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\begin{document}

\title{A3 Rings and Modules}
\author{Richard Earl}
\date{Hilary Term 2017}
\maketitle

{\LARGE Syllabus}

\noindent Recap on rings (not necessarily commutative or with an identity)
and examples: $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$, fields, polynomial rings (in more than one variable), matrix rings.
Zero-divisors, integral domains. Units. The characteristic of a ring.
Discussion of fields of fractions and their characterization (proofs
non-examinable) [2]\medskip

\noindent Homomorphisms of rings. Quotient rings, ideals and the first
isomorphism theorem and consequences, e.g. Chinese remainder theorem.
Relation between ideals in $R$ and $R/I$. Prime ideals and maximal ideals,
relation to fields and integral domains. Examples of ideals. [3]\medskip

\noindent Euclidean Domains. Examples. Principal Ideal Domains. EDs are
PIDs. Application of quotients to constructing fields by adjunction of
elements; examples to include $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\cong 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]/\langle x^{2}+1\rangle $ and some finite fields. Degree of a
field extension, the tower law. [2.5]\medskip

\noindent Unique factorisation for PIDs. Gauss's Lemma and Eisenstein's
Criterion for irreducibility. [2]\medskip

\noindent Modules: Definition and examples: vector spaces, abelian groups,
vector spaces with an endomorphism. Submodules and quotient modules and
direct sums. The first isomorphism theorem. [2]\medskip

\noindent Row and column operations on matrices over a ring. Equivalence of
matrices and canonical forms of matrices over a Euclidean Domain.
[1.5]\medskip

\noindent Free modules and presentations of finitely generated modules.
Structure of finitely generated modules of a Euclidean domain. [2]\medskip

\noindent Application to rational canonical form for matrices, and structure
of finitely generated Abelian groups. [1]\medskip

{\LARGE Recommended Texts}

\begin{itemize}
\item P. B. Bhattacharya, S. K. Jain, S. R. Nagpaul, Basic Abstract Algebra,
CUP (1994) (Covers all of the basic algebra material most undergraduate
courses have).

\item Joseph Gallian, Contemporary Abstract Algebra (9th edition, CENGAGE
2016) (Excellent text covering material on groups, rings and fields).

\item B. Hartley, T. O. Hawkes, Chapman and Hall, Rings, Modules and Linear
Algebra. (Possibly out of print, but many library should have it. Relatively
concise and covers all the material in the course).

\item Neils Lauritzen, Concrete Abstract Algebra, CUP (2003) (Excellent on
groups, rings and fields, and covers topics in the Number Theory course
also. Does not cover material on modules).

\item Michael Artin, Algebra (2nd ed. Pearson, (2010). (Excellent but highly
abstract text covering everything in this course and much more besides).
\end{itemize}

\pagebreak

\begin{center}
\textbf{EXAMPLES OF\ ALGEBRAIC\ STRUCTURES}
\end{center}

\noindent \textbf{RINGS}

\noindent $%
%TCIMACRO{\U{2124} }%
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\mathbb{Z}
%EndExpansion
$ the integers under $+$ and $\times .$

\noindent $%
%TCIMACRO{\U{2124} }%
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\mathbb{Z}
%EndExpansion
_{n}$ -- the integers, modulo $n$, under $+$ and $\times .$

\noindent $\mathbb{H}$ -- the quaternions under $+$ and $\times .$

\noindent $R[x]$ -- polynomials in $x$ with coefficients in the ring $R$
under $+$ and $\times .$

\noindent $R[[x]]$ -- formal power series in $x$ with coefficients in the
ring $R$ under $+$ and $\times .$

\noindent $R[A]$ -- polynomials in a square matrix $A$ with entries in a CRI 
$R.$

\noindent $R^{S}=\left\{ f:S\rightarrow R\right\} $ -- the set of maps from
a set $S$ to a ring $R$ under pointwise $+$ and $\times .$

\noindent $C(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)$ -- the ring of continuous functions $f:%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ under pointwise $+$ and $\times .$

\noindent $C^{1}(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)$ -- the ring of continuously differentiable $f:%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ under pointwise $+$ and $\times .$

\noindent $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ -- the Gaussian integers, i.e. the subring of complex numbers $%
a+bi$ where $a,b\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$

\noindent $M_{n}(F)$ -- the $n\times n$ matrices with entries in the field $%
F $ under matrix addition and mutliplication.

\noindent $\mathrm{End}(V)$ -- endomorphisms (i.e. linear maps $V\rightarrow
V$) of a vector space $V$ under $+$ and $\circ $.

\noindent $\mathcal{P}(X)$ -- the power set of $X$ under $\Delta $ and $\cap 
$.

\noindent $R/I$ -- the quotient (or factor) ring of a ring $R$ by an ideal $%
I $ of $R$.\medskip

\noindent \textbf{FIELDS}

\noindent $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
,$ $%
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%BeginExpansion
\mathbb{R}
%EndExpansion
,$ $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ -- the rationals/reals/complex numbers under $+$ and $\times .$

\noindent $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}$ -- the integers modulo $p$ (a prime) under $+$ and $\times .$

\noindent $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack i]$ -- $\{q_{1}+iq_{2}:q_{1},q_{2}\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\}$ under $+$ and $\times .$

\noindent $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ -- $\{q_{1}+q_{2}\sqrt{2}:q_{1},q_{2}\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\}.$

\noindent $F(x)$ -- rational functions in $x$ with coefficients in the field 
$F$.

\noindent $\mathbb{F}_{q}$ -- the finite field with $q$ elements; $q=p^{n}$
for a prime $p$ and $n\geqslant 1.$\pagebreak

\chapter{A Recap on Rings}

\noindent The material in this chapter that is covered in A0 Linear Algebra
will be only briefly revisited.

\begin{definition}
A \textbf{ring }$(R,+,\times )$ consists of a set $R$ with $+$ and $\times $
being two binary operations on $R$%
\begin{eqnarray*}
+ &:&R\times R\rightarrow R,\qquad \left( a,b\right) \mapsto a+b; \\
\times &:&R\times R\rightarrow R,\qquad \left( a,b\right) \mapsto a\times b,
\end{eqnarray*}%
such that $(R,+)$ is an abelian group, that is:

\begin{itemize}
\item $(A1)$ associativity: for all $a,b,c\in R$ we have $a+(b+c)=(a+b)+c;$

\item $(A2)$ commutativity: for all $a,b\in R$ we have $a+b=b+a;$

\item $(A3)$ zero element: there exists $0_{R}\in R$ such that for all $a\in
R$ we have $a+0_{R}=a=0_{R}+a;$

\item $(A4)$ inverses: for any $a\in R$ there exists $-a\in R$ such that $%
a+(-a)=(-a)+a=0;$
\end{itemize}

\noindent and further that:

\begin{itemize}
\item $(M1)$ for all $a,b,c\in R$ we have $a\times (b\times c)=(a\times
b)\times c;$

\item $(D)$ for all $a,b,c\in R$ we have $a\times (b+c)=a\times b+a\times c$
and $(a+b)\times c=a\times c+b\times c.$
\end{itemize}
\end{definition}

\begin{notation}
We will often suppress the $\times $ symbol and simply write $ab$ for $%
a\times b.$
\end{notation}

\begin{remark}
If the operations $+$ and $\times $ are clear then we will often refer
simply to a ring $R.$
\end{remark}

\begin{definition}
We say that a ring $R$ is \textbf{commutative} if multiplication is
commutative -- that is if the following axiom holds

\begin{itemize}
\item $(M2)$ commutativity: for all $a,b\in R$ we have $ab=ba$.
\end{itemize}
\end{definition}

\begin{definition}
We say that a ring $R$ has a $1$ or an \textbf{identity} if there is an
element $1_{R}\in R$ such that

\begin{itemize}
\item $(M3)$ existence of an identity: for all $a\in R$ we have $a\times
1_{R}=a=1_{R}\times a;$

\item $(Z)$ avoiding collapse: $1_{R}\neq 0_{R}.$
\end{itemize}

\noindent If $R$ has an identity then it is necessarily unique.
\end{definition}

Basic rules of algebra following from the ring axioms are:

\begin{proposition}
Let $R$ be a ring with a $1$ and $a,b\in R$.

(a) If $a+b=a+c$ then $b=c.$

(b) $-(-a)=a.$

(c) $a0_{R}=0_{R}=0_{R}a.$

(d) $-\left( ab\right) =\left( -a\right) b=a\left( -b\right) .$

(e) $\left( -1_{R}\right) a=-a=a\left( -1_{R}\right) .$
\end{proposition}

\begin{proof}
Left as exercises.
\end{proof}

\begin{notation}
Most, but not all, of the rings we will study are commutative and also have
an identity; we will use the acronym CRI as shorthand for "commutative ring
with an identity".
\end{notation}

\noindent A ring, then, is a half-way house between a group and a field, in
that the second operation $\times $ is not "fully-developed": it need not be
commutative, there need not be an identity, and there need not be inverses.
In a ring, multiplication is permissible but division may well not be.

\begin{example}
$%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,$ $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
,$ $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
,$ $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ with $+$ and $\times $ are all CRIs. In each case $0_{R}=0$ and $1_{R}=1$.
\end{example}

\begin{example}
$%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}$ -- the ring of integers modulo $n$ -- is a CRI with standard $+$ and $%
\times $ and $0_{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}}=\bar{0}$ and $1_{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}}=\bar{1}.$
\end{example}

\begin{example}
If $R$ is a ring, then we can consider $R[x]$ the ring of polynomials with
coefficients in $R.$ If $R$ is commutative, then so is $R\left[ x\right] $
and if $R$ has an identity then so does $R[x],$ namely the constant
polynomial $1_{R}$. Explicitly, given two polynomials $p,q$ over $R,$ say%
\begin{equation*}
p(x)=\sum_{i}a_{i}x^{i};\qquad q(x)=\sum_{i}b_{i}x^{i}
\end{equation*}%
then the sum $p+q$ and product $pq$ in $R[x]$ are defined by%
\begin{equation*}
p(x)+q(x)=\sum_{i}(a_{i}+b_{i})x^{i};\qquad p(x)q(x)=\sum_{k}\left(
\sum_{i+j=k}a_{i}b_{j}\right) x^{k}.
\end{equation*}%
We can more generally define polynomial rings over several variables, $%
R[x_{1},\ldots ,x_{n}].$ These rings can be inductively defined by%
\begin{equation*}
R[x_{1},\ldots ,x_{n}]=R[x_{1},\ldots ,x_{n-1}][x_{n}].
\end{equation*}
\end{example}

\begin{example}
Note that the rule $\deg (fg)=\deg f+\deg g$ will not apply generally in
rings. For example in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}[x]$ we have%
\begin{equation*}
(\bar{2}x+1)(\bar{2}x+1)=\bar{4}x^{2}+\bar{4}x+\bar{1}=\bar{1}
\end{equation*}%
Also in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}[x]$ we see that the quadratic $x^{2}-\bar{1}$ has four roots $\bar{1},%
\bar{3},\bar{5},\bar{7}$ and has two distinct factorizations%
\begin{equation*}
x^{2}-\bar{1}=(x-\bar{1})(x+\bar{1})=(x-\bar{3})(x-\bar{5}).
\end{equation*}
\end{example}

\begin{example}
Given a ring $R$ and a square matrix $A$ with entries in $R$, we can
consider the ring $R[A]$ with addition and multiplication defined similarly
to those in $R[x].$ Recall that if%
\begin{equation*}
p(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}\qquad \text{then}\qquad
p(A)=a_{0}I+a_{1}A+\cdots +a_{n}A^{n}.
\end{equation*}%
So with $R=%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ and%
\begin{eqnarray*}
A &=&\left( 
\begin{array}{cc}
1 & 0 \\ 
0 & 1%
\end{array}%
\right) ,\qquad R[A]=%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
I_{2}\cong 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
. \\
B &=&\left( 
\begin{array}{cc}
0 & 1 \\ 
0 & 0%
\end{array}%
\right) ,\qquad R[B]=\left\{ \left( 
\begin{array}{cc}
x & y \\ 
0 & x%
\end{array}%
\right) :x,y\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} . \\
C &=&\left( 
\begin{array}{cc}
0 & 1 \\ 
-1 & 0%
\end{array}%
\right) \qquad R[C]=\left\{ \left( 
\begin{array}{cc}
x & y \\ 
-y & x%
\end{array}%
\right) :x,y\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} \cong 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
\end{eqnarray*}%
The isomorphism with $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ in the third example is shown in Exercise Sheet 1, \#2. Note that $%
C^{2}=-I_{2}.$
\end{example}

\begin{example}
Given two rings $R$ and $S$ we can form their \textbf{direct sum} $R\oplus S$
which has $R\times S$ as its underlying set, and operations%
\begin{equation*}
(r_{1},s_{1})+(r_{2},s_{2})=(r_{1}+_{R}r_{2},s_{1}+_{S}s_{2}),\qquad
(r_{1},s_{1})\times (r_{2},s_{2})=(r_{1}\times _{R}r_{2},s_{1}\times
_{S}s_{2}).
\end{equation*}
\end{example}

\begin{example}
$C(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)$, the set of continuous functions from $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ to $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ forms a CRI under pointwise addition and multiplication with $0_{C(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)}$ and $1_{C(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)}$ being the constant functions $0$ and $1$. With the same operations\ and
identities, $C^{1}(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)$, the set of differentiable functions from $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ to $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ likewise forms a CRI.
\end{example}

\begin{example}
The even integers form a ring $2%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ under $+$ and $\times $ which is commutative but which has no identity.
\end{example}

\begin{example}
The $n\times n$ real matrices $M_{n}(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)$ form a non-commutative ring with an identity $I_{n}$. We can similarly
consider $M_{n}(%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
)$, $M_{n}(%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
)$ or $M_{n}(F)$ where $F$ is any field or indeed any ring.
\end{example}

\begin{example}
If $V$ is a vector space then $\mathrm{End}(V),$ the set of linear maps $%
V\rightarrow V$, forms a ring under addition and composition. The zero and
identity maps are the additive and multiplicative identities of $\mathrm{End}%
(V).$
\end{example}

\begin{example}
The power set $\mathcal{P}(X)$ of a set $X$, that is, the set of subsets of $%
X$, forms a CRI under symmetric difference $\Delta $ and intersection $\cap $%
. That is, for $A,B\subseteq X$ we have%
\begin{equation*}
A+B:=A\,\Delta \,B=(A\backslash B)\cup (B\backslash A);\qquad A\times
B:=A\cap B.
\end{equation*}%
We further have $0_{\mathcal{P}(X)}=$ $\varnothing $ and $1_{\mathcal{P}%
(X)}=X$. The additive inverse of $A$ is itself $A.$ Notice that every
element satisfies $A^{2}=A$.

If $X=\{x_{1},\ldots x_{n}\}$ then we can identify $\mathcal{P}(X)$ with $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}^{n}$ by $S\leftrightarrow (e_{1},\ldots ,e_{n})$ with $e_{i}=\bar{1}$
if and only if $x_{i}\in S.$
\end{example}

\begin{example}
The following \textbf{sequence spaces} all form commutative rings under
coordinate-wise addition and multiplication:

\begin{itemize}
\item $l^{\infty }$ the space of bounded real sequences;

\item $l^{1}$ the space of real sequences which are absolutely summable;

\item $c$ the space of convergent real sequences;

\item $c_{0}$ the space of real sequences which converge to $0$.
\end{itemize}

\noindent Both $l^{\infty }$ and $c$ have an identity in the constant
sequence $(1,1,1,\ldots )$.
\end{example}

\begin{example}
The \textbf{Gaussian integers} $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ are the complex numbers $\{a+bi:a,b\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\}$ and form a CRI.
\end{example}

\begin{example}
If $R$ is a ring and $S$ is a set, then we can form a ring $%
R^{S}=\{f:S\rightarrow R\}$ with the set of maps from $S$ to $R$ by taking $%
+ $ and $\times $ to be pointwise addition and multiplication. $R^{S}$ has
an identity if and only if $R$ has and is commutative if and only if $R$ is.
\end{example}

\begin{example}
A \textbf{quaternion }is a "four-dimensional" number of the form%
\begin{equation*}
\mathbf{q}=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}
\end{equation*}%
where $a,b,c,d$ are real numbers. The ring of quaternions is denoted $%
\mathbb{H}$ after the mathematician William Rowan Hamilton (1805-1865) who
discovered them in 1843. Two quaternions add component-wise as one would
expect and multiply according to the rules%
\begin{equation*}
\mathbf{i}^{2}=\mathbf{j}^{2}=\mathbf{k}^{2}=\mathbf{ijk}=-1.
\end{equation*}%
From these rules one can further show that%
\begin{equation*}
\mathbf{ij}=\mathbf{k}=-\mathbf{ji},\qquad \mathbf{jk}=\mathbf{i}=-\mathbf{kj%
},\qquad \mathbf{ki}=\mathbf{j}=\mathbf{-ik},
\end{equation*}%
so we see that $\mathbb{H}$ is non-commutative. The elements%
\begin{equation*}
\left\{ 1,-1,\mathbf{i},-\mathbf{i},\mathbf{j},-\mathbf{j},\mathbf{k},-%
\mathbf{k}\right\}
\end{equation*}%
form a group which is denoted $Q_{8}.$
\end{example}

\begin{definition}
Let $R$ be a ring. A subset $S\subseteq R$ is a \textbf{subring }of $R$ if
the operations $+$ and $\times $ restrict to make a ring of $S$. That is, $S$
is a subring of $R$ if:

(a) $0_{R}\in S$;

(b) whenever $r_{1},r_{2}\in S$ then $r_{1}\pm r_{2}\in S$.

(c) whenever $r_{1},r_{2}\in S$ then $r_{1}r_{2}\in S$.
\end{definition}

\begin{remark}
Note that the axioms $A1,A2,M1,D$ automatically apply in $S$ as they apply
to all elements of $R,$ including those of $S.$
\end{remark}

\begin{example}
(a) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ is a subring of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].$

(b) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ is a subring of $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.$

(c) $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$ is a subring of $C(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
).$

(d) $2%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ is a subring of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$

(e) $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
=\{a+b\mathbf{i}:a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\}$ is a subring of $\mathbb{H}.$

(e) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}$ is \textbf{not} a subring of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ as it is not a subset of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$
\end{example}

\begin{example}
$\{0_{R}\}$ and $R$ are always subrings of $R.$ Given a subset $X$ of $R$
the subring generated by $X$ is the smallest subring containing $X$. It is
the intersection of all subrings which contain $X.$
\end{example}

\begin{definition}
In a ring $R,$ we say that a non-zero element $a\in R$ is a \textbf{%
zero-divisor} if there exist non-zero $b,c\in R$ such that $ab=0=ca.$ In a
commutative ring, if $a$ is a zero-divisor, then we can assume $b=c.$
\end{definition}

\begin{definition}
An \textbf{integral domain} $R$ is a CRI which has no zero-divisors.
\end{definition}

\begin{definition}
In a ring $R,$ with an identity $1_{R},$ we say that $a\in R$ is a \textbf{%
unit} if there exists $b\in R$ such that $ab=1_{R}=ba$. If such $b$ exists
then it is unique and we will denote it $a^{-1}.$
\end{definition}

\begin{example}
(a) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,\,%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
,\,%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
,\,%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ are integral domains. In $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ the units are only $1$ and $-1$ whilst every non-zero element is a unit in 
$%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
,\,%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
,\,%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.$

(b) $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$ is an integral domain and the units are the non-zero constant
polynomials.

(c) In $M_{n}(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
),$ the non-zero singular matrices are the zero-divisors and the invertible
matrices are the units.

(d) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ is an integral domain and the units are $1,-1,i,-i$ only.

(e) In $\mathcal{P}(X),$ every non-empty proper set is a zero-divisor and
the only unit is $X$.

(f) In $\mathbb{H}$, there are no zero divisors (but, not being commutative, 
$\mathbb{H}$ is not an integral domain) and every non-zero element is a unit.
\end{example}

\begin{proposition}
(\textbf{Cancellation Law) }Let $R$ be an integral domain and $a,b,c\in R$
with $a\neq 0$. If $ab=ac$ then $b=c.$
\end{proposition}

\begin{proof}
As $ab=ac$ then $a(b-c)=0_{R}$. As $R$ is an integral domain then $a$ is not
a zero-divisor and so $b-c=0_{R}.$
\end{proof}

\begin{proposition}
Let $R$ be a ring with a $1$. Then $R^{\ast }$, the set of units in $R$,
forms a group under multiplication.
\end{proposition}

\begin{proof}
Suppose now that $u,v$ are units in $R.$ Then%
\begin{eqnarray*}
(uv)(v^{-1}u^{-1}) &=&u(vv^{-1})u^{-1}=uu^{-1}=1; \\
(v^{-1}u^{-1})(uv) &=&v^{-1}(u^{-1}u)v=v^{-1}v=1,
\end{eqnarray*}%
so that $uv$ is a unit. So multiplication is a binary operation on $R^{\ast
} $ and is associative by $M1.$ Further $1_{R}$ is the identity in $R^{\ast
} $. Also%
\begin{equation*}
(u^{-1})u=1=u(u^{-1})
\end{equation*}%
so that $u^{-1}$ is a unit and the inverse of $u$ in $R^{\ast }.$
\end{proof}

\begin{example}
(a) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{\ast }=\{1,-1\}.$

(b) $(M_{n}(F))^{\ast }=GL(n,F)$ for any field $F.$

(c) $(F[x])^{\ast }=F^{\ast }$ when identified with the non-zero constant
polynomials.

(d) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}^{\ast }=\{\bar{1},\bar{3},\bar{5},\bar{7}\}$ and is isomorphic to $%
C_{2}\times C_{2}.$
\end{example}

\begin{proposition}
Let $R$ be a ring with an identity.

(a) An element cannot be both a zero-divisor and a unit.

(b) If $R$ is finite and commutative then every non-zero element is a
zero-divisor or a unit.

(c) More generally, a non-zero element may be neither a zero-divisor or a
unit.
\end{proposition}

\begin{proof}
(a) Suppose that $a\in R$ and that $b,c,d$ are non-zero and such that $%
ab=0_{R}=ca,$ and $ad=1_{R}=da.$ Then%
\begin{equation*}
b=1_{R}b=(da)b=d(ab)=d0_{R}=0_{R}
\end{equation*}%
which is a contradiction.

\noindent (b) For $x\neq 0$ the map $r\mapsto xr$ is either 1-1 or not. If
it is 1-1 then it is onto as $R$ is finite, and hence there exists $r$ such
that $xr=1_{R}$ and we have that $x$ is a unit. If the map is not 1-1 then
there exist $r_{1}\neq r_{2}$ such that $xr_{1}=xr_{2}.$ Hence $%
x(r_{1}-r_{2})=0$ and we see that $x$ is a zero-divisor.

\noindent (c) In $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ we see $2$ is neither a unit nor a zero-divisor.
\end{proof}

In the linear algebra courses, we defined a field to be a triple $%
(R,+,\times )$ satisfying the field axioms $A1,A2,A3,A4,M1,M2,M3,M4,Z,D.$ An
integral domain satisfies each of these except (M4): existence of
multiplicative inverses. So we can reformulate the field axioms as:

\begin{definition}
A \textbf{field }is an integral domain in which every non-zero element is a
unit.
\end{definition}

\begin{theorem}
A finite integral domain is a field.
\end{theorem}

\begin{proof}
Let $R$ be a finite integral domain and $a\in R$ with $a\neq 0$. Consider
the map $m_{a}:R\rightarrow R$ given by $r\mapsto ar$. By the cancellation
law, $m_{a}$ is 1-1 and hence is onto as $R$ is finite. In particular, there
is $r\in R$ such that $ar=1_{R}$. Hence $a$ is a unit and $R$ is a field.
\end{proof}

We know that $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p},$ where $p$ is prime, is a finite field. There are in fact examples of
other finite fields. Below is an example of a finite field with 4 elements.

\begin{example}
The field of 4 elements can be written as follows. Let $F_{4}=\{0,1,2,3\}$
where $0$ is the additive identity and $1$ is the multiplicative identity.
It must be the case that, as an additive group, $F\cong C_{2}\times C_{2}$
for if $F\cong C_{4}$ we would find we had zero-divisors. And it must be the
case that $F^{\ast }\cong C_{3}$ as this is the only group of order 3. The
addition and multiplication tables for $F$ are then%
\begin{equation*}
\begin{tabular}{|l|llll|}
\hline
$+$ & $0$ & $1$ & $2$ & $3$ \\ \hline
$0$ & $0$ & $1$ & $2$ & $3$ \\ 
$1$ & $1$ & $0$ & $3$ & $2$ \\ 
$2$ & $2$ & $3$ & $0$ & $1$ \\ 
$3$ & $3$ & $2$ & $1$ & $0$ \\ \hline
\end{tabular}%
\qquad \qquad 
\begin{tabular}{|l|llll|}
\hline
$\times $ & $0$ & $1$ & $2$ & $3$ \\ \hline
$0$ & $0$ & $0$ & $0$ & $0$ \\ 
$1$ & $0$ & $1$ & $2$ & $3$ \\ 
$2$ & $0$ & $2$ & $3$ & $1$ \\ 
$3$ & $0$ & $3$ & $1$ & $2$ \\ \hline
\end{tabular}%
.
\end{equation*}
\end{example}

\begin{definition}
Given any ring $R$ we define its characteristic $\mathrm{char}R$ as follows:

\begin{itemize}
\item If such a positive integer exists, we define the $\mathrm{char}R$ to
be the smallest positive integer $n$ such that $nr=0$ for all $r\in R.$

\item If no such $n$ exists, then we define $\mathrm{char}R=0.$
\end{itemize}

\noindent In the event that $R$ has an identity, then $\mathrm{char}R$ is
the additive order of $1_{R}$ when this is finite, and $0$ when that order
is infinite.
\end{definition}

\begin{example}
(a) $\mathrm{char}%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}=n.$

(b) $\mathrm{char}%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
=\mathrm{char}%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
=\mathrm{char}%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
=\mathrm{char}%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
=0.$

(c) $F_{4}$ has characteristic $2.$

(d) For the ring $R$ of $4$ elements, with $(R,+)\cong C_{4}$ and
multiplication defined by $ab=0$ for all $a,b,$ then $\mathrm{char}R=4.$
\end{example}

\begin{proposition}
If $R$ is an integral domain then $\mathrm{char}R$ is prime or zero.
\end{proposition}

\begin{proof}
If the additive order $n$ of $1_{R}$ is finite then we have%
\begin{equation*}
\underset{n\text{ times}}{\underbrace{1_{R}+\cdots +1_{R}}}\text{\thinspace }%
=0_{R},
\end{equation*}%
and if $n=uv,$ where neither $u$ nor $v$ were $1$, we'd have%
\begin{equation*}
\underset{u\text{ times}}{\underbrace{1_{R}+\cdots +1_{R}}\,}\underset{v%
\text{ times}}{\times \,\underbrace{1_{R}+\cdots +1_{R}}}\,=0_{R}.
\end{equation*}%
As $R$ is an ID then we'd have either that%
\begin{equation*}
\underset{u\text{ times}}{\underbrace{1_{R}+\cdots +1_{R}}\,}=0_{R}\qquad 
\text{or}\qquad \underset{v\text{ times}}{\underbrace{1_{R}+\cdots +1_{R}}%
\,=0_{R}},
\end{equation*}%
both of which would contradict the minimality of $n.$
\end{proof}

\begin{proposition}
Let $F$ be a field.

(a) If $\mathrm{char}F=p$, a prime, then there is a smallest subfield of $F$
isomorphic to $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}.$

(b) If $\mathrm{char}F=0$, then there is a smallest subfield of $F$
isomorphic to $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
.$

This subfield is called the \textbf{prime subfield}.
\end{proposition}

\begin{proof}
In both cases, as $0_{F}$ and $1_{F}$ both need to be in any subfield, then
the smallest subfield of $F$ is the one generated by $1_{F}.$ In the case
that $\mathrm{char}F=p$ then the map%
\begin{equation*}
\phi :%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}\rightarrow F\text{\qquad given by\qquad }\phi (\bar{k})=\underset{k%
\text{ times}}{\underbrace{1_{F}+\cdots +1_{F}}}
\end{equation*}%
is well-defined and an isomorphism onto its image. In the case that $\mathrm{%
char}F=0$ then the map%
\begin{equation*}
\phi :%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\rightarrow F\text{\qquad given by\qquad }\phi \left( \frac{m}{n}\right)
=\left( \underset{m\text{ times}}{\underbrace{1_{F}+\cdots +1_{F}}}\right)
\left( \underset{n\text{ times}}{\underbrace{1_{F}+\cdots +1_{F}}}\right)
^{-1}
\end{equation*}%
is well-defined and an isomorphism onto its image.
\end{proof}

\begin{proposition}
Let $R$ be an integral domain. Then we can create a \textbf{field of
fractions} $F$ for $R$ as follows. We define an equivalence relation on $%
R^{2}$ by%
\begin{equation*}
(m,n)\sim (M,N)\qquad \iff \qquad mN=Mn,
\end{equation*}%
and define $F=R^{2}/\sim $ to be the set of $\sim $-equivalence classes with
the operations%
\begin{equation*}
\left[ \left( m_{1},n_{1}\right) \right] +\left[ \left( m_{2},n_{2}\right) %
\right] =\left[ \left( m_{1}n_{2}+m_{2}n_{1},n_{1}n_{2}\right) \right]
,\qquad \left[ \left( m_{1},n_{1}\right) \right] \times \left[ \left(
m_{2},n_{2}\right) \right] =\left[ \left( m_{1}m_{2},n_{1}n_{2}\right) %
\right] .
\end{equation*}%
Then

(a) $F$ is a well-defined field with $0_{F}=[(0_{R},1_{R})]$ and $%
1_{F}=[(1_{R},1_{R})].$

(b) $F$ contains a copy of $R$ as an isomorphic subring, namely the elements 
$R\equiv \{[(r,1_{R})]:r\in R\}.$

(c) $F$ is uniquely characterized by the "universal property" that any 1-1
homomorphism $\phi :R\rightarrow K$ to a field $K$, can be uniquely extended
to a 1-1 homorphism $\tilde{\phi}:F\rightarrow K$.
\end{proposition}

\begin{remark}
The above formulae may be a little less surprising when we realize in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ that%
\begin{equation*}
\frac{m_{1}}{n_{1}}+\frac{m_{2}}{n_{2}}=\frac{m_{1}n_{2}+m_{2}n_{1}}{%
n_{1}n_{2}}\qquad \text{and}\qquad \frac{m_{1}}{n_{1}}\times \frac{m_{2}}{%
n_{2}}=\frac{m_{1}m_{2}}{n_{1}n_{2}}.
\end{equation*}%
We can also see for non-zero $n_{1},n_{2}$ that%
\begin{equation*}
m_{1}n_{2}=m_{2}n_{1}\qquad \iff \qquad \frac{m_{1}}{n_{1}}=\frac{m_{2}}{%
n_{2}}.
\end{equation*}%
Thus the above construction would yield $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ as the field of fractions of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ and would identify $m/n$ with the equivalence class $[(m,n)].$
\end{remark}

\begin{proof}
(Non-examinable) We will not prove the details here, but rather just give a
list of all the facts that need checking, none of which are particularly
difficult.

(i) $\sim $ is an equivalence relation.

(ii) $+$ and $\times $ are well-defined binary operations on the set of
equivalence classes.

(iii) the field axioms hold amongst the equivalence classes with $%
0_{F}=[(0_{R},1_{R})]$ and $1_{F}=[(1_{R},1_{R})].$

(iv) $r\mapsto \lbrack (r,1_{R})]$ is a 1-1 homomorphism from $R$ to $F.$

(v) defining $\phi \left[ \left( m,n\right) \right] =\phi (m)\phi (n)^{-1}$
is a well-defined 1-1 homomorphism from $F$ to $K$ which extends $\phi .$

(vi) For any other field $\tilde{F}$ with this universal property, extending
inclusion $\iota :R\rightarrow F$ yields an isomorphosm $\tilde{\iota}:%
\tilde{F}\rightarrow F.$
\end{proof}

\begin{example}
(a) The field of fractions of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ is $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
.$

(b) The field of fractions of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ is $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
(x).$

(c) The field of fractions of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{2}]$ is $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}].$

(d) The field of fractions of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \pi ]$ is $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
(\pi ).$
\end{example}

\chapter{The Isomorphism Theorems}

\begin{definition}
Let $R$ be a ring. A non-empty subset $I\subseteq R$ is said to be a \textbf{%
ideal} of $R$ if%
\begin{equation*}
\text{whenever }i_{1},i_{2}\in I\text{ then }i_{1}+i_{2}\in I\text{;\qquad
whenever }i\in I,r\in R\text{ then }ri\in I\text{ and }ir\in I.
\end{equation*}%
This is then written $I\lhd R$. In particular, ideals are subrings though
the converse is not generally true. [N.B. any text which insists that rings
have 1s will then have a convention that ideals are not generally subrings.]
\end{definition}

\begin{definition}
Let $R$ be a ring and $a\in R.$ Then the \textbf{principal ideal} $\langle
a\rangle $ generated by $a$ is the smallest ideal to contain $a.$ If $R$ has
an identity $1_{R}$ then%
\begin{equation*}
\langle a\rangle =\left\{ ras:r,s\in R\right\} .
\end{equation*}%
$\langle a\rangle $ is also commonly written as $(a).$ An ideal $I$ of $R$
is said to be \textbf{principal} if there exists $a\in R$ such that $%
I=\langle a\rangle .$
\end{definition}

\begin{example}
$\left\{ 0_{R}\right\} $ and $R$ are always ideals of $R.$
\end{example}

\begin{example}
Let $R=C(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)$ and $S=%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$. Then $S$ is a subring of $R$ but is not an ideal of $R,$ as $%
x\in S,\,e^{x}\in R$ but $xe^{x}\notin S.$
\end{example}

\begin{example}
Let $R=C(%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
)$ and $I=\left\{ f\in R:f\left( 0\right) =0\right\} .$ Then $I$ is an ideal
of $R.$
\end{example}

\begin{example}
Let $R$ be a ring with a $1_{R}$ and $a_{1},\ldots ,a_{k}\in R$. Then the%
\textbf{\ ideal generated by} $\left\{ a_{1},\ldots ,a_{k}\right\} $ is%
\begin{equation*}
I=\langle a_{1},\ldots ,a_{k}\rangle =\left\{ r_{1}a_{1}s_{1}+\cdots
+r_{k}a_{k}s_{k}:r_{i},s_{i}\in R\right\}
\end{equation*}%
is the smallest ideal of $R$ containing $a_{1},\ldots ,a_{k}.$
\end{example}

\begin{example}
Let $R=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ and $I=\langle 2,x\rangle .$ Then $I$ is an example of an ideal
which is not principal. Note%
\begin{equation*}
I=\langle 2,x\rangle =\{p(x)\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]:p(x)\text{ has an even constant term}\}.
\end{equation*}%
If $I$ were principal then, for some $f(x)$, we would have $I=\langle
f(x)\rangle =f(x)%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ and then $f(x)$ would divide both $2$ and $x$. The only such
polynomials are $f(x)=\pm 1$ but $\langle \pm 1\rangle =%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]\neq I.$
\end{example}

\begin{proposition}
The ideals of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ are $n%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
=\langle n\rangle $ where $n\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ So every ideal of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ is principal.
\end{proposition}

\begin{proof}
For any $n\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$, we have $n%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
=\langle n\rangle \lhd 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ Conversely, suppose that $I\lhd 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$. Then, in particular, $I$ is a subgroup of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ and $I=n%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ for some $n\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$
\end{proof}

\begin{proposition}
Let $R$ be a CRI. Then $R$ is a field if and only if the only ideals of $R$
are $\left\{ 0_{R}\right\} $ and $R$.
\end{proposition}

\begin{proof}
Suppose that $R$ is a field and $I\lhd R$. If $I\neq \left\{ 0_{R}\right\} $
then there exists non-zero $a\in I$. As $R$ is a field, then $a$ is a unit
and so there exists $b\in R$ such that $ba=1_{R}$. Then, for any $r\in R$,%
\begin{equation*}
\left( rb\right) a=r\in I
\end{equation*}%
and we see $I=R$. Conversely, suppose that the only ideals are $\left\{
0_{R}\right\} $ and $R$ and that $a\neq 0$. Then $1_{R}\in R=\langle
a\rangle $ and there exists $r\in R$ such that $ar=1_{R}$. So $a$ is a unit
and $R$ is a field.
\end{proof}

\begin{remark}
As with the case of normal subgroups for groups, ideals are the natural
subsets a ring might be "modulo"-ed by. If we are to introduce new rules
into the algebra of a ring $R$ and set $i,j\sim 0_{R}$ (where $\sim $ is an
equivalence relation) then for reasonable algebra in $R/\sim ,$ we must
expect to have%
\begin{equation*}
i\pm j\sim 0_{R}\qquad \text{and\qquad }ri\sim 0_{R}\text{\quad and\quad }%
ir\sim 0_{R}\text{\quad for any }r\in R.
\end{equation*}
\end{remark}

\begin{theorem}
Let $R$ be a ring\ and let $I\lhd R$. The \textbf{coset} of $r\in R$ is%
\begin{equation*}
r+I=\{r+i:i\in I\}.
\end{equation*}%
The operations 
\begin{eqnarray*}
\left( r_{1}+I\right) \oplus \left( r_{2}+I\right) &=&\left(
r_{1}+r_{2}\right) +I; \\
\left( r_{1}+I\right) \otimes \left( r_{2}+I\right) &=&\left(
r_{1}r_{2}\right) +I;
\end{eqnarray*}%
lead to well-defined binary operations on the set $R/I$ of cosets of $I$
with $(R/I,\oplus ,\otimes )$ being a CRI known as the \textbf{quotient ring}%
. If $R$ is commutative then $R/I$ is commutative, and if $R$ has an
identity then so does $R/I$ with $1_{R/I}=1_{R}+I.$
\end{theorem}

\begin{notation}
The notation $r+I$ will become cumbersome and so we will write $\bar{r}=r+I$%
. This very much ties in with the notion that $R/I$ is "$R\func{mod}I$" and
that $r+I$ contains all those $s$ such that $r\equiv s\func{mod}I.$
\end{notation}

\begin{proof}
Suppose that $I\lhd R.$ Note for $r_{1},r_{2}\in R$ that%
\begin{equation*}
\overline{r_{1}}=\overline{r_{2}}\quad \iff \quad r_{1}-r_{2}\in I.
\end{equation*}%
So suppose that $\overline{r_{1}}=\overline{r_{2}}$ and $\overline{s_{1}}=%
\overline{s_{2}}$. Then%
\begin{equation*}
\left( r_{1}+s_{1}\right) -\left( r_{2}+s_{2}\right) =\left(
r_{1}-r_{2}\right) +\left( s_{1}-s_{2}\right) \in I
\end{equation*}%
showing that%
\begin{equation*}
\overline{r_{1}}\oplus \overline{s_{1}}=\overline{r_{1}+s_{1}}=\overline{%
r_{2}+s_{2}}=\overline{r_{2}}\oplus \overline{s_{2}}.
\end{equation*}%
That is $\oplus $ is well-defined. Similarly%
\begin{equation*}
r_{1}s_{1}-r_{2}s_{2}=r_{1}\left( s_{1}-s_{2}\right) +s_{2}\left(
r_{1}-r_{2}\right) \in I
\end{equation*}%
showing that%
\begin{equation*}
\overline{r_{1}} \otimes \overline{s_{1}}=\overline{r_{1}s_{1}}=\overline{%
r_{2}s_{2}}=\overline{r_{2}}\otimes \overline{s_{2}}
\end{equation*}%
and that $\otimes $ is well-defined on the set of cosets $R/I.$

$(R/I,\oplus ,\otimes )$ meets the axioms of a CRI because these properties
are inherited from the fact that $\left( R/I,+,\times \right) $ is a CRI.%
\begin{eqnarray*}
(A1) &:&\left( \bar{r}\oplus \bar{s}\right) \oplus \bar{t}=\overline{(r+s)+t}%
=\overline{r+(s+t)}=\bar{r}\oplus \left( \bar{s}\oplus \bar{t}\right) . \\
(A2) &:&\bar{r}\oplus \bar{s}=\overline{r+s}=\overline{s+r}=\bar{s}\oplus 
\bar{r} \\
(A3) &:&0_{R/I}=\overline{0_{R}}=I. \\
(A4) &:&-\bar{r}=\overline{-r} \\
(M1) &:&\left( \bar{r}\otimes \bar{s}\right) \otimes \bar{t}=\overline{(rs)t}%
=\overline{r(st)}=\bar{r}\otimes \left( \bar{s}\otimes \bar{t}\right) . \\
(D) &:&\bar{r}\otimes \left( \bar{s}\oplus \bar{t}\right) =\overline{r(s+t)}=%
\overline{rs+rt}=\left( \bar{r}\otimes \bar{s}\right) \oplus (\bar{r}\otimes 
\bar{t}). \\
(M2) &:&\bar{r}\otimes \bar{s}=\overline{rs}=\overline{sr}=\bar{s}\otimes 
\bar{r} \\
(M3) &:&1_{R/I}=\overline{1_{R}}=1_{R}+I.
\end{eqnarray*}
\end{proof}

\begin{example}
Let $R=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,\,I=\langle n\rangle =n%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$. As a ring, we can naturally identify $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
/n%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}.$
\end{example}

\begin{example}
Let $R=%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ and $I=\langle x^{2}\rangle $. Then $R/I$ is not an integral
domain as $\bar{x}$ is a zero-divisor (because $\bar{x}\neq \bar{0}$ and $%
\bar{x}\bar{x}=\bar{0}$). To better understand the ring $R/I$ note by the
division algorithm, that every polynomial $p(x)$ can uniquely be written%
\begin{equation*}
p(x)=a+bx+x^{2}p(x)\text{ \ for some }a,b\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
,f(x)\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].
\end{equation*}%
So every coset in $R/I$ can uniquely be represented as $a+b\bar{x}$ where $%
a,b\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
.$ These cosets than add and multiply as%
\begin{eqnarray*}
(a+b\bar{x})+\left( c+d\bar{x}\right) &=&\left( a+c\right) +(b+d)\bar{x}; \\
(a+b\bar{x})\left( c+d\bar{x}\right) &=&ac+(bc+ad)\bar{x},
\end{eqnarray*}%
as $\bar{x}^{2}=\bar{0}$. From this we can see that the only zero-divisors
in $R/I$ are of the form $b\bar{x}$ where $b\neq 0.$
\end{example}

\begin{example}
Write down the addition and multiplication tables of $R_{1}$ and $R_{2}$
where%
\begin{equation*}
R_{1}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]}{\langle x^{2}\rangle },\qquad R_{2}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]}{\langle x^{2}+1\rangle }.
\end{equation*}%
Show that the rings are isomorphic.
\end{example}

\begin{solution}
By the division algorithm, there are four cosets of $\langle x^{2}\rangle $
in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]$ and of $\langle x^{2}+1\rangle $ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]$, and we can represent these cosets in both rings with the elements%
\begin{equation*}
\{\bar{0},\bar{1},\bar{x},\bar{x}+\bar{1}\}.
\end{equation*}%
The addition and mutliplication tables of $R_{1}$ are then%
\begin{equation*}
\begin{tabular}{|c|c|c|c|c|}
\hline
$+$ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{0}$ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{1}$ & $\bar{1}$ & $\bar{0}$ & $\bar{x}+\bar{1}$ & $\bar{x}$ \\ \hline
$\bar{x}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ & $\bar{0}$ & $\bar{1}$ \\ \hline
$\bar{x}+\bar{1}$ & $\bar{x}+\bar{1}$ & $\bar{x}$ & $\bar{1}$ & $\bar{0}$ \\ 
\hline
\end{tabular}%
\,,\qquad 
\begin{tabular}{|c|c|c|c|c|}
\hline
$\times $ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{0}$ & $\bar{0}$ & $\bar{0}$ & $\bar{0}$ & $\bar{0}$ \\ \hline
$\bar{1}$ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{x}$ & $\bar{0}$ & $\bar{x}$ & $\bar{0}$ & $\bar{x}$ \\ \hline
$\bar{x}+\bar{1}$ & $\bar{0}$ & $\bar{x}+\bar{1}$ & $\bar{x}$ & $\bar{1}$ \\ 
\hline
\end{tabular}%
\,,
\end{equation*}%
and those for $R_{2}$ are%
\begin{equation*}
\begin{tabular}{|c|c|c|c|c|}
\hline
$+$ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{0}$ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{1}$ & $\bar{1}$ & $\bar{0}$ & $\bar{x}+\bar{1}$ & $\bar{x}$ \\ \hline
$\bar{x}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ & $\bar{0}$ & $\bar{1}$ \\ \hline
$\bar{x}+\bar{1}$ & $\bar{x}+\bar{1}$ & $\bar{x}$ & $\bar{1}$ & $\bar{0}$ \\ 
\hline
\end{tabular}%
\,,\qquad 
\begin{tabular}{|c|c|c|c|c|}
\hline
$\times $ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{0}$ & $\bar{0}$ & $\bar{0}$ & $\bar{0}$ & $\bar{0}$ \\ \hline
$\bar{1}$ & $\bar{0}$ & $\bar{1}$ & $\bar{x}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{x}$ & $\bar{0}$ & $\bar{x}$ & $\bar{1}$ & $\bar{x}+\bar{1}$ \\ \hline
$\bar{x}+\bar{1}$ & $\bar{0}$ & $\bar{x}+\bar{1}$ & $\bar{x}+\bar{1}$ & $%
\bar{0}$ \\ \hline
\end{tabular}%
\,.
\end{equation*}%
We can see that the first set of tables match the second set of tables when
we swap every $\bar{x}$ with $\bar{x}+\bar{1}$ and vice-versa. This is
perhaps less of a surprise when we note%
\begin{equation*}
x^{2}+1=\left( x+1\right) ^{2}\text{ \ \ in }%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x].
\end{equation*}
\end{solution}

\begin{definition}
Let $R$ and $S$ be rings. A \textbf{(ring) homomorhism }$\phi \colon
R\rightarrow S$ is a map satisfying%
\begin{equation*}
\phi (r_{1}+_{R}r_{2})=\phi (r_{1})+_{S}\phi (r_{2}),\qquad \phi (r_{1}\ast
_{R}r_{2})=\phi (r_{1})\ast _{S}\phi (r_{2})\qquad \text{for all }%
r_{1},r_{2}\in R.
\end{equation*}%
If further $\phi $ is bijective then $\phi $ is called a \textbf{(ring)\
isomorphism.}
\end{definition}

\begin{proposition}
Let $R$ and $S$ be rings and $\phi \colon R\rightarrow S$ a homomorphism.
Let $r\in R$ and $n$ be an integer. Then:

(a) $\phi (0_{R})=0_{S}.$

(b) if $S$ is an integral domain and $R$ has an identity, then either $\phi
\equiv 0$ or $\phi (1_{R})=1_{S}.$

(c) $\phi (nr)=n\phi (r)$.

(d) if $n\geqslant 1$ then $\phi (r^{n})=\phi (r)^{n}.$
\end{proposition}

\begin{proof}
Left as exercises.
\end{proof}

\begin{example}
Let $R$ be a CRI and $a\in R$. Then the map $\phi \colon R\left[ x\right]
\rightarrow R$ given by $\phi (p(x))=p(a)$ is a homomorphism because -- by
definition -- we have%
\begin{equation*}
(\phi +\psi )(a)=\phi (a)+\psi (a),\qquad (\phi \psi )(a)=\phi (a)\psi (a).
\end{equation*}
\end{example}

\begin{example}
In a similar vein, let $A$ be an $n\times n$ real matrix. Then the map $\phi
:%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\left[ A\right] ,$ defined by $p(x)\mapsto p(A),$ is a homomorphism.
\end{example}

\begin{example}
Let $n\geqslant 2$ be an integer. Then $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}$ given by $x\mapsto \bar{x}$ is a ring homomorphism as%
\begin{equation*}
\overline{x+y}=\bar{x}+\bar{y},\qquad \overline{xy}=\bar{x}\bar{y}.
\end{equation*}%
More generally if $I\lhd R$ then the quotient maps $\pi :R\rightarrow R/I$
given by $\pi (r)=\bar{r}$ is a homomorphism.
\end{example}

\begin{example}
Let $n\geqslant 2.$ The only ring homomorphism $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ is the zero map because $0=\phi \left( \bar{0}\right) =\phi (\bar{n}%
)=n\phi (\bar{1})$ and so $\phi (\bar{1})=0$ and $\phi \equiv 0.$
\end{example}

\begin{example}
The only ring homomorphisms $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ are $x\mapsto 0$ for all $x$ and $x\mapsto x.$
\end{example}

\begin{example}
Find all the ring homomorphisms $\phi \colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$.
\end{example}

\begin{solution}
If $\phi (1)=1$ then $\phi (m)=m$ for any integer $m$. Likewise, for $n>0,$
we have $n\phi (m/n)=\phi (m)=m$ and so $\phi (m/n)=m/n.$ Hence the only
ring homomorphisms are%
\begin{equation*}
\phi (x)=0\qquad \text{and\qquad }\phi (x)=x.
\end{equation*}
\end{solution}

\begin{example}
A ring homomorphism $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}$ is of the form $\phi (\bar{x})=\bar{a}\bar{x}$ where $\bar{a}^{2}=\bar{%
a}.$

\noindent Hence the ring homomorphisms $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}$ are%
\begin{equation*}
\phi (\bar{x})=\bar{0},\qquad \phi (\bar{x})=\bar{x},\qquad \phi (\bar{x})=%
\bar{3}\bar{x},\qquad \phi (\bar{x})=\bar{4}\bar{x},
\end{equation*}%
and the only ring homomorphisms $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}$ (where $p$ is a prime) are%
\begin{equation*}
\phi (\bar{x})=\bar{0},\qquad \phi (\bar{x})=\bar{x}.
\end{equation*}
\end{example}

\begin{solution}
Certainly any homomorphism $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}$ is of the form $\phi (\bar{x})=\bar{a}\bar{x}$ where $\bar{a}=\phi (%
\bar{1})$. For such maps we immediately have%
\begin{equation*}
\phi (\overline{x+y})=\phi (\bar{x})+\phi (\bar{y}).
\end{equation*}%
Futher we need%
\begin{equation*}
\bar{a}=\phi (\bar{1})=\phi (\bar{1})\phi (\bar{1})=\bar{a}^{2}.
\end{equation*}%
Finally if $\bar{a}^{2}=\bar{a}$ then we have%
\begin{equation*}
\phi (\bar{x}\bar{y})=\bar{a}\bar{x}\bar{y}=\left( \bar{a}\bar{x}\right)
\left( \bar{a}\bar{y}\right) =\phi (\bar{x})\phi (\bar{y}).
\end{equation*}%
We can check that $\bar{0},\bar{1},\bar{3},\bar{4}$ are the only roots of $%
x\left( x-1\right) =0$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}$. In $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}$, which is a field when $p$ is prime, there are no zero-divisors and so 
$0$ and $1$ are the only roots of this quadratic.
\end{solution}

\begin{example}
Show that $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is not (ring) isomorphic to $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.$
\end{example}

\begin{solution}
Suppose that $\phi \colon 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is a ring isomorphism. Then $\phi (1)=1$ and so%
\begin{equation*}
\phi (i)^{2}=\phi (i^{2})=\phi (-1)=-1
\end{equation*}%
but there is no element $x\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ which satisfies $x^{2}=-1.$
\end{solution}

\begin{definition}
Let $\phi \colon R\rightarrow S$ be a homomorphism between rings. Then the 
\textbf{kernel }of $\phi $, written $\ker \phi $, is%
\begin{equation*}
\ker \phi =\left\{ r\in R:\phi (r)=0_{S}\right\} \subseteq R.
\end{equation*}%
The \textbf{image }of $\phi $, written $\func{Im}\phi $, is%
\begin{equation*}
\func{Im}\phi =\left\{ \phi (r):r\in R\right\} \subseteq S.
\end{equation*}
\end{definition}

\begin{proposition}
Let $\phi \colon R\rightarrow S$ be a homomorphism between rings. Then:

(a) $\ker \phi $ is an ideal of $R.$

(b) $\func{Im}\phi $ is a subring of $S.$
\end{proposition}

\begin{proof}
(a) Suppose that $i_{1},i_{2}\in \ker \phi $ and $r\in R.$ Then%
\begin{eqnarray*}
\phi (i_{1}\pm i_{2})=\phi (i_{1})\pm \phi (i_{2})=0_{S}\pm 0_{S}=0_{S}; &&
\\
\phi (ri_{1})=\phi (ri_{1})=\phi (r)\phi (i_{1})=\phi (r)0_{S}=0_{S}, &&
\end{eqnarray*}%
showing that $i_{1}\pm i_{2}\in \ker \phi $ and that $ri_{1}\in \ker \phi .$
Similarly $i_{1}r\in \ker \phi .$

(b) Suppose that $s_{1},s_{2}\in \func{Im}\phi $. Then there exist $r_{i}\in
R$ such that $\phi (r_{i})=s_{i}$ and so%
\begin{eqnarray*}
s_{1}\pm s_{2}=\phi (r_{1})\pm \phi (r_{2})=\phi (r_{1}\pm r_{2})\in \func{Im%
}\phi ; && \\
s_{1}s_{2}=\phi (r_{1})\phi (r_{2})=\phi (r_{1}r_{2})\in \func{Im}\phi . &&
\end{eqnarray*}
\end{proof}

\begin{example}
Let $R$ be a CRI and $a\in R$. The kernel and image of the map $\phi \colon R%
\left[ x\right] \rightarrow R$ given by $\phi (p(x))=p(a)$ are%
\begin{equation*}
\ker \phi =\langle x-a\rangle ;\qquad \func{Im}\phi =R.
\end{equation*}%
Note for the former that we cannot presume the division algorithm applies
(which would be the case if $R$ were a field) but we can work with
identities such as%
\begin{equation*}
x^{n}=(x^{n-1}+ax^{n-2}+\cdots +a^{n-2}x+a^{n-1})(x-a)+a^{n}.
\end{equation*}
\end{example}

\begin{example}
The kernel and image of the map $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}$ given by $x\mapsto \bar{x}$ are%
\begin{equation*}
\ker \phi =n%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
;\qquad \func{Im}\phi =%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}.
\end{equation*}%
More generally $\pi \colon R\rightarrow R/I$ given by $\pi (r)=\bar{r}$ has
kernel $I$ and image $R/I.$
\end{example}

\begin{example}
The kernel and image of the map $\phi \colon 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ given by $p(x)\mapsto p(i)$ are%
\begin{equation*}
\ker \phi =\langle x^{2}+1\rangle ;\qquad \func{Im}\phi =%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
\end{equation*}%
This is because any real polynomial which has $i$ as a root also has $-i$ as
a root (conjugate pairs).
\end{example}

\begin{example}
The kernels and images of the ring homomorphisms $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}$ are%
\begin{equation*}
\begin{tabular}{|c|c|c|c|c|}
\hline
$\phi $ & $\phi (\bar{x})=\bar{0}$ & $\phi (\bar{x})=\bar{x}$ & $\phi (\bar{x%
})=\bar{3}\bar{x}$ & $\phi (\bar{x})=\bar{4}\bar{x}$ \\ \hline
$\ker \phi $ & $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}$ & $\{\bar{0}\}$ & $\{\bar{0},\bar{2},\bar{4}\}$ & $\left\{ \bar{0},%
\bar{3}\right\} $ \\ \hline
$\func{Im}\phi $ & $\{\bar{0}\}$ & $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}$ & $\{\bar{0},\bar{3}\}$ & $\{\bar{0},\bar{2},\bar{4}\}$ \\ \hline
\end{tabular}%
\end{equation*}
\end{example}

\begin{example}
\label{313}Let%
\begin{equation*}
X=\left( 
\begin{array}{cc}
0 & 1 \\ 
0 & 0%
\end{array}%
\right) \text{ \ \ and \ \ }Y=\left( 
\begin{array}{cc}
0 & 1 \\ 
1 & 0%
\end{array}%
\right) \text{ \ \ and \ \ }Z=\left( 
\begin{array}{cc}
0 & 1 \\ 
-1 & 0%
\end{array}%
\right) .
\end{equation*}%
Note that $X^{2}=0_{2},\,Y^{2}=I_{2},\,Z^{2}=-I_{2}.$ The homomorphisms%
\begin{eqnarray*}
\alpha &:&%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\left[ X\right] \text{ \ given by \ }\alpha (p(x))=p(X); \\
\beta &:&%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\left[ Y\right] \text{ \ given by \ }\beta (p(x))=p(Y); \\
\gamma &:&%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\left[ Z\right] \text{ \ given by \ }\gamma (p(x))=p(Z),
\end{eqnarray*}%
can be shown to have kernels%
\begin{equation*}
\ker \alpha =\langle x^{2}\rangle ,\qquad \ker \beta =\langle x^{2}-1\rangle
,\qquad \ker \gamma =\langle x^{2}+1\rangle .
\end{equation*}
\end{example}

As with the case of groups we then have an equivalent version of the
isomorphism theorem.

\begin{theorem}
(\textbf{First Isomorphism Theorem for Rings) }Let $\phi :R\rightarrow S$ be
a homomorphism between CRIs. Then:

(a) $\ker \phi \lhd R.$

(b) $\func{Im}\phi \leqslant S.$

(c)\ The map 
\begin{equation*}
\bar{\phi}\colon \frac{R}{\ker \phi }\rightarrow \func{Im}\phi \text{ \ \
given by \ \ }\bar{\phi}(r+\ker \phi )=\phi (r)
\end{equation*}%
is a (ring)\ isomorphism.
\end{theorem}

\begin{proof}
(a)\ and (b)\ were proven earlier. To show (c) we note (in a similar fashion
to groups)\ that%
\begin{eqnarray*}
\bar{\phi}\left( r_{1}+\ker \phi \right) =\bar{\phi}\left( r_{2}+\ker \phi
\right) &\iff &\phi (r_{1})=\phi (r_{2}) \\
&\iff &\phi (r_{1}-r_{2})=0_{S} \\
&\iff &r_{1}-r_{2}\in \ker \phi \\
&\iff &r_{1}+\ker \phi =r_{2}+\ker \phi .
\end{eqnarray*}%
The above implications right-to-left show that $\bar{\phi}$ is well-defined
and those left-to-right that $\bar{\phi}$ is injective. We also note $\bar{%
\phi}$ clearly maps onto $\func{Im}\phi $. Finally we note%
\begin{eqnarray*}
\bar{\phi}\left( \left( r_{1}+\ker \phi \right) +\left( r_{2}+\ker \phi
\right) \right) &=&\bar{\phi}\left( \left( r_{1}+r_{2}\right) +\ker \phi
\right) \\
&=&\phi \left( r_{1}+r_{2}\right) \\
&=&\phi (r_{1})+\phi (r_{2}) \\
&=&\bar{\phi}\left( r_{1}+\ker \phi \right) +\bar{\phi}\left( r_{2}+\ker
\phi \right)
\end{eqnarray*}%
and%
\begin{eqnarray*}
\bar{\phi}\left( \left( r_{1}+\ker \phi \right) \left( r_{2}+\ker \phi
\right) \right) &=&\bar{\phi}\left( \left( r_{1}r_{2}\right) +\ker \phi
\right) \\
&=&\phi \left( r_{1}r_{2}\right) \\
&=&\phi (r_{1})\phi (r_{2}) \\
&=&\bar{\phi}\left( r_{1}+\ker \phi \right) \bar{\phi}\left( r_{2}+\ker \phi
\right) .
\end{eqnarray*}
\end{proof}

If we apply the isomorphism theorem to the following homomorphisms we find:

\begin{example}
Let $a\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ and $\phi \colon 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ where $p(x)\mapsto p(a)$. The isomorphism theorem then says that%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-a\rangle }\cong 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
.
\end{equation*}
\end{example}

\begin{example}
The isomorphism theorem applied to the map $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}$ where $n\mapsto \bar{n}$ shows that%
\begin{equation*}
\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}{n%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{n}.
\end{equation*}
\end{example}

\begin{example}
The isomorphism theorem applied to the map $\phi \colon 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ where $p(x)\mapsto p(i)$ shows that%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{2}+1\rangle }\cong 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
\end{equation*}
\end{example}

\begin{example}
The isomorphism theorem applied to the map $\phi \colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ where $p(x)\mapsto p(\sqrt{2})$ shows that%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle x^{2}-2\rangle }\cong 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}].
\end{equation*}
\end{example}

\begin{example}
The isomorphism theorem applied to the map $\phi \colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]\rightarrow 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ where $p(x)\mapsto p\left( \frac{1}{2}\right) $ shows that%
\begin{equation*}
\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]}{\langle 2x-1\rangle }\cong \left\{ \frac{m}{2^{k}}:m\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,\text{\thinspace }k\geqslant 0\right\} =%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\left[ 1/2\right] .
\end{equation*}
\end{example}

\begin{example}
Let%
\begin{equation*}
X=\left( 
\begin{array}{cc}
0 & 1 \\ 
0 & 0%
\end{array}%
\right) \text{ \ \ and \ \ }Y=\left( 
\begin{array}{cc}
0 & 1 \\ 
1 & 0%
\end{array}%
\right) \text{ \ \ and \ \ }Z=\left( 
\begin{array}{cc}
0 & 1 \\ 
-1 & 0%
\end{array}%
\right) ,
\end{equation*}%
(as in Example \ref{313}) so that%
\begin{eqnarray*}
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack X] &=&\left\{ aI+bX:a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} =\left\{ \left( 
\begin{array}{cc}
a & b \\ 
0 & a%
\end{array}%
\right) :a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} . \\
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack Y] &=&\left\{ aI+bY:a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} =\left\{ \left( 
\begin{array}{cc}
a & b \\ 
b & a%
\end{array}%
\right) :a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} . \\
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack Z] &=&\left\{ aI+bZ:a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} =\left\{ \left( 
\begin{array}{cc}
a & b \\ 
-b & a%
\end{array}%
\right) :a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} .
\end{eqnarray*}%
The isomorphism theorem applied to the maps%
\begin{equation*}
\alpha (p(x))=p(X),\qquad \beta (p(x))=p(Y),\qquad \gamma (p(x))=p(Z),
\end{equation*}%
shows that%
\begin{equation*}
\left\{ \left( 
\begin{array}{cc}
a & b \\ 
0 & a%
\end{array}%
\right) :a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} \cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{2}\rangle },
\end{equation*}%
that%
\begin{equation*}
\left\{ \left( 
\begin{array}{cc}
a & b \\ 
b & a%
\end{array}%
\right) :a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} \cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{2}-1\rangle }
\end{equation*}%
which further we know to be isomorphic to $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}$ (as in Sheet 1, Question 7), and that%
\begin{equation*}
\left\{ \left( 
\begin{array}{cc}
a & b \\ 
-b & a%
\end{array}%
\right) :a,b\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\right\} \cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\left[ x\right] }{\langle x^{2}+1\rangle }\cong 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
,
\end{equation*}%
(as in Sheet 1, Question 2).
\end{example}

We conclude this discussion with an important application of the isomorphism
theorem, namely the Chinese remainder theorem. Specific instances of its use
date back as far as the 3rd century, and a general alogorithm for its
solution in the integers to the 6th century. Historically it applies to
problems such as the following.

\begin{example}
Say that an integer satisfies $x\equiv 3\func{mod}7$ and $x\equiv 7\func{mod}%
13$, what is $x\func{mod}91$?
\end{example}

\begin{solution}
A slightly ad hoc approach is to note that if $x\equiv r\func{mod}91,$ where 
$0\leqslant r<91,$ then $x=r+91k$ and so%
\begin{equation*}
r\func{mod}13=x\func{mod}13=7\quad \implies \quad r=7,20,33,46,59,72,85.
\end{equation*}%
But these numbers $r$ satisfy%
\begin{equation*}
7,20,33,46,59,72,85\equiv 0,6,5,4,3,2,1\func{mod}7
\end{equation*}%
so that we see $r=59$ is the only one satisfying $r\equiv 3\equiv \func{mod}%
7.$ So we see $x\equiv 59\func{mod}91.$
\end{solution}

More generally the Chinese remainder theorem is a means of decomposing a
quotient ring $R/(I\cap J)$ where $I$ and $J$ are \textit{coprime} ideals.

\begin{definition}
Given ideals $I$ and $J$ of a ring $R$, then we can define the following
ideals in terms of them.

(a) Their \textbf{sum} $I+J$ equals%
\begin{equation*}
I+J=\{i+j:i\in I,\,j\in J\}.
\end{equation*}%
We say that $I$ and $J$ are \textbf{coprime} if $I+J=R.$

(b)\ Their\textbf{\ intersection} $I\cap J$ equals%
\begin{equation*}
I\cap J=\{r:r\in I\text{ \ and \ }r\in J\}.
\end{equation*}%
(c) Their \textbf{product} $IJ$ equals%
\begin{equation*}
IJ=\left\{ \sum_{k=1}^{n}i_{k}j_{k}:i_{k}\in I,\,j_{k}\in I\right\} .
\end{equation*}
\end{definition}

\begin{example}
If $m$ and $n$ are integers, note that%
\begin{eqnarray*}
\langle m\rangle +\langle n\rangle &=&\langle h\rangle \qquad \text{where \ }%
h=\mathrm{hcf}(m,n). \\
\langle m\rangle \cap \langle n\rangle &=&\langle l\rangle \qquad \text{%
where \ }l=\mathrm{\func{lcm}}(m,n). \\
\langle m\rangle \langle n\rangle &=&\langle mn\rangle .
\end{eqnarray*}%
Note that $\langle m\rangle $ and $\langle n\rangle $ are coprime ideals if
and only if $m$ and $n$ are coprime integers.
\end{example}

\begin{theorem}
\label{Chinese}(\textbf{Chinese Remainder\ Theorem) }Let $I$ and $J$ be
coprime ideals of a ring\ $R$ with an identity $1_{R}.$ Then the map%
\begin{equation*}
\phi \colon \frac{R}{I\cap J}\rightarrow \frac{R}{I}\oplus \frac{R}{J}\qquad 
\text{given by}\qquad r+I\cap J\mapsto (r+I,r+J)
\end{equation*}%
is a well-defined isomorphism.
\end{theorem}

\begin{proof}
Firstly suppose that $r_{1}+I\cap J=r_{2}\cap I+J.$ Then $r_{1}-r_{2}=k$ for
some integer $k$ in both $I$ and $J.$ Hence $r_{1}+I=r_{2}+I$ and $%
r_{1}+J=r_{2}+J.$ Hence $\phi $ is well-defined. Further%
\begin{eqnarray*}
\phi \left( (r_{1}+r_{2})+I\cap J\right) &=&\left(
(r_{1}+r_{2})+I,(r_{1}+r_{2})+J\right) \\
&=&\left( r_{1}+I,r_{1}+J\right) +\left( r_{2}+I,r_{2}+J\right) \\
&=&\phi \left( r_{1}+I\cap J\right) +\phi \left( r_{2}+I\cap J\right) .
\end{eqnarray*}%
Similarly%
\begin{eqnarray*}
\phi \left( r_{1}r_{2}+I\cap J\right) &=&\left(
r_{1}r_{2}+I,r_{1}r_{2}+J\right) \\
&=&\left( r_{1}+I,r_{1}+J\right) \left( r_{2}+I,r_{2}+J\right) \\
&=&\phi \left( r_{1}+I\cap J\right) \phi \left( r_{2}+I\cap J\right) ,
\end{eqnarray*}%
and so $\phi $ is a ring homorphism. Finally we need to check that $\phi $
is 1-1 and onto. If%
\begin{equation*}
\phi \left( r_{1}+I\cap J\right) =\phi \left( r_{2}+I\cap J\right)
\end{equation*}%
then $r_{1}+I=r_{2}+I$ and $r_{1}+J=r_{2}+J$. So $r_{1}-r_{2}$ lies in both $%
I$ and $J$ and we have%
\begin{equation*}
r_{1}+I\cap J=r_{2}+I\cap J.
\end{equation*}%
Hence $\phi $ is 1-1. Finally as $I$ and $J$ are coprime ideals then we know
that there exist $i_{0}\in I$ and $j_{0}\in J$ such that $i_{0}+j_{0}=1_{R}.$
So for any $r_{1},r_{2}\in R$ we have%
\begin{eqnarray*}
\phi (r_{1}j_{0}+r_{2}i_{0}+I\cap J)
&=&(r_{1}j_{0}+r_{2}i_{0}+I,r_{1}j_{0}+r_{2}i_{0}+J) \\
&=&(r_{1}j_{0}+I,r_{2}i_{0}+J) \\
&=&(r_{1}(1-i_{0})+I,r_{2}(1-j_{0})+J) \\
&=&(r_{1}+I,r_{2}+J)
\end{eqnarray*}%
and we see $\phi $ is also onto. [Note in the above how $i_{0}$ corresponds
to $(0,1)$ and $j_{0}$ corresponds to $(1,0)$ under $\phi .$]
\end{proof}

\begin{example}
Determine the inverse $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{11}\times 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{13}\rightarrow 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{143}$ of the map%
\begin{equation*}
r\,\func{mod}143\mapsto (r\,\func{mod}11,r\,\func{mod}13).
\end{equation*}
\end{example}

\begin{solution}
We see from the above proof that we first need to determine integers $u$ and 
$v$ such that $11u+13v=1$. By inspection we see that $u=6$ and $v=-5$ work.
Under the above map%
\begin{equation*}
11u=66\mapsto (0,1),\qquad 13v=-65\mapsto (1,0)
\end{equation*}%
so that%
\begin{equation*}
-65x+66y\mapsto (x,y)
\end{equation*}%
and the map's inverse is given by%
\begin{equation*}
(x\,\func{mod}11,y\,\func{mod}13)\mapsto 66y-65x\,\func{mod}143.
\end{equation*}
\end{solution}

\begin{example}
Note%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{3}-x^{2}+x-1\rangle }=\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle (x-1)(x^{2}+1)\rangle }=\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle \cap \langle x^{2}+1\rangle }\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{2}+11\rangle }\equiv 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
.
\end{equation*}
\end{example}

\begin{example}
How many units are there in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{120}$?
\end{example}

\begin{solution}
Using the Chinese remainder theorem note%
\begin{equation*}
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{120}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}{120%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}\cong \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}{8%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\cap 15%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}\cong \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}{8%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
}{(3%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\cap 5%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
)}\equiv 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{5}.
\end{equation*}%
There are $4$ units in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8},$ $2$ units in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}$ and $4$ units in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{5}$ and so there are $4\times 2\times 4=32$ units in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{120}$. In fact we can see that%
\begin{equation*}
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{120}^{\ast }\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}^{\ast }\times 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}^{\ast }\times 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{5}^{\ast }\cong C_{2}\times C_{2}\times C_{2}\times C_{4}.
\end{equation*}
\end{solution}

\chapter{More on Ideals and Quotient Rings. Field Extensions.}

We now move from the isomorphism theorem to the more general connection
between ideals and quotient rings.

\begin{proposition}
\label{corr}Let $R$ be a ring and $I$ an ideal of $R.$ Then there is a 1-1
correspondence between the ideals of $R$ that contain $I$ and the ideals of $%
R/I.$
\end{proposition}

\begin{proof}
Let $\pi \colon R\rightarrow R/I$ denote the quotient map and let $J$ be an
ideal of $R$ which contains $I$. Then $I$ is an ideal of $J$ as well as it
is still closed under addition and multiplication by elements of $J.$
Further $\bar{J}=\pi (J)=J/I$ is an ideal of $R/I$. Conversely say that $%
\bar{J}$ is an ideal of $R/I.$ If we define%
\begin{equation*}
J=\pi ^{-1}(\bar{J})=\left\{ r\in R:\bar{r}\in \bar{J}\right\} ,
\end{equation*}%
then this is an ideal of $R$ and contains $I$ as $\bar{0}\in \bar{J}.$ The
above maps $J\mapsto \pi (J)$ and $\bar{J}\mapsto \pi ^{-1}(\bar{J})$ are
inverse processes of one another and the result follows.
\end{proof}

\begin{example}
The ideals of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
/6%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ are%
\begin{equation*}
\{\bar{0}\},\qquad \{\bar{0},\bar{2},\bar{4}\},\qquad \{\bar{0},\bar{3}%
\},\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}.
\end{equation*}%
These correspond to the ideals $6%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,2%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,3%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ which contain $6%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$
\end{example}

\begin{definition}
(a) A proper ideal $I$ of a ring $R$ is said to be \textbf{prime} if
whenever $ab\in I$ then $a\in I$ or $b\in I$ (or both).

(b) A proper ideal $I$ of a ring $R$ is said to be \textbf{maximal} if the
only ideal of $R$ to strictly contain $I$ is $R$ itself. (Note therefore
that $R$ is not maximal.)
\end{definition}

\begin{proposition}
Let $I$ be a proper ideal of a CRI $R.$ Then

(a) $I$ is prime if and only if $R/I$ is an integral domain.

(b) $I$ is maximal if and only if $R/I$ is a field.
\end{proposition}

\begin{solution}
(a) Assume $I$ is prime and say that $\bar{r}\bar{s}=\bar{0}.$ Then $rs\in I$
and so $r\in I$ or $s\in \dot{I}.$ But this is equivalent to $\bar{r}=\bar{0}
$ or $\bar{s}=\bar{0}.$

Conversely, say that $R/I$ is an integral domain. If $rs\in I$ then $%
\overline{rs}=\bar{r}\bar{s}=\bar{0}$ and, as $R/I$ is an integral domain,
then $\bar{r}=\bar{0}$ or $\bar{s}=\bar{0}$ or equivalentlt $r\in I$ or $%
s\in I.$ Hence $I$ is prime.

(b) Say that $I$ is maximal. Say that $\bar{r}\neq \bar{0}\in R/I.$ Then the
ideal $J=\langle I,r\rangle $ strictly contains $I$ and so by maximality $%
J=R.$ Consequently we can write $1=sr+i$ for some $s\in R$ and $i\in I.$ We
then have $\bar{s}\bar{r}=\bar{1}$. So $\bar{r}$ is a unit in $R/I$ and $R/I$
is a field.

Conversely say that $R/I$ is a field. Then the only ideals are $\left\{ \bar{%
0}\right\} $ and $R/I.$ These correspond (under Proposition \ref{corr}) to $%
I $ and $R$ being the only ideals of $R$ containing $I$ and hence $I$ is
maximal.
\end{solution}

\begin{corollary}
Maximal ideals are prime. (This is something that can also be proved
directly in a relatively straightforward manner.)
\end{corollary}

\begin{example}
The prime ideals of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ are $p%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ where $p$ is prime or $p=0.$

The maximal ideals of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ are $p%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ where $p$ is prime.
\end{example}

\begin{example}
(See Sheet 1, \#7) The ideal $\langle x^{2}-1\rangle $ is not prime in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ as $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]/\langle x^{2}-1\rangle \cong 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
^{2}$ is not an integral domain.

The ideal $\langle x^{2}-2\rangle $ is maximal in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ as $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]/\langle x^{2}-2\rangle \cong 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ is a field.
\end{example}

\begin{example}
(See Sheet 1, \#6) The ideal $\langle x^{2}+x+1\rangle $ is maximal in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]$ as $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]/\langle x^{2}+x+1\rangle $ is a field.
\end{example}

\begin{proposition}
Let $F$ be a field.

(a) Every ideal of $F[x]$ is principal.

(b) An ideal $\langle p(x)\rangle $ is maximal if and only if $p(x)$ is
irreducible (that is $f(x)$ cannot be written as a product of two
non-constant polynomials).
\end{proposition}

\begin{proof}
(a) Let $I$ be an ideal of $F[x].$ If $I=\{0\}$ then we have $I=\langle
0\rangle .$ Otherwise there is a non-zero polynomial $f(x)$ in $I$ of least
degree. By the division algorithm, for any $g(x)$ in $I$ we can write%
\begin{equation*}
g(x)=q(x)f(x)+r(x)\qquad \text{where\quad }\deg r<\deg f\text{ \ or \ }r=0.
\end{equation*}%
As $r=g-qf\in I$ then we must have $r=0$ and so $g(x)=q(x)f(x)\in \langle
f(x)\rangle .$ But $\langle f(x)\rangle \subseteq I$ by definition and hence 
$I=\langle f(x)\rangle .\medskip $

(b) Say that $p(x)$ is irreducible. Then for any $q(x)\notin \langle
p(x)\rangle ,$ $p(x)$ and $q(x)$ are coprime and so by B\'{e}zout's lemma
there exist $u(x),v(x)\in F[x]$ such that%
\begin{equation*}
u(x)p(x)+v(x)q(x)=1.
\end{equation*}%
But then $\langle p(x),q(x)\rangle =F[x]$ and so $\langle p(x)\rangle $ is
maximal. Conversely say that $p(x)$ is not irreducible and $p(x)=a(x)b(x)$
where neither $a$ nor $b$ is a constant polynomial.\ Then $\langle
p(x)\rangle $ is strictly contained in $\langle a(x)\rangle \neq F[x]$ and
so $\langle p(x)\rangle $ is not maximal.
\end{proof}

\begin{proposition}
In an integral domain $R$ in which every ideal is principal, then non-zero
prime ideals are maximal.
\end{proposition}

\begin{proof}
Let $I\lhd R$ with $I$ prime. Then there exists $x\neq 0$ such that $%
I=\langle x\rangle .$ For $r\notin I$ then $\langle x,r\rangle =\langle
y\rangle $ for some $y$ as every ideal is principal$.$ This in particular
means that $x=yu.$ As $yu\in I$ and $I$ is prime then either $y\in I$ or $%
u\in I.$ Were it the case that $y\in I=\langle x\rangle $ then we'd have $%
y=xv$ for some $v$ and so $x=xvu;$ by the cancellation law $u$ would be a
unit and we'd have $I=\langle y\rangle =\langle x,r\rangle $ which is a
contradiction as $\langle x,r\rangle $ strictly contains $I.$ So $u\in I$
and $y$ is a unit and hence $\langle x,r\rangle =R$. We have therefore shown
that $I$ is maximal.
\end{proof}

\begin{example}
As $x^{3}+x+1$ has no roots in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ then the polynomial is irreducible. (Any reduction of a cubic
polynomial involves a linear factor.) So%
\begin{equation*}
\mathbb{F}_{8}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]}{\langle x^{3}+x+1\rangle }
\end{equation*}%
is a field with $8$ elements. Every coset has a representative $a+bx+cx^{2}$
where $a,b,c\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}.$

Similarly $x^{2}+1$ is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}$ and%
\begin{equation*}
\mathbb{F}_{9}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}[x]}{\langle x^{2}+1\rangle }
\end{equation*}%
is a field with $9$ elements.

Note in the first case that we can identify $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ with $\{0,1\}\subseteq \mathbb{F}_{8}$, its prime subfield and
similarly we can $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}$ with $\{0,1,2\}\subseteq \mathbb{F}_{9}.$
\end{example}

The above is part of a more general approach to \textit{adjoining a root.}

\begin{proposition}
Let $F$ be a field and $f(x)$ be an irreducible polynomial in $F[x].$ Then%
\begin{equation*}
K=\frac{F[x]}{\langle f(x)\rangle }
\end{equation*}%
contains an isomorphic copy of $F$ and there is a root of $f(x)$ in $K.$
\end{proposition}

\begin{proof}
$\langle f(x)\rangle $ is maximal and so $K$ is a field. $K$ contains an
isomorphic copy of $F$ when we identify $F$ with the cosets%
\begin{equation*}
a\leftrightarrow a+\langle f(x)\rangle =\bar{a}.
\end{equation*}%
Consequently we can make sense of $p(k)$ where $k\in K$ and $p(x)\in F[x].$
In particular we have that%
\begin{equation*}
f(\bar{x})=f(x+\langle f(x)\rangle )=f(x)+\langle f(x)\rangle =\langle
f(x)\rangle =0_{K}
\end{equation*}%
and we see that $f$ has a root $\bar{x}$ in $K.$
\end{proof}

\begin{example}
The polynomial $x^{2}+x+1$ is irreducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]$. If we set%
\begin{equation*}
\mathbb{F}_{4}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[y]}{\langle y^{2}+y+1\rangle }
\end{equation*}%
then we note that $y$ is a root of $x^{2}+x+1$ in $\mathbb{F}_{4}.$ We then
note%
\begin{equation*}
x^{2}+x+1=(x-y)(x+y+1)
\end{equation*}%
as expansion gives%
\begin{equation*}
x^{2}+(y+1-y)x+(-y^{2}-y)=x^{2}+x+1.
\end{equation*}
\end{example}

\begin{example}
The polynomial $x^{3}-2$ is irreducible over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
.$ If we set $\alpha =\sqrt[3]{2}$ then we see in%
\begin{equation*}
K=%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \alpha ]=\left\{ q_{1}+q_{2}\alpha +q_{3}\alpha ^{2}:q_{i}\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right\}
\end{equation*}%
that in $K[x]$ we can factorize as%
\begin{equation*}
x^{3}-2=x^{3}-\alpha ^{3}=(x-\alpha )(x^{2}+\alpha x+\alpha ^{2}).
\end{equation*}%
However we cannot factorize further over $K$ -- the other two roots are
complex and $K$ is a subfield of $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
.$
\end{example}

\begin{definition}
Let $F$ be a field. A \textbf{field extension} $K$ of $F$ is a field which
contains $F$ as a subfield. We denote this extension as $K\colon F.$

$K$ can then be considered as a vector space over $F.$ The \textbf{degree}
of the field extension is denoted $\left\vert K\colon F\right\vert $ and
equals $\dim _{F}K.$
\end{definition}

\begin{example}
(a) $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ is a field extension of $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ of degree $2$. A basis of $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ over $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is $\{1,i\}.$

(b) $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is a field extension of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ of infinite degree. To see this note that any finite degree field
extension of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ is isomorphic as a vector space to $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
^{n}$ for some $n$ and in particular is finite. As $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is uncountable this cannot be the case.

(c) $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ is a field extension of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ of degree $2$. A basis of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ over $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is $\{1,\sqrt{2}\}.$

(d) $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt[3]{2}]$ is a field extension of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ of degree $3$. A basis of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt[3]{2}]$ over $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ is $\{1,\sqrt[3]{2},\sqrt[3]{4}\}.$

(e) $\mathbb{F}_{8}=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]/\langle x^{3}+x+1\rangle $ is a field extenison over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ of degree $3$ with a basis being $\{1,x,x^{2}\}.$

(f) $\mathbb{F}_{9}=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}[x]/\langle x^{2}+1\rangle $ is a field extenison over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}$ of degree $2$ with a basis being $\{1,x\}.$
\end{example}

\begin{proposition}
A finite field has order $p^{n}$ for some prime $p$ and a positive integer $%
n.$
\end{proposition}

\begin{proof}
Let $F$ be a finite field. As a field of characteristic $0$ contains a prime
subfield isomorphic to $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ then $F$ has characteristic $p$ for some prime $p$ and contains a copy of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}$ as its prime subfield. $F$ is the a field extension of some degree $n$
over its prime subfield and so $F$ is isomorphic to $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}^{n}$ \textit{as a vector space}. Hence $\left\vert F\right\vert
=\left\vert 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}^{n}\right\vert =p^{n}.$
\end{proof}

\begin{remark}
(\textbf{Off-syllabus}) There exists a field of order $p^{n}$ for each prime 
$p$ and positive integer $n.$ One can show this by demonstrating that, for
each $n$, there is an irreducible polynomial over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}$ of degree $n$. It can further be shown that, up to isomorphism, there
is a unique field $\mathbb{F}_{p^{n}}$ of order $p^{n}.$ Finite fields are
commonly called \textbf{Galois fields} in honour of \'{E}variste Galois who
first proved these results. The multiplicative group of any finite field is
cyclic though it can be computationally difficult to find a generator, and
consequently finite fields can be used in this way in crytopgraphy.
\end{remark}

\begin{example}
\label{F16}Give an example of a field $\mathbb{F}_{16}$ with $16$ elements.
How many generators does its multiplicative group have?
\end{example}

\begin{solution}
We can produce such a field by finding an irreducible quartic over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}.$ Consider $x^{4}+x+1.$ Certainly neither $0$ nor $1$ are roots; it
remains to show that $x^{4}+x+1$ cannot be written as a product of two
irreducible quadratics. The only irreducible quadratic over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ is $x^{2}+x+1$ and we note that%
\begin{equation*}
(x^{2}+x+1)^{2}=x^{2}+2x^{3}+3x^{2}+2x+1=x^{4}+x^{2}+1,
\end{equation*}%
which is not equal to our polynomial. Hence%
\begin{equation*}
\mathbb{F}_{16}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[x]}{\langle x^{4}+x+1\rangle }
\end{equation*}%
is a field of $16$ elements. Given what we've been told above, $\mathbb{F}%
_{16}^{\ast }\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{15}$ which has $8$ generators, $1,2,4,7,8,11,13,14.$ Note in $\mathbb{F}%
_{16}$ that $x^{5}=x^{2}+x\neq 1$ and so $x$ does not have order $3$ or $5$
and so must have order $15.$ So the $8$ generators of $\mathbb{F}_{16}^{\ast
}$ are%
\begin{equation*}
x,\quad x^{2},\quad x^{4},\quad x^{7},\quad x^{8},\quad x^{11},\quad
x^{13},\quad x^{14},
\end{equation*}%
whose cosets have "preferred" representatives%
\begin{equation*}
x,\quad x^{2},\quad x+1,\quad x^{3}+x+1,\quad x^{2}+1,\quad
x^{3}+x^{2}+x,\quad x^{3}+x^{2}+1,\quad x^{3}+1.
\end{equation*}
\end{solution}

\begin{definition}
Given a field extension $K\colon F$ and $\alpha \in K$, we say that $\alpha $
is \textbf{algebraic} over $F$ if there exists a polynomial $f(x)\in F[x]$
such that $f(\alpha )=0.$
\end{definition}

\begin{proposition}
Let $K\colon F$ be a field extension and $\alpha \in K$ be algebraic over $%
F. $

(a) There exists a unique monic polynomial $m(x)$ in $F[x]$ such that $%
m(\alpha )=0.$ The polynomial $m(x)$ is known as the \textbf{minimal
polynomial} of $\alpha .$

(b) For $f(x)\in F[x],$ we have $f(\alpha )=0$ if and only if $m(x)$ divides 
$f(x).$ Further $m(x)$ is irreducible.

(c) $F[\alpha ]$ is a subfield of $K$ of degree equal to the degree of $%
m(x). $
\end{proposition}

\begin{proof}
(a) As there exists a polynomial $f(x)$ such that $f(\alpha )=0$ then there
exists such a polynomial $m(x)$ of least degree. By dividing by its leading
coefficient if necessary we can assume $m(x)$ to be monic. If $m_{1}(x)$ and 
$m_{2}(x)$ were monic polynomials of least degree such that $m_{1}(\alpha
)=m_{2}(\alpha )=0$ then $m_{1}-m_{2}$ would be a polynomial of strictly
lower degree with $\alpha $ as a root -- a contradiction.

(b) By the division algorithm we can write $f(x)=q(x)m(x)+r(x)$ where $\deg
r<\deg m$ or $r=0.$ We then have $0=f(\alpha )=r(\alpha )$ and by the
minimality of $m$'s degree it must be that $r=0.$ Hence $f(x)=q(x)m(x)$ and
we see $m(x)$ divides $f(x).$ The converse is obvious. Finally if $m(x)$
were reducible as $m(x)=g(x)h(x)$ then we'd have $g(\alpha )h(\alpha
)=m(\alpha )=0$. As $K$ is a field, and so an integral domain, $g(\alpha )=0$
or $h(\alpha )=0$ and either would contradict the minimality of $m.$

(c) We already Certainly $F[\alpha ]$ is a CRI, it only remains to show that
non-zero elements are units. Say that $g(\alpha )\in F[\alpha ]$ and $%
g(\alpha )\neq 0.$ Then $m(x)$ does not divide $g(x)$ and as $m(x)$ is
irreducible then it follows that $g(x)$ and $m(x)$ are coprime. By B\'{e}%
zout's Lemma in $F[x]$ we know that there are polynomials $u(x)$ and $v(x)$
such that%
\begin{equation*}
u(x)g(x)+v(x)m(x)=1.
\end{equation*}%
But then $u(\alpha )g(\alpha )=1$ and we see that $g(\alpha )$ is a unit.
Finally, if $n=\deg m(x)$ and $f(\alpha )\in F[\alpha ]$ we see from the
division algorithm that $f(\alpha )=r(\alpha )$ for some $r(\alpha )$ in the
span of $1,\alpha ,\ldots ,\alpha ^{n-1}.$ Further $1,\alpha ,\ldots ,\alpha
^{n-1}$ are independent by the minimality of $m(x)$ and so a basis for $%
F[\alpha ]$ over $F.$ Hence $n=\left\vert F[\alpha ]\colon F\right\vert $ as
well.
\end{proof}

\begin{example}
Let $\mathbb{F}_{16}$ be as in Example \ref{F16}. Find the minimal
polynomials of (i) $x$, (ii) $x+1,$ (iii) $x^{2}+x$?
\end{example}

\begin{solution}
(i) We know that $x$ is a root of satisfies $y^{4}+y+1$ which is
irreducible, so this must be its minimal polynomial.

(ii) We similarly then have that $x+1$ satisfies 
\begin{equation*}
(y+1)^{4}+(y+1)+1=y^{4}+4y^{3}+6y^{2}+4y+1+y+1+1=y^{4}+y+1
\end{equation*}%
which again is irreducible and so $x+1$ has this as its minimal polynomial.

(iii) Note with $\alpha =x^{2}+x$ we have%
\begin{equation*}
\alpha ^{2}=x^{4}+2x^{3}+x^{2}=x+1+0+x^{2}=\alpha +1.
\end{equation*}%
Hence the minimal polynomial of $\alpha $ is $y^{2}+y+1.$ In particular this
means that%
\begin{equation*}
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}[\alpha ]=\{0,1,\alpha ,\alpha +1\}=\{0,1,x^{2}+x,x^{2}+x+1\}
\end{equation*}%
is a subfield of $\mathbb{F}_{16}.$
\end{solution}

\begin{proposition}
(\textbf{Tower Law}) Let $L,K,F$ be fields with $F\subseteq K\subseteq L$.
Then $L$ has finite degree over $F$ if and only if $\left\vert L\colon
K\right\vert $ and $\left\vert K\colon F\right\vert $ are finite. In this
case%
\begin{equation*}
\left\vert L\colon F\right\vert =\left\vert L\colon K\right\vert \left\vert
K\colon F\right\vert .
\end{equation*}
\end{proposition}

\begin{proof}
Say that $\left\vert L\colon K\right\vert =m$ and $\left\vert K\colon
F\right\vert =n$ are finite and $l_{1},\ldots ,l_{m}$ are a basis for $L$
over $K$ and $k_{1},\ldots ,k_{n}$ are a basis for $K$ over $F.$ Then we
will show that%
\begin{equation*}
\left\{ l_{i}k_{j}:1\leqslant i\leqslant m,1\leqslant j\leqslant n\right\}
\end{equation*}%
is a basis for $L$ over $F.$ \textit{Independence}: say that%
\begin{equation*}
0=\sum_{i.j}f_{ij}l_{i}k_{j}=\sum_{i}\left( \sum_{j}f_{ij}k_{j}\right)
l_{i}\qquad \text{for some }f_{ij}\in F.
\end{equation*}%
As the $l_{i}$ are independent over $K$ then it follows that%
\begin{equation*}
\sum_{j}f_{ij}k_{j}=0.
\end{equation*}%
And as the $k_{j}$ are independent over $F$ then it follows that $f_{ij}=0$
for all $i,j.$ Consequently the $l_{i}k_{j}$ are linearly independent
elements of $L$ over $F.$ \textit{Spanning:} say that $l\in L.$ Then there
exists $\kappa _{i}\in K$ such that%
\begin{equation*}
l=\sum_{i}\kappa _{i}l_{i}.
\end{equation*}%
Similarly there exist $\phi _{oj}$ such that%
\begin{equation*}
\kappa _{i}=\sum_{j}\phi _{ij}k_{j}
\end{equation*}%
so that%
\begin{equation*}
l=\sum_{i}\sum_{j}\phi _{ij}k_{j}l_{i}
\end{equation*}%
and we see that the $l_{i}k_{j}$ are spanning.

Conversely if $\dim _{F}L$ is finite then so is $\dim _{F}K$ as $K$ is a
subspace of $L$ (over $F$) and $\dim _{K}L\leqslant \dim _{F}L$ as any set
that spans $L$ over $F$ also spans $L$ over $K.$
\end{proof}

\begin{example}
Find the degrees of the following extensions.%
\begin{equation*}
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2},i]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\qquad \text{and}\qquad 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2},\sqrt{3}].
\end{equation*}
\end{example}

\begin{solution}
The former is somewhat easier. The minimal polynomial of $\sqrt{2}$ over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ is $x^{2}-2.$ It cannot be linear as $\sqrt{2}$ is irrational. So $%
\left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right\vert =2.$ Similarly the minimal polynomial of $i$ over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ is $x^{2}+1$; it cannot be linear as $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ is a real subfield and $i$ is not real. So%
\begin{equation*}
\left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2},i]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right\vert =\left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2},i]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]\right\vert \left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right\vert =2\times 2=4.
\end{equation*}%
Again $\left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right\vert =2.$ We need to take a little care to show that%
\begin{equation*}
\left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2},\sqrt{3}]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]\right\vert =2.
\end{equation*}%
Certainly $x^{2}-3$ has $\sqrt{3}$ as a root, but is this polynomial
irreducible over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$? A general element of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ is $q_{1}+q_{2}\sqrt{2}$ where $q_{1},q_{2}\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$. If $x^{2}-3$ reduced in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ then we'd have%
\begin{equation*}
(q_{1}^{2}-2q_{2}^{2})+2q_{1}q_{2}\sqrt{2}=(q_{1}+q_{2}\sqrt{2})^{2}=3
\end{equation*}%
for some $q_{1},q_{2}.$ As $1$ and $\sqrt{2}$ are independent over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ then we have $q_{1}=0$ or $q_{2}=0$ both of which lead to contradictions.
Hence $x^{2}-3$ is the minimal polynomial of $\sqrt{3}$ over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ and we have%
\begin{equation*}
\left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2},\sqrt{3}]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right\vert =\left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2},\sqrt{3}]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]\right\vert \left\vert 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]\colon 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\right\vert =2\times 2=4.
\end{equation*}
\end{solution}

\begin{example}
Show that $\mathbb{F}_{16}$ does not contain a subfield of order $8.$
\end{example}

\begin{solution}
If $\mathbb{F}_{16}$ has a subfield $K$ of order $8$ we would have $%
\left\vert F\colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\right\vert =3$ and%
\begin{equation*}
4=\left\vert \mathbb{F}_{16}\colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\right\vert =\left\vert \mathbb{F}_{16}\colon K\right\vert \left\vert
K\colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\right\vert =3\times \left\vert K\colon 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\right\vert
\end{equation*}%
which is a contradiction as $3$ does not divide $4.$
\end{solution}

\chapter{Factorization. EDs. PIDs. UFDs.}

\textit{Throughout this section, rings will be assumed to be CRIs
(commutative and having an identity) unless otherwise stated.}

\begin{definition}
Let $R$ be a ring and let $a,b,c\in R$.

\noindent (i) If $a\neq 0,$ we say that $a$ \textbf{divides} $b$ if there
exists $r\in R$ such that $ra=b$. This is written%
\begin{equation*}
a|b.
\end{equation*}%
Equivalently we say that $a$ is a \textbf{factor} of $b$ or that $b$ is a 
\textbf{multiple }of $a.$

\noindent (ii) If $c\neq 0$, we say that $c$ is a \textbf{common factor} of $%
a$ and $b$ if $c|a$ and $c|b.$

\noindent (iii) If $c$ is a common factor of $a$ and $b$, then we say that $%
c $ is a \textbf{highest common factor }(or hcf) of $a$ and $b$ if whenever $%
d$ is a common factor of $a$ and $b$ then $d|c$.

\noindent (iv) If $a,b\neq 0$ we say that $c$ is a \textbf{common multiple }%
of $a$ and $b$ if $a|c$ and $b|c.$

\noindent (v) If $c$ is a common multiple of $a$ and $b$, then we say that $%
c $ is a \textbf{least common multiple }(or lcm) of $a$ and $b$ if whenever $%
d$ is a common multiple of $a$ and $b$ then $c|d.$
\end{definition}

\begin{proposition}
\label{hcf}In an integral domain, the hcf of two elements, if it exists, is
unique up to multiplication by a unit. The corresponding result also holds
for lcms.
\end{proposition}

\begin{proof}
Let $R$ be an integral domain and $a,b$ be non-zero elements of $R$. Say
that $h_{1}$ and $h_{2}$ are two hcfs of $a$ and $b.$ In particular $h_{1}$
and $h_{2}$ are common factors of $a$ and $b,$ so that $h_{1}|h_{2}$ as $%
h_{2}$ is an hcf and $h_{2}|h_{1}$ as $h_{1}$ is an hcf . So there exist $%
r_{1},r_{2}\in R$ such that%
\begin{equation*}
h_{1}r_{1}=h_{2},\qquad h_{2}r_{2}=h_{1}.
\end{equation*}%
Hence $h_{1}\left( r_{1}r_{2}-1_{R}\right) =0$ and, as $h_{1}\neq 0,$ by the
cancellation law $r_{1}r_{2}=1_{R}$ and $r_{1}$ and $r_{2}$ are both units
in $R.$
\end{proof}

\begin{example}
\noindent (i) Let $R=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ Let $a=105$ and $b=441.$ Then $21$ and $-21$ are hcfs of $a$ and $b$.

\noindent (ii) Let $R=%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$. Let $a=x^{2}-3x+4$ and $b=x^{2}-2x+1$. Then the hcfs of $a$ and 
$b$ are of the form $c\left( x-1\right) $ where $c$ is a non-zero rational.

\noindent (iii) Let $R=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$ Let $a=2x-2$ and $b=4x^{2}-2x+6$. Then the hcfs are $2$ and $-2$%
.

\noindent (iv) Let $R=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]=\{x+yi:x,y\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\}.$ Let $a=2$ and $b=-1+3i$. Then the hcfs are $1+i,$ $-1-i,$ $-1+i,$ $1-i$.

\noindent (v) Let $R=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{-3}]=\left\{ x+y\sqrt{-3}:x,y\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\right\} .$ Then $a=4$ and $b=2+2\sqrt{-3}$ have no highest common factor.
To appreciate this:%
\begin{equation*}
\text{if}\quad 4=(\alpha +\beta \sqrt{-3})(\gamma +\delta \sqrt{-3})\quad 
\text{then}\quad 16=(\alpha ^{2}+3\beta ^{2})(\gamma ^{2}+3\delta ^{2})
\end{equation*}%
and, knowing the factorizations of $16$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ we see that the factors of $4$ are $\pm 1,\pm 1\mp \sqrt{-3},\pm 2,\pm 4$
and%
\begin{equation*}
\text{if}\quad 2+2\sqrt{-3}=(\alpha +\beta \sqrt{-3})(\gamma +\delta \sqrt{-3%
})\quad \text{then}\quad 16=(\alpha ^{2}+3\beta ^{2})(\gamma ^{2}+3\delta
^{2})
\end{equation*}%
and so we see that the factors of $2+2\sqrt{-3}$ are $\pm 1,\pm 2,\pm 1\mp 
\sqrt{-3}.$ So the common factors of $a$ and $b$ are%
\begin{equation*}
\pm 1,\quad \pm 2,\quad \pm (1+\sqrt{-3}).
\end{equation*}%
$1$ cannot be a highest common factor as it divides the other two. However
neither $2$ nor $1+\sqrt{-3}$ is a highest common factor as neither divides
the other.
\end{example}

\begin{definition}
In an integral domain, non-zero elements are said to be \textbf{coprime }if $%
1_{R}$ (or equivalently a unit) is an hcf of theirs.
\end{definition}

\begin{proposition}
\label{bezout1}Let $R$ be an integral domain and $a,b\in R$. If there exist $%
u,v\in R$ such that $ua+vb=1_{R}$ then $a$ and $b$ are coprime.
\end{proposition}

\begin{proof}
(a) Let $c$ be a common factor of $a$ and $b$. Then there exist $r,s\in R$
such that $a=cr$ and $b=cs$. Hence%
\begin{equation*}
1_{R}=ua+vb=(ur+vs)c
\end{equation*}%
and we see that $c$ is a unit. Finally, by definition, units are factors of
all elements in a ring.
\end{proof}

\begin{definition}
Given a ring $R$ we say that a non-zero, non-unit $x\in R$ is

(a) a \textbf{prime} element if whenever $x|yz$ then $x|y$ or $x|z.$

(b) an \textbf{irreducible }element if whenever $x=yz$ then $y$ is a unit or 
$z$ is a unit.
\end{definition}

\begin{proposition}
(a) Given a non-zero, non-unit $x$, then the principal ideal $\langle
x\rangle $ is prime if and only if $x$ is a prime element.

(b) In an integral domain, prime elements are irreducible.

(c) $2$ is irreducible but not prime in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{-3}].$
\end{proposition}

\begin{proof}
(a) Say that $x$ is prime and $ab\in \langle x\rangle .$ Then $x|ab$ and as $%
x$ is prime we have $x|a$ or $x|b.$ That is $a\in \langle x\rangle $ or $%
b\in \langle x\rangle .$ Conversely say that $\langle x\rangle $ is prime
with $x$ non-zero and a non-unit. If $x|ab$ then $ab\in \langle x\rangle $
and so $a\in \langle x\rangle $ or $b\in \langle x\rangle $ i.e. $x|a$ or $%
x|b.$ As $x$ is a non-unit then $x$ is a prime element.

(b) Let $R$ be an integral domain and $x$ be prime. If $x=yz$ then $x|yz$
and so $x|y$ or $x|z.$ If $x|y$ then $y=ux$ for some $u$ and then $x=uxz.$
By the cancellation rule $uz=1$ and so $z$ is a unit, showing that $x$ is
irreducible.

(c) Note that $2|(1+\sqrt{-3})(1-\sqrt{-3})=4$ but that $2$ divides neither $%
1\pm \sqrt{-3}.$ So $2$ is not prime. But if we could write%
\begin{equation*}
2=(a+b\sqrt{-3})(c+d\sqrt{-3}).
\end{equation*}%
Taking the modulus squared we see%
\begin{equation*}
4=(a^{2}+3b^{2})(c^{2}+3d^{2}).
\end{equation*}%
If we have $a^{2}+3b^{2}=1$ then $a+b\sqrt{-3}$ is a unit. So a genuine
reduction of $2$ would mean%
\begin{equation*}
a^{2}+3b^{2}=c^{2}+3d^{2}=2
\end{equation*}%
and these equations have no solution. Hence $2$ is irreducible.
\end{proof}

\section{Euclidean Domains (EDs)}

\begin{definition}
A \textbf{Euclidean domain }$R$ is essentially an integral domain which
admits the division algorithm. That is $R$ is an integral domain together
for which there exists a function, often called a \textbf{norm}, $d\colon
R\backslash \{0\}\rightarrow 
%TCIMACRO{\U{2115} }%
%BeginExpansion
\mathbb{N}
%EndExpansion
=\{0,1,2,3\ldots \}$ such that

(a) $d(a)\leqslant d(ab)$ for all $a,b\in R\backslash \{0\}.$

(b) given $a,b\in R$ with $b\neq 0$, there exist $q,r\in R$ such that $%
a=qb+r $ with $d(r)<d(b)$ or $r=0.$
\end{definition}

\begin{example}
(a) The integers $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ form an ED with $d(x)=|x|$ for $x\neq 0.$

(b) Given a field $F,$ the polynomial ring $F[x]$ forms an ED with $%
d(f)=\deg f$ for any $f\neq 0.$
\end{example}

\begin{proposition}
The Gaussian integers $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ form an ED with%
\begin{equation*}
d(a+bi)=\left\vert a+bi\right\vert ^{2}=a^{2}+b^{2}\qquad \text{where }%
a+bi\neq 0.
\end{equation*}
\end{proposition}

\begin{proof}
We immediately have for non-zero $\alpha ,\beta $ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ that%
\begin{equation*}
d(\alpha \beta )=\left\vert \alpha \beta \right\vert ^{2}=\left\vert \alpha
\right\vert ^{2}\left\vert \beta \right\vert ^{2}\geqslant \left\vert \alpha
\right\vert ^{2}=d(\alpha ).
\end{equation*}%
Further we have that $\alpha /\beta \in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ and -- noting the Gaussian integers form a grid of unit squares in $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$ -- there exists $q\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ such that%
\begin{equation*}
\left\vert \frac{\alpha }{\beta }-q\right\vert \leqslant \frac{1}{\sqrt{2}}.
\end{equation*}%
If we set $r=\alpha -q\beta \in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]$ then we have%
\begin{equation*}
d(r)=\left\vert \alpha -q\beta \right\vert ^{2}=\left\vert \frac{\alpha }{%
\beta }-q\right\vert ^{2}\left\vert \beta \right\vert ^{2}\leqslant \frac{1}{%
2}\left\vert \beta \right\vert ^{2}=\frac{1}{2}d(\beta )<d(\beta ).
\end{equation*}
\end{proof}

\begin{example}
The ring $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{2}]$ is an ED with%
\begin{equation*}
d(a+b\sqrt{2})=\left\vert a^{2}-2b^{2}\right\vert .
\end{equation*}%
Show that the equations $x^{2}-2y^{2}=1$ and $x^{2}-2y^{2}=-1$ each have
infinitely many integer solutions.
\end{example}

\begin{solution}
Firstly note that $d$ is multiplication. If $\alpha =a_{1}+a_{2}\sqrt{2}$
and $\beta =b_{1}+b_{2}\sqrt{2}$ then%
\begin{eqnarray*}
d((a_{1}+a_{2}\sqrt{2})(b_{1}+b_{2}\sqrt{2}))
&=&d((a_{1}b_{1}+2a_{2}b_{2})+(a_{2}b_{1}+a_{1}b_{2})\sqrt{2}) \\
&=&\left\vert
(a_{1}b_{1}+2a_{2}b_{2})^{2}-2(a_{2}b_{1}+a_{1}b_{2})^{2}\right\vert \\
&=&\left\vert
a_{1}^{2}b_{1}^{2}+4a_{1}a_{2}b_{1}b_{2}+4a_{2}^{2}b_{2}{}^{2}-2a_{2}^{2}b_{1}^{2}-2a_{1}^{2}b_{2}^{2}-4a_{1}a_{2}b_{1}b_{2}\right\vert
\\
&=&\left\vert a_{1}^{2}-2a_{2}^{2}\right\vert \left\vert
b_{1}^{2}-2b_{2}^{2}\right\vert \\
&=&d(a_{1}+a_{2}\sqrt{2})d(b_{1}+b_{2}\sqrt{2}).
\end{eqnarray*}%
Let $\alpha ,\beta \in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{2}]\backslash \{0\}.$ So we have $d(\alpha )\leqslant d(\alpha
)d(\beta )=d(\alpha \beta )$ as $d(\beta )\geqslant 1.$ Also we have $\alpha
/\beta =q_{1}+q\sqrt{2}\in 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack \sqrt{2}]$ and so we can find $\gamma =c_{1}+c_{2}\sqrt{2}\in 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{2}]$ such that $\left\vert q_{i}-c_{1}\right\vert \leqslant
1/2 $ so that%
\begin{eqnarray*}
d\left( \frac{\alpha }{\beta }-\gamma \right) &=&\left\vert
(q_{1}-c_{1})^{2}-2(q_{2}-c_{2})^{2}\right\vert \\
&\leqslant &(q_{1}-c_{1})^{2}+2(q_{2}-c_{2})^{2} \\
&\leqslant &\frac{1}{4}+\frac{2}{4}=\frac{3}{4}.
\end{eqnarray*}%
If we set $\delta =\alpha -\beta \gamma $ and so%
\begin{equation*}
d(\delta )=d\left( \alpha -\beta \gamma \right) =d\left( \frac{\alpha }{%
\beta }-\gamma \right) d(\beta )\leqslant \frac{3}{4}d(\beta )<d(\beta ).
\end{equation*}%
If instead we had defined $D(a+b\sqrt{2})=a^{2}-2b^{2}$ then we see as above
that $D$ is multiplicative. As $D(1+\sqrt{2})=1^{2}-2\times 1^{2}=-1$ it
then follows that%
\begin{equation*}
D(x_{n}+\sqrt{2}y_{n})=D((1+\sqrt{2})^{n})=(-1)^{n}
\end{equation*}%
and so $(x_{2n},y_{2n})$ is a solution of $x^{2}-2y^{2}=1$ and $%
(x_{2n+1},y_{2n+1})$ is a solution of $x^{2}-2y^{2}=-1$.
\end{solution}

\begin{proposition}
Let $R$ be an ED and $I\lhd R.$ Then $I$ is principal.
\end{proposition}

\begin{proof}
If $I=\{0_{R}\}$ then we are done as $I=\langle 0_{R}\rangle .$ Otherwise
there exists $x\in R$ with $x\neq 0$ and $d(x)$ minimal. Certainly $\langle
x\rangle \subseteq I.$ Conversely if $y\in I$ then there exist $q,r\in R$
such that $y=qx+r$ where $d(r)<d(x)$ or $r=0.$ As $r=y-qx\in I$ then $r=0$
by the minimality of $d(x).$ So $y=qx$ and hence $I\subseteq \langle
x\rangle .$ Hence $I$ is principal.
\end{proof}

\begin{corollary}
In an ED, irreducible elements are prime.
\end{corollary}

\begin{proof}
Say that $x$ is irreducible and $\langle x\rangle \subseteq I\lhd R.$ There
exists $y$ such that $\langle y\rangle =I$ and so there exists $z$ such that 
$x=yz.$ By the irreducibility of $x$ we either have $y$ is a unit and $%
I=\langle y\rangle =R$ or $z$ is a unit and we have $I=\langle y\rangle
=\langle x\rangle .$ Hence $\langle x\rangle $ is maximal, and so prime, and
hence $x$ is a prime element.
\end{proof}

\begin{example}
$%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ and $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x,y]$ are not Euclidean domains
\end{example}

\begin{solution}
In the first case note that $\langle 2,x\rangle $ is not principal, and in
the second case $\langle x,y\rangle $ is not principal.
\end{solution}

\section{Aside -- the Euclidean Algorithm (Off-syllabus)}

The Euclidean algorithm employs the division algorithm repeatedly to find
the hcf of two integers $a$ and $b$. The algorithm first appeared in
Euclid's \textit{Elements} Book VII in around $300$ B.C.. First we describe
an example in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ to see how the algorithm works.

\begin{example}
\label{EAex}Find the highest common factor of $53714$ and $30281.$
\end{example}

\begin{solution}
Let $n_{1}=53714$ (the greater of the two numbers) and $n_{2}=30281.$

$n_{1}/n_{2}=1.77\ldots $ and so $q_{1}=1$ and we set $n_{3}=r_{1}=n_{1}-1%
\times n_{2}=23433$;

$n_{2}/n_{3}=1.29\ldots $ and so $q_{2}=1$ and $n_{4}=r_{2}=n_{2}-1\times
n_{3}=6848$;

$n_{3}/n_{4}=3.42\ldots $ and so $q_{3}=3$ and $n_{5}=r_{3}=n_{3}-3\times
n_{4}=2889$;

$n_{4}/n_{5}=2.37\ldots $ and so $q_{4}=2$ and $n_{6}=r_{4}=n_{4}-2\times
n_{5}=1070$;

$n_{5}/n_{6}=2.7$ and so $q_{5}=2$ and $n_{7}=r_{5}=n_{5}-2\times n_{6}=749$;

$n_{6}/n_{7}=1.42\ldots $ and so $q_{6}=1$ and $n_{8}=r_{6}=n_{6}-1\times
n_{7}=321$;

$n_{7}/n_{8}=2.33\ldots $ and so $q_{7}=2$ and $n_{9}=r_{7}=n_{7}-2\times
n_{8}=107;$

$n_{8}/n_{9}=3$ and so $q_{8}=3$ and $n_{10}=n_{9}-3\times n_{8}=0.$

\noindent At this point the algorithm terminates and the output is the last
positive number namely $107$.
\end{solution}

\begin{algorithm}
(\textbf{Euclidean algorithm}) Let $a_{1}$ and $a_{2}$ be non-zero elements
of an ED with $d(a_{1})\geqslant d(a_{2})$. Then the Euclidean algorithm
uniquely defines two sequences $q_{i}$ and $a_{i}$ of integers by%
\begin{eqnarray*}
a_{1} &=&q_{1}a_{2}+a_{3}\qquad \text{where }d(a_{3})<d(a_{2}) \\
a_{2} &=&q_{2}a_{3}+a_{4}\qquad \text{where }d(a_{4})<d(a_{3}) \\
a_{3} &=&q_{3}a_{4}+a_{5}\qquad \text{where }d(a_{5})<d(a_{4})
\end{eqnarray*}%
and so on. The algorithm terminates if $a_{k}=0$ for some $k.$
\end{algorithm}

\begin{theorem}
The Euclidean algorithm always terminates with $a_{k}=0$ for some $k$ and%
\begin{equation*}
a_{k-1}=\mathrm{hcf}\left( a_{1},a_{2}\right) .
\end{equation*}
\end{theorem}

\begin{proof}
The sequence of positive integers $d(a_{i})$ is strictly decreasing and
bounded below, and so can only be finite in length. If $d(a_{i})>0$ then it
possible to run (at least)\ one further application of the division
algorithm and so the Euclidean algorithm terminates when $a_{k}=0$ for some $%
k.$

Firstly, I claim $a_{k-1}$ divides $a_{i}$ for $1\leqslant i\leqslant k.$
The proof follows by reverse induction. Certainly $a_{k-1}|a_{k-1}$ and $%
a_{k-1}|a_{k}=0$. Suppose (as an inductive hypothesis) that $a_{k-1}$
divides $a_{r}$ and $a_{r+1}.$ Then%
\begin{equation*}
a_{r-1}=q_{r-1}a_{r}+a_{r+1}
\end{equation*}%
is also divisible by $a_{k-1}$. Hence by induction $a_{k-1}$ is a common
factor of $a_{1}$ and $a_{2}.$

Secondly, we need to show that $a_{k-1}=\mathrm{hcf}\left(
a_{1},a_{2}\right) .$ Suppose that $m$ is a common factor of $a_{1}$ and $%
a_{2}$. I\ claim that $m$ is also a factor of $a_{k-1}$ with the proof
following by induction. Suppose (as an inductive hypothesis)\ that $m$
divides $a_{r}$ and $a_{r+1}$. Then $m$ divides%
\begin{equation*}
a_{r+2}=a_{r}-q_{r}a_{r+1}
\end{equation*}%
also. By induction $m$ divides $a_{k-1}$ also completing the proof.
\end{proof}

\begin{theorem}
(\textbf{B\'{e}zout's Lemma}) Let $a,b$ be non-zero elements of an ED $R$
with a highest common factor $h$. Then there exist $u$ and $v$ in $R$ such
that%
\begin{equation*}
ua+vb=h.
\end{equation*}
\end{theorem}

\begin{proof}
Set $a_{1}=a$ and $a_{2}=b.$ The proof uses \textit{reverse} \textit{%
induction} working backwards through the calculations performed in the
Euclidean Algorithm. We will show that, for each $i=1,2,\ldots ,k,$ there
exist $u_{i}$ and $v_{i}$ such that%
\begin{equation}
u_{i}a_{i}+v_{i}a_{i+1}=h.  \label{indhyp}
\end{equation}%
As $h=a_{k-1}$ then we can set $u_{k-2}=0$ and $v_{k-2}=1$ to see that (\ref%
{indhyp}) is true for $i=k-2.$ Now suppose, as an inductive hypothesis, that
(\ref{indhyp}) holds true for $i=I$, where $1<I\leqslant k-2$, and we shall
aim to show it's true for $i=I-1.$ We have%
\begin{eqnarray*}
h &=&u_{I}a_{I}+v_{I}a_{I+1} \\
&=&u_{I}a_{I}+v_{I}\left( a_{I-1}-q_{I-1}a_{I}\right) \\
&=&v_{I}a_{I-1}+\left( u_{I}-q_{I-1}v_{I}\right) a_{I}
\end{eqnarray*}%
thus proving that (\ref{indhyp}) holds true for $i=I-1$ with $u_{I-1}=v_{I}$
and $v_{I-1}=u_{I}-q_{I-1}v_{I}$. By induction (\ref{indhyp}) holds true
when $i=1$ which is the required result for the case of $a$ and $b.$
\end{proof}

\begin{example}
\label{BLZ}Find integers $u$ and $v$ such that $53714u+30281v=107.$
\end{example}

\begin{solution}
Recall from our earlier calculations that%
\begin{eqnarray*}
n_{3} &=&n_{1}-n_{2};\qquad n_{4}=n_{2}-n_{3};\qquad n_{5}=n_{3}-3n_{4}; \\
n_{6} &=&n_{4}-2n_{5};\qquad n_{7}=n_{5}-2n_{6};\qquad n_{8}=n_{6}-n_{7};
\end{eqnarray*}%
with $n_{9}=n_{7}-2n_{8}=107$ being the highest common factor. We will
"reverse" the above equations to write $n_{9}$ in terms of $n_{7}$ and $%
n_{8} $, then in terms of $n_{6}$ and $n_{7}$, and so on repeatedly until we
have $n_{8}$ in terms of $n_{1}$ and $n_{2}$. We see%
\begin{eqnarray*}
107 &=&n_{7}-2n_{8} \\
&=&n_{7}-2(n_{6}-n_{7})=-2n_{6}+3n_{7} \\
&=&-2n_{6}+3(n_{5}-2n_{6})=3n_{5}-8n_{6} \\
&=&3n_{5}-8(n_{4}-2n_{5})=-8n_{4}+19n_{5} \\
&=&-8n_{4}+19(n_{3}-3n_{4})=19n_{3}-65n_{4} \\
&=&19n_{3}-65(n_{2}-n_{3})=-65n_{2}+84n_{3} \\
&=&-65n_{2}+84(n_{1}-n_{2})=84n_{1}-149n_{2}
\end{eqnarray*}
\end{solution}

\begin{example}
Find the multiplicative inverse of $2167$ in $\func{mod}65537$ arithmetic.
\end{example}

\begin{solution}
We first find integers $u$ and $v$ such that $2167u+65537v=1$. Applying
first the Euclidean algorithm we set:

Let $n_{1}=65537$ and $n_{2}=2167.$

As $n_{1}/n_{2}=30.24\ldots $ and so $q_{1}=30$ and we set $%
n_{3}=r_{1}=n_{1}-30\times n_{2}=527.$

As $n_{2}/n_{3}=4.11\ldots $ then $q_{2}=4$ and $n_{4}=r_{2}=n_{2}-4\times
n_{3}=59.$

As $n_{3}/n_{4}=8.93\ldots $ then $q_{3}=8$ and $n_{5}=r_{3}=n_{3}-8\times
n_{4}=55.$

As $n_{4}/n_{5}=1.07\ldots $ then $q_{4}=1$ and $n_{6}=r_{4}=n_{4}-n_{5}=4.$

As $n_{5}/n_{6}=13.75$ then $q_{5}=13$ and $n_{7}=r_{5}=n_{5}-13\times
n_{6}=3.$

As $n_{6}/n_{7}=1.25$ then $q_{6}=1$ and $n_{8}=r_{6}=n_{6}-n_{7}=1.$

\noindent Hence $\mathrm{hcf}\left( 65537,2167\right) =1$ as required and we
can work backwards to find%
\begin{eqnarray*}
1 &=&n_{6}-n_{7} \\
&=&n_{6}-\left( n_{5}-13n_{6}\right) =-n_{5}+14n_{6} \\
&=&-n_{5}+14\left( n_{4}-n_{5}\right) =14n_{4}-15n_{5} \\
&=&14n_{4}-15\left( n_{3}-8n_{4}\right) =-15n_{3}+134n_{4} \\
&=&-15n_{3}+134\left( n_{2}-4n_{3}\right) =134n_{2}-551n_{3} \\
&=&134n_{2}-551\left( n_{1}-30n_{2}\right) =-551n_{1}+16664n_{2}.
\end{eqnarray*}%
Hence we have that%
\begin{equation*}
\left( -551\right) \times \left( 65537\right) +\left( 16664\right) \times
\left( 2167\right) =1.
\end{equation*}%
So in $\func{mod}65537$ arithmetic we have%
\begin{equation*}
16664\times 2167=1\,\func{mod}65537\text{ \ or equally \ }2167^{-1}=16664\,%
\func{mod}65537.
\end{equation*}
\end{solution}

\section{Unique Factorization in PIDs.}

\begin{definition}
An integral domain $R$ is said to be a \textbf{principal ideal domain} (or
PID) if every ideal is principal.
\end{definition}

\begin{example}
We know that every Euclidean domain is a PID. These include (with $F$ a
field)%
\begin{equation*}
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
,\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i],\qquad F[x],\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{2}]
\end{equation*}%
but not $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ nor $F[x,y].$
\end{example}

\begin{definition}
An integral domain $R$ is said to be a \textbf{unique factorization domain}
(or UFD) if every non-zero, non-unit element $x$ can be written%
\begin{equation*}
x=p_{1}p_{2}\cdots p_{r}
\end{equation*}%
where $p_{1},\ldots ,p_{r}$ are irreducible elements and further this
factorization is unique in the sense that if%
\begin{equation*}
x=p_{1}p_{2}\cdots p_{r}=q_{1}q_{2}\ldots q_{s}
\end{equation*}%
are two factorizations into irreducible elements then $r=s$ and (with a
possible reordering of the factors) we have%
\begin{equation*}
p_{i}=u_{i}q_{i}
\end{equation*}%
for each $i$ where each $u_{i}$ is a unit.
\end{definition}

\begin{remark}
Though we shall not prove it explicitly, the above definition is equivalent
to requiring that (i) non-zero elements can be written as the product of
irreducible elements and (ii) irreducible elements are prime.
\end{remark}

\begin{example}
(i) $4=(1+\sqrt{-3})(1-\sqrt{-3})=2\times 2$ are two essentially different
factorizations in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{-3}].$

(ii) $5=(1+2i)(1-2i)=(2-i)(2+i)$ are essentially the same factorizations in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i].$

(iii) $2x^{2}-6x+4=(2x-2)(x-2)=(x-1)(2x-4)$ are essentially the same
factorization in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].$

(iv) $x^{2}-1=(x-1)(x+1)=(x-3)(x-5)$ are two essentially different
factorizations in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}[x].$
\end{example}

\begin{example}
Factorize $17-i$ into irreducibles in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i].$
\end{example}

\begin{solution}
Note that $d(17-i)=17^{2}+1^{2}=290=2\times 5\times 29$ so that any
factorization into irreducibles can involve at most three factors. Up to
multiplication by units, the only elements with a norm of $2$ are $1\pm i.$
Now note that%
\begin{equation*}
\frac{17-i}{1+i}=\frac{(17-i)(1-i)}{2}=\frac{16-18i}{2}=8-9i.
\end{equation*}%
Likewise the only Gaussian integers with a norm of $5$ are $1\pm 2i$ up to
units. Note that%
\begin{equation*}
\frac{8-9i}{1+2i}=\frac{(8-9i)(1-2i)}{5}=\frac{-10-25i}{5}=-2-5i.
\end{equation*}%
Hence%
\begin{equation*}
17-i=(1+i)(1+2i)(-2-5i)
\end{equation*}%
is a factorization into irreducibles.
\end{solution}

\begin{proposition}
In a UFD two non-zero elements have a unique hcf.
\end{proposition}

\begin{proof}
Let $x,y$ be non-zero elements of a UFD $R$. By considering the irreducible
elements which divide $a$ or $b$ we may write%
\begin{equation*}
x=up_{1}^{\alpha _{1}}\cdots p_{r}^{\alpha _{r}}\qquad \text{and}\qquad
y=vp_{1}^{\beta _{1}}\cdots p_{r}^{\beta _{r}}
\end{equation*}%
where $u$ and $v$ are units and $\alpha _{i},\beta _{i}\geqslant 0.$ Then it
can be shown that%
\begin{equation*}
\mathrm{hcf}(x,y)=p_{1}^{\gamma _{1}}\cdots p_{r}^{\gamma _{r}}\qquad \text{%
where }\gamma _{i}=\min \{\alpha _{i},\beta _{i}\}
\end{equation*}%
and that this hcf is unique up to multiplication by a unit. Details are
omitted here.
\end{proof}

We know that in an ID\ a prime element is irreducible, but that the converse
need not hold (e.g. $2$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack \sqrt{-3}]$). We see now that in UFDs and PIDs that irreducibles are
prime. As we shall see, PIDs are UFDs and so the latter should not be
surprising. But we shall need the equivalence of primes and irreducibles in
PIDs to actually show that PIDs are UFDs.

\begin{proposition}
(a) In a UFD an irreducible element is prime.

(b) In a PID an irreducible element is prime.
\end{proposition}

\begin{proof}
(a) Say that $x$ is irreducible and that $x|yz.$ We then have that $xv=yz$
for some $v$. The elements $v,y,z$ can each be factorized into irreducibles,
and by the uniquenes of the factorizations of $xv=yz$ it much be that $x$
(up to a unit) is present in the factorizations of $y$ or $z$ (or both). So $%
x|y$ or $x|z$ and we see that $x$ is prime.

(b) Say that $x$ is irreducible and $\langle x\rangle \subseteq I\lhd R.$
There exists $y$ such that $\langle y\rangle =I$ and so there exists $z$
such that $x=yz.$ By the irreducibility of $x$ we either have $y$ is a unit
and $I=\langle y\rangle =R$ or $z$ is a unit and we have $I=\langle y\rangle
=\langle x\rangle .$ Hence $\langle x\rangle $ is maximal, and so prime, and
hence $x$ is a prime element.
\end{proof}

\begin{proposition}
PIDs are Noetherian. That is given an increasing sequence of ideals%
\begin{equation*}
I_{1}\subseteq I_{2}\subseteq I_{3}\subseteq \cdots
\end{equation*}%
in a PID $R$, there exists $N$ such that $I_{n}=I_{N}$ for $n\geqslant N.$
\end{proposition}

\begin{proof}
This is left to Sheet 2, Exercise 5(i).
\end{proof}

\begin{theorem}
A PID\ is a UFD.
\end{theorem}

\begin{proof}
\textit{Existence}: Let $R$ be a PID and $x$ be non-zero. If $x$ is
irreducible then we are done; otherwise we may write $x=yz$ where $y$ and $z$
are not units. If $y$ and $z$ are irreducible then we are done, and
otherwise we may continue factorizing the composite elements that arise and,
say, write next $y=ab$. If the process terminates (through lack of any
genuinely composite remaining factors) then we have written $x$ as a product
of irreducibles and otherwise we are done. However if the process did not
terminate then we could produce an infinite strictly increasing sequence of
ideals%
\begin{equation*}
\langle x\rangle \subseteq \langle y\rangle \subseteq \langle a\rangle
\subseteq \cdots .
\end{equation*}%
However as PIDs are Noetherian then this cannot occur and the above
factorization will terminate.

\textit{Uniqueness}: Say that we have%
\begin{equation*}
x=p_{1}p_{2}\cdots p_{r}=q_{1}q_{2}\cdots q_{s}
\end{equation*}%
are two factorizations of $x$ into irreducible elements. As $p_{1}$ is
irreducible, and so prime, then $p_{1}|q_{1}q_{2}\cdots q_{s}$ implies that $%
p_{1}|q_{i}$ for some $i.$ As $q_{i}$ is irreducible and $p_{1}$ is not a
unit we then have $q_{i}=up_{1}$ for some unit $u$. By renumbering $%
q_{1},\ldots ,q_{s}$ and incorporating the unit $u$ into one of the other
factors we may assume that $p_{1}=q_{1}$ and then by the cancellation law we
have that%
\begin{equation*}
p_{2}\cdots p_{r}=q_{2}\cdots q_{s}
\end{equation*}%
and may proceed along similar lines again, ultimately showing that $r=s$ and
that the two factorizations are essentially the same.
\end{proof}

\section{Factorization in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ and $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$}

Factorization in $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]$ is, in principal at least, very straightforward. The fundamental
theorem of algebra tells us that any complex polynomial can be uniquely
written as%
\begin{equation*}
qc(x-\alpha _{1})(x-\alpha _{2})\cdots (x-\alpha _{n})
\end{equation*}%
for some $\alpha _{1},\ldots ,\alpha _{n},c\in 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
$. These linear factors are irreducible and so no further factorization is
possible.

The above also helps us appreciate how factorization works in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$ when one recalls that a real polynomial's non-real roots will
arise as conjugate pairs. Hence we can factorize any real polynomial can be
unqiuely factorized as%
\begin{equation*}
c(x-a_{1})(x-a_{2})\cdots (x-a_{r})q_{1}(x)q_{2}(x)\cdots q_{s}(x)
\end{equation*}%
where $a_{1},\ldots ,a_{s},c$ are real and $q_{1},\cdots ,q_{s}$ are
irreducible monic quadratics.

We know from our earlier discussion that $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ is a UFD but we are yet to prove this for $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ (though this is indeed the case)$.$ But we have no general
theorem like that above to help us determine these rings' irreducible
elements. We begin with a naive treatment of the following example.

\begin{example}
Show that the cubic $x^{3}-2x+3$ is irreducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$
\end{example}

\begin{solution}
If the cubic did factorize then its factors would include a linear factor $%
ax+b$. Note that $a$ would have to divide $1$ and that $b$ would have to
divide $3$. So up to units, the only linear factors could be $%
x-1,x+1,x-3,x+3.$ However as none of $1,-1,3,-3$ is a root of the cubic then
we can see that the cubic is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$
\end{solution}

How might we have approached this problem in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$? Certainly not so naively treating a finite number of
possibilities. Reassuringly Gauss showed these problems to be largely
equivalent. We first note the following useful and quite general approach
involving the polynomial rings $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}[x].$

\begin{proposition}
Let $f(x)$ be an integer polynomial whose leading coefficient is not
divisible by the prime $p$. If $\overline{f(x)}$ is irreducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}[x]$ then $f(x)$ is irreducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$
\end{proposition}

\begin{proof}
Suppose that $f=gh$ be a proper factorization in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$. Then $g$ and $h$ are both of positive degree and we also have $%
\deg f=\deg f+\deg g.$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}[x]$ we have $\overline{f}=\overline{g}\overline{h}$ and%
\begin{equation*}
\deg f=\deg \overline{f}=\deg \overline{g}+\deg \overline{h}\leqslant \deg
g+\deg h=\deg f
\end{equation*}%
as the leading coefficient of $f$ is not divisible by $p,$ as $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}$ is a field and as $\deg \overline{p}\leqslant \deg p$ in general.
Hence it must be the case that $\deg \overline{g}$ and $\deg \overline{h}$
are both positive and $\overline{f}$ is reducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}[x].$
\end{proof}

\begin{example}
Show that the following polynomials are irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$%
\begin{equation*}
f_{1}(x)=17x^{3}+7x+3,\qquad f_{2}(x)=2x^{3}+3x^{2}+x-2,\qquad
f_{3}(x)=7x^{4}+5x-3.
\end{equation*}
\end{example}

\begin{solution}
$f_{1}(x)$ $\func{mod}2$ equals $x^{3}+x+1$ which is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ as it has no roots in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}.$ Hence $f_{1}(x)$ is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$

We cannot use $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ for $f_{2}(x)$ as $2$ divides the leading coefficient. However $\func{%
mod}3$ we obtain the polynomial%
\begin{equation*}
2x^{3}+x+1
\end{equation*}%
which does not have a root in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}$ and so $f_{2}(x)$ is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$

$f_{3}(x)$ $\func{mod}2$ equals $x^{4}+x+1$ which is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ as $\overline{f_{3}(x)}$ has no roots in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ and as $x^{2}+x+1$, the only irreducible quadratic $\func{mod}2$ is
not a factor. So $f_{3}(x)$ is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$
\end{solution}

Whilst thinking along these lines we will also introduce the following
criterion for irreducibility. The criterion may seem somewhat contrived but
can in fact be very useful particularly for cyclotomic polynomials.

\begin{proposition}
(\textbf{Eisenstein's Criterion}) Let $f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots
+a_{1}x+a_{0}$ be an integer polynomial and $p$ be a prime such that%
\begin{equation*}
p\text{ does not divide }a_{n};\qquad p\text{ divides }a_{0},a_{1},\ldots
,a_{n-1};\qquad p^{2}\text{ does not divide }a_{0}.
\end{equation*}%
Then $f(x)$ is irreducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$
\end{proposition}

\begin{proof}
Note that $\func{mod}p$ we have%
\begin{equation*}
\overline{f(x)}=\overline{a_{n}}x^{n}
\end{equation*}%
where $\overline{a_{n}}\neq \bar{0}.$ If $f(x)=g(x)h(x)$ is a genuine
reduction into integer polynomials, then, as $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}[x]$ is a UFD, we have%
\begin{equation*}
\overline{g(x)}=\bar{b}x^{r}\qquad \qquad \overline{h(x)}=\bar{c}x^{s}
\end{equation*}%
where $r+s=n$ and $r,s\geqslant 1.$ So $p$ divides the constant coefficients
of $g(x)$ and $h(x)$ and hence $p^{2}$ divides the constant constant of $%
f(x),$ but this is the required contradiction.
\end{proof}

\begin{example}
Show that $x^{4}+x^{3}+x^{2}+x+1$ is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$
\end{example}

\begin{solution}
A standard trick for such "cyclotomic" polynomials (irreducible factors of $%
x^{n}-1$) is to set $x=u+1.$ The new polynomial in $u$ will be irreducible
if and only if the original one in $x$ is. Now%
\begin{equation*}
(u+1)^{4}+(u+1)^{3}+(u+1)^{2}+(u+1)+1=u^{4}+5u^{3}+10u^{2}+10u+5.
\end{equation*}%
We see that the new polynomial is irreducible -- by Eisenstein with $p=5$ --
and so the original polynomial also is.
\end{solution}

We return now to discussion relating factorization over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ to factorization over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$

\begin{definition}
Let $f(x)$ be a polynomial over the integers. The \textbf{content} $c(f)$ is
the hcf of the coefficients of $f$. A polynomial is said to be \textbf{%
primitive} is it has content equal to $1.$
\end{definition}

Note that any integer polynomial can be written uniquely as $f(x)=cp(x)$
where $c=c(f)$ and $p(x)$ is a primitive polynomial.

\begin{lemma}
(\textbf{Gauss Lemma}) The product of two primitive polynomials is
primitive. Consequently%
\begin{equation*}
c(fg)=c(f)c(g)
\end{equation*}%
for any two integer polynomials $f$ and $g.$
\end{lemma}

\begin{proof}
Say that $f$ and $g$ are primitive polynomials and suppose for a
contradiction that $fg$ is not primitive. Let $p$ be a prime factor of $%
c(fg).$ We then have that%
\begin{equation*}
\overline{f(x)}\overline{g(x)}=0\func{mod}p.
\end{equation*}%
As $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}[x]$ is an ID\ then either $\overline{f(x)}=0$ or $\overline{g(x)}=0$ in 
$%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{p}[x]$. But then $p$ divides the coefficients of $f$ or divides the
coefficients of $g$. This would contradict $f$ and $g$ being primitive.

More generally for integer polynomials $f$ and $g$ we have%
\begin{equation*}
f=c(f)p_{f},\qquad g=c(g)p_{g},\qquad fg=c(fg)p_{fg}.
\end{equation*}%
Then%
\begin{equation*}
c(fg)p_{fg}=c(f)c(g)p_{f}p_{g}.
\end{equation*}%
As $p_{f}p_{g}$ is primitive, then by the uniqueness of the above
expressions we have $c(fg)=c(f)c(g)$ and $p_{fg}=p_{f}p_{g}.$
\end{proof}

Gauss' lemma plays a key role in relating factorization in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ with factorization in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].$ Note

\begin{itemize}
\item $2x-6$ is irreducible in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ but not in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ as $2x-6=2(x-3)$ and $2$ is not a unit in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$

\item But if $f(x)$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ reduces into positive degree polynomials in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ then it is reducible in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ as $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]\subseteq 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].$

\item In general polynomials in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ are not in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ but can be multiplied by a non-zero integer to form a polynomial
in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$
\end{itemize}

Given the example that started this section we would much rather work with
irreducibility in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ rather than in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ if at all possible. The following result addresses precisely
this point.

\begin{theorem}
\label{thmgauss}Let $f$ be a primitive non-constant polynomial in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$ Then $f(x)$ is irreducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ if and only if $f(x)$ is irreducible in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].$
\end{theorem}

\begin{proof}
If $f$ is irreducible over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ then it is irreducible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$, given the comments above. Say now that $f(x)$ is genuinely reducible in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x],$ say $f(x)=q_{1}(x)q_{2}(x)$ for polynomials in $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].$ We then have $g_{1}=d_{1}q_{1}$ and $g_{2}=d_{2}q_{2}$ are
integer polynomials where $d_{i}$ is the lcm of the denominators of the
coefficients in $q_{i}.$ Finally we have $g_{i}=c(g_{i})p_{i}$ where $p_{i}$
are primitive integer polynomials. So%
\begin{equation*}
d_{1}d_{2}f=g_{1}g_{2}=c(g_{1})c(g_{2})p_{1}p_{2}.
\end{equation*}%
As $f$ and $p_{1}p_{2}$ are both primitive then we have, by uniqueness, that 
$f=p_{1}p_{2}$ and so $f$ is reducible in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x].$
\end{proof}

\begin{proposition}
$%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ is a UFD.
\end{proposition}

\begin{proof}
Any $f(x)$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack x]$ can be written uniquely as $f(x)=c(f)p(x)$ where $p$ is
primitive. As $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ is a UFD then $c(f)$ can be uniquely written as a product of prime
(integers). As $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ is a UFD then $p(x)$ can be written as a product of irreducible
rational polynomials. As seen in the proof of Theorem \ref{thmgauss} any
reduction of $p(x)$ into rational polynomials can be replaced with a
reduction of $p(x)$ into integer polynomials, again in a unique fashion (up
to the units of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$).
\end{proof}

In fact the above theorems generalize quite naturally to demonstrate the
following result. As UFDs have well defined hcfs then we can introduce the
notion of content to polynomials in $R[x].$ Replacing $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ with $R$ and $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ with the field of fractions of $R$ then we can rewrite the above arguments
to show:

\begin{proposition}
If $R$ is a UFD then so is $R[x].$
\end{proposition}

\chapter{Modules}

Modules are essentially the equivalent of vector spaces when the scalars
come from a ring rather than a field. Throughout we will consider only
modules over Euclidean domains. (The results that follow can all be proved
over PIDs, though some of the proofs can be somewhat more laborious and in
any case the important examples we will meet are all over EDs.)

\begin{definition}
Let $R$ be an ED. An $R$-\textbf{module} $M$ is an abelian group together
with scalar multiplication $R\times M\rightarrow M,$ denoted as $%
(r,m)\mapsto rm,$ such that

\begin{itemize}
\item $r(m_{1}+m_{2})=rm_{1}+rm_{2}$ for all $r\in R$ and $m_{1},m_{2}\in M.$

\item $(r_{1}+r_{2})m=r_{1}m+r_{2}m$ for all $r_{1},r_{2}\in R$ and $m\in M.$

\item $(r_{1}r_{2})m=r_{1}(r_{2}m)$ for all $r_{1},r_{2}\in R$ and $m\in M.$

\item $1_{R}m=m$ for all $m\in M.$
\end{itemize}
\end{definition}

\begin{example}
\textbf{Modules over} $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.$ A $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-module is an abelian group and vice versa an abelian group is a $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-module. Scalar multiplication by integers is entirely determined by the
abelian group structure as for any positive integer $n$ we have%
\begin{equation*}
n.m=(1+\cdots +1).m=(1.m)+\cdots +(1.m)=m+\cdots +m
\end{equation*}%
and $(-n).m=(-1).(n.m)=-(n.m).$
\end{example}

\begin{example}
\textbf{Modules over fields}. A module over a field is a vector space.
\end{example}

\begin{example}
\textbf{Modules over polynomial rings}. Say that $M$ is a module over $F[x]$
where $F$ is a field. Then $M$ is a vector space over $F$ when considered as
the constant polynomials in $F[x].$

Multiplication by $x$, that is $T(m)=xm$ has the effect of an $F$-linear map 
$T\colon M\rightarrow M.$

Note that the entire $F[x]$-module structure on $V$ is entirely determined
by this map $T$ as by the module axioms we have%
\begin{equation*}
p(x).m=p(T)m
\end{equation*}%
for any polynomial $p(x)$ in $F[x].$

Conversely given a vector space $M$ over $F$ and a linear map $T\colon
M\rightarrow M$ we can define $M$ as a $F[x]$-module by defining scalar
multiplication as%
\begin{equation*}
x.m=T(m).
\end{equation*}
\end{example}

\begin{example}
Given a field $F$ and a square matrix $A$ over $F$ then $F[A]$ is an $F[x]$%
-module with%
\begin{equation*}
x.p(A)=(xp)(A).
\end{equation*}%
\textbf{N.B. }In general though $F[A]$ is a different module to the $F[x]$%
-module defined by $A.$

For example, consider the real matrix%
\begin{equation*}
A=\left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 1 \\ 
0 & 0 & 1%
\end{array}%
\right) .
\end{equation*}%
The $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$-module defined by $A$ is a $3$-dimensional real vector space. By
comparison%
\begin{equation*}
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack A]\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle (x-1)^{2}\rangle },
\end{equation*}%
because $m_{A}(x)=(x-1)^{2},$ and is a two-dimensional real vector space. $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack A]$ is spanned by $I,A$.
\end{example}

\begin{example}
Given two $R$-modules $M$ and $N$ then we can form the \textbf{direct sum} $%
M\oplus N$ as an $R$-module in the natural way by component-wise addition
and scalar multiplication.

Generally for a positive integer $n$ we can define the $R$-module $%
R^{n}=R\oplus \cdots \oplus R$ ($n$ times). These are the \textbf{free
modules} over $R.$
\end{example}

\begin{example}
If $R$ is an ED and $I$ is an ideal of $R$ then $R/I$ is naturally an $R$%
-module. For example with $R=%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ and $I=\langle x^{2}+1\rangle $ then%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle x^{2}+1\rangle }
\end{equation*}%
also has the structure of a two-dimensional vector space over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
$ with basis $1,x.$ Scalar multiplication is defined by%
\begin{equation*}
x.1=x,\qquad x.x=x^{2}=-1
\end{equation*}%
so that multiplication by $x$ is represented by the matrix%
\begin{equation*}
B=\left( 
\begin{array}{cc}
0 & -1 \\ 
1 & 0%
\end{array}%
\right)
\end{equation*}%
with respect to the above basis. Hopefully unsurprisingly%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle x^{2}+1\rangle }\qquad \text{\ and}\qquad 
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack B]
\end{equation*}%
are isomorphic as $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$-modules, with such an isomorphism being $a+bx\mapsto a+bB.$
\end{example}

Here we make rigorous the idea of being isomorphic as $R$-modules and also
introduce the idea of a module homomorphism.

\begin{definition}
Let $M$ and $N$ be $R$-modules. A map $\phi \colon M\rightarrow N$ is a 
\textbf{module homomorphism} if%
\begin{equation*}
\phi (m_{1}+m_{2})=\phi (m_{1})+\phi (m_{2}),\qquad \phi (rm_{1})=r\phi
(m_{1})
\end{equation*}%
where $r\in R,m_{1},m_{2}\in M.$ We say that $\phi $ is a \textbf{module
isomorphism} if $\phi $ is a bijection.
\end{definition}

\begin{example}
A module homomorphism between $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-modules is a group homomorphism.
\end{example}

\begin{example}
If $R$ is a field then the module homomorphisms are precisely the linear
maps.
\end{example}

\begin{example}
Let%
\begin{equation*}
C=\left( 
\begin{array}{cc}
1 & 0 \\ 
0 & 2%
\end{array}%
\right) ,\qquad D=\left( 
\begin{array}{cc}
1 & 0 \\ 
0 & 3%
\end{array}%
\right) .
\end{equation*}%
Find all module homomorphisms from $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C]$ to $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack D].$
\end{example}

\begin{solution}
Every element of $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C]$ can be written as $\alpha I+\beta C$ and every of $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack D]$ can be written $\alpha I+\beta D.$ So if we take initial basis $%
\left\{ I,C\right\} $ and final basis $\left\{ I,D\right\} $ then a module
homomorphism $\phi $ can must be represented by a $2\times 2$ real matrix.
Say%
\begin{equation*}
\phi =\left( 
\begin{array}{cc}
a & b \\ 
c & d%
\end{array}%
\right)
\end{equation*}%
so that $\phi (I)=aI+cD$ and $\phi (C)=bI+dD.$ Our further requirements on $%
\phi $ are that%
\begin{eqnarray*}
bI+dD &=&\phi (C)=\phi (x.I)=x.\phi (I)=D(aI+cD)=-3cI+(a+4c)D; \\
(-2a+3b)I+(-2c+3d)D &=&\phi (3C-2I)=\phi (C^{2}) \\
&=&\phi (x.C)=x.\phi (C)=D(bI+dD)=-3dI+(b+4d)D,
\end{eqnarray*}%
noting that $C^{2}=3C-2I$ and $D^{2}=4D-3I.$ Comparing coefficients we have
that%
\begin{equation*}
b=-3c,\qquad d=a+4c,\qquad -2a+3b=-3d,\qquad -2c+3d=b+4d.
\end{equation*}%
Solving these we see there is a one-parameter family of solutions with%
\begin{equation*}
a=b=-3c\qquad \text{and}\qquad d=c.
\end{equation*}%
So%
\begin{equation*}
\phi =c\left( 
\begin{array}{cc}
-3 & -3 \\ 
1 & 1%
\end{array}%
\right) .
\end{equation*}%
We might have done the above calculation in a slightly more systematic way.
Under the identification $\left( \alpha ,\beta \right) ^{T}\leftrightarrow
\alpha I+\beta C$ of $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}$ with $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C]$ we note%
\begin{equation*}
x.\left( 
\begin{array}{c}
\alpha \\ 
\beta%
\end{array}%
\right) \leftrightarrow x.(\alpha I+\beta C)=\alpha C+\beta C^{2}=-2\beta
I+(\alpha +3\beta )C\leftrightarrow \left( 
\begin{array}{c}
\alpha \\ 
\beta%
\end{array}%
\right) =\left( 
\begin{array}{cc}
0 & -2 \\ 
1 & 3%
\end{array}%
\right) \left( 
\begin{array}{c}
\alpha \\ 
\beta%
\end{array}%
\right)
\end{equation*}%
so that multiplication by $x$ in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C]$ is given by the matrix%
\begin{equation*}
\left( 
\begin{array}{cc}
0 & -2 \\ 
1 & 3%
\end{array}%
\right)
\end{equation*}%
with respect to our basis, and similarly multiplication by $x$ in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack D]$ is given by the matrix%
\begin{equation*}
\left( 
\begin{array}{cc}
0 & -3 \\ 
1 & 4%
\end{array}%
\right)
\end{equation*}%
with respect to $\{I,D\}.$ So the matrices for $\phi $ that we found are
precisely those that satisfy%
\begin{equation*}
\left( 
\begin{array}{cc}
0 & -3 \\ 
1 & 4%
\end{array}%
\right) \left( 
\begin{array}{cc}
a & b \\ 
c & d%
\end{array}%
\right) =\left( 
\begin{array}{cc}
a & b \\ 
c & d%
\end{array}%
\right) \left( 
\begin{array}{cc}
0 & -2 \\ 
1 & 3%
\end{array}%
\right)
\end{equation*}%
as a consequence of requiring $x.\phi (v)=\phi (x.v)$ for all $v$ in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C].$ (We will see in due course that these two matrices are the
companion matrices for $m_{C}(x)$ and $m_{D}(x).$)

Why are there limited module homomorphisms here? And how might we better
understand them? The second question will have a clearer answer once we
better understand the structure of modules. However, to answer the first
question, note that scalar multiplication by $x^{2}-3x+2$ is the same as
multiplication by zero in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C]$ as $C^{2}-3C+2I=0$. However this is not the case in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack D]$ as $D^{2}-3D+2I\neq 0.$
\end{solution}

\begin{definition}
A non-empty subset $N$ of a module $M$ is a \textbf{submodule} if $N$ is
closed under addition and scalar multiplication.
\end{definition}

\begin{example}
When we consider $R$ as an $R$-module the submodules of $R$ are the ideals.
\end{example}

\begin{example}
The submodules of a $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-module are the subgroups.
\end{example}

\begin{example}
Let $V$ be a vector space over a field $F$ and $T\colon V\rightarrow V$ be a
linear map defining $V$ as an $F[x]$-module. Then the submodules of $V$ as
the $T$-invariant subspaces -- i.e. those subspaces $U$ of $V$ such that $%
T(U)\subseteq U.$
\end{example}

\begin{example}
Given a module homomorphism $\phi \colon M\rightarrow N$ then%
\begin{eqnarray*}
\ker \phi &=&\left\{ m\in M:\phi (m)=0\right\} \qquad \text{\ is a submodule
of }M; \\
\func{Im}\phi &=&\left\{ \phi (m):m\in M\right\} \qquad \text{\ is a
submodule of }N.
\end{eqnarray*}
\end{example}

\begin{example}
The rational matrix%
\begin{equation*}
F=\left( 
\begin{array}{cc}
1 & 1 \\ 
1 & 1%
\end{array}%
\right)
\end{equation*}%
defines a $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$ module structure on $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
^{2}$. The submodules of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
^{2}$ are the $F$-invariant subspaces and so include the eigenspaces%
\begin{equation*}
E_{2}=\langle (1,1)^{T}\rangle ,\qquad E_{0}=\langle (1,-1)^{T}\rangle
\end{equation*}%
and we see that, as a direct sum of $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]$-modules,%
\begin{equation*}
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
^{2}=E_{2}\oplus E_{0},\qquad \left( 
\begin{array}{c}
a \\ 
b%
\end{array}%
\right) =\left( \frac{a+b}{2}\right) \left( 
\begin{array}{c}
1 \\ 
1%
\end{array}%
\right) +\left( \frac{a-b}{2}\right) \left( 
\begin{array}{c}
1 \\ 
-1%
\end{array}%
\right)
\end{equation*}
\end{example}

\begin{definition}
(a) We say that elements $x_{1},x_{2},\ldots ,x_{n}$ of an $R$-module $M$
are \textbf{linearly independent} if the only solution of the equation%
\begin{equation*}
r_{1}x_{1}+r_{2}x_{2}+\cdots +r_{n}x_{n}=0,\qquad r_{i}\in R
\end{equation*}%
is $r_{1}=r_{2}=\cdots =r_{n}=0.$

(b) We say that elements $x_{1},x_{2},\ldots ,x_{n}$ of an $R$-module $M$ 
\textbf{generate }or \textbf{span} $M$ if every element $x$ of $M$ can be
written%
\begin{equation*}
x=r_{1}x_{1}+r_{2}x_{2}+\cdots +r_{n}x_{n}
\end{equation*}%
for some $r_{1},r_{2},\ldots ,r_{n}$ in $R.$

(c) We say that elements $x_{1},x_{2},\ldots ,x_{n}$ of an $R$-module $M$
form a basis for $M$ if the elements are linearly independent and span $M.$

(d) We say that a module $M$ is \textbf{finitely generated} if there is a
finite subset of $M$ that generates $M.$

(e) A module with a basis is called a \textbf{free module}. \textbf{N.B.
Most modules don't have bases.}
\end{definition}

\begin{example}
The elements $e_{1}=(1,0,\ldots ,0),e_{2}=(0,1,0,\ldots ,0),\ldots
,e_{n}=(0,\ldots ,0,1)$ are a basis for $R^{n}.$
\end{example}

\begin{example}
The set $\{2\}$ in $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ is linearly independent but not a basis.
\end{example}

\begin{example}
$%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ is not free as a $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-module as $2.\bar{1}=\bar{0}$ but $2\neq 0.$ But $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$ is free as a $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}$-module -- of course it remains the case that $\bar{2}.\bar{1}=\bar{0}$
but now the scalar $\bar{2}$ is the zero scalar.
\end{example}

\begin{proposition}
Any basis of $R^{n}$ contains $n$ elements. $n$ is known as the \textbf{rank 
}of $R^{n}.$ Note an $R$-module with a basis containing $n$ elements is
isomorphic as an $R$-module to $R^{n}.$
\end{proposition}

\begin{proof}
Say that $b_{1},\ldots ,b_{m}$ is a basis for $R^{n}.$ Then we may write
each $b_{i}$ as%
\begin{equation*}
b_{i}=b_{1i}e_{1}+b_{2i}e_{2}+\cdots +b_{ni}e_{n}
\end{equation*}%
so that the $n\times m$ matrix $B=(b_{ij})$ is the change of basis matrix
from the $e_{i}$ to the $b_{i}.$ However as the $b_{i}$ are also a basis, we
can write the $e_{i}$ in terms of the $b_{i}$ and hence there is an $m\times
n$ change of basis matrix $A$. We then have that%
\begin{equation*}
BA=I_{n}\qquad \text{and\qquad }AB=I_{m}.
\end{equation*}%
If $m\neq n$ then we may assume without loss of generality that $m>n.$ We
can introduce the $m\times m$ matrices%
\begin{equation*}
A^{\prime }=(A\,|\,0_{m(m-n)}),\qquad B^{\prime }=\left( 
\begin{array}{c}
B \\ 
0_{(m-n)m}%
\end{array}%
\right)
\end{equation*}%
which still satisfy $A^{\prime }B^{\prime }=I_{m}.$ However we would then
have%
\begin{equation*}
1=\det I_{m}=\det A^{\prime }\det B^{\prime }=0\times 0=0.
\end{equation*}%
A contradiction. (Note that the veracity of the determinant product rule
relies only on the commutativity of the matrices' entries.)

If a module $M$ has a basis $b_{1},\ldots ,b_{n}$ then every element $x$ of $%
M$ can be uniquely written as%
\begin{equation*}
x=r_{1}b_{1}+r_{2}b_{2}+\cdots +r_{n}b_{n}
\end{equation*}%
and the identification $x\leftrightarrow (r_{1},r_{2},\ldots ,r_{n})$ is a
module isomorphism with $R^{n}.$
\end{proof}

\begin{proposition}
A submodule $M$ of the free module $R^{n}$ is finitely generated.
\end{proposition}

\begin{proof}
We shall prove this by induction on $n$. The submodule of $R$ are the ideals
which are principal, and so generated by a single element, proving the case $%
n=1.$ Say now that the result holds for $R^{n-1}$ and let $W$ be a submodule
of $R^{n}.$ Then%
\begin{equation*}
W_{0}=\left\{ \mathbf{w}=(w_{1},\ldots ,w_{n})\in W:w_{n}=0\right\}
\end{equation*}%
is isomorphic to a submodule of $R^{n-1}$ and so is finitely generated. If $%
W=W_{0}$ then we are done. Otherwise there is $\mathbf{x}\in W$ with $%
x_{n}\neq 0$ and with norm $d(x_{n})$ being minimal. Say that $\mathbf{y}%
_{1},\ldots ,\mathbf{y}_{k}$ generate $W_{0}$ By minimality of $d(x_{n})$
for any $\mathbf{w}\in W$ we have $w_{n}=rx_{n}$ for some $r$ and hence $%
\mathbf{w}-r\mathbf{x}\in W_{0}$. It follows that $\mathbf{y}_{1},\ldots ,%
\mathbf{y}_{k},\mathbf{x}$ generate $W$ and so $W$ is finitely generated and
the result follows by induction.
\end{proof}

\begin{corollary}
A submodule $M$ of $R^{n}$ is free and $\mathrm{rank}(M)\leqslant n.$
\end{corollary}

\begin{proof}
From the proof of the proposition, we can also see how one might produce a
basis for $W$ with $n$ or fewer elements. We can create a nested sequence of
submodules%
\begin{equation*}
W_{0}^{(1)}=W_{0},\qquad W_{0}^{(k)}=(W_{0}^{(k-1)})_{0},\qquad \text{and}%
\qquad \{\mathbf{0\}}=W_{0}^{(n)}\subseteq W_{0}^{(n-1)}\subseteq \cdots
\subseteq W_{0}^{(1)}\subseteq W,
\end{equation*}%
so that $W_{0}^{(k)}$ consists of those elements of $W$ whose last $k$
entries are $0.$ From the previous proof we see that at each stage we need
add at most one generator to the generators of $W_{0}^{(k)}$ to produce a
set of generators for $W_{0}^{(k-1)}$ and won't need to add any generator if 
$W_{0}^{(k)}=W_{0}^{(k-1)}.$ Further by the nature of how these generators
are constructed they are linearly independent. Hence we can produce a basis
for $W$ containing at most $n$ elements.
\end{proof}

\begin{remark}
Do make note of the important differences between the theory of vector
spaces and the more general theory of modules.

\begin{itemize}
\item Most modules don't have bases -- only the free modules.

\item A linearly independent set cannot always be extended to a basis (even
in a free module).

\item A spanning set need not contain a basis (even in a free module)..

\item A proper submodule of a free module can have the same rank.
\end{itemize}
\end{remark}

\begin{definition}
A module $M$ is said to be \textbf{cyclic }if it is generated by a single
element. For an $F[x]$-module such a generator is called a \textbf{cyclic
vector}.
\end{definition}

\begin{example}
The cyclic $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-modules are the cyclic groups.
\end{example}

\begin{example}
Given a square matrix $A$ over a field $F$ then $F[A]$ is a cyclic module as
it is generated by $I.$
\end{example}

\begin{example}
The $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$-module defined on $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3}$ by%
\begin{equation*}
A=\left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 1 \\ 
0 & 0 & 1%
\end{array}%
\right)
\end{equation*}%
is not cyclic. Note that $m_{A}(x)=(x-1)^{2}$ and so for any $\mathbf{v}\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3}$ we see $\mathbf{v}$ generates $\langle \mathbf{v},A\mathbf{v}\rangle
\neq 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3}$ as $A^{2}\mathbf{v}=2A\mathbf{v}-\mathbf{v}$ and so there is no cyclic
vector.

The $x$-axis $M_{1}$ and $yz$-plane $M_{2}$ however are submodules (they are 
$A$-invariant subspaces) and are cyclic as they are respectively generated
by $\mathbf{i}$ and $\mathbf{k}$. So we can decompose the module%
\begin{equation*}
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3}=M_{1}\oplus M_{2}
\end{equation*}%
as the direct sum of cyclic submodules.
\end{example}

\begin{proposition}
Let $V$ be an $n$-dimensional vector space over $F$ with an $F[x]$-module
structure defined by $T\colon V\rightarrow V.$ Say that $V$ is cyclic as an $%
F[x]$-module, and that $v\in V$ is a cyclic vector. Then the vectors%
\begin{equation*}
v,Tv,T^{2}v,\ldots ,T^{n-1}v
\end{equation*}%
form a basis for $V$ as a vector space. With respect to this basis $T$ has
matrix%
\begin{equation*}
C(f)=\left( 
\begin{array}{ccccc}
0 & 0 & \cdots & 0 & -a_{0} \\ 
1 & 0 & \cdots & 0 & -a_{1} \\ 
0 & 1 & \ddots & \vdots & -a_{2} \\ 
\vdots & \ddots & \ddots & 0 & \vdots \\ 
0 & \cdots & 0 & 1 & -a_{n-1}%
\end{array}%
\right)
\end{equation*}%
where the minimal and characteristic polynomials of $T$ both equal%
\begin{equation*}
f(x)=x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}.
\end{equation*}%
The matrix $C(f)$ is called the \textbf{companion matrix }of $f(x).$
\end{proposition}

\begin{proof}
As $v$ generates $V$ as an $F[x]$-module then the set%
\begin{equation*}
v,Tv,T^{2}v,\ldots
\end{equation*}%
spans $V$ as a vector space. If $k$ is the first occasion that%
\begin{equation*}
v,Tv,T^{2}v,\ldots ,T^{k}v
\end{equation*}%
are linearly dependent as vectors, then we have that $v,Tv,T^{2}v,\ldots
,T^{k-1}v$ are independent and spanning and so $k=\dim V=n.$

Hence $v,Tv,T^{2}v,\ldots ,T^{n-1}v$ is a basis for $V$ as a vector space.
That $T(T^{k}v)=T^{k+1}v$ for $0\leqslant k\leqslant n-2$ accounts for the
first $n-1$ columns of $C(f).$ Then for some $a_{0},a_{1},\ldots ,a_{n-1}\in
F$ we have%
\begin{equation*}
T^{n}v=-a_{0}v-a_{1}Tv-\cdots -a_{n-1}T^{n-1}v.
\end{equation*}%
If we define $f(x)$ as above then we have $f(T)v=0.$ But as polynomials in $%
T $ commute we also have $f(T)T^{k}v=0$ for all $k$ and hence $f(T)=0.$ It
must be the case that $\det m_{T}\geqslant n$ as $v,Tv,T^{2}v,\ldots
,T^{n-1}v$ are independent and also the case that $m_{T}|f.$ As $f$ is monic
then $m_{T}=f.$ As $c_{T}$ has degree $n$ and $m_{T}|c_{T}$ then we also
have $m_{T}=c_{T}.$
\end{proof}

\begin{definition}
Given an $R$-module $M$ and submodule $N$ then we can form the \textbf{%
quotient }$R$\textbf{-module} $M/N$ by%
\begin{equation*}
M/N=\left\{ m+N:m\in M\right\}
\end{equation*}%
with%
\begin{equation*}
(m_{1}+N)+(m_{2}+N)\colon =(m_{1}+m_{2})+N,\qquad r.(m+N)\colon =r.m+N.
\end{equation*}
\end{definition}

\begin{example}
Whilst somewhat cumbersome in notation expressing a module as a quotient
module can make transparent the module structure. For example%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x+1\rangle }
\end{equation*}%
is isomorphic to the $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$-module defined on $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$ by $-I_{1}=(-1).$ All this information is captured in writing the module
as above -- namely that scalar multiplication by $x$ acts in the same way as
multiplication by $-1$.
\end{example}

\begin{example}
Note that for any polynomial $p(x)$ in $F[x]$ that%
\begin{equation*}
\frac{F[x]}{\langle p(x)\rangle }
\end{equation*}%
is a cyclic $F[x]$-module as it is generated by $x.$
\end{example}

\begin{example}
The Chinese Remainder Theorem still holds for modules and so we have for
example that%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{2}-1\rangle }\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x+1\rangle }
\end{equation*}%
with a module isomorphism being given by%
\begin{equation*}
a+bx+\langle x^{2}-1\rangle \mapsto (a+bx+\langle x-1\rangle ,a+bx+\langle
x+1\rangle )=(a+b+\langle x-1\rangle ,a-b+\langle x+1\rangle ).
\end{equation*}
\end{example}

\begin{theorem}
(\textbf{First Isomorphism Theorem}) Let $\phi \colon M\rightarrow N$ be a
module homomorphism. Then the map%
\begin{equation*}
\tilde{\phi}\colon \frac{M}{\ker \phi }\rightarrow \func{Im}\phi \qquad 
\text{where\quad }\tilde{\phi}(m+\ker \phi )=\phi (m)
\end{equation*}%
is a module isomorphism.
\end{theorem}

\begin{proof}
This proof is almost identical to the first isomorphism theorem for rings,
with an extra line to check the $R$-module structures align.
\end{proof}

\begin{example}
We return here to the modules $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C]$ and $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack D]$ introduced earlier where%
\begin{equation*}
C=\left( 
\begin{array}{cc}
1 & 0 \\ 
0 & 2%
\end{array}%
\right) ,\qquad D=\left( 
\begin{array}{cc}
1 & 0 \\ 
0 & 3%
\end{array}%
\right) .
\end{equation*}%
We can decompose these modules in such a way that should make clearer the
nature of the homomorphisms we found between them. We shall show that%
\begin{equation*}
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C]\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{2}-3x+2\rangle }\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-2\rangle }.
\end{equation*}%
These isomorphisms are given by%
\begin{equation*}
p(C)\mapsto p(x)+\langle x^{2}-3x+2\rangle \qquad \text{and by}\qquad
p(x)+\langle x^{2}-3x+2\rangle \mapsto \left( p(x)+\langle x-1\rangle
,p(x)+\langle x-2\rangle \right) .
\end{equation*}%
In a similar fashion%
\begin{equation*}
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack D]\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-3\rangle }.
\end{equation*}%
Whilst these may look somewhat cumbersome decompositions they contain
explicitly and transparently the $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$-module structure. Thus finding the module homomorphisms between
these two modules is a lot more straightforward and the answer a lot
clearer. Note for example that%
\begin{equation*}
x.(\alpha ,\beta )=(\alpha ,2\beta )\qquad \text{in \ }\frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-2\rangle }.
\end{equation*}

A module homomorphism%
\begin{equation*}
\phi \colon \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-2\rangle }\rightarrow \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-3\rangle }
\end{equation*}%
is, in particular, an $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
$-linear map and so is determined by $\phi (1,0)$ and $\phi (0,1).$ Say that 
$\phi (1,0)=(a,b)$ and $\phi (0,1)=(c,d)$; then, as $\phi $ is a module
homomorphism, we have 
\begin{eqnarray*}
(a,2b) &=&x.(a,b)=x.\phi (1,0)=\phi (x.(1,0))=\phi (1,0)=(a,b); \\
(c,3d) &=&x.(c,d)=x.\phi (0,1)=\phi (x.(0,1))=\phi (0,2)=(2c,2d).
\end{eqnarray*}%
So we see that $b=c=d=0$ and that the only homomorphisms are of the form%
\begin{equation*}
\left( \alpha ,\beta \right) \mapsto (a\alpha ,0).
\end{equation*}%
These module homomorphisms correspond to a scaling in the $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]/\langle x-1\rangle $ factor and a collapsing of the $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]/\langle x-2\rangle $, hopefully not surprisingly as scalar
multiplication by $x$ in $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack D]$ never corresponds to multiplication by $2$ as it does in the
second summand of $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack C].$
\end{example}

\begin{example}
We return to the $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$-module $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3}$ defined by the matrix%
\begin{equation*}
A=\left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 1 \\ 
0 & 0 & 1%
\end{array}%
\right) .
\end{equation*}%
We have already seen that this ideal is not cyclic but can be decomposed as
the direct sum of the $x$-axis $M_{1}$ and the $yz$-plane $M_{2}$ which are
cyclic submodumles. Somewhat more transparently we can represent the module
structure by writing%
\begin{equation*}
M_{1}\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\cong 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack I_{1}]\qquad \text{and}\qquad M_{2}\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle (x-1)^{2}\rangle }\cong 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\left[ \left( 
\begin{array}{cc}
1 & 1 \\ 
0 & 1%
\end{array}%
\right) \right] .
\end{equation*}
\end{example}

\chapter{Smith Normal Form. Presentations.}

Throughout let $R$ denote a Euclidean domain.

\begin{definition}
(a) A square matrix $P$ with entries in $R$ is said to be \textbf{invertible}
if there is a matrix $Q$ with entries in $R$ such that $PQ=I=QR.$

(b) Two $m\times n$ matrices $A$ and $B$ are said to be \textbf{equivalent}
if there exist an invertible $m\times m$ matrix $P$ and an invertible $%
n\times n$ matrix $Q$ such that $PAQ=B.$ A simple check shows that
equivalence is indeed an equivalence relation.
\end{definition}

\begin{example}
The matrix $\mathrm{diag}(1,2)$ is not invertible over $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$ but would be, say, over $%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x].$
\end{example}

\begin{theorem}
(\textbf{Smith Normal Form}) Let $A$ be an $m\times n$ matrix with entries
in $R.$ Then there exist elements $d_{1},d_{2},\ldots ,d_{r}$, known as 
\textbf{invariant factors}, and unique up to multiplication by units, such
that $A$ is equivalent to%
\begin{equation*}
\left( 
\begin{array}{cc}
\mathrm{diag}(d_{1},d_{2},\ldots ,d_{r}) & 0_{r(n-r)} \\ 
0_{(m-r)r} & 0_{(m-r)(n-r)}%
\end{array}%
\right)
\end{equation*}%
where%
\begin{equation*}
d_{1}|d_{2}|d_{3}|\cdots |d_{r}.
\end{equation*}
\end{theorem}

\begin{proof}
\textit{Existence}: Our first aim is to employ EROs and ECOs to show that $A$
is equivalent to%
\begin{equation*}
\left( 
\begin{array}{cc}
d_{1} & 0_{1(n-1)} \\ 
0_{(m-1)1} & M%
\end{array}%
\right)
\end{equation*}%
where $d_{1}$ divides every entry of $M$.

\textit{Step One}: By permuting rows and columns we can move an entry $d$ of
least norm to the first-row-first-column entry. We now aim to clear out the
first row: note every entry of the first row can be written in the form $%
a_{1j}=q_{j}d+r_{j}$ and if a remainder $r_{j}$ of smaller norm is produced
we permute it to the top left entry and begin again. This process must
terminate as the norm we are continually decreasing takes a positive
integral value.

\textit{Step Two}: We then proceed similarly seeking to clear out the first
column. Should we produce an entry of smaller norm then we move that row to
the first row and return to Step One.

\textit{Step Three}: Eventually the first row and first column have been
cleared and it follows that the matrix $A$ can be put in the above form but
it may not be the case that $d$ divides every entry of $M.$ If all of the
entries of $M$ are not divisible by $d$ then we are done. If not then by
EROs and ECOs we can again produce an element of smaller norm than $d$ and
we return to Step One.

The process must ultimately terminate as at each state we are producing
entries of strictly smaller norm. Thus we have demonstrated that $A$ is
equivalent to a matrix of the above form and by repeating this process on $M$
and so on we have shown the existence of the Smith normal form.

\textit{Uniqueness}: To prove uniqueness, we introduce the notion of \textit{%
determinantal divisors}. The $i$th determinantal divisor $D_{i}(A)$ is the
highest common factor of the determinants of all $i\times i$ submatrices of $%
A$. Note that $D_{i}(A)$ is only defined up to multiplication by units and
is invariant under EROs and ECOs. When the matrix is in Smith normal form,
as above, then we see that%
\begin{equation*}
D_{i}(A)=d_{1}d_{2}\cdots d_{i}\qquad \text{for }i\leqslant r\qquad \text{and%
}\qquad D_{i}(A)\qquad \text{for }i>r.
\end{equation*}%
Hence $d_{k}=D_{k}(A)/D_{k-1}(A)$ is invariant under EROs and ECOs and so
the invariant factors are unique.
\end{proof}

\begin{corollary}
(\textbf{Submodules of Free Modules}) Let $M$ be a submodule of $R^{n}.$
Then there exist elements $d_{1},d_{2},\ldots ,d_{r}$, where $r$ is the rank
of $M,$ and a basis $f_{1},\ldots ,f_{n}$ for $R^{n}$ such that $%
d_{1}f_{1},d_{2}f_{2},\ldots ,d_{r}f_{r}$ is a basis for $M$ and $%
d_{1}|d_{2}|d_{3}|\cdots |d_{r}.$
\end{corollary}

To see why this corollary follows we will apply the method to the following
matrix.

\begin{example}
Put the following matrix in Smith normal form%
\begin{equation*}
\left( 
\begin{array}{cccc}
12 & 6 & 4 & 8 \\ 
3 & 9 & 6 & 12 \\ 
2 & 16 & 14 & 28 \\ 
20 & 10 & 10 & 20%
\end{array}%
\right) .
\end{equation*}
\end{example}

\begin{solution}
We will do more than simply put the matrix into Smith normal form and keep
track in particular of the ECOs that we are using. We find%
\begin{eqnarray*}
\left( 
\begin{array}{cccc}
e_{1} & e_{2} & e_{3} & e_{4} \\ 
12 & 6 & 4 & 8 \\ 
3 & 9 & 6 & 12 \\ 
2 & 16 & 14 & 28 \\ 
20 & 10 & 10 & 20%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
e_{1} & e_{2} & e_{3}+2e_{4} & e_{4} \\ 
12 & 6 & 4 & 0 \\ 
3 & 9 & 6 & 0 \\ 
2 & 16 & 14 & 0 \\ 
20 & 10 & 10 & 0%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
e_{1} & e_{2} & e_{3}+2e_{4} & e_{4} \\ 
12 & 6 & 4 & 0 \\ 
1 & -7 & -8 & 0 \\ 
2 & 16 & 14 & 0 \\ 
20 & 10 & 10 & 0%
\end{array}%
\right) && \\
\sim \left( 
\begin{array}{cccc}
e_{1} & e_{2} & e_{3}+2e_{4} & e_{4} \\ 
0 & 90 & 100 & 0 \\ 
1 & -7 & -8 & 0 \\ 
0 & 30 & 30 & 0 \\ 
0 & 150 & 170 & 0%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
e_{1} & e_{2} & e_{3}+2e_{4} & e_{4} \\ 
0 & 0 & 10 & 0 \\ 
1 & -7 & -8 & 0 \\ 
0 & 30 & 30 & 0 \\ 
0 & 0 & 20 & 0%
\end{array}%
\right) && \\
\sim \left( 
\begin{array}{cccc}
e_{1} & e_{2} & e_{3}+2e_{4} & e_{4} \\ 
0 & 0 & 10 & 0 \\ 
1 & -7 & -8 & 0 \\ 
0 & 30 & 0 & 0 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
e_{1}-7e_{2}-8e_{3}-16e_{4} & e_{3}+2e_{4} & e_{2} & e_{4} \\ 
1 & 0 & 0 & 0 \\ 
0 & 10 & 0 & 0 \\ 
0 & 0 & 30 & 0 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right) &&
\end{eqnarray*}%
Hence the Smith normal form is%
\begin{equation*}
\left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & 10 & 0 & 0 \\ 
0 & 0 & 30 & 0 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right) .
\end{equation*}
\end{solution}

But we are also able to answer the following.

\begin{example}
Find a basis $f_{1},f_{2},f_{3},f_{4}$ for $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{4}$ such that $d_{1}f_{1},\ldots ,d_{r}f_{r}$ is a basis for%
\begin{equation*}
M=\langle (12,6,4,8),\,(3,9,6,12),\,(2,16,14,28),(20,10,10,20)\rangle
\end{equation*}%
where $r=\mathrm{rank}(M)$ and $d_{1}|d_{2}|\cdots |d_{r}.$
\end{example}

\begin{solution}
In our previous calculation we saw that%
\begin{equation*}
\left( 
\begin{array}{cccc}
e_{1} & e_{2} & e_{3} & e_{4} \\ 
12 & 6 & 4 & 8 \\ 
3 & 9 & 6 & 12 \\ 
2 & 16 & 14 & 28 \\ 
20 & 10 & 10 & 20%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
e_{1}-7e_{2}-8e_{3}-16e_{4} & e_{3}+2e_{4} & e_{2} & e_{4} \\ 
1 & 0 & 0 & 0 \\ 
0 & 10 & 0 & 0 \\ 
0 & 0 & 30 & 0 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right) .
\end{equation*}%
The column headings are still a basis for $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{4}$ as we began with a basis $e_{1},e_{2},e_{3},e_{4}$ and each ECO is
invertible. The rows (with co-ordinates understood wrt this new basis) still
span $M$ and are now clearly independent. So we have shown that%
\begin{equation*}
f_{1}=e_{1}-7e_{2}-8e_{3}-16e_{4},\qquad f_{2}=e_{3}+2e_{4},\qquad
f_{3}=e_{2},\qquad f_{4}=e_{4}
\end{equation*}%
is a basis for $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{4}$ and $f_{1},10f_{2},30f_{3}$ is a basis for $M.$
\end{solution}

Before we move on to the Structure Theorem we all need to introduce the
ideas of generators, relations and presentations. We begin with some
motivational examples.

\begin{example}
Consider the following modules.%
\begin{equation*}
M_{1}=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{5},\qquad M_{2}=%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x^{2}+1\rangle }.
\end{equation*}%
The $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-module $M_{1}$ is generated by%
\begin{equation*}
a=(1,0,0),\qquad b=(0,1,0),\qquad c=(0,0,1).
\end{equation*}%
The module is not free and we see for example the "relations" $2a=4b=5c=0.$
We could have chosen other generators for $M_{1}$ such as $a$ and $b+c$
which satisfy the relations $2a=20(b+c)=0$. These facts are represented in
the isomorphisms%
\begin{equation*}
M_{1}\cong \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{3}}{\langle (2,0,0),(0,4,0),(0,0,5)\rangle }\cong \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{2}}{\langle (2,0),(0,20)\rangle }.
\end{equation*}%
Note that every relation involving $a$ and $b+c$ can be deduced from $%
2a=20(b+c)=0$ with these relations generating all relations.

For $M_{2}$ we note again that $a,b,c$ are generators with $%
(x-1)b=(x^{2}+1)c=0$ or again might have been generated by $a$ and $b+c$.
Again we have%
\begin{equation*}
M_{2}\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]^{3}}{\langle (0,x-1,0),(0,0,x^{2}+1)\rangle }\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]^{2}}{\langle (0,0),(0,(x-1)(x^{2}+1))\rangle }.
\end{equation*}%
For each module, and for each different set of generators, the relations
these generators satisfy can be captured in a "presentation" matrix. In the
case of $M_{1}$ these would be the matrices%
\begin{equation*}
\left( 
\begin{array}{ccc}
2 & 0 & 0 \\ 
0 & 4 & 0 \\ 
0 & 0 & 5%
\end{array}%
\right) \qquad \text{and}\qquad \left( 
\begin{array}{cc}
2 & 0 \\ 
0 & 20%
\end{array}%
\right)
\end{equation*}%
where the columns in the first matrix relate to $a,b,c$ and in the second to 
$a,b+c.$ For $M_{2}$ the two presentation matrices would be%
\begin{equation*}
\left( 
\begin{array}{ccc}
0 & x-1 & 0 \\ 
0 & 0 & x^{2}+1%
\end{array}%
\right) \qquad \text{and}\qquad \left( 
\begin{array}{cc}
0 & 0 \\ 
0 & (x-1)(x^{2}+1)%
\end{array}%
\right) .
\end{equation*}
\end{example}

\begin{definition}
(a) As we have defined already the elements $x_{1},\ldots ,x_{n}$ in the $R$%
-module $M$ are \textbf{generators} if every element $x$ of $M$ can be
written in the form%
\begin{equation*}
x=r_{1}x_{1}+r_{2}x_{2}+\cdots +r_{n}x_{n}.
\end{equation*}%
(b) A \textbf{relation} in the generators $x_{1},\ldots ,x_{n}$ is any
(trivial or non-trivial) combination that adds to $0$, that is%
\begin{equation*}
r_{1}x_{1}+r_{2}x_{2}+\cdots +r_{n}x_{n}=0.
\end{equation*}
\end{definition}

\begin{proposition}
Given a finitely generated $R$-module $M$ with generators $x_{1},\ldots
,x_{n}$ there is an onto module homomorphism $\phi \colon R^{n}\rightarrow M$
and the relations in $x_{1},\ldots ,x_{n}$ correspond to $\ker \phi $.
\end{proposition}

\begin{proof}
There is a module homomorphism%
\begin{equation*}
\phi \colon R^{n}\rightarrow M\qquad \text{given by\quad }e_{i}=(0,\ldots
,0,1,0,\ldots ,0)\mapsto x_{i}.
\end{equation*}%
As the $x_{i}$ generate $M$ then $\phi $ is onto and we have, by the first
isomorphism theorem, that%
\begin{equation*}
\frac{R^{n}}{\ker \phi }\cong \func{Im}\phi =M.
\end{equation*}%
Further%
\begin{equation*}
r_{1}x_{1}+r_{2}x_{2}+\cdots +r_{n}x_{n}=0
\end{equation*}%
is a relation if and only if $(r_{1},r_{2},\ldots ,r_{n})\in \ker \phi .$
\end{proof}

\begin{corollary}
A finitely generated module is isomorphic to a quotient of a free module.
\end{corollary}

\begin{proof}
This is immediate from the above by the First Isomorphism Theorem as $M\cong
R^{n}/\ker \phi .$
\end{proof}

\begin{corollary}
A submodule of a finitely generated module is finitely generated.
\end{corollary}

\begin{proof}
If $M$ is finitely generated then $M\cong R^{n}/K$ for some submodule $K$ of 
$R^{n}.$ There is a correspondence between the submodules of $R^{n}/K$ and
the submodules of $R^{n}$ which contain $K$ (as with rings). So a submodule $%
N$ of $M$ is of the form $P/K$ where $P$ is a submodule of $R^{n}$. As
submodules of free modules are free, and so finitely generated, then $N=P/K$
is finitely generated.
\end{proof}

\begin{definition}
Let $M$ be an $R$-module generated by $x_{1},\ldots ,x_{n}$ and $\phi \colon
R^{n}\rightarrow M$ be as above. Then the \textbf{relation module} $\ker
\phi $ is finitely generated, say by the relations%
\begin{equation*}
a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n}=0,\qquad \cdots \qquad
a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n}=0.
\end{equation*}%
Then the \textbf{presentation matrix} for $M$ with respect to these
generators and relations is $A=(a_{ij}).$ As $\ker \phi $ is free the
relations can be chosen to be a basis (but do not need to be).
\end{definition}

\begin{example}
Say that $A$ is the abelian group generated by $a,b,c$ subject to the
relations%
\begin{equation*}
6a+18b+12c=0,\qquad 12a-9b+15c=0,\qquad 9a-12b-24c=0.
\end{equation*}%
Show that $A$ is isomorphic to $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{414}$ and find, in terms of $a,b,c$ all elements of order $3.$
\end{example}

\begin{solution}
This module is%
\begin{equation*}
A=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{3}}{\langle (6,18,12),(12,-9,15),(9,-12,-24)\rangle }
\end{equation*}%
though this is not a particularly informative or transparent presentation.
Rather we put the presentation matrix into Smith normal form as follows:%
\begin{eqnarray*}
\left( 
\begin{array}{ccc}
a & b & c \\ 
6 & 18 & 12 \\ 
12 & -9 & 15 \\ 
9 & -12 & 24%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
a & b & c \\ 
6 & 18 & 12 \\ 
0 & -45 & -9 \\ 
3 & -30 & 12%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
a & b & c \\ 
3 & -30 & 12 \\ 
0 & 45 & 9 \\ 
0 & 78 & -12%
\end{array}%
\right) \sim && \\
\left( 
\begin{array}{ccc}
a-10b+4c & b & c \\ 
3 & 0 & 0 \\ 
0 & 45 & 9 \\ 
0 & 78 & -12%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
a-10b+4c & b & 5b+c \\ 
3 & 0 & 0 \\ 
0 & 0 & 9 \\ 
0 & 138 & -12%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
a-10b+4c & 5b+c & b \\ 
3 & 0 & 0 \\ 
0 & -3 & 138 \\ 
0 & 9 & 0%
\end{array}%
\right) && \\
\sim \left( 
\begin{array}{ccc}
a-10b+4c & 5b+c & b \\ 
3 & 0 & 0 \\ 
0 & 3 & -138 \\ 
0 & 0 & 414%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
a-10b+4c & c-41b & b \\ 
3 & 0 & 0 \\ 
0 & 3 & 0 \\ 
0 & 0 & 414%
\end{array}%
\right) &&
\end{eqnarray*}%
Hence if we set $\alpha =a-10b+4c,\beta =c-41b,\gamma =b$ then we see%
\begin{equation*}
A=\langle \alpha ,\beta ,\gamma :3\alpha =3\beta =414\gamma =0\rangle \cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{414}.
\end{equation*}%
We can see then that there are $26$ elements of order three and that these
are of the form%
\begin{equation*}
\varepsilon _{1}\alpha +\varepsilon _{2}\beta +138\varepsilon _{3}\gamma
\qquad \text{where }\varepsilon _{1},\varepsilon _{2},\varepsilon _{3}\in
\{0,1,2\}\text{ without all being zero.}
\end{equation*}
\end{solution}

\begin{example}
Consider the $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$-module $M$ defined on $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{3}$ by%
\begin{equation*}
A=\left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 1 \\ 
0 & 0 & 1%
\end{array}%
\right) .
\end{equation*}%
We have already noted that this module is not cyclic; but we can see that $%
e_{1},e_{3}$ are generators as $x.e_{3}=Ae_{3}=e_{2}+e_{3}.$

Now note that $x.e_{1}=e_{1}$ so that $(x-1)e_{1}=0$ and we also have%
\begin{equation*}
x^{2}.e_{3}=x.(e_{2}+e_{3})=2e_{2}+e_{3}=2x.e_{3}-e_{3}
\end{equation*}%
so that $(x-1)^{2}.e_{3}=0.$ So%
\begin{equation*}
(x-1).e_{1}=0,\qquad (x-1)^{2}.e_{3}=0
\end{equation*}%
are relations. In fact these relations are sufficient to generate all
relations. To appreciate this note that $\dim _{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
}M=3$ and that the map%
\begin{equation*}
\phi \colon M\rightarrow \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]^{2}}{\langle ((x-1),0),(0,(x-1)^{2})\rangle }\cong \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]}{\langle (x-1)^{2}\rangle }
\end{equation*}%
given by $\phi (e_{1})=(1,0),$ $\phi (e_{2})=(0,x-1),\,\phi (e_{3})=(0,1)$
is an $%
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
\lbrack x]$-homomorphism and bijective as the LHS and RHS\ are $3$%
-dimensional real vector spaces.
\end{example}

Given a vector space $V$ over a field $F$ with basis $e_{1},\ldots ,e_{n}$
and defined as a $F[x]$-module via a matrix $A$, then $V$ is generated by $%
e_{1},e_{2},\ldots ,e_{n}$ and we have the relations $xe_{i}-Ae_{i}=0$ for
each $i$. In fact these relations generate the relation module as the entire
module structure is a consequence of linearity and inductive use of these
relations. Thus the presentation matrix for $V$ with respect to generators $%
e_{1},e_{2},\ldots ,e_{n}$ and the relations $xe_{i}-Ae_{i}=0$ is the matrix%
\begin{equation*}
xI_{n}-A.
\end{equation*}

\begin{example}
Let $V=%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{3}$ and%
\begin{equation*}
T=\left( 
\begin{array}{ccc}
1 & -1 & 1 \\ 
0 & 0 & 1 \\ 
0 & 1 & 0%
\end{array}%
\right) .
\end{equation*}%
Put the presentation matrix $xI-T$ into Smith normal form and find the
invariant factors. Find generators for $V.$ What is the minimal polynomial
of $T$?
\end{example}

\begin{solution}
We see%
\begin{eqnarray*}
\left( 
\begin{array}{ccc}
e_{1} & e_{2} & e_{3} \\ 
x-1 & 1 & -1 \\ 
0 & x & -1 \\ 
0 & -1 & x%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
e_{2} & e_{1} & e_{3} \\ 
1 & x-1 & -1 \\ 
x & 0 & -1 \\ 
-1 & 0 & x%
\end{array}%
\right) && \\
\sim \left( 
\begin{array}{ccc}
e_{2} & e_{1} & e_{3} \\ 
1 & x-1 & -1 \\ 
0 & x-x^{2} & x-1 \\ 
0 & x-1 & x-1%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
e_{2}+(x-1)e_{1}-e_{3} & e_{1} & e_{3} \\ 
1 & 0 & 0 \\ 
0 & x-1 & x-1 \\ 
0 & x-x^{2} & x-1%
\end{array}%
\right) && \\
\sim \left( 
\begin{array}{ccc}
e_{2}+(x-1)e_{1}-e_{3} & e_{1} & e_{3} \\ 
1 & 0 & 0 \\ 
0 & x-1 & x-1 \\ 
0 & 0 & x^{2}-1%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
e_{2}+(x-1)e_{1}-e_{3} & e_{1}+e_{3} & e_{3} \\ 
1 & 0 & 0 \\ 
0 & x-1 & 0 \\ 
0 & 0 & x^{2}-1%
\end{array}%
\right) &&
\end{eqnarray*}%
Hence the invariant factors are $x-1$ and $x^{2}-1.$ This also means that
the module defined by $T$ is isomorphic to%
\begin{equation*}
V=\frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x-1\rangle }\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x^{2}-1\rangle }
\end{equation*}%
with generators being $e_{1}+e_{3}$ and $e_{3}$ and relations being%
\begin{equation*}
(x-1).(e_{1}+e_{3})=0,\qquad (x^{2}-1).e_{3}=0.
\end{equation*}%
It also follows that the minimal polynomial of $T$ is $x^{2}-1.$
\end{solution}

\chapter{Structure Theorem. Applications.}

\begin{definition}
Let $M$ be a finitely-generated $R$-module. An element $r\in M$ is said to
be a\textbf{\ torsion element}\emph{, }if $r.m=0$ for some $r\neq 0.$ $M$ is
said to have \textbf{torsion} if it has any non-zero torsion elements. $M$
is said to be torsion-free if $0$ is the only torsion element.
\end{definition}

\begin{proposition}
(a) Let $M$ be a finitely-generated $R$-module. The torsion elements form a
submodule $T$ of $M.$

(b) A finitely generated torsion-free $R$-module is free.

(c) The module $F=M/T$ is a free $R$-module and $M\cong F\oplus T.$
\end{proposition}

\begin{proof}
(a) Note that $0\in T.$ If $m_{1},m_{2}\in T$ then there exist non-zero $%
r_{1},r_{2}$ such that $r_{1}.m_{1}=r_{2}.m_{2}=0.$ Then $r_{1}r_{2}=0$ and
we have%
\begin{equation*}
r_{1}r_{2}.(m_{1}+m_{2})=0.
\end{equation*}%
Further for $r\in R$ we have that $rm_{1}\in T$ as%
\begin{equation*}
r_{1}.(rm)=r_{1}r.m=rr_{1}.m=r.r_{1}m=0.
\end{equation*}%
Hence $T$ is a submodule.

(b) Let $N$ be a finitely generated torsion-free $R$-module. Say that $%
x_{1},\ldots ,x_{n}$ generate $N$ and that (by reordering if necessary) $%
x_{1},\ldots ,x_{m}$ is a maximal linearly independent subset of the $x_{i}.$
We set $F$ to be the span of $x_{1},\ldots ,x_{m}$, noting that $F$ is free.
Note that there exist $r_{i},r_{ij}\in R$ not all zero and such that%
\begin{equation*}
r_{i}x_{i}+\sum_{j=1}^{m}r_{ij}x_{j}=0.
\end{equation*}%
This is clear for $i\leqslant m$ and follows by the maximality of $m$ for $%
i>m.$ Further as $x_{1},\ldots ,x_{m}$ are independent then $r_{i}\neq 0$
for all $i.$ We set%
\begin{equation*}
r=r_{1}r_{2}\cdots r_{n}\neq 0.
\end{equation*}%
Note that $r_{i}x_{i}$ is in $F$ for all $i$ and hence $rx_{i}\in F$ for all 
$i.$ But then $rN\subseteq F.$ Note that the map $n\mapsto r.n$ is an
injective module homomomorphism as $N$ is torsion-free. So $N$ is isomorphic
to its image which is a submodule of the free module $F$ and so free itself.

(c) As $M$ is finitely generated then $F=M/T$ is finitely generated and is
torsion free, and so free, by construction. Let $x_{1},\ldots ,x_{n}$ be a
basis for $F.$ Then any element of $M/T$ can be uniquely written as%
\begin{equation*}
r_{1}x_{1}+\cdots +r_{n}x_{n}+T.
\end{equation*}%
Then the map $\phi \colon M/T\oplus T\rightarrow M$ given by%
\begin{equation*}
\phi (r_{1}x_{1}+\cdots +r_{n}x_{n}+T,t)=r_{1}x_{1}+\cdots +r_{n}x_{n}+t
\end{equation*}%
is a module isomorphism.
\end{proof}

\begin{example}
A linear map $T\colon V\rightarrow V$ on a finite dimensional vector space $%
V $ over $F$ defines an $F[x]$-module. In this module every element is a
torsion element as $m_{T}(x).v=0$ for all $v$ yet $m_{T}(x)\neq 0\in F[x].$
\end{example}

\begin{theorem}
(\textbf{Structure Theorem for Finitely Generated Modules}) Let $M$ be a
finitely-generated $R$-module. Then there exists a non-negative integer $r$,
called the (torsion-free) \textbf{rank} of $M$ and non-zero, non-unit
elements $d_{i}\in R,$ known as the \textbf{invariant factors} such that%
\begin{equation*}
d_{1}|d_{2}|d_{3}|\cdots |d_{k}
\end{equation*}%
and such that%
\begin{equation*}
M\cong R^{r}\oplus \frac{R}{\langle d_{1}\rangle }\oplus \frac{R}{\langle
d_{2}\rangle }\oplus \cdots \oplus \frac{R}{\langle d_{k}\rangle }.
\end{equation*}%
The rank $r$ is unique and $d_{1},\ldots ,d_{k}$ unique up to multiplication
by units.
\end{theorem}

\begin{proof}
Say that $M$, which is finitely generated, is generated by $x_{1},\ldots
,x_{n}.$ There is then a module homomorphism%
\begin{equation*}
\phi \colon R^{n}\rightarrow M\qquad \text{given by\quad }e_{i}=(0,\ldots
,0,1,0,\ldots ,0)\mapsto x_{i}.
\end{equation*}%
As the $x_{i}$ generate $M$ then $\phi $ is onto and we have, by the first
isomorphism theorem, that%
\begin{equation*}
\frac{R^{n}}{\ker \phi }\cong \func{Im}\phi =M.
\end{equation*}%
Now by the corollary to the Smith normal form, we know that there is a basis 
$f_{1},\ldots ,f_{n}$ for $R^{n}$ and $d_{i}$ as above with%
\begin{equation*}
\ker \phi =0\oplus \cdots \oplus 0\oplus \langle d_{1}\rangle \oplus \cdots
\oplus \langle d_{k}\rangle
\end{equation*}%
and hence we have%
\begin{equation*}
M\cong \frac{R^{n}}{0\oplus \cdots \oplus 0\oplus \langle d_{1}\rangle
\oplus \cdots \oplus \langle d_{k}\rangle }\cong R^{n-k}\oplus \frac{R}{%
\langle d_{1}\rangle }\oplus \frac{R}{\langle d_{2}\rangle }\oplus \cdots
\oplus \frac{R}{\langle d_{k}\rangle }.
\end{equation*}%
Had any of the $d_{i}$ been units then we would have $R/\langle d_{i}\rangle
=R/R\cong 0$ and we can just omit such factors.

Now note that $d_{k}M=d_{k}R^{r}$ is a free module and so the rank of $M$ is
uniquely determined. However we will postpone for now proving the uniqueness
of invariant factors until we have introduced the notion of elementary
divisors.
\end{proof}

\begin{example}
Present each of the following modules as described in the structure theorem.

(a) $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{12}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{16}\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{16}\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{12}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{48}.$

(b)%
\begin{equation*}
\frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle (x^{2}-4)(x^{3}-8)\rangle }.
\end{equation*}%
This is already cyclic (it is generated by $1$) and so is in the required
form.

(c)%
\begin{equation*}
\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 2\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 4\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 5\rangle }\cong \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 2\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 20\rangle }.
\end{equation*}%
Note that all of the above three modules all have zero rank.
\end{example}

There is an alternative form of the structure theorem. The above
decomposition might be viewed as a minimal decomposition into cyclic
submodules. Alternatively we can decompose some of the summands yet further
by using the Chinese remainder theorem applied to the coprime factors of the
invariant factors. Let $p_{1},\ldots ,p_{n}$ be the prime factors of at
least one of the $d_{i}.$ Then we may write%
\begin{equation*}
d_{i}=p_{1}^{\alpha _{1i}}p_{2}^{\alpha _{2i}}\cdots p_{n}^{\alpha _{ni}}
\end{equation*}%
where $0\leqslant \alpha _{ki}\leqslant \alpha _{(k+1)i}.$ Then we have%
\begin{equation*}
\frac{R}{\langle d_{i}\rangle }\cong \frac{R}{\langle p_{1}^{\alpha
_{1i}}\rangle }\oplus \frac{R}{\langle p_{2}^{\alpha _{2i}}\rangle }v\cdots
\oplus \frac{R}{\langle p_{n}^{\alpha _{ni}}\rangle }
\end{equation*}%
so that%
\begin{equation*}
M\cong \left( \dbigoplus\limits_{i=1}^{k}\frac{R}{\langle p_{1}^{\alpha
_{1i}}\rangle }\right) \oplus \left( \dbigoplus\limits_{i=1}^{k}\frac{R}{%
\langle p_{2}^{\alpha _{2i}}\rangle }\right) \oplus \cdots \oplus \left(
\dbigoplus\limits_{i=1}^{k}\frac{R}{\langle p_{n}^{\alpha _{ni}}\rangle }%
\right) .
\end{equation*}%
The elements $p_{j}^{\alpha _{ji}},$ where $\alpha _{ji}>0$, are known as
the \textbf{elementary divisors}.

\begin{example}
Applying this alternative decomposition to the previous three examples we
would write%
\begin{eqnarray*}
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{12}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{16} &\cong &%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{16}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}. \\
\frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle (x^{2}-4)(x^{3}-8)\rangle } &\cong &\frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle (x-2)^{2}\rangle }\oplus \frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle x+2)\rangle }\oplus \frac{%
%TCIMACRO{\U{211a} }%
%BeginExpansion
\mathbb{Q}
%EndExpansion
\lbrack x]}{\langle x^{2}+2x+4\rangle }. \\
\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 2\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 4\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 5\rangle } &\cong &\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 1+i\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle (1+i)^{2}\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 1-i\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle (1-i)^{2}\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 1+2i\rangle }\oplus \frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
\lbrack i]}{\langle 1-2i\rangle }.
\end{eqnarray*}
\end{example}

\begin{proposition}
The elementary divisors and invariant factors are unique (up to
multiplication by units).
\end{proposition}

\begin{proof}
We can separately consider the different irreducibles $p$ for which there is
some non-zero element $x$ of $M$ satisfying $p^{n}x=0$ for a power of $p,$
because different (non-associate) irreducibles will be coprime. So say that $%
M$ is a module with $p^{n}M=0$ for some positive integer $n$, and choose $n$
to be the least such $n.$ Then we have that $p^{n-1}M$ is a non-zero vector
space over the field $R/\langle p\rangle $ for if $\overline{r_{1}}=%
\overline{r_{2}}$ in $R/\langle p\rangle $ then $\ r_{1}-r_{2}=rp$ for some $%
r$ and we have%
\begin{equation*}
r_{1}.p^{n-1}m=(r_{2}+rp).p^{n-1}m=r_{2}.p^{n-1}m.
\end{equation*}%
The dimension of $p^{n-1}M$ as an $R/\langle p\rangle $-vector space is the
number of copies of $R/\langle p^{n-1}\rangle $ in the decomposition and so
recoverable from $M.$ In a similar fashion $p^{n-2}M/p^{n-1}M$ is an $%
R/\langle p\rangle $-vector space and its dimension is the total number of
copies of $R/\langle p^{n-2}\rangle $ and $R/\langle p^{n-1}\rangle $ in the
decomposition, and subtracting our previously found dimension we now know
the number of summands of $R/\langle p^{n-2}\rangle $ in the decomposition.
Continuing in this fashion we are able to determine the number of each
different summand. The invariant factors are then recoverable from the
elementary divisors in a straightforward, but notationally painful manner.
Begin by putting the highest power of each irreducible among the elementary
divisors into $d_{k}$ and keep repeating this process to produce all the
invariant factors.
\end{proof}

\begin{example}
To help understand the previous proposition, here is the argument made for a
specific $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-module 
\begin{equation*}
M=%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}.
\end{equation*}%
Note that $8M=0$ and that%
\begin{equation*}
4M=0\oplus 0\oplus 0\oplus \langle 4\rangle \oplus \langle 4\rangle \cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}^{2}.
\end{equation*}%
This dimension, $2$ is the number of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}$ summands. Now%
\begin{equation*}
\frac{2M}{4M}\cong \frac{0\oplus 0\oplus \langle 2\rangle \oplus \langle
2\rangle \oplus \langle 2\rangle }{0\oplus 0\oplus 0\oplus \langle 4\rangle
\oplus \langle 4\rangle }\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}^{3}.
\end{equation*}%
This dimension, $3$, is the number of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}$ and $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}$ summands in total. Finally%
\begin{equation*}
\frac{M}{2M}=\frac{%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus \langle 1\rangle \oplus \langle 1\rangle \oplus \langle 1\rangle 
}{0\oplus 0\oplus \langle 2\rangle \oplus \langle 2\rangle \oplus \langle
2\rangle }\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}^{5}
\end{equation*}%
and this dimension $5$ is the total number of $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2},%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4},%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{8}$ summands. So we see that the number of each elementary divisor is
recoverable from these dimensions.
\end{example}

\begin{corollary}
(\textbf{Classification Theorem for Finitely Generated Abelian Groups}) Let $%
A$ be a finitely generated abelian group. Then there exist unique
non-negative $r$ and integers $d_{i}\geqslant 2$ with $d_{1}|d_{2}|\cdots
|d_{k}$ such that%
\begin{equation*}
A\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
^{r}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{d_{1}}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{d_{2}}\oplus \cdots \oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{d_{k}}.
\end{equation*}
\end{corollary}

\begin{proof}
This is simply a statement of the Structure Theorem for $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
$-modules.
\end{proof}

\begin{example}
Find, up to isomorphism, all abelian groups of order $360.$ What are their
elementary divisors?

If not explicitly on your list, explain which of your groups is isomorphic
to $%
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{90}$?
\end{example}

\begin{solution}
Note that $360=2^{3}3^{2}5.$ Hence $k\leqslant 3$ and we must have $2\times
3\times 5=30|d_{k}.$ We see that the only abelian groups, up to isomorphism,
are%
\begin{equation*}
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{360},,\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{180},\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{3}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{120},\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{60},\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{90},\qquad 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{6}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{30}.
\end{equation*}%
The elementary divisors of these groups are respectively%
\begin{equation*}
8,9,5,\qquad 2,4,9,5,\qquad 8,3,3,5,\qquad 2,4,3,3,5,\qquad 2,2,2,9,5,\qquad
2,2,2,3,3,5.
\end{equation*}%
Now note that%
\begin{equation*}
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{90}\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{4}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{45}\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{2}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{180}
\end{equation*}%
or we might have noted its elementary divisors to be $4,2,5,9$ and so it is
the second listed group.
\end{solution}

\begin{example}
Identify the abelian group generated by four elements $a,b,c,d$ subject to
the relations%
\begin{equation*}
12a+6b+4c+8d=0,\quad 3a+9b+6c+12d=0,\quad 2a+16b+15c+28d=0,\quad
20a+10b+10c+20d=0.
\end{equation*}
\end{example}

\begin{solution}
We know that we can use EROs and ECOs to put the matrix into Smith normal
form, and from our previous calculation we have%
\begin{equation*}
\left( 
\begin{array}{cccc}
a & b & c & d \\ 
12 & 6 & 4 & 8 \\ 
3 & 9 & 6 & 12 \\ 
2 & 16 & 14 & 28 \\ 
20 & 10 & 10 & 20%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
a-7b-8c-16d & c+2d & b & d \\ 
1 & 0 & 0 & 0 \\ 
0 & 10 & 0 & 0 \\ 
0 & 0 & 30 & 0 \\ 
0 & 0 & 0 & 0%
\end{array}%
\right) .
\end{equation*}%
Hence if we set%
\begin{equation*}
\alpha =a-7b-8c-16d,\qquad \beta =c+2d,\qquad \gamma =b,\qquad \delta =d
\end{equation*}%
then the described abelian group is%
\begin{equation*}
\langle \alpha ,\beta ,\gamma ,\delta :\alpha =10\beta =30\gamma =0\rangle
\cong 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{10}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
_{30}\oplus 
%TCIMACRO{\U{2124} }%
%BeginExpansion
\mathbb{Z}
%EndExpansion
.
\end{equation*}%
There are two elements of order $3$ for example, namely $\left( 0,\pm
10,0\right) $ or in terms of the generators these are%
\begin{equation*}
\pm 10\beta =\pm (10c+20d).
\end{equation*}
\end{solution}

\begin{corollary}
(\textbf{Rational Canonical Form}) Let $A$ be an $n\times n$ matrix over a
field $F.$ Then there $A$ is similar to a matrix in the form%
\begin{equation*}
\mathrm{diag}(C(d_{1}),C(d_{2}),\ldots C(d_{k}))
\end{equation*}%
where $d_{i}\in F[x]$ are monic polynomials, $C(d_{i})$ denotes the
companion matrix of $d_{i}$ and%
\begin{equation*}
d_{1}|d_{2}|\cdots |d_{k}.
\end{equation*}%
The $d_{i}$ are unique up to multiplication by units. Note further that 
\begin{equation*}
m_{A}(x)=d_{k}(x)\qquad \text{and that}\qquad
c_{A}(x)=d_{1}(x)d_{2}(x)\cdots d_{k}(x).
\end{equation*}%
The above matrix representative of $A$ is known as its \textbf{rational
canonical form} or \textbf{Frobenius normal form}.
\end{corollary}

\begin{remark}
Equivalently, in terms of the $F[x]$-module structure defined on $F^{n}$ by $%
A$, the above says that%
\begin{equation*}
F^{n}\cong F[C(d_{1})]\oplus F[C(d_{2})]\oplus \cdots \oplus F[C(d_{k})],
\end{equation*}%
hence decomposing the $F[x]$-module defined by $A$, which in general will
not be cyclic, into cyclic $F[x]$-modules defined by the above companion
matrices.
\end{remark}

\begin{proof}
Consider the $F[x]$-module structure defined on $F^{n}$ by the linear map $T(%
\mathbf{v})=A\mathbf{v}.$ This module has rank zero and every element is a
torsion element. By the Structure Theorem we know that there exist invariant
factors $d_{i}\in F[x]$ such that%
\begin{equation*}
d_{1}|d_{2}|\cdots |d_{k}
\end{equation*}%
with%
\begin{equation*}
F^{n}\cong \frac{F[x]}{\langle d_{1}(x)\rangle }\oplus \frac{F[x]}{\langle
d_{2}(x)\rangle }\oplus \cdots \oplus \frac{F[x]}{\langle d_{k}(x)\rangle }.
\end{equation*}%
If $\deg d_{1}=n_{1}$ then $1,x,\ldots ,x^{n_{1}-1}$ is a basis for $%
F[x]/\langle d_{1}(x)\rangle $ as a vector space and with respect to this
basis multiplication by $x,$ or equivalently by $A$, on $F[x]/\langle
d_{1}(x)\rangle $ is given by the companion matrix $C(d_{1}).$ Taking such a
basis for each summand, the union of these bases is a basis for $F^{n}$ and
with respect to this basis $T$ has matrix representative%
\begin{equation*}
\mathrm{diag}(C(d_{1}),C(d_{2}),\ldots C(d_{k}))
\end{equation*}%
to which $A$ is similar. (In choosing the above basis we have found a change
of basis matrix $P$ such that $P^{-1}AP$ equals the above matrix
representative.) Conversely any such matrix representation leads to a
decomposition of the $F[x]$-module as above and we know that the invariant
factors are unique up to multiplication by units.

For elements in the final summand $F[x]/\langle d_{k}(x)\rangle $ we know
that the minimal polynomial is $d_{k}(x).$ As $d_{i}|d_{k}$ for all $i$ then 
$d_{k}(A)$ also annihilates the other summands and we see that $%
m_{A}(x)=d_{k}(x).$ We also have that the characteristic polynomial of $%
C(d_{i})$ is $d_{i}(x);$ as the characteristic polynomial of the above
matrix representative equals the product of the characteristic polynomials
of the blocks we have%
\begin{equation*}
c_{A}(x)=d_{1}(x)d_{2}(x)\cdots d_{k}(x).
\end{equation*}
\end{proof}

\begin{corollary}
(\textbf{Jordan Normal Form}) Let $A$ be a complex $n\times n$ matrix with
distinct eigenvalues $\lambda _{1},\lambda _{2},\ldots ,\lambda _{k}$. Then $%
A$ is similar to a matrix of the form%
\begin{equation*}
\mathrm{diag}(J(\lambda _{1},r_{11}),J(\lambda _{1},r_{12}),\ldots J(\lambda
_{1},r_{1n_{1}}),\ldots J(\lambda _{k},r_{k1}),J(\lambda _{1},r_{k2}),\ldots
J(\lambda _{k},r_{kn_{k}}))
\end{equation*}%
where $J(\lambda ,r)$ denotes the $r\times r$ Jordan block matrix%
\begin{equation*}
J(\lambda ,r)=\left( 
\begin{array}{ccccc}
\lambda  & 0 & 0 & \cdots  & 0 \\ 
1 & \lambda  & 0 & \ddots  & \vdots  \\ 
0 & 1 & \lambda  & \ddots  & 0 \\ 
\vdots  & \ddots  & \ddots  & \ddots  & 0 \\ 
0 & \cdots  & 0 & 1 & \lambda 
\end{array}%
\right) 
\end{equation*}%
and%
\begin{equation*}
r_{i1}\leqslant r_{i2}\leqslant \cdots \leqslant r_{in_{i}}\qquad \text{for
each }i.
\end{equation*}%
Note that%
\begin{equation*}
\mathrm{nullity}(A-\lambda _{i}I)^{j}-\mathrm{nullity}(A-\lambda
_{i}I)^{j-1}=\text{number of }J(\lambda _{i},r)\text{ blocks with }%
r\geqslant j.
\end{equation*}
\end{corollary}

\begin{remark}
Equivalently, in terms of the $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]$-module structure defined on $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{n}$ by $A$, the above says that%
\begin{equation*}
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{n}\cong 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack J(\lambda _{1},r_{11})]\oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack J(\lambda _{1},r_{12})]\oplus \cdots \oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack J(\lambda _{k},r_{kn_{k}})],
\end{equation*}%
hence decomposing the $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
[x]$-module defined by $A$, which in general will not be cyclic, into cyclic 
$%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
[x]$-modules defined by the above Jordan block matrices.
\end{remark}

\begin{proof}
Consider the $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]$-module structure defined on $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{n}$ by $A$. We have that%
\begin{equation*}
c_{A}(x)=d_{1}(x)d_{2}(x)\cdots d_{k}(x)
\end{equation*}%
and by the Fundamental Theorem of Algebra the elementary divisors are all of
the form $(x-\lambda )^{r}$ for some $r$ and $\lambda $ an eigenvalue. So
the alternative statement of the Structure Theorem gives us%
\begin{equation*}
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{n}\cong \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle (x-\lambda _{1})^{r_{11}}\rangle }\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle (x-\lambda _{1})^{r_{12}}\rangle }\oplus \cdots \oplus 
\frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle (x-\lambda _{1})^{r_{1n_{1}}}\rangle }\oplus \cdots
\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle (x-\lambda _{1})^{r_{k1}}\rangle }\oplus \cdots \oplus 
\frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle (x-\lambda _{1})^{r_{kn_{k}}}\rangle }.
\end{equation*}%
Note that is a basis for $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]/\langle (x-\lambda )^{r}\rangle $ as a vector space, and with
respect to this basis we see that multiplication by $x$ is represented by
the matrix $J(\lambda ,r).$ To see this note that%
\begin{equation*}
x(x-\lambda )^{s}=\lambda (x-\lambda )^{s}+(x-\lambda )^{s+1}\qquad \text{%
for }0\leqslant s<r-1
\end{equation*}%
and that $x(x-\lambda )^{r-1}=\lambda (x-\lambda )^{r-1}$ as $(x-\lambda
)^{r}=0.$ Hence with respect to the union of these bases multiplication by $A
$ is represented by the above matrix of Jordan blocks.

We know that the elementary divisors are unique (up to units) and so the
Jordan normal form is unique. However we can further appreciate that the
number of $J(\lambda ,r)$ blocks equals $\mathrm{nullity}(A-\lambda I)$, a
basis for $\ker (A-\lambda I)$ consisting of the last basis vector
associated with each such block. A basis for $\ker (A-\lambda I)^{2}$
consists of those vectors just described and the penultimate basis vectors
of those $J(\lambda ,r)$ blocks for which $r\geqslant 2$. etc.
\end{proof}

\begin{remark}
If we instead worked with the (ordered) basis $(x-\lambda )^{r-1},\ldots
,(x-\lambda )^{2},(x-\lambda ),1,$ then the Jordan blocks would have the form%
\begin{equation*}
\left( 
\begin{array}{ccccc}
\lambda  & 1 & 0 & \cdots  & 0 \\ 
0 & \lambda  & 1 & \ddots  & \vdots  \\ 
0 & 0 & \lambda  & \ddots  & 0 \\ 
\vdots  & \ddots  & \ddots  & \ddots  & 1 \\ 
0 & \cdots  & 0 & 0 & \lambda 
\end{array}%
\right) 
\end{equation*}%
which is a form just as commonly used.
\end{remark}

\begin{example}
Find the RCFs and JNFs of the following complex matrices.%
\begin{equation*}
X=\left( 
\begin{array}{ccc}
0 & 1 & 1 \\ 
0 & 0 & 1 \\ 
0 & 0 & 0%
\end{array}%
\right) \hspace{-0.04in},\qquad Y=\left( 
\begin{array}{ccc}
0 & 1 & 0 \\ 
0 & 0 & 1 \\ 
1 & 0 & 0%
\end{array}%
\right) \hspace{-0.04in},\qquad Z=\left( 
\begin{array}{ccc}
2 & 0 & 0 \\ 
0 & 2 & 0 \\ 
0 & 1 & 2%
\end{array}%
\right) \hspace{-0.04in}.
\end{equation*}
\end{example}

\begin{solution}
Note that $m_{X}(x)=c_{X}(x)=x^{3}$. Hence the rational canonical form is $%
C(x^{3}).$ The geometric multiplicity of $0$ is $1$ and so we have just one
Jordan block $J(0,3)$. So the RCF and JNF of $X$ are both%
\begin{equation*}
\left( 
\begin{array}{ccc}
0 & 0 & 0 \\ 
1 & 0 & 0 \\ 
0 & 1 & 0%
\end{array}%
\right) \hspace{-0.04in}.
\end{equation*}%
Note that $m_{Y}(x)=c_{Y}(x)=x^{3}-1.$ This has three distinct complex roots 
$1,\,\omega $ and $\omega ^{2}$ where $\omega =\mathrm{cis}(2\pi /3).$ So
the RCF equals $C(x^{3}-1)$ and the JNF equals $\mathrm{diag}%
(J(1,1),J(1,\omega ),J(1,\omega ^{2}))$. Explicitly these are%
\begin{equation*}
\left( 
\begin{array}{ccc}
0 & 0 & 1 \\ 
1 & 0 & 0 \\ 
0 & 1 & 0%
\end{array}%
\right) ,\qquad \left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & \omega  & 0 \\ 
0 & 0 & \omega ^{2}%
\end{array}%
\right) \hspace{-0.04in}.
\end{equation*}%
Finally note that $C$ is already in JNF equalling $\mathrm{diag}%
(J(2,1),J(2,2)).$ We have $m_{Z}(x)=(x-2)^{2}$ and $c_{Z}(x)=(x-2)^{3}$ so
that the invariant factors are $(x-2),(x-2)^{2}$ and the RCF equals $\mathrm{%
diag}(C(x-2),C((x-2)^{2})\dot{)}.$ Explicitly the RCF and JNF are%
\begin{equation*}
\left( 
\begin{array}{ccc}
2 & 0 & 0 \\ 
0 & 0 & -4 \\ 
0 & 1 & 4%
\end{array}%
\right) ,\qquad \left( 
\begin{array}{ccc}
2 & 0 & 0 \\ 
0 & 2 & 0 \\ 
0 & 1 & 2%
\end{array}%
\right) .
\end{equation*}
\end{solution}

\begin{example}
(a) Find the RCF\ and JNF of%
\begin{equation*}
T=\left( 
\begin{array}{cccc}
1 & 1 & 1 & 1 \\ 
1 & 1 & 1 & 1 \\ 
1 & 1 & 1 & 1 \\ 
1 & 1 & 1 & 1%
\end{array}%
\right) 
\end{equation*}%
by finding the minimal and characteristic polynomials.

(b) What are the decompositions of the module defined by $T$ that are
associated with the RCF and JNF?

(c) Rederive the invariant factors by finding the Smith normal form of $xI-T$
\end{example}

\begin{solution}
(a) We have $c_{T}(x)=x^{3}(x-4)$ and $m_{T}(x)=x(x-4).$ So the invariant
factors are%
\begin{equation*}
x,\qquad x,\qquad x(x-4).
\end{equation*}%
Hence the RCF and JNF are respectively 
\begin{equation*}
\mathrm{diag}(C(x),C(x),C(x(x-4))\qquad \text{and}\qquad \mathrm{diag}%
(J(0,1),J(0,1),J(0,1),J(4,1)).
\end{equation*}%
Explicitly these are%
\begin{equation*}
\left( 
\begin{array}{cccc}
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 \\ 
0 & 0 & 1 & 4%
\end{array}%
\right) ,\qquad \left( 
\begin{array}{cccc}
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 0 \\ 
0 & 0 & 0 & 4%
\end{array}%
\right) .
\end{equation*}%
(b) Two bases that the above matrix representatives for $T$ are with respect
to are%
\begin{eqnarray*}
&&\{(1,-1,0,0)^{T},(0,1,-1,0)^{T},(0,0,1,0)^{T},(1,1,1,1)^{T}\}; \\
&&\{(1,-1,0,0)^{T},(0,1,-1,0)^{T},(0,0,1,-1)^{T},(1,1,1,1)^{T}\}.
\end{eqnarray*}%
These correspond respectively to the following decompositions of $%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{4}$ into submodules as%
\begin{eqnarray*}
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{4} &=&\langle (1,-1,0,0)^{T}\rangle \oplus \langle (0,1,-1,0)^{T}\rangle
\oplus \langle (0,0,1,0)^{T},(1,1,1,1)^{T}\rangle  \\
&=&\langle (1,-1,0,0)^{T}\rangle \oplus \langle (0,1,-1,0)^{T}\rangle \oplus
\langle (0,0,1,0)^{T}\rangle \oplus \langle (1,1,1,1)^{T}\rangle .
\end{eqnarray*}%
or we might write these as%
\begin{equation*}
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{4}\cong \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x\rangle }\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x\rangle }\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x(x-4)\rangle }\cong \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x\rangle }\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x\rangle }\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x\rangle }\oplus \frac{%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\lbrack x]}{\langle x-4\rangle }
\end{equation*}%
or as further alternatives as%
\begin{equation*}
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{4}\cong 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\left[ 0_{1}\right] \oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\left[ 0_{1}\right] \oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\left[ \left( 
\begin{array}{cc}
0 & 0 \\ 
1 & 4%
\end{array}%
\right) \right] \cong 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\left[ 0_{1}\right] \oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\left[ 0_{1}\right] \oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\left[ 0_{1}\right] \oplus 
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
\left[ 4I_{1}\right] .
\end{equation*}%
(c) If we put $xI-T$ into Smith normal form then we find%
\begin{eqnarray*}
\left( 
\begin{array}{cccc}
x-1 & -1 & -1 & -1 \\ 
-1 & x-1 & -1 & -1 \\ 
-1 & -1 & x-1 & -1 \\ 
-1 & -1 & -1 & x-1%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
1 & 1 & 1 & 1-x \\ 
-1 & x-1 & -1 & -1 \\ 
-1 & -1 & x-1 & -1 \\ 
x-1 & -1 & -1 & -1%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
1 & 1 & 1 & 1-x \\ 
0 & x & 0 & -x \\ 
0 & 0 & x & -x \\ 
0 & -x & -x & x^{2}-2x%
\end{array}%
\right)  && \\
\sim \left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & x & 0 & -x \\ 
0 & 0 & x & -x \\ 
0 & -x & -x & x^{2}-2x%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & x & 0 & -x \\ 
0 & 0 & x & -x \\ 
0 & 0 & 0 & x^{2}-4x%
\end{array}%
\right) \sim \left( 
\begin{array}{cccc}
1 & 0 & 0 & 0 \\ 
0 & x & 0 & 0 \\ 
0 & 0 & x & 0 \\ 
0 & 0 & 0 & x^{2}-4x%
\end{array}%
\right) . &&
\end{eqnarray*}
\end{solution}

\begin{example}
Let $V=%
%TCIMACRO{\U{2102} }%
%BeginExpansion
\mathbb{C}
%EndExpansion
^{3}$ and%
\begin{equation*}
T=\left( 
\begin{array}{ccc}
1 & -1 & 1 \\ 
0 & 0 & 1 \\ 
0 & 1 & 0%
\end{array}%
\right) .
\end{equation*}%
Find the RCF and JNF of $T.$
\end{example}

\begin{solution}
We will determine this by putting $xI-T$ into Smith normal form. We see%
\begin{eqnarray*}
\left( 
\begin{array}{ccc}
x-1 & 1 & -1 \\ 
0 & x & -1 \\ 
0 & -1 & x%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
1 & x-1 & -1 \\ 
x & 0 & -1 \\ 
-1 & 0 & x%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
1 & x-1 & -1 \\ 
0 & x-x^{2} & x-1 \\ 
0 & x-1 & x-1%
\end{array}%
\right)  && \\
\sim \left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & x-1 & x-1 \\ 
0 & x-x^{2} & x-1%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & x-1 & x-1 \\ 
0 & 0 & x^{2}-1%
\end{array}%
\right) \sim \left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & x-1 & 0 \\ 
0 & 0 & x^{2}-1%
\end{array}%
\right) . &&
\end{eqnarray*}%
Hence the invariant factors of $T$ are $x-1$ and $x^{2}-1$ and the
elementary divisors are $x-1,x-1,x+1.$ The RCF and JNF are therefore
respectively%
\begin{equation*}
\left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 0 & 1 \\ 
0 & 1 & 0%
\end{array}%
\right) ,\qquad \left( 
\begin{array}{ccc}
1 & 0 & 0 \\ 
0 & 1 & 0 \\ 
0 & 0 & -1%
\end{array}%
\right) .
\end{equation*}
\end{solution}

\end{document}
