Question

Find an Dedekind finite ring that is not IC (other than examples 5.10 & 5.12 given in K.R. Goodearl - Von Neumann Regular Rings)

Answer 1

Solution:

Proof.

Let D be any noncommutative division ring (e.g., Hamilton's ring of real quaternions), and let $R=\mathbb{M}_{n}(D[x])$ , where $n \geqslant 2$ . Since D[x] is a PRID (principal right ideal ring), all modules in $\mathcal{P}(D[x])$ are free. This implies that D[x] is stably IC, so R is IC. Since

$\mathbb{M}_{r}(R)=\mathbb{M}_{r}\left(\mathbb{M}_{n}(D[x])\right) \cong \mathbb{M}_{r n}(D[x])$

it follows that R is in fact stably IC. But

$R[y]=\mathbb{M}_{n}(D[x])[y] \cong \mathbb{M}_{n}(D[x, y])$ ...(i)

and by a result of Ojanguren and Sridharan Lemma, the polynomial ring S:=D[x, y] has a non-principal right ideal P such that $P \oplus S \cong S \oplus S\ in\ \mathcal{P}(S)$ . This implies that $S_{S}^{2}$ is not internally cancellable. Since $n \geqslant 2$ , the isomorphism in (i) implies that the ring R[y] is not IC. (Of course, for n=1, R[y]=D[x, y] would have been IC, since it is a domain.)

Thus, There exists a stably IC ring R such that the polynomial ring R[y] is not IC.

Proved.