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Let $G$ and $H$ be groups, and $R$ a commutative ring. Then elements of $RG$ look like finite sums $\sum\limits_{g\in G}r_g\,g$, and similarly for $RH$. So $RG$ and $RH$ are $R$-modules with bases $G$ and $H$, respectively.

Does it follow that $RG\otimes_R RH$ has basis given by simple tensors $g\otimes h$?

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Note that the fact that each element of $RG$ is of the form $\sum_{g\in G}r_gg$ does not on its own say that the elements of $G$ form a basis for $RG$. It just says that they generate $RG$ as an $R$-module. That they generate $RG$ as an $R$-module together with the fact that the coefficients $r_g$ are unique, i.e., the elements of $G$ are $R$-linearly independent, shows that they form a basis. – Keenan Kidwell Sep 16 '12 at 22:52
You're right; sorry, I didn't mean for it read as an implication. I just meant them to be separate declarative statements – Bey Sep 16 '12 at 23:02
up vote 3 down vote accepted

Yes. If $R$ is any commutative ring and $M$ and $N$ are free $R$-modules with bases $\{m_i:i\in I\}$ and $\{n_j:j\in J\}$, then $M\otimes_RN$ is free with basis $\{m_i\otimes n_j:i\in I,j\in J\}$.

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Thank you. I'll try to write out a proof – Bey Sep 16 '12 at 23:01
@Bey And once you have the proof, you'll see why $R[G]\otimes R[H]\cong R[G\times H]$! – rschwieb Sep 17 '12 at 0:21
Question for you rschwieb: I've noticed that some folks denote group rings/algebras with brackets around the group, as you did in your comment; is there any particular significance to this? – Bey Sep 17 '12 at 4:07

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