Let \(G = (G, \cdot)\) be a group and let \(R\) be a commutative ring with \(1\neq 0\) (often \(R\) is taken to be a field). Let \(R[G]\) be the collection of formal expressions of the form indicated below, where the \(+\) operation is assumed to be commutative:
Equivalently, a typical element of \(R[G]\) can be written as \(\sum_{g \in G} r_g g\text{,}\) where \(r_g \in R\) for all \(g\) and \(r_g = 0\) for all but a finite number of \(g\)’s.
With these definitions, \(R[G]\) is a ring, called the group ring of \(G\) with coefficients in \(R\text{.}\) It is a unital ring with identity \(1_Re_G\text{.}\) If \(R=F\) is a field then \(F[G]\) is an \(F\)-vector space.
Remark8.40.
Some authors write \(R[G]\) as \(RG\text{,}\) but the notation \(R[G]\) is more standard.
Proposition8.41.
Let \(R\) be any commutative ring and \(G\) a group. Then \(R[G]\) is commutative if and only if \(G\) is abelian.
Example8.42.
Let \(Q_8 = \{\pm 1, \pm i, \pm j, \pm k\}\) denote the group of quaterinons and \({\mathbb{R}}\) the field of real numbers, and let us consider the group ring \({\mathbb{R}}[Q_8]\text{.}\) Actually, the notation here is not so good since \((-1) k\) is easily confused with \(1 (-k)\text{,}\) and, even worse, things like \(1 \cdot 1\text{,}\)\(1 \cdot (-1)\) are highly confusing. So, let us rename the elements of \(Q\text{,}\) so that
so that \(e\) is what we were writing as \(1\text{,}\)\(e'\) is what we were writing as \(-1\text{,}\)\(i'\) is what we were writing as \(-i\text{,}\) etc. So, for example, we now have \(i^2 = e'\) in this group.
\({\mathbb{R}}[Q_8]\) is a non-commutative ring, and you might guess that it is the same as the quaterinons \(\mathbb{H}\) defined above, but it can’t be: \({\mathbb{R}}[Q_8]\) is \(8\)-dimensional as a \({\mathbb{R}}\)-vector space whereas \(\mathbb{H}\) is \(4\) dimensional. In fact \({\mathbb{R}}[Q_8]\) is not a division ring, since it has zero divisors: \((e - i)(e + i + i^2 + i^3) =0\) and so neither of the two factors can be units.
The problem is that \((-1) e \in R[Q_8]\) is not the same thing as \(1 e'\in R[Q]\text{,}\) but we want them to be the same in \(\mathbb{H}\text{.}\) Once we learn about quotient rings, we will be able to show that \(\mathbb{H}\) is the quotient of \(R[Q_8]\) by the ideal generated by \(e' + e\text{.}\) Roughly this means we mod out by the relation \(e' \sim -e\) and all consequences of this relation. For example, once one imposes this equivalence relation, the element
\begin{equation*}
e + i + i^2 + i^3 = e + i + e' + e' i = (e + e') + i(e + e')
\end{equation*}
becomes the zero element.
Proposition8.43.
For any commutative ring \(R\text{,}\) the inclusion \(i:G \hookrightarrow R[G]\) given by \(i(g)= 1_Rg\) lands in \(R[G]^\times\) and induces a homomorphism of abelian groups \(G \to R[G]^{\times}\text{.}\)
Note that \((1_R g)(1_R h) = 1_R (gh)\) by definition of multiplication in \(R[G]\text{.}\)
Lemma 2.18. If \(F\) is a field, \(F[G]\) is an \(F\)-vector space and \(G\) is a basis, so that \(\operatorname{dim}_F(F[G]) = \#G\text{.}\)
Example 2.5. Take \(G = S_3\) and \(R = {\mathbb{R}}\text{.}\) Then \({\mathbb{R}}[S_3]\) is a six dimensional real vector space with basis \(\{e, (1 \, 2), (1 \, 3), (2 \, 3), (1 \, 2 \, 3), (1 \, 3 \, 2) \}\text{.}\) An element is any expression of the form
where \(r_1, \dots, r_6\) are real numbers. This ring has some zero divisors — for example
\begin{equation*}
(e - (1 \, 2))(e + (1 \, 2)) = e - (1 \, 2) + (1 \, 2) - (1 \, 2)^2 = e - e = 0.
\end{equation*}
Here we are abusing notation a bit — for example, \(- (1 \, 2)\) is really \((-1_R) (1 \, 2)\text{.}\) In general, \(1_R g\) is just written as \(g\) in \(R[G]\) and \((-r)g\) is just written as \(- rg\text{,}\) since \((-r)g\) is the additive inverse of \(rg\text{.}\)
Exercise 2.19.
We identify \(G\) as a subset of \(R[G]\) in the obvious way (by identifying \(1_R g\) with \(g\)).
thus \(i(g)\) is a unit in \(R[G]\) with inverse \(i(g^{-1})\text{.}\) This shows that \(\operatorname{Im}(i)\subseteq R[G]^\times\text{.}\)
The formula \((1_R g)(1_R h) = 1_R (gh)\) also gives that the map \(g\mapsto 1_Rg\) is group homomorphism. ◻
SubsectionWhat is a Polynomial Ring?
Example8.44.
Group rings give lots of cool examples of rings, but we will now just focus on the boring case when \(G\) is a free abelian group (written with multiplicative notation) generated by \(x_1, x_2, \dots, x_n\text{.}\) In this case an element of \(G\) may be written uniquely as \(x_1^{e_1}, \dots, x_n^{e_n}\) for \(e_1, \dots, e_n \in\mathbb{Z}\text{.}\) For any commutative ring \(R\) a typical element of \(R[G]\) is thus
This is a Laurent polynomial in the variables \(x_1, \dots, x_n\) with \(R\)-coefficients. Say \(n = 1\) and let \(x = x_1\text{,}\) so that \(G = \langle x \rangle\text{.}\) Then \(a x^{-3} + b x^{1} + c + dx^5\) with \(a,b,c,d \in R\) is a representative example of an element of \(R[G]\text{.}\) Addition is by combining like powers of \(x\text{.}\) Multiplication is uniquely determined by the fact that it must satisfy the distributive law and \(x^i x^j = x^{i +j}\) for \(i, j \in \mathbb{Z}\text{.}\)
It is clear that from the rules for \(+\) and \(\cdot\) that if we consider those elements with \(e_i \geq 0\) for all \(i\) in a Laurent polynomial ring, we obtain a subring:
Definition8.45.
Let \(G\) be a free abelian groups with generators \(x_1, \dots, x_n\text{.}\) For any commutative ring \(R\text{,}\) the polynomial ring in \(x_1, \dots, x_n\), written \(R[x_1, \dots, x_n]\text{,}\) is the subring of \(R[G]\) consisting of (finite) sums of the form
Example 2.8. If \(n = 1\text{,}\) letting \(x = x_1\text{,}\) then \(R[x]\) consists of all expressions of the form \(\sum_{i=0}^\infty r_i x^i\) with \(r_i = 0\) for all but a finite number of \(i\text{.}\)
Definition 2.22. Let \(R\) be a commutative ring with \(1 \neq 0\) and let \(M\) be a monoid (set endowed with a binary operation that is associative and has an identity). The monoid ring\(R[M]\) is the set of formal expressions
\begin{equation*}
R[m]=\left\{\sum_{m \in M} r_mm \mid r_m \in R, m \in ,M r_m = 0 \text{ for all but a finite number of }m\right\},
\end{equation*}
Definition 2.23. The polynomial ring on \(n\) variables \(x_1,\ldots,x_n\) with coefficients in \(R\) is the monoid ring \(R[x_1,...,x_n] = R(\mathbb{N}^n)\) on the free abelian monoid \(\mathbb{N}^n\) where each \(x_i\) is identified with \((0,\ldots, 0,1,0,\ldots, 0)\) with the 1 in the \(i\)-th position.
Remark 2.24. When \(R = {\mathbb{R}}\text{,}\) one is tempted to think of \({\mathbb{R}}[x]\) as being a subring of the ring \(\mathcal C({\mathbb{R}})\) of continuous, real-valued functions that are defined on all of \({\mathbb{R}}\text{.}\) This is not technically true: elements of \({\mathbb{R}}[x]\) are just formal expressions, not functions. But there is an injective ring homomorphism (see below)
