Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 17 September 2014

Lecture 5: Rings and Fields

An abelian group is a set $A$ with a function (addition) $$\begin{array}{ccc}A\times A& ⟶& A\\ (a,b)& ⟼& a+b\end{array}$$ such that

(a)	If $a_1,a_2,a_3\in A$ then $(a_1+a_2)+a_3=a_1+(a_2+a_3)\text{.}$
(b)	There exists $0\in A$ such that if $a\in A$ then $0+a=a$ and $a+0=a\text{.}$
(c)	If $a\in A$ then there exists $-a\in A$ such that $a+(-a)=0$ and $(-a)+a=0\text{.}$

(a)	$\mathbb{Z}$ with addition.
(b)	$M_{1\times 5}\left(ℝ\right)=\left\{\text{column vectors of length 5 with entries in}\hspace{0.17em}ℝ\right\}$ with $$\left(\begin{array}{c}{a}_1\\ a_2\\ a_3\\ a_4\\ a_5\end{array}\right)+\left(\begin{array}{c}{b}_1\\ b_2\\ b_3\\ b_4\\ a_{\mathrm{b}}\end{array}\right)=\left(\begin{array}{c}{a}_1+b_1\\ a_2+b_2\\ a_3+b_3\\ a_4+b_4\\ a_5+b_5\end{array}\right)\text{.}$$

A ring is an abelian group $R$ with a function (multiplication) $$\begin{array}{ccc}R\times R& ⟶& R\\ (a,b)& ⟼& ab\end{array}$$ such that

(d)	If $r_1,r_2,r_3\in R$ then $\left(r_1{r}_2\right)r_3=r_1\left(r_2{r}_3\right),$
(e)	There exists $1\in R$ such that if $r\in R$ then $r·1=r$ and $1·r=r\text{.}$
(f)	If $r_1,r_2,r_3\in R$ then $$r_1(r_2+r_3)=r_1{r}_2+r_1{r}_3\text{and}(r_1+r_2)r_3=r_1{r}_3+r_2{r}_3$$ (the distributive properties).

(a)	$\mathbb{Z}$ with addition and multiplication.
(b)	$\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$m\mathbb{Z}$}\right.$ with addition and multiplication.
(c)	$ℂ\left[t\right],$ polynomials, with addition and multiplication.
(d)	$M_{n\times n}\left(ℝ\right),$ square matrices, with addition and multiplication.

$$ℂ\left[t\right]=\{a_0+a_{1}t+a_2{t}^2+\cdots \hspace{0.17em}|\hspace{0.17em}{a}_i\in ℂ\hspace{0.17em}\text{and all but a finite number of the}\hspace{0.17em}{a}_i\hspace{0.17em}\text{are}\hspace{0.17em}0\}$$ $$\begin{array}{ccc}{\displaystyle (5t^2+3t+7)(3t^3+5)}& =& {\displaystyle 15t^5+25t^2+9t^4+15t+21t^3+35}\\ & =& {\displaystyle 15t^5+9t^4+2t^3+25t^2+15t+35\text{.}}\end{array}$$

A commutative ring is a ring $R$ such that

(g)	if $r_1,r_2\in R$ then $r_1{r}_2=r_2{r}_1\text{.}$

A field is a commutative ring $𝔽$ such that

(h)	If $r\in 𝔽$ and $r\ne 0$ then there exists $r^{-1}\in 𝔽$ such that $r·r^{-1}=1$ and $r^{-1}·r=1\text{.}$

(a)	$\mathbb{Q}=\left\{\frac{a}{b}\hspace{0.17em}\right|\hspace{0.17em}a,b\in \mathbb{Z},b\ne 0\}$ with $\frac{a}{b}=\frac{c}{d}$ if $ad=bc\text{.}$
(b)	$ℝ=\left\{\text{decimal expansions}\right\}\text{.}$
(c)	$ℂ=\{a+bi\hspace{0.17em}|\hspace{0.17em}a,b\in ℝ\}$ with $i^2=-1\text{.}$

Clocks

$$\begin{array}{ccccccccc}\begin{array}{c}1\end{array}& & \begin{array}{c}1\\ 2\end{array}& & \begin{array}{ccc}& 1& \\ 3& & 2\end{array}& & \begin{array}{ccc}& 1& \\ 4& & 2\\ & 3& \end{array}& & \begin{array}{ccc}& 1& \\ 5& & 2\\ 4& & 3\end{array}\\ \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$1\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$2\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$3\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$4\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$5\mathbb{Z}$}\right.\end{array}$$ Better to write $$\begin{array}{ccccccccc}\begin{array}{c}0\end{array}& & \begin{array}{c}0\\ 1\end{array}& & \begin{array}{ccc}& 0& \\ 2& & 1\end{array}& & \begin{array}{ccc}& 0& \\ 3& & 1\\ & 2& \end{array}& & \begin{array}{ccc}& 0& \\ 4& & 1\\ 3& & 2\end{array}\\ \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$1\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$2\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$3\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$4\mathbb{Z}$}\right.& & \raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$5\mathbb{Z}$}\right.\end{array}$$ All of these are commutative rings. Which are fields? In $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$5\mathbb{Z}$}\right.\text{:}$ $1·1=1,$ $2·3=1,$ $3·2=1,$ $4·4=1\text{.}$
In $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$4\mathbb{Z}$}\right.\text{:}$ $1·1=1,$ $3·3=1,$ but $2·x$ is never $1\text{.}$ So $2$ is not invertible in $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$4\mathbb{Z}$}\right.,$ and $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$4\mathbb{Z}$}\right.$ is not a field.

$M_{2\times 2}\left(ℂ\right)$ is not a commutative ring since $$\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\in M_{2\times 2}\left(ℂ\right)\text{and}$$ $$\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\text{and}\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)$$ so that $$\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\ne \left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\text{.}$$

Let $A$ be an abelian group. Let $a\in A\text{.}$ Show that $-(-a)=a\text{.}$

		Proof.
	Assume $a\in A\text{.}$ To show: $-(-a)=a\text{.}$ We know: $-a$ is an element (call it $b\text{)}$ such that $$\begin{array}{cc}b+a=0\text{and}a+b=0\text{.}& \text{(1)}\end{array}$$ We know: $-(-a)=-b$ is an element (call it $c\text{)}$ such that $$\begin{array}{cc}c+b=0\text{and}b+c=0\text{.}& \text{(2)}\end{array}$$ To show: $c=a\text{.}$ $$c=c+0=c+(b+a)=(c+b)+a=0+a=a,$$ by properties (b), (1), (a), (2), (b), respectively. $\square$

Let $A$ be an abelian group. Show that $0\in A$ is unique.

		Proof.
	To show: $0\in A$ is unique. $0$ is an element (call it $a\text{)}$ such that $$\begin{array}{cc}\text{if}\hspace{0.17em}x\in A\hspace{0.17em}\text{then}\hspace{0.17em}a+x=x\hspace{0.17em}\text{and}\hspace{0.17em}x+a=x\text{.}& \text{(3)}\end{array}$$ Let $b$ be another element such that $$\begin{array}{cc}\text{if}\hspace{0.17em}x\in A\hspace{0.17em}\text{then}\hspace{0.17em}b+x=x\hspace{0.17em}\text{and}\hspace{0.17em}x+b=x\text{.}& \text{(4)}\end{array}$$ To show: $a=b$ $$a=a+b=b,$$ by (4) and (3), respectively. $\square$

Let $R$ be a ring. Let $a\in R\text{.}$ Show that $0·a=0\text{.}$

		Proof.
	Assume $a\in R\text{.}$ To show: $0·a=0\text{.}$ $$0·a=(0+0)·a=0·a+0·a\text{(by (b) and the distributive law).}$$ Add $-(0·a)$ to each side to get $$0=0·a\text{(because}\hspace{0.17em}0·a+(-(0·a))=0\text{).}$$ $\square$

Let $R$ be a ring. Let $a\in R\text{.}$ Show that $(-1)·a=-a\text{.}$

		Proof.
	Assume $a\in R\text{.}$ To show: $(-1)·a=-a\text{.}$ We know: $-a$ is an element (call it $b\text{)}$ such that $$b+a=0\text{and}a+b=0\text{.}$$ We know: $-1$ is an element (call it $c\text{)}$ such that $$c+1=0\text{and}1+c=0\text{.}$$ To show: $c·a=b\text{.}$ $$\begin{array}{ccc}{\displaystyle c·a+a}& =& {\displaystyle (c+1)a=0·a=0,\text{and}}\\ {\displaystyle a+c·a}& =& {\displaystyle (1+c)a=0·a=0\text{.}}\end{array}$$ Then $$b=b+0=b+a+c·a=0+c·a=c·a\text{.}$$ $\square$

Notes and References

These are a typed copy of Lecture 5 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on August 3, 2011.

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