Let $R$ be a ring. Since $R$ is also a group then we can talk about the group ring $R[R]$. I want to understand this group ring $R[R]$.

An element $x\in R[R]$ is written as a finite formal sum $$x=r_1s_1+r_2s_2+\cdots+r_ns_n$$ where both $r_i$ and $s_i$ are in $R$ but since the ring $R$ is closed under sum and multiplication, then it is clear that $x\in R$. So can't we just say that the group ring $R[R]$ is equal to $R$?

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    Because it isn't. It's better to use different notation, e.g., a typical element would be $\sum r_i [s_i]$ with multiplication $(r[s])(r'[s'])=rr'[s+s']$ etc. – Lord Shark the Unknown Oct 18 at 18:25
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    I'm not sure what motivated the downvoter in this case. There is a point of confusion expressed, sufficient context for us to see clearly what the problem is, and some clear evidence of prior thought. I couldn't find a duplicate either. Seems fine to me. +1 – rschwieb Oct 18 at 18:37
  • @LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ? – palio Oct 18 at 18:54
  • @palio No, I did not. – Lord Shark the Unknown Oct 18 at 20:27
  • @LordSharktheUnknown Then shouldn't $(r[s])(r'[s'])=(rr')([s]+[s'])$? where $rr'$ is the product in $R$ and $[s]+[s']$ is the sum in $R$ ? – palio Oct 19 at 17:16
up vote 7 down vote accepted

It would be better to write elements of $R[R]$ in the form $$x=\sum_{s\in R}r_se^s.$$ Since $e^se^t=e^{s+t}$, multiplication in $R[R]$ captures the group structure of $(R,+)$. It also avoids the confusion you are having.

  • in $x=\sum_{s\in R}r_se^s.$ what is $r$ and what is $e$ ? Thanks! – palio Oct 18 at 18:41
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    $r_s\in R$ is the coefficient. The symbol $e$ is just that, a formal symbol (which you should think of like the natural base of exponentiation). – David Hill Oct 18 at 18:45
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    here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character – David Hill Oct 18 at 18:47
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    I think we can use this formal exponentiation trick anytime we have a group ring $R[G]$ where $G$ is an abelian group which is denoted additively and this trick makes multiplication in the group ring clear, is this correct ? – palio Oct 18 at 18:51
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    yes, that is correct. – David Hill Oct 18 at 19:01

It's important to remember that the linear combinations are formal in that the way we write it distinguishes coefficients from generators: the $r_i$'s are coefficients, and the $s_i$'s are basis elements.

Their juxtaposition does not denote multiplication in $R$, but rather that $r_i$ is the coefficient at the base element $s_i$.

One can form a group ring over the additive group $(R,+)$ or a monoid ring over the monoid $(R,\cdot)$, so the notation above is a little ambiguous. It would perhaps be beneficial to just forget that $R$ is a ring and talk about its underlying abelian group $A$ (or use $M$ if you're doing the monoid instead.)

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