# The group ring of a ring.

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 $$R$$?

• 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 '18 at 18:25
• 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 '18 at 18:37
• @LordSharktheUnknown Here you identified $[s+s']$ with $[s]+[s']$ ? – palio Oct 18 '18 at 18:54
• @palio No, I did not. – Lord Shark the Unknown Oct 18 '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 '18 at 17:16

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 '18 at 18:41
• $r_s\in R$ 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 '18 at 18:45
• here is an example where the notation is used: en.wikipedia.org/wiki/Algebraic_character – David Hill Oct 18 '18 at 18:47
• 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 '18 at 18:51
• yes, that is correct. – David Hill Oct 18 '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.)