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If a ring $R$ satisfies the property $P$, but the matrix rings $M_n(R)$ do not satisfy $P$, then $P$ is not a Morita property; that is the definition.

But that is not fix this situation. I want to know the reason that Invariant Basis Number (IBN) is not a Morita property. (IBN means that $R^m=R^n$ implies $m=n$).

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In other words, you want an example of a ring $R$ that has IBN, but with a matrix ring $M_n(R)$ that does not have IBN? (What does "does not fix this situation" mean?) – Arturo Magidin Apr 5 '11 at 18:26
up vote 5 down vote accepted

Lam's Lectures on Rings and Modules (GTM 189, Springer-Verlag 1999) contains in the exercises an example attributed to George Bergman that shows that IBN is not a Morita invariant property (exercise 11, page 502).

(Lam also mentions earlier in the book that Cohn's book, Free Rings and their Relations, contained an exercise asking the reader to prove that IBN was a Morita invariant property...)

Anyway, here's the sketch: Given a ring $R$, let $\mathcal{P}(R)$ be the monoid of isomorphism classes of finitely generated right $R$-modules under the direct sum operation.

It can be shown (using coproducts) that there is a ring $R$ for which $\mathcal{P}(R)$ is generated as a monoid by $[R]$ and $[M],[N]$, together with the defining relations $[M]+[N]=[R]=[R]+[R]$.

Assuming this holds, $S = \mathrm{End}_R(M\oplus N)$ is Morita equivalent to $R$ and has IBN, but $R$ does not (since $R\cong R\oplus R$). To show that $S$ has IBN, the suggestion is to use the following exercise:

Let $S=\mathrm{End}(P_R)$, where $P_R$ is a progenerator over the ring $R$; then $S$ has IBN if and only if $P^n\cong P^m$ as right $R$-modules implies $n=m$.

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Yes,I just read this book,but the question is not self-contain,suppose a claim without proof,maybe that is really complicative. – Strongart May 2 '11 at 7:24
@Strongart: I'm having a bit of trouble parsing your comment. Are you saying that the answer is not self-contained (given the reference to to a paper by George Bergman), that the existence of the ring $R$ with the given property is claimed but not proven, and that it is possibly very complicated to give the explicit example? – Arturo Magidin May 2 '11 at 15:32
Yes,I hope maybe there is another – Strongart May 4 '11 at 5:56
Yes,I can not find the paper and hope maybe there are some other simple way,so I ask the question because that is really amazing. – Strongart May 4 '11 at 6:03

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