Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $g\otimes h$?

share|improve this question
2  
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

1 Answer 1

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 $\{m_i\otimes n_j:i\in I,j\in J\}$.

share|improve this answer
    
Thank you. I'll try to write out a proof –  Bey Sep 16 '12 at 23:01
2  
@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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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