Non-trivial stably free, projective right ideals I am working on _Stably free, projective right ideals_ (1985) by J.T. Stafford. Trying to get through Theorem 1.2 on page 66 [4]. From the beginning of the paper we are supposed to know: taking two regular non commutative elements $a, b \in S$, then $S=aS+bS$ (would be nice to know why). In case of the theorem we choose $a=r$ and $b=x+s$. $K$ is supposed to be $aS \cap bS$. From the first exact sequence of the theorem's proof: $0 \rightarrow K \rightarrow aS \oplus bS \rightarrow aS+bS \rightarrow 0$ follows that $K \oplus S \cong S \oplus S$. Why is that the case? Since we know, that $S=aS+bS$ and S is free (hence projective, hence sequence is splitting), we see that: $K \oplus S \cong aS \oplus bS$. So the crux of the question would be to know why: $aS \oplus bS=S \oplus S$?

Since $a$ is regular, the map $r\mapsto ar$ is an isomorphism. The same applies to b too.

The author isn't claiming that all such noncommuting a and b have that property, he is just saying such a pair can be found.