# Ideal generated by two elements that is not free

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?

## 1 Answer

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.

• Actually I am working with Lam's "Serres problem on projekctive modules" where he is presenting that example. But it is very interesting to see the original paper. I understand why $P$ can't be free if it is generated by two elements. Anyway I don't get why we can conclude that it has to be generated by two elemnts if we just consider the argumentation of Lam. I will try to post the link of the book later. Thanks a lot already!! – Heffalump Aug 3 '13 at 11:06
• @Heffalump if you give the page I might be able to see it in googlebooks – rschwieb Aug 4 '13 at 11:55
• It's on page 74 but I think this page does not appear on googlebooks. Lam says "Since $P\oplus R\cong R^2$, we know $J$ has two generators". Before he shows that the Ideal $J$ can be expressed by $(x+a)R\cap (y+b)R$ where the kernel of the homomorphisms (x+a,y+b): R^2\rightarrow R defines the stably free module $P$. – Heffalump Aug 5 '13 at 8:22
• so I understand if we show that $P$ can't be generated by one element that we are done. Anyway I would like to understand that other argumentation as it's seems to me that it is kind of a basic construction. In the paper ob Stafford which you posted (I think) they are doing the same thing: take an ideal that is generated by the intersection of to principal ideals, show that this implys that the ideal resp. module is staby free of type one and then show that it is generated by two elements... – Heffalump Aug 5 '13 at 8:31