Is it true that $(R\times S)[G]\cong R[G]\times S[G]$?

Question

I know for two groups $G, H$ (not necessarily finite) we have $R[G\times H]\cong (R[G])[H]$, but I was wondering if we had a similar statement for rings $R,\,S$. In other words, if $R,\,S$ are two (possibly noncommutative rings), is it true that $(R\times S)[G]\cong R[G]\times S[G]$?

Answer 1

Denote $T=R\times S$. Let $e=(1_R,0)1_G$ and $f=(0,1_S)1_G$ so that $(e+f)\cdot 1_G=1_{T[G]}$ .

Then $e$ and $f$ are central idempotents and $eT[G]=R[G]$, $fT[G]=S[G]$ and $T[G]$ is the direct product of the two.

Answer 2

Since $R[G]$, is defined by the adjuction $Hom_{R-alg}(R[G], R')=Hom_{Grp}(G, R'^*)$. In particular the equality chain $$Hom_{R\oplus S-alg}((R\oplus S)[G], R')=Hom_{Grp}(G, R'^*)=Hom_{Grp}(G, R'^*)\oplus Hom_{Grp}(G, R'^*)=Hom_{R-alg}(R[G], R')\oplus Hom_{S-alg}(S[G], R')=Hom_{R\oplus S-alg}((R[G]\oplus S[G], R')$$ Thus $R[G]\oplus S[G]=(R\oplus S)[G]$!