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

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 $(R\times S)[G]\cong R[G]\times S[G]$?

• is it true if we just take $R=S=1$ ? – seeker Oct 26 '15 at 18:16
• Did you try to write down some maps between them? – Espen Nielsen Oct 26 '15 at 18:16
• @seeker Do you mean the zero ring? If so, then yes, since $0[G]=0$. – Espen Nielsen Oct 26 '15 at 18:18
• @EspenNielsen :- ya i am sorry...i misinterpreted it. – seeker Oct 26 '15 at 18:19
• What do you mean by $R[G]$ if $R$ is noncommutative? – Qiaochu Yuan Oct 27 '15 at 18:35

## 2 Answers

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.

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 $R[G]\oplus S[G]=(R\oplus S)[G]$!

• I'm not fluent in such computations, so I ask a naive question or two now. Is the coproduct of G with itself really necessary? Perhaps I overlooked something in my solution, but maybe it hasn't been long enough to realize why. – rschwieb Oct 27 '15 at 17:51
• That coproduct looks wrong to me. For example, if $R = S$ is a field $k$ and $G$ is finite, then the LHS is finite-dimensional over \$k but the RHS usually won't be. – Qiaochu Yuan Oct 27 '15 at 18:34
• @QiaochuYuan Yeap, you guys are right. – Pax Kivimae Oct 27 '15 at 19:41