I$$ is a two-sided ideal. If $$I$$ is a maximal left ideal, the result of the quotient is just a simple left module (one with only trivial submodules.)

Now, for your actual text, it is true that there is a ring $$R$$ with a maximal two-sided ideal $$I$$ such that $$R/I$$ isn't a division ring. If you take any field $$F$$ and make the matrix ring $$R=M_n(F)$$ for any $$n> 1$$, then the zero ideal is the only maximal two-sided ideal, the quotient by the zero ideal is isomorphic to $$R$$, and clearly $$R$$ isn't a division ring.

Is a "maximal ideal" even a well defined notion if $$R$$ is not commutative?

If $$I\lhd R$$ is a proper ideal, then $$I$$ is said to be maximal if there is no other ideal properly between $$I$$ and $$R$$. Nothing about this refers to commutativity, and so it's perfectly well-defined for all rings, even ones without identity.

## Is $$R/I$$ not a division ring really the end of the approach?

You observed that not getting a division ring is the first place where the usual commutative proof stops applying. Does that spell the end for the argument? The issue becomes whether or not this can be patched up by somehow showing simple rings have IBN.

The answer is, however, no, there exist simple rings which do not have the IBN. Nothing about simple rings has the power that dimensionality provided for division rings. That means the approach has broken down completely.

Backing up, though, one can see that the reason division rings worked was that they too had the IBN. So easily you can see that the commutative proof is a corollary of another proposition:

If there exists a proper ideal such that $$R/I$$ has IBN, then $$R$$ has IBN.

• Great response, thanks Nov 3 '15 at 22:53