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If $R$ is a ring with 1 which satisfies $R^r=R^s$ for some $r\neq s$.

Are there any explicit calculation of $K_0(R)$ for such $R$?

I want to know such examples because I think that such $R$ may not have $\mathbb{Z}$-summands.

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    @Rasmus: the free module $R\oplus\cdots\oplus R$ with $r$ summands (I don't say *of rank $r$* because in the situation of the question the rank is not well difined :) )2011-03-30

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For each $n\geq2$, the Leavitt algebra $V_{1,n}$ is a universal example of an algebra $A$ such that $A\cong A^n$ as modules, but $K_0(V_{1,n})\cong\mathbb Z/(n-1)\mathbb Z$. This was proved by Pere Ara, Ken Goodearl and E. Pardo; see http://arXiv.org/abs/math/0111066v1

On the other hand, if $U$ is an infinite dimensional vector space over a field and $B=\operatorname{End}(V)$ is its endomorphism algebra, then $K_0(B)=0$ and $B\cong B^2$. This is done in Rosenberg's book on Algebraic K-theory, if I recall correctly.