# Rank of Free Module over a Noncommutative Ring

There's a fairly standard proof that the rank of a free module $F$ over a commutative ring $R$ is well defined. We take a maximal ideal $I$ and note that $R/I$ is a field. Taking $R/I \otimes_R F$ gives a vector space of dimension rank of $F$, which gives the result.

I was wondering where the proof breaks down in the non-commutative case (let's assume we have a unit). According to the wikipedia article on Division Rings, every module over a division ring is free with well-defined rank, and I don't see any issue with taking a maximal ideal (do we need to be careful with selecting maximal left/right/two-sided ideals?) or with taking the tensor product. If someone could point out what's wrong that would be greatly appreciated.

• Well, there are tons of noncommutative rings without any nonzero bilateral ideal at all, maximal or not —and onesided ones will do you no good, as then $R/I$ is not even a ring. Oct 10, 2016 at 17:21
• Let's suppose we take our ring to be unital, then we always have a maximal ideal. There are examples of unital non-commutative rings where the rank of a free module is not well defined. I edited the question to reflect this.
– user353491
Oct 10, 2016 at 18:11
• Only if you included zero ideals... There are rings which are not division rings and which have no nonzero proper ideals, and with those your argument is useless. Oct 10, 2016 at 19:46
• I see, could you give me an example of such a ring? Or perhaps point me in the direction of one?
– user353491
Oct 10, 2016 at 23:57
• The Weyl algebra is a simple ring, for example. See en.wikipedia.org/wiki/Weyl_algebra or many other better sources —google gives me web.maths.unsw.edu.au/~danielch/thesis/dcock.pdf which looks nice. Oct 11, 2016 at 0:07

The flaw is that if $I$ is a maximal ideal in a noncommutative ring, $R/I$ need not be a division ring (here "ideal" should mean two-sided ideal, since otherwise $R/I$ won't even be a ring at all). The usual proof for commutative rings is that if $r\in R$ is not a unit, then the ideal generated by $r$ is a proper ideal (since it just consists of the multiples of $r$ and $1$ is not a multiple of $r$), and thus if a ring has no nonzero proper ideals it is a field. But for noncommutative rings, the (two-sided) ideal generated by a single element $r$ is much more complicated: it is the set of all sums of elements of the form $arb$ for $a,b\in R$. Note that for instance, there is no way to write a sum like $arb+crd$ as a single two-sided multiple of $r$ in general. So even if $1$ is not a multiple of $r$, $1$ might be in the ideal generated by $r$.