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

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

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

  • $\begingroup$ 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. $\endgroup$ – rschwieb Oct 27 '15 at 17:51
  • $\begingroup$ 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. $\endgroup$ – Qiaochu Yuan Oct 27 '15 at 18:34
  • $\begingroup$ @QiaochuYuan Yeap, you guys are right. $\endgroup$ – Pax Kivimae Oct 27 '15 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.