@darji, regarding your example, isn't it the case that if the characteristic of $K$ does not divide $|G|$, then $K[G]$ contains a copy of each irreducible representation, and that you need algebraic closure to ensure that every representation $V$ occurs $\dim V$ times and hence that $K[G]$ is isomorphic to the direct sum of the endomorphism rings? So if $K$ is not algebraically closed, you have no assurance that any of the endomorphism rings that $K$ is a sum of are more than $1$-dimensional?