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Abstract algebra A Kursobjekt for Math 331

Aaron Greicius

Section 2.3 Group rings, polynomials, power series

In this section and the next we introduce three ring constructions: group rings, polynomial (and power series) rings, and Hamilton quaternion rings. Each of these uses as an underlying set a collection of “formal sums” with coefficients in a base ring \(R\text{.}\) This somewhat ambiguous language of formal sums comes up frequently in mathematics. As we make official with Fiat 2.3.1 a formal sum is really just a tuple that we have decided to write using summation notation. The definition of the complex numbers is a typical example of this phenomenon: complex numbers are usually defined as a formal sum \(a+bi\) where \(a,b\in \R\text{,}\) when in essence a complex number is just a pair \((a,b)\in \R^2\text{.}\) Why not just use tuple notation? We tend to reach for the formal sum notation in situations where there is a (well-behaved) multiplication defined on our tuples, which further more respects the component-wise addition of these tuples. In these situations algebraic manipulation often proceeds more smoothly with the formal sum representation of objects, as opposed to the tuple representation.
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Subsection 2.3.1 Group rings

Specimen 23. Group rings.

Let \(R\) be a nontrivial commutative ring, and let \(G\) be a finite group. We define \(R\, G\) to be the set of all formal sums
\begin{equation*} \sum_{g\in G}a_g g\text{,} \end{equation*}
where \(a_g\in R\) for all \(g\in G\text{,}\) and we define binary operations \(+\) and \(\cdot\) as follows:
\begin{align*} \sum_{g\in G}a_g g+\sum_{g\in G}b_g g \amp =\sum_{g\in G}(a_g+b_g) g \\ \sum_{g\in G}a_g g\, \cdot\sum_{g\in G}b_g g \amp = \sum_{g\in G}c_g g, \amp c_g=\sum_{\substack{h,k\in G\\ hk=g}}a_h b_k \text{.} \end{align*}
The triple \((R\, G, +, \cdot)\) is a ring called the group ring of \(G\) with coefficients in \(R\text{.}\) The additive and multiplicative identities of \(R\) are given as
\begin{align*} 0 \amp = \sum_{g\in G}0g\\ 1 \amp =1e+\sum_{g\ne e}0g\text{.} \end{align*}
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Remark 2.3.2. Group rings.

  1. The subset \(\{re\mid r\in R\}\) is easily seen to be a subring of \(R\, G\text{.}\) Indeed, using the definition of the group ring operations, we see that
    \begin{align*} re + se \amp = (r+s) e \\ re\cdot se \amp =(rs) e\text{.} \end{align*}
    Thus we can think of the subring \(\{re\mid r\in R\}\) as a copy of \(R\) living as a subset of \(R\, G\text{.}\) (Later we will be able to say that this subring is isomorphic to \(R\text{.}\)) By slight abuse of notation, we will identify \(R\) with this subring of \(R\, G\text{,}\) and sometimes drop the \(e\) from the notation \(r\, e\text{.}\)
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  2. From the group ring operations definition, it follows that
    \begin{align*} r\, e \cdot \left(\sum_{g\in G}a_g\, g \right)\amp =\sum_{g\in G}ra_g\, g\\ \amp =\left(\sum_{g\in G}a_g\, g\right) \cdot r\, e\text{.} \end{align*}
    Identifying \(R\) with \(\{re\mid r\in R\}\text{,}\) we see that \(R\leq Z(R)\text{.}\)
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  3. We can also think of the group \(G\) as sitting inside \(R\, G\text{,}\) using the identification
    \begin{align*} g \amp =1\, g+\sum_{h\ne g}0\, h\text{.} \end{align*}
    We call such elements of \(R\, G\) group elements. With this identification in place, we that the ring multiplication on \(R\, G\) extends the group operation of \(G\text{.}\) That is, given \(g,h\in G\text{,}\) identifying these as the ring elements \(1\, g\) and \(1\, h\) and taking their product, we see that
    \begin{align*} g\cdot h \amp = gh \text{.} \end{align*}
    As a result, it is easy to see that a group element \(g\in R\, G\) is invertible with inverse the group element \(g^{-1}\text{.}\) It follows that we can think of \(G\) as a subgroup of the group of units of \(R\, G\) (with respect to our identification): i.e.,
    \begin{align*} G\leq (R\, G)^* \amp \text{.} \end{align*}
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  4. Combining the previous observations, we can now give a decent motivation for the multiplication operation on \(R\, G\text{.}\) Namely, it is the unique binary operation on \(R\, G\) satisfying the following properties:
    1. it distributes over the given addition operation on \(R\, G\text{;}\)
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    2. it extends the group operation in the sense that \(g\cdot h=gh\text{;}\)
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    3. we have \(r\, e\in Z(R\, G)\) for all \(r\in R\) with respect to this multiplication.
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    In this sense our multiplication operation is the unique way of extending the group operation of \(G\) to a ring operation on \(R\, G\) in such a way that our copy of \(R\leq R\, G\) lies in the center of \(R\, G\text{.}\)
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Example 2.3.3. Group ring of \(D_3\).

Let \(D_3=\angvec{r,s\mid r^3=s^2=e, srs=r^{-1}}\text{,}\) and consider the group ring \(\F_3\, D_3\text{.}\)
  1. Let \(a=2rs+ r^2s\) and \(b=s+2rs\text{.}\) Compute \(ab\) and \(ba\text{.}\)
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  2. Show that \(r+r^2\) is a unit in \(\F_3\,D_3\)
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Solution.
When computing an explicit product in a group ring, instead of using the formal definition in Specimen 23, it is usually easiest to expand out using distributivity, use the fact that the ring coefficients commute with everything and that the ring product of group elements is given by the group product, and the collect like terms.
  1. We have
    \begin{align*} ab \amp =(2rs+r^2s)(s+2rs)\\ \amp =2rs^2+4rsrs+r^2s^2+2r^2srs\\ \amp =2r+1e+r^2+2r \amp ((rs)^2=e, 4=1, r^2srs=r)\\ \amp =1e+4r+r^2\\ \amp =e+r+r^2\\ ba \amp =(s+2rs)(2rs+r^2s)\\ \amp =2srs+sr^2s+4rsrs+2rsr^2s\\ \amp =2r^2+r+4e+r^2\\ \amp =4e+r+3r^2\\ \amp =e+r \amp (4=1, 3=0)\text{.} \end{align*}
    Observe that \(ab\ne ba\text{:}\) no great surprise in light of Example 2.3.5.
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  2. First observe that \(r\) is a unit in \(\F_3\, D_3\) with inverse \(r^2\text{.}\) (See Remark 2.3.2.) Since \(r+r^2=r(e+r) \text{,}\) it suffices to show that \(e+r\) is a unit of \(\F_3\, D_3\text{.}\) (If \(a\) and \(b\) are units, then so is \(ab\text{.}\)) Playing around with some expressions, we find that
    \begin{align*} (e+r)(2e+r+2r^2) \amp =(2e+r+2r^2)(e+r)=1e=1\text{.} \end{align*}
    Thus \((e+r)\) is a unit. We conclude that \(r+r^2\) is a unit, with inverse
    \begin{align*} (e+r)^{-1}r^{-1} \amp =(2e+r+2r^2)r^2=e+2r+2r^2\text{.} \end{align*}
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Remark 2.3.4. Group rings over a field.

Let \(K\) be a field, and let \(G=\{g_1,g_2,\dots, g_n\}\) be a group of cardinality \(n\text{.}\) In this case the group ring \(K\, G\) has the additional structure of a \(K\)-vector space, where vector addition is just the ring addition, and scalar multiplication by an element \(c\in K\) is defined as ring multiplication by the element \(c\, e\in K\, G\text{.}\) In fact, identifying elements of \(K\, G\) as tuples via the correspondence
\begin{align*} \sum_{i=1}^n a_i\, g_i \amp \leftrightarrow (a_1,a_2,\dots, a_n) \text{,} \end{align*}
we see that as a vector space, \(K\, G\) is literally equal to the vector space \(K^n\) of \(n\)-tuples with coefficients in \(K\) with its usual vector operations:
\begin{align*} K\, G \amp \longleftrightarrow K^n\\ \sum_{i=1}^na_i\, g_i \amp \longleftrightarrow (a_i)_{i=1}^n\\ \sum_{i=1}^na_i\,g_i+ \sum_{i=1}^nb_i\,g_i \amp \longleftrightarrow (a_i)_{i=1}^n+(b_i)_{i=1}^n\\ ce\cdot \sum_{i=1}^na_i\, g_i \amp \longleftrightarrow c(a_i)_{i=1}^n\text{.} \end{align*}
As you will see on your homework, this added vector space structure gives us additional tools for studying group rings over fields.
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Example 2.3.5. Group ring properties.

Let \(R\) be a nontrivial commutative ring, and let \(G\) be a finite group.
  1. Prove that \(R\, G\) is commutative if and only if \(G\) is abelian.
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  2. Consider the element \(r=\sum_{g\in G} g\in R\, G\text{.}\) Compute \(h\cdot r\text{,}\) where \(h\) is an element of \(G\text{.}\)
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  3. Let \(g\in G\) be a nontrivial element. Show that \(e-g\) is a zero divisor.
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  4. When is a group ring \(R\, G\) a division ring?
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Solution.
Let \(g_1,g_2,\dots, g_n\) be the distinct elements of \(G\text{.}\) To distinguish between the group operation and ring multiplication, we will use concatenation for the group operation and \(\cdot\) for the ring operation.
  1. Assume that \(R\, G\) is commutative. For all \(g,h\in G\text{,}\) we have
    \begin{align*} gh \amp = g\cdot h \amp (\knowl{./knowl/xref/rem_group_rings.html}{\text{2.3.2}}) \\ \amp =h\cdot g \amp (R\, G \text{ commutative})\\ \amp =hg \amp (\knowl{./knowl/xref/rem_group_rings.html}{\text{2.3.2}})\text{.} \end{align*}
    Thus \(G\) is abelian.
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    Now assume \(G\) is abelian. Given
    \begin{align*} r \amp =\sum_{g\in G}a_g\, g \amp s\amp= \sum_{g\in G}b_g\, g\text{,} \end{align*}
    expanding out using distributivity and commutativity of ring coefficients, we have
    \begin{align*} rs \amp =\sum_{(g,h)\in G^2}(a_g b_h)gh \amp \\ \amp =\sum_{(g,h)\in G^2}(b_h a_g)hg \amp (R \text{ comm.}, G \text{ abelian})\\ \amp = sr \text{.} \end{align*}
    Thus \(R\, G\) is commutative.
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  2. We have
    \begin{align*} h\cdot r \amp = h\cdot \sum_{g\in G} g \amp \\ \amp =\sum_{g\in G} hg \\ \amp =\sum_{g'\in G} g' \amp (hg=g') \\ \amp =r \text{,} \end{align*}
    where the penultimate line follows since as \(g\) runs through all elements of \(G\text{,}\) so too does \(g'=hg\text{.}\)
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    Observe that if \(G\) is nontrivial, the taking \(h\ne e\text{,}\) the equality \(hr=r\) implies that \(r\) is noninvertible: if it were, we could cancel it to conclude \(h=e\text{,}\) a contradiction.
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  3. We have
    \begin{align} (e-g)\sum_{i=0}^{n-1}g^i \amp =\sum_{i=0}g^i-\sum_{i=0}^{n-1}g^{i+1}\tag{2.3.1}\\ \amp =e-g^n\tag{2.3.2}\\ \amp =e-e=0\text{.}\tag{2.3.3} \end{align}
    Since \(e-g\) and \(\sum_{i=0}^{n-1}g^i=e+g+\cdots +g^{n-1}\) are both nonzero elements of \(R\, G\text{,}\) we conclude that \(e-g\) is a zero divisor (as is \(1+g+\cdots +g^{n-1}\)).
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  4. Both (2) and (3) imply that \(R\, G\) is a division ring if and only if \(G=\{e\}\) is trivial and \(R\) is a field.
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Subsection 2.3.2 Polynomials and power series

Whereas normally power series are introduced as a sort of generalization of polynomials, we will take the approach of defining the notion of a ring of power series, and identify a subring as the ring of polynomials.
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Specimen 24. Power series ring.

Let \(R\) be a nontrivial commutative ring. We define the set \(R[[x]]\) to be the set of all formal sums of the form
\begin{align*} \sum_{n=0}^\infty a_n x^n \amp \text{,} \end{align*}
where \(a_n\in R\) for all \(n\in \Z_{\geq 0}\text{,}\) and we define binary operations \(+\) and \(\cdot\) as follows:
\begin{align*} \sum_{n=0}^\infty a_nx^n+\sum_{n=0}^\infty b_nx^n \amp = \sum_{n=0}^\infty (a_n+b_n)x^n\\ \sum_{n=0}^\infty a_nx^n\cdot \sum_{n=0}^\infty b_nx^n \amp = \sum_{n=0}^\infty c_nx^n \amp c_n=\sum_{k=0}^n a_k b_{n-k}\text{.} \end{align*}
The triple \((R[[x]], +,\cdot)\) is a commutative ring called the power series ring over (or with coefficients in) \(R\) in the unknown (or indeterminate) \(x\). The additive and multiplicative identities of \(R[[x]]\) are given as
\begin{align} 0 \amp =\sum_{n=0}^\infty 0x^n = 0+0x+0x^2+\cdots \tag{2.3.4}\\ 1 \amp =1x^0+\sum_{n=1}^\infty 0x^n =1+0x+0x^2+\cdots \text{.}\tag{2.3.5} \end{align}
Given a power series \(\sum_{n=0}^\infty\text{,}\) we call \(a_nx^n\) its \(n\)-th term, and \(a_n\) its \(n\)-th coefficient.
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We will use function-like notation to name power series. In more detail, we will refer to power series using function letters like \(f\) and \(g\text{,}\) and will write \(f(x)=\sum_{n=0}^\infty a_nx_n\) when explicitly introducing names of coefficients.
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Remark 2.3.6. Power series equality.

When are two power series \(f(x)=\sum_{n=0}^\infty a_nx^n\) and \(g(x)=\sum_{n=0}^\infty b_nx^n\) equal? Identifying these formal sums as tuples \((a_n)_{n=0}^\infty\text{,}\) \((b_n)_{n=0}^\infty\text{,}\) following Fiat 2.3.1, we see that
\begin{gather*} f=g \iff a_n=b_n \text{ for all } n\in \Z_{\geq 0}\text{.} \end{gather*}
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We now identify the ring \(R[x]\) of polynomials over \(R\) as the subring of \(R[[x]]\) consisting of all power series whose coefficients are “eventually” equal to zero.
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Specimen 25. Polynomial rings.

Let \(R\) be a nontrivial commutative ring. Define the polynomial ring over \(R\) in the indeterminate \(x\) to be the subset \(R[x]\subseteq R[[x]]\) of all power series whose coefficients are eventually equal to zero: i.e.,
\begin{align*} R[x] \amp = \{f(x)=\sum_{n=0}^\infty a_nx^n\in R[[x]]\mid a_n=0 \text{ for all } n> N \text{ for some } N\in \Z_{\geq 0} \} \\ \amp =\{f\in R[[x]]\mid f(x)=\sum_{n=0}^Na_nx^n \text{ for some } N\in \Z_{\geq 0}, a_n\in R\}\text{.} \end{align*}
It is easy to see that \(R[x]\) is a subring of \(R[[x]]\text{,}\) and thus a ring in its own right. Restricted to \(R[x]\text{,}\) the power series multiplication rule becomes the familiar polynomial multiplication rule:
\begin{align} (\anpoly)\amp \cdot (\bmpoly) =\sum_{k=0}^{m+n}c_kx^k\tag{2.3.6}\\ c_k\amp =\sum_{j=0}^k a_j b_{k-j}\text{.}\tag{2.3.7} \end{align}
As illustrated above, when denoting polynomials we will usually drop all the terms with zero coefficients: e.g., \(f(x)=a_0+a_1x+\cdots +a_nx^n\text{,}\) or \(f(x)=\anpoly\text{.}\)
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Given a nonzero polynomial \(f\in R[x]\text{,}\) we define its degree, \(\deg f\text{,}\) to be the largest \(n\) satisfying \(a_n\ne 0\text{.}\) If \(f = 0\text{,}\) we define \(\deg f=-\infty\text{.}\)
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If \(f(x)=\anpoly\) is a polynomial of degree \(n\ne -\infty\text{,}\) we call \(a_{n}\) the leading coefficient of \(f\text{.}\) A polynomial with leading coefficient equal to \(1\in R\) is called monic.
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Remark 2.3.7. Polynomial and power series rings.

When working with a power series ring \(R[[x]]\text{,}\) we will identify the ring \(R\) with power series of the form
\begin{align*} r+\sum_{n=1}^\infty 0x^n \amp =r+0x+0x^2+\cdots\text{.} \end{align*}
Since this identification preserves the ring operations of \(R\text{,}\) we end up with a chain of subring inclusions
\begin{align*} R \amp \subseteq R[x] \subseteq R[[x]]\text{.} \end{align*}
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The degree operation on \(R[x]\) turns out to be very useful, thanks in part to the properties articulated in the following proposition.
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Proof.

  1. Assume \(\deg f=n\) and \(\deg g=m\text{,}\) so that
    \begin{align*} f(x) \amp =\anpoly \amp \\ g(x)\amp=\bmpoly \amp \text{,} \end{align*}
    where \(a_n\ne 0\) and \(b_m\ne 0\text{.}\) From the polynomial multiplication rule given in (2.3.6), we see that
    \begin{align*} fg(x) \amp =a_nb_mx^{n+m}+\text{terms of lower degree}\text{.} \end{align*}
    It follows that \(\deg fg\leq n+m=\deg f+\deg g\text{.}\) If \(R\) is an integral domain, then \(a_nb_m\ne 0\) and thus \(\deg fg=m+n=\deg f+\deg g\text{.}\)
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  2. The proof is left to the reader.
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Example 2.3.9. Polynomial degree.

Find polynomials \(f,g\in \Z/4\Z[x]\) satisfying \(\deg fg< \deg f+\deg g\text{.}\)
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Solution.
Let \(f(x)=2x+1\) and \(g(x)=2x\text{.}\) We have \(\deg f=\deg g=1\) and
\begin{align} \deg fg \amp =\deg(4x^2+2x)\tag{2.3.8}\\ \amp =\deg(2x)\tag{2.3.9}\\ \amp = 1\tag{2.3.10}\\ \amp < 1+1\tag{2.3.11}\\ \amp =\deg f+\deg g\text{.}\tag{2.3.12} \end{align}
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As a first illustration of the utility of the degree operation, we show that the polynomial ring construction preserves the integral domain property.
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Proof.

Since clearly a subring of an integral domain is an integral domain, we have (3)\(\implies\)(2)\(\implies\)(1). It remains only to show (1)\(\implies\)(3).
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Assume \(R\) is an integral domain. Given nonzero elements \(f,g\in R[[x]\text{,}\) we can write
\begin{align} f(x) \amp =a_nx^n+a_{n+1}x^{n+1}+\cdots \tag{2.3.13}\\ g(x) \amp =b_mx^m+b_{m+1}x^{m+1}+\cdots \text{,}\tag{2.3.14} \end{align}
where \(a_n,b_m\ne 0\text{.}\) But then
\begin{align} fg (x) \amp = a_nb_mx^{n+m}+\text{higher degree terms}\text{.}\tag{2.3.15} \end{align}
Since \(R\) is an integral domain, and since \(a_n, b_m\ne 0\text{,}\) we conclude that \(fg\ne 0\text{,}\) as desired.
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Example 2.3.11. Units of polynomial and power series rings.

Let \(R\) be a nontrivial commutative ring.
  1. Show that \(f(x)=\sum_{n=0}^\infty a_nx^n\) is a unit if and only if \(a_0\in R^*\text{.}\)
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  2. Prove: if \(R\) is an integral domain, then \((R[x])^*=R^*\text{.}\)
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Solution.
Left as an exercise.
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