Does Mashke's theorem hold for semisimple rings?

Question

I saw a proof of Mashke's theorem using the theory of modules. However, it seems like it works in much more generality than it was stated in the text; can anyone confirm if it does, or if there's something I'm missing?

Theorem: If $R$ is a semisimple ring and $G$ is a finite group such that $|G|$ is invertible in $R,$ then $R[G]$ is semisimple.

Proof: It is equivalent to prove that every short exact sequence of $R[G]$ modules splits. Let $0 \to N \to M\to M/N\to 0$ be such a short exact sequence. Since $R$ is semisimple, there is a map $s : M/N \to M$ which is $R$-linear and splits the sequence, e.g. $\pi(s(x)) = x$ for $\pi : M\to M/N$ the projection.

Then the map $S : M/N \to M$ given by $$S(x) = \frac{1}{|G|}\sum_{g\in G} g \cdot s(g^{-1}x)$$

is $R[G]$-linear, since its clear $R$-linear and for every $h \in G,$ we have $$S(hx) = \frac{1}{|G|}\sum_{g\in G} hg \cdot s((g^{-1}h^{-1})hx) = hS(x).$$

Also, clearly $\pi(S(x)) = x,$ and so $S$ splits the short exact sequence in the category of $R[G]$-modules, proving that $R[G]$ is semisimple.

Answer 1

Yes, it is given so in Lam's First course in noncommutative rings page 80, for example:

Let $k$ be any ring and $G$ be a finite group. Then $R=kG$ is semisimple iff $k$ is semisimple and $|G|\cdot 1$ is a unit in $k$.