TITLE: Where does this proof of $R^n\cong R^m\Rightarrow m=n$ break down for noncommutative rings?
QUESTION [5 upvotes]: It can be that $R\cong R^2$ for some ring as wikipedia says, something like the column finite ring. In this the author proved in lemma 1.1 something different. But I could not find any mistakes in the proof. What is happening?

REPLY [8 votes]: The difference is that maximal ideals in noncommutative rings do not always produce a field when you use them to make the quotient ring.
The proof given there relies entirely on the fact that $R/M$ is at least a division ring, so that ordinary linear algebra takes over.
In general, $R/M$ is merely a simple ring, and there is no corresponding dimensionality that you can use to make the same argument.