Is there a finite dimensional Banach algebra $A$ for which $K\_{0}(A)$ is a finite group?


I asked this question in MSE but I received no answer


<https://math.stackexchange.com/questions/1624250/k-theory-of-finite-dimenional-banach-algebras>



$K$-Theory of finite dimensional Banach algebras
The definition of K$\_0 (A)$ (where $A$ is a finite dimensional Banach algebra), that it be the kernel of $K\_0(\tilde A) \to K\_0(C) \cong Z$ leads immediately to it being a free abelian group, possibly zero (free on no generators); the latter occurs iff $A$ is nilpotent; $\tilde A$ denotes the unitification. 


Since $\tilde A$ is a finite dimensional unital algebra, $\overline{A}:= \tilde A/rad(\tilde A)$ is finite-dimensional semisimple, hence $K\_0(\overline A) $ is free finitely generated; since the radical of a finite dimensional algebra is nilpotent, $K\_0(\tilde A) \cong K\_0(\overline A) \cong Z^d$ for some $d$. 


As $K\_0(\tilde A) \to K\_0(C) = Z$ is onto, $K\_0(A)$ is free on $d-1$ generators. So $K\_0(A)$ is either zero or torsion free. If $A$ is nilpotent, then $\tilde A$ has a unique maximal ideal, and then $d=1$. Conversely, if $d=1$, then $\overline A = C$, forcing $\tilde A$ again to have a unique maximal (two-sided) ideal. For finite dimensional algebras (over any field), $\tilde A$ having unique maximal ideal is equivalent to $A$ being nilpotent (otherwise, $A$ would have a nontrivial idempotent, etc). 


We only need that $A$ be a finite dimensional algebra over some field for this to work, not a Banach algebra. The outcome is that $K\_0(A)$ is finite iff it is zero iff $A$ is nilpotent. 


