TITLE: Ideal generated by two elements that is not free
QUESTION [5 upvotes]: I am having problems understanding an example constructed by M. Ojanguren and R. Sridharan showing that over the polynomial ring in two variables over a division ring (which is not a field) there exists a stably free module which is not free. 
Let $k$ be a division ring that is not a field and $R=k[x,y]$. It can be easily shown that there exists a stably free $R$-module $P$ for which $P\oplus R\cong R^2$. Further we can show that there is a (right) ideal $J$ which is isomorphic to $P$ generated by the intersection of two principal ideals of $R$.
My question is why we can conclude from the above that $J$ must be generated by two elements and why this means that $J$, resp. $P$ is not free.
Can anybody help me with this?

REPLY [2 votes]: It appears you are reading this paper, or else something else very much like it: Cancellation of Azumaya algebras. I can't follow the last part where they show it isn't free, but I think it can be proven in the way I describe below.
We keep in mind that $R$ is a Noetherian ring. In particular it has the IBN property and $P$ is finitely generated.
The authors appear to give a detailed explanation of why $P$ is not generated by a single element, so I trust you believe it's generated by more than one element.
Suppose $P$ were free and thus isomorphic to $R^n$. Since then $P\oplus R\cong R^{n+1}\cong R^2$, we're forced to conclude that $n=1$ since $R$ has the IBN property. But then $P$ is generated by a single element, a contradiction. Consequently, $P$ isn't free.
By the way, while looking for this, I had the opportunity to skim what looks like a very nice paper by J.T. Stafford: Stably free, projective right ideals (1985), in case you are interested.