Morita equivalent about group algebra and representation ring

Question

Let $\mathbb{K}G$ be a group algebra and $R$ be the representation ring(Grothendieck Ring) of $\mathbb{K}G$, that means the product of $R$ is the tensor product of representations, add is the direct sum of representations. So are $\mathbb{K}G$ and $R$ Morita equivalent(morita equivalent for rings rather than tensor categories)? If true, I need to find a right $\mathbb{K}G$-module $M$, how do I to find it?

Answer 1

Well typically $R$ is just a ring, whereas $\mathbb{K}[G]$ is a $\mathbb{K}$-algebra. So they don't really have a chance of being Morita equivalent. We could look at $R \otimes \mathbb{K}$ though, so let's do that. For simplicity I will assume $\mathbb{K}$ is algebraically closed, but I don't think it changes much.

If $\mathbb{K} = \mathbb{C}$ (or some other algebraically closed field of characteristic zero) then $\mathbb{C}G$ and $R \otimes \mathbb{C}$ are both semisimple with the same number of Artin-Weddurburn blocks ($\#$conjugacy classes many), so they are Morita equivalent. Alternatively: A semisimple $\mathbb{K}$-algebra is Morita equivalent to its center, and character theory provides an isomorphism between $Z(\mathbb{C}[G])$ and $R \otimes \mathbb{C}$.

If $\mathbb{K}$ is algebraically closed of characteristic $p$ dividing $|G|$ then $\mathbb{K}G$ is not semisimple. Its semisimplification $\mathbb{K}G/J(\mathbb{K}G)$ has ($\#$ $p'$ conjugacy classes) blocks. $R \otimes \mathbb{K}$ on the other hand is a commutative algebra of dimension ($\#$ $p'$ conjugacy classes). If its semisimple then it can't be Morita equivalent to something non-semisimple, and if it isn't semisimple then its semisimplification $R \otimes \mathbb{K}/ J(R \otimes \mathbb{K})$ has strictly smaller dimension so it can't be equivalent to $\mathbb{K}G/J(\mathbb{K}G)$. Either way we see they are not Morita equivalent in this case.