I think I can prove the following using the compactness of first order logic and I am wondering what a purely algebraic proof would look like. 



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Of course, if $G$ is a finite group and $R=\mathbb ZG$, then this is true by Maschke's theorem. I realized this is true for finite monoids the other day. I believe the the above statement is true because the semisimplicity of $K\otimes R$ can be written as a statement in the first order theory of algebraically closed fields of characteristic 0 and hence what I want can be obtained from standard compactness arguments in the model theory of algebraically closed fields. 


My question is what does a purely algebraic proof look like?



Lefschetz Principle for semisimplicity
Over a perfect field, or one with characteristic larger than $\dim\_F A$, semisimple is the same as [separable](https://en.wikipedia.org/wiki/Separable_algebra). 
Over a field $F$ of characteristic larger than $\dim\_F A$, separable is the same as strongly separable. So, discarding finitely many characteristics, we can assume we care about strong separability. Choose $e\_1$, $e\_2$, ..., $e\_N$ in $R$ a basis for $R \otimes \mathbb{Q}$; discarding finitely many characteristics, we can assume $e\_i$ is a basis for the additive group of $R \otimes F$.


Let $M$ be the integer matrix $M\_{ij} = Tr(e\_i e\_j)$, where $Tr(a)$ means trace of the linear map $R \to R$ given by multiplication by $a$. Then $R \otimes F$ is strongly separable [if and only if](http://ncatlab.org/nlab/show/separable+algebra) $\det M\_{ij}$ is nonzero in $F$. In particular, this only depends on the characteristic of $F$ and, if it is true in characteristic zero, it is true in all but finitely many fields.


