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\title{{\sc Algebra Notes}}
\author{Ralph Freese and William DeMeo}
%\date{14 December 2010}
\date{\today}
\begin{document}
\maketitle
\tableofcontents
\pagestyle{fancy}
\newpage
\noindent
{\bf Primary Textbook:} Jacobson, {\it Basic Algebra}~\cite{Jacobson:1985}.
\\[8pt]
{\bf Supplementary Textbooks:} Hungerford,
{\it Algebra}~\cite{Hungerford:1974};
Dummitt and Foote. {\it Abstract Algebra}~\cite{DummitFoote:2004};
\\[8pt]
{\bf Primary Subject:} Classical algebra systems: groups, rings, fields,
modules (including vector spaces). Also a little universal algebra and lattice
theory.
\\[8pt]
\underline{List of Notation}
\begin{itemize}
\item $A^A$, the set of maps from a set $A$ into itself.
\item $\Aut(\bA)$, the group of automorphisms of an algebra $\bA$.
\item $\End(\bA)$, the set of endomorphisms an algebra $\bA$.
\item $\Hom(\bA, \bB)$, the set of homomorphism from an algebra $\bA$ into an algebra $\bB$.
\item $\Con(\bA)$, the set of congruence relations of an algebra $\bA$.
\item $\bCon \bA$, the lattice of congruence relations of an algebra $\bA$.
\item $\Eq(A)$, the set of equivalence relations of a set $A$.
\item $\bEq \bA$, the lattice of equivalence relations of a set $A$.
\item $\Sub(\bA)$, the set of subalgebras of an algebra $\bA$.
\item $\bSub \bA$, the lattice of subalgebras of an algebra $\bA$.
\item $\Sg_{\bA}(X)$, the subuniverse generated by a set $X\subseteq A$.
\item $\N = \{1, 2, \dots\}$, the set of natural numbers.
\item $\Z = \{\dots, -1, 0, 1, \dots\}$, the ring of integers.
\item $\R = (-\infty, \infty)$, the real number field.
\item $\C$, the complex number field.
\item $\Q$, the rational number field.
\end{itemize}
\newpage
\part{Fall 2010: Universal Algebra \& Group Theory}
\newpage
\section{Universal Algebra}
\subsection{Basic concepts}
\index{algebra!definition|textit}
A (universal) \defn{algebra} is a pair
\begin{equation}
\label{eq:algebra}
\bA = \algebra{ A; F}
\end{equation}
where $A$ is a nonempty set and $F = \{f_i : i\in I\}$ is a set of finitary
operations on $A$; that is, $f_i : A^n \rightarrow A$ for some $n\in \N$.
A common shorthand notation for (\ref{eq:algebra}) is $\< A; f_i \>_{i\in I}$.
The number $n$ is called the \defn{arity} of the operation $f_i$.
\index{arity|textit}
Thus, the arity of an operation is the number of operands upon which it
acts, and we say that $f\in F$ is an \defn{$n$-ary} operation on $A$
if $f$ maps $A^n$ into $A$. An operation is called \emph{nullary}
(or constant) if its arity is zero. \emph{Unary}, \emph{binary}, and
\emph{ternary} operations have arities 1, 2, and 3, respectively.
\index{unary operation|textit} \index{binary operation|textit}
\index{ternary operation|textit}
\begin{example}
If $A=\R$ and $f: \R\times \R \rightarrow \R$ is the map
$f(a,b) = a+b$, then $\algebra{A; f}$ is an algebra with a
single binary operation.
Many more examples will be given below.
\end{example}
\index{algebra!unary|textit}
An algebra \bA\ is called \defn{unary} if all of its operations are unary. An
algebra \bA\ is \defn{finite} if $|A|$ is finite and \defn{trivial} if $|A| = 1$.
\index{algebra!trivial|textit}
Given two algebras $\bA$ and $\bB$, we say that $\bB$ is a
\index{reduct|textit}
\defn{reduct} of $\bA$ if both algebras have the same universe and $\bA$
(resp.~$\bB$) can be obtained from $\bB$ (resp.~$\bA$) by adding (resp.~removing) operations.
\\[6pt]
\index{operation symbol|textit}
\underline{A better approach:} An \defn{operation symbol} $f$ is an object that
has an associated arity, which we'll denote $\arity(f)$. A set of operation
symbols $F$ is called a \defn{similarity type}.
An algebra of similarity type $F$ is a pair $\bA = \$, where
$F^\bA = \{f^\bA : f\in F\}$
and $f^\bA$ is an operation on $A$ of arity $\arity(f)$.
\begin{example}
Consider the set of integers $\Z$ with operation
sysbols $F = \{+, \cdot, -, 0, 1\}$, which have respective
arities $\{2, 2, 1, 0, 0\}$. The operation
$+^\Z$ is the usual binary addition, while $-^\Z$ is negation: $a \mapsto -a$.
The constants $0^\Z$ and $1^\Z$ are nullary operations.
Of course we usually just write $+$ for $+^\Z$, etc.
\end{example}
\subsection{Subalgebras and Homomorphisms}
Suppose $\bA = \algebra{A; F^\bA}$ is an algebra. We call the nonempty
set $A$ the \defn{universe} of $\bA$. \index{algebra!universe of}
If a subset $B\subseteq A$ is \emph{closed} under all
operations in $F^\bA$, we call $B$ a \defn{subuniverse} of $A$.
\index{subuniverse!defined|textit} By closed under all operations we
mean the following: for each $f\in F^\bA$ (say $f$ is $n$-ary), we have
$f(b_0, \dots, b_{n-1}) \in B$, for all $b_0, \dots, b_{n-1}\in B$.
If $B\ne \emptyset$ is a subuniverse of $\algebra{A; F^\bA}$, and if we
let\footnote{Here $f\res B$ denotes restriction of the
function $f$ to the set $B$ (see Appendix Sec.~\ref{sec:functions}).}
$F^\bB = \{f\res B : f\in F^\bA\}$, then the algebra $\bB = \algebra{B;
F^\bB}$ is called a \defn{subalgebra} of $\bA$. \index{subalgebra|textit}
If $\bB$ is a subalgebra of $\bA$, we denote this fact by $\bB\leq \bA$.
Similarly, we write $B\leq A$ if $B$ is a subuniverse of $A$. We denote
the set of all subalgebras of $\bA$ by $\Sub(\bA)$.
Note that universe of an algebra is not allowed to be empty.
However, the empty set can be a subuniverse (if
$\bA$ has no nullary operations)..
\begin{theorem}
If $\bA_i \leq \bA$, $i\in I$, then $\bigcap A_i$ is a
subuniverse.
\end{theorem}
\index{subuniverse!generated by a set|textit}
If $S$ is a nonempty subset of $A$, the \defn{subuniverse generated by} $S$, denoted
$\Sg_\bA(S)$ or $\$ is the smallest subuniverse of $\bA$ containing the set $S$.
When $\ = A$, we say that $S$ generates $A$.
\begin{theorem}
If $S \subseteq A$,
then $\Sg_\bA(S) = \< S \> = \bigcap \{ B\leq A : S\subseteq B \}$.
\end{theorem}
Define $\{S_i\}$ recursively as follows:
\begin{align*}
S_0 &= S;\\
S_{i+1} &= \{f(a_1, \dots, a_k) : \text{$f$ is a $k$-ary basic operation of $A$ and $a_i\in S_i$}\}.
\end{align*}
\begin{theorem}
$\Sg_\bA(S) = \< S \> = \bigcup_{i=0}^\infty S_i$.
\end{theorem}
Let $\bA = \< A; F^\bA \>$ and $\bB = \$ be algebras of the same
type $F$, and let $F_n$ denote the set of $n$-ary operation symbols in $F$.
Consider a mapping $\varphi : A \rightarrow B$ and
operation symbol $f\in F_n$, and
suppose that for %any operation symbol $f\in F_n$ and for
all $a_0, \dots a_{n-1}\in A$ the following equation holds:
\[
\varphi(f^\bA(a_0, \dots, a_{n-1})) = f^\bB(\varphi(a_0), \dots, \varphi(a_{n-1})).
\]
Then $\varphi$ is said to \emph{respect the interpretation of} $f$.
If $\varphi$ respects the interpretation of every $f\in F$,
\index{homomorphism|textit}
then we call $\varphi$ a \defn{homomorphism} from $\bA$ into $\bB$, and we write
$\varphi \in \Hom(\bA, \bB)$, or simply, $\varphi: \bA \rightarrow \bB$.
\subsection{Direct Products}
\index{direct product!of sets}
The \defn{direct product} of two sets $A_0$ and $A_1$, denoted $A_0 \times A_1$,
is the set of all ordered pairs\footnote{For the definition of {\it ordered
pair}, consult the appendix.} $\< a_0, a_1 \>$ such that $a_0\in A_0$ and
$a_1 \in A_1$. That is, we define
\[
A_0 \times A_1 := \{\langle a_0, a_1 \rangle : a_0\in A_0, a_1 \in A_1\}.
\]
More generally, $A_0 \times \cdots \times A_{n-1}$ is the set of all sequences of length
$n$ with $i^{th}$ element in $A_i$. That is,
\[
A_0 \times \cdots \times A_{n-1}
:= \{\langle a_0,\ldots, a_{n-1}\rangle : a_0\in A_0, \ldots, a_{n-1} \in A_{n-1}\}.
\]
Equivalently, $A_0 \times \cdots \times A_{n-1}$ it is the set of all functions
with domain $\{0, 1, \dots, n-1\}$ and range $\bigcup\limits_{i=1}^{n-1}A_i$.
More generally still, let $\{A_i : i\in I\}$ be an indexed family of sets.
Then the \defn{direct product} of the $A_i$ is
\[
\prod_{i\in I} A_i := \{f \mid f: I \rightarrow \bigcup\limits_{i\in I} A_i \text{ with }
f(i) \in A_i\}.
\]
When $A_0 = A_1 = \cdots = A$, we often use the shorthand notation $A^2 := A
\times A$ and $A^n := A \times \cdots \times A$ ($n$ terms).
\\[4pt]
{\it Question:}
How do you know $\prod\limits_{i\in I} A_i \neq \emptyset$, even supposing
$I\neq \emptyset$ and $A_i \neq \emptyset$ for all
$i\in I$.\footnote{{\it Answer:} Each $f$ ``chooses'' an element from each $A_i$, but
when the $A_i$ are all different and $I$ is infinite, we may not be able to do
this. The Axiom of Choice (AC) \index{Axiom of Choice}
says you can.
G\"{o}del proved that the
AC is consistent with the other axioms of set theory. Cohen proved that the
negation of the AC is also consistent.
}
\\[8pt]
% [WJD] moved the section on relations ahead of the section on homomorphism and
% congruence relations (the definitions of the former are needed for the latter).
\subsection{Relations}
\label{subsec:relations}
\index{relation|textit}
A \defn{$k$-ary relation} $R$ on a set $A$ is a subset of the cartesian product
$A^k$.
\begin{examples}~
\begin{enumerate}[(a)]
\item $A = \R$ (the line) and $R = \{a\in \R^2: \text{$a$ is rational}\}$.
\item $A = \R^2$ (the plane) and $R = \{(a,b,c) \in \R^2\times \R^2\times
\R^2: \text{$a$, $b$, $c$ lie on a line}\}$; i.e. tripples of points which
are {\it colinear}.
\item $A = \R^2$ and $R = \{(a, b) \in \R^2\times \R^2 : a = (a_1,
a_2), \; b = (b_1, b_2), \; a_1^2+ a_2^2 = b_1^2+ b_2^2 \}.$
This is an equivalence relation. The equivalence classes are circles centered
at $(0,0)$.
\item $A = \R^2$ and $R =$ ``$\leq$ on each component''
$= \{(a, b) \in \R^2\times \R^2 : a_1 \leq b_1, \; a_2\leq b_2 \}$.
\end{enumerate}
\end{examples}
\index{partial order|textit}
\noindent The relation in the last example above is a \defn{partial order}; that is,
$\leq$ satisfies, for all $a, b, c$,
\begin{enumerate}
\item $a\leq a$ \hspace{54mm} \emph{(reflexive)}
\item $a\leq b, \; b\leq a \quad \Rightarrow \quad a=b$ \hspace{2cm} \emph{(anti-symmetric)}
\item $a\leq b, \; b\leq c \quad \Rightarrow \quad a\leq c$ \hspace{2cm} \emph{(transitive)}
\end{enumerate}
\index{equivalence relation|textit}
A relation $R$ on a set $A$ is an \defn{equivalence relation} if it satisfies, for all $a, b, c$,
\begin{enumerate}
\item $a \rel{R} a$ \hspace{54mm} \emph{(reflexive)}
\item $a \rel{R} b \quad \Rightarrow \quad b\rel{R} a$ \hspace{32mm} \emph{(symmetric)}
\item $a\rel{R} b, \; b\rel{R} c \quad \Rightarrow \quad a\rel{R} c$ \hspace{21mm} \emph{(transitive)}
\end{enumerate}
We denote the set of all equivalence relations on a set $A$ by $\Eq(A)$.
\index{partition|textit}
A \defn{partition} of a set $A$ is a collection $\Pi = \{A_i : i\in I\}$
of non-empty subsets of $A$ such that
\[
\bigcup_{i\in I} A_i = A \quad \text{ and } \quad A_i \cap A_j =
\emptyset \text{ for all pairs $i\neq j$ in $I$.}
\]
\underline{Facts and notation:}
\begin{enumerate}
\item The $A_i$ are called ``blocks.''
\item Each partition $\Pi$ determines an equivalence relation -- namely, the relation $\theta$ defined by $a\rel{\theta} b$ if and
only if $a$ and $b$ are in the same block of $\Pi$.
\item Conversely, if $\theta$ is an equivalence relation on $A$ ($\theta\in
\Eq(A)$), we denote the equivalence class of $\theta$ containing $a$ by
$a/\theta := \{b\in A : a \rel{\theta} b\}$
and the set $A/\theta := \{a/\theta : a\in A\}$ of all $\theta$ classes is a
partition of $A$.
\item ``an SDR'' (add something here)
\end{enumerate}
\subsection{Congruence Relations}
Let $A$ and $B$ be sets and let $\varphi : A\rightarrow B$ be any
mapping.
%\footnote{Recall (Appendix~\ref{sec:functions}), a mapping
%$\varphi : A\rightarrow B$ is a relation $\varphi \subseteq A\times B$
%such that for each $a$ in $\dom \varphi = A$ there is only one $b$ such that $a
%\rel{\varphi} b$.}
\index{kernel!of a relation}
We say that a pair $\\in A^2$ belongs to the \defn{kernel} of $\varphi$, and we
write $\langle a_0, a_1\rangle \in \ker \varphi$, just in case
$\varphi(a_0)=\varphi(a_1)$. Thus, to every map $\varphi$ there corresponds a relation
$\sim_\varphi = \ker \varphi$, defined by
\[
a \sim_\varphi a' \qquad \Leftrightarrow \qquad \varphi(a) = \varphi(a').
\]
We leave it as an exercise to prove
%\footnote{Equivalence relation is defined in the appendix, Sec.~\ref{sec:relations}.
% We denote the set of all equivalence relations on a set $A$ by $\Eq(A)$.}
\begin{prop}\label{prop:1}
$\ker \varphi$ is an equivalence relation.
\end{prop}
%\begin{exercise} Prove proposition \ref{prop:1}. \end{exercise}
If $\sim$ is an equivalence relation on a set $A$, then $a/{\sim}$ denotes the equivalence class
containing $a$; that is, $a/{\sim} := \{ a' \in A : a' \sim a\}$. The set of all
equivalence classes of $\sim$ in $A$ is denoted $A/{\sim}$. That is,
$A/{\sim} = \{a/{\sim} : a\in A\}$.
Let $\bA = \< A; F^\bA \>$ and $\bB = \$ be algebras of the same type
and let $\varphi : A \rightarrow B$ be any mapping. As above,
$\varphi$ determines the equivalence relation $\sim_\varphi = \ker \varphi$ on $A$.
\index{congruence relation|textit}
A \defn{congruence relation} on $\bA$ is an equivalence relation $\sim_\varphi$
arising from a \emph{homomorphism} $\varphi : \bA \rightarrow \bB$, for some
$\bB$. We denote the set of all congruence relations on $\bA$ by $\Con \bA$.
To reiterate, $\theta \in \Con \bA$ iff $\theta = \ker \varphi$ for some
homomorphism $\varphi$ of $\bA$. It is easy to check that this is
equivalent to:
$\theta \in \Con \bA$ iff $\theta \in \Eq(A)$ and
\begin{equation}
\label{eq:cong-re}
\ \in \theta \quad (0\leq i < n) \qquad \Rightarrow \qquad
\ \in \theta,
\end{equation}
for all $f\in F_n$ and all $a_0, \dots, a_{n-1}, a_0', \dots, a_{n-1}' \in A$.
Equivalently,\footnote{The direct product algebra $\bA \times \bA$ is
defined below in section~\ref{sec:directproducts}.}
\[
\Con \bA = \Eq(A) \cap \Sub(\bA\times \bA).
\]
We record this fact as
\begin{theorem}
\label{thm:cong-rel}
Suppose $\bA = \algebra{A; F}$ is an algebra and $\theta$ is an equivalence
relation on the set $A$. Then $\theta$ is congruence relation on the algebra
$\bA$ if and only if %(for every $n$ in the similarity type of $\bA$)
for every $n$-ary basic operation $f\in F$ and for all
$a_0, \dots, a_{n-1}, a_0', \dots, a_{n-1}' \in A$,
\[
a_i \rel{\theta} a_i' \quad (0\leq i < n) \qquad \Rightarrow \qquad
f(a_0, \dots, a_{n-1}) \rel{\theta} f(a_0', \dots, a_{n-1}').
\]
\end{theorem}
In fact, we can replace the final implication in the theorem with: for each $0
\leq i < n$,
\[
a_i\rel{\theta} a_i' \qquad \Rightarrow \qquad
f(a_0, \dots, a_i, \dots a_{n-1})\rel{\theta} f(a_0, \dots,
a_i', \dots, a_{n-1}).
\]
\underline{Proof sketch:} Recall that $a/\theta$ denotes the equivalence class of
$\theta$ which contains $a$, and $A/\theta$ denotes the full set of $\theta$
classes: $A/\theta= \{a/\theta : a\in A\}$. For every
operation $f^\bA$ on $A$ (say it is $n$-ary), we \emph{define}
\[
f^{\bA/\theta}(a_1/\theta, \dots, a_n/\theta) := f^\bA(a_1, \dots, a_n)/\theta.
\]
One easily checks that this is well-defined and that the map $a \mapsto
a/\theta$ is a homomorphism of $\bA$ onto the algebra $\bA/\theta :=
\$, where $F^{\bA/\theta} := \{f^{\bA/\theta} :
f^\bA \in F\}$. Finally, note that the kernel of this homomorphism is
$\theta$.
\subsection{Quotient Algebras}
\index{quotient algebra}
Let $\bA = \$ be an algebra. Recall, $F$ denotes the set of operation symbols,
and to each operation symbol $f\in F$ there corresponds an arity, and the set of
arities determines the similarity type of the algebra. (We might do better to
denote the algebra by $\$, since the algebra is not defined
until we have associated to each operation symbol $f\in F$ an actual operation
$f^\bA$ on $A$, but this is a technical point, and we will often denote two
algebras of the same similarity type as $\$ and $\$ with the
understanding that the meaning of $F$ depends on the context.)
Let $\theta \in \Con\bA$ be a congruence relation. The \defn{quotient algebra}
$\bA/\theta$ is an algebra with the same similarity type as $\bA$, with universe
$A/\theta = \{a/\theta : a\in A\}$, and operation symbols $F$, where for each
($k$-ary) symbol $f\in F$ the operation $f^{\bA/\theta}$ is defined as follows:
for $(a_1/\theta, \dots, a_k/\theta) \in (A/\theta)^k$,
\[
f^{\bA/\theta}(a_1/\theta, \dots, a_k/\theta) =
f^{\bA}(a_1,\dots, a_k)/\theta. % \; \in \; A/\theta.
\]
(As mentioned above, $F$ denotes the set of operation symbols of the
similarity type of the algebra, and it is ``overloaded'' in the sense that we
write $\bA = \$ and $\bA/\theta = \$, and for each
$f\in F$ the corresponding operation in these algebras is interpreted
appropriately -- i.e., as $f^\bA$ or $f^{\bA/\theta}$.)
\subsection{Direct Products of Algebras}
\label{sec:directproducts}
\index{direct product!of algebras}
Above we defined direct products of sets. We now define direct products of
algebras. Let $\bA = \$ and $\bB = \$ be two algebras of the same
similarity type. The \defn{direct product} $\bA \times \bB$ is an
algebra of the same type as $\bA$ and $\bB$, with universe $A \times
B = \{(a, b) : a\in A, b\in B\}$, and operation symbols $F$. To each ($k$-ary)
symbol $f\in F$ corresponds an operation $f^{\bA \times \bB}$ defined as follows:
for $((a_1, b_1), \dots, (a_k, b_k)) \in
(A\times B)^k$,
\begin{equation}
\label{eq:prod}
f^{\bA\times \bB}
((a_1, b_1), \dots, (a_k, b_k)) =
(f^{\bA}(a_1, \dots, a_k),f^{\bB}(b_1, \dots, b_k)).
\end{equation}
This definition can be easily extended to the direct product $\prod \bA_i$ of
any collection of algebras $\{\bA_i : i\in I\}$, and we leave it to the reader
to write down the defining property of the operations, which is completely
analogous to (\ref{eq:prod}).
\index{direct power|textit}
When all the algebras are the same -- that is,
when $\bA_i \cong \bA$, for some $\bA$ -- we call
$\prod \bA_i$ the \defn{direct power} of $\bA$. When the set $I$ is
finite, say, $I = {1, 2, \dots, n}$,
%$\{\bA_i : i\in I\} = \{\bA_1, \bA_2, \dots, \bA_n\}$,
we have alternative notations for the direct power, namely,
\[
\prod \bA_i = \bA_1 \times \bA_2 \times \cdots \times \bA_n = \bA^n.
\]
The constructions of this subsection section and the preceeding one
are often combined to give direct products of quotient algebras. For instance, if
$\theta_1$ and $\theta_2$ are two congruences of $\bA$, the algebra
$\bA/\theta_1\times \bA/\theta_2$ has the same similarity type as $\bA$, and its
universe is
\[
A/\theta_1\times A/\theta_2 = \{
(a/\theta_1, b/\theta_2) : a, b \in A\}.
\]
The operation symbols are again $F$, and to each ($k$-ary) symbol $f\in F$
corresponds an operation $f^{\bA/\theta_1\times \bA/\theta_2}$ defined as follows:
for $((a_1/\theta_1, b_1/\theta_2), \dots, (a_k/\theta_1, b_k/\theta_2)) \in
(A/\theta_1\times A/\theta_2)^k$,
\begin{align*}
f^{\bA/\theta_1\times \bA/\theta_2}
((a_1/\theta_1, b_1/\theta_2), \dots, (a_k/\theta_1, b_k/\theta_2))
&=
(f^{\bA/\theta_1}
(a_1/\theta_1, \dots, a_k/\theta_1),
f^{\bA/\theta_2}
(b_1/\theta_2, \dots, b_k/\theta_2))\\
&=
(f^{\bA}(a_1, \dots, a_k)/\theta_1,
f^{\bA}(b_1, \dots, b_k)/\theta_2). %\; \in \; A/\theta_1 \times A/\theta_2.
\end{align*}
\subsection{Lattices}
\index{relational structure|textit}
A \defn{relational structure} is a set $A$ and a collection of (finitary)
relations on $A$.
\index{partially ordered set|textit} \index{poset|see{partially ordered set}}
A \defn{partially ordered set}, or \defn{poset}, is a set $A$ together
with a partial order (Sec.~\ref{subsec:relations}) $\leq$ on it, denoted $\$.
Let $\$ be a poset and let $B$ be a subset of the set $A$. An
element $a$ in $A$ is an upper bound for $B$ if $b \leq a$ for every $b$ in $B$. An
element $a$ in $A$ is the \defn{least upper bound} of $B$, denoted $\Join\!B$,
\index{supremum|textit}
or \defn{supremum} of $B$ ($\sup B$), if $a$ is an upper bound of $B$, and
$b\leq c$ for every $b$ in $B$ implies $a \leq c$ (i.e., $a$ is the smallest among
the upper bounds of $B$). Similarly,
$a$ is a lower bound of $B$ provided $a\leq b$ for all $b$ in $B$, and $a$
is the \defn{greatest lower bound} of $B$ ($\Meet\!B$), or \defn{infimum} of $B$
\index{infimum|textit}
($\inf B$) if $a$ is a lower bound and is above every other lower bound of $B$.
Let $a, c$ be two elements in the poset $A$. We say $c$ \defn{covers} $a$, or $a$ is covered
by $c$ provided $a \leq c$ and whenever $a \leq b \leq c$ it follows that $a =
b$ or $b = c$. We use the notation $a \prec c$ to denote that $c$ covers $a$.
\index{lattice|textit}
A \defn{lattice} is a
partially ordered set $\$ such that for each pair $a, b \in L$ there
is a least upper bound, denoted $a \join b := \lub \{a, b\}$,
and a greatest lower bound, denoted $a \meet b := \glb \{a, b\}$, contained in
$L$. A lattice can also be viewd as an algebra $\algebra{L; \join, \meet}$
where $\join$, called ``join,'' and $\meet$, ``meet,'' are binary operations
satisfying
\index{join operation}
\index{meet operation}
\begin{enumerate}
\item $x\join x = x$ and $x \meet x = x$ \hspace{32mm} \emph{(idempotent)}
\item $x\join y = y\join x$ and $x\meet y = y\meet x$ \hspace{2cm} \emph{(commutative)}
\item $x\join (y\join z) = (x \join y) \join z$\hspace{34mm} \emph{(associative)}
\item $x\join (y\meet x) = x$ and $x \meet (y \join x) = x$\hspace{16mm} \emph{(absorbtive)}
\end{enumerate}
Posets in general, and lattices in particular, can be visualized using a
so-called \defn{Hasse diagram}.
\index{Hasse diagram|textit}
The Hasse diagram of a poset $\$
is a graph in which each element of the set $A$ is denoted by a vertex, or
``node'' of the graph.
If $a \prec b$ then we draw the node for $b$ above the node for $a$, and join
them with a line segment. The resulting diagram gives a visual description of the
relation $\leq$, since $a \leq b$ holds iff for some
finite sequence of elements $c_1, \dots, c_n$ in $A$ we have $a = c_1 \prec c_2
\prec \cdots \prec c_n = b$. Some examples appear in the figures below.
\begin{figure}[h,centering]
\caption{Hasse diagrams}
\label{fig:hasse}
\begin{center}
\begin{tikzpicture}[scale=0.6]
\node (b1) at (-4,-2) [fill,circle,inner sep=1.2pt] {};
\node (t1) at (-4,2) [fill,circle,inner sep=1.2pt] {};
\node (b2) at (0,-2) [fill,circle,inner sep=1.2pt] {};
\node (t2) at (0,2) [fill,circle,inner sep=1.2pt] {};
\node (b3) at (4,-2) [fill,circle,inner sep=1.2pt] {};
\node (t3) at (4,2) [fill,circle,inner sep=1.2pt] {};
\node (l11) at (-5,-1) [fill,circle,inner sep=1.2pt] {};
\node (l12) at (-3,-1) [fill,circle,inner sep=1.2pt] {};
\node (l21) at (-1,-1) [fill,circle,inner sep=1.2pt] {};
\node (l22) at ( 1,-1) [fill,circle,inner sep=1.2pt] {};
\node (l31) at ( 3,-1) [fill,circle,inner sep=1.2pt] {};
\node (l32) at ( 5,-1) [fill,circle,inner sep=1.2pt] {};
\node (u11) at (-5,1) [fill,circle,inner sep=1.2pt] {};
\node (u12) at (-3,1) [fill,circle,inner sep=1.2pt] {};
\node (u21) at (-1,1) [fill,circle,inner sep=1.2pt] {};
\node (u22) at ( 1,1) [fill,circle,inner sep=1.2pt] {};
\node (u31) at ( 3,1) [fill,circle,inner sep=1.2pt] {};
\node (u32) at ( 5,1) [fill,circle,inner sep=1.2pt] {};
\draw (2.6,1.2) node {$a$};
\draw (5.4,1.2) node {$b$};
\draw (2.6,-1.2) node {$c$};
\draw (5.4,-1.2) node {$d$};
\draw[semithick] (b1) to (l11) to (u11) to (t1) to (u12) to (l12) to (b1);
\draw[semithick] (b2) to (l21) to (u22) to (t2) to (u21) to (l22) to (b2);
\draw[semithick] (b3) to (l31) to (u31) to (t3) to (u32) to (l32) to (b3);
\draw[semithick] (l31) to (u32);
\draw[semithick] (l32) to (u31);
\end{tikzpicture}
\end{center}
\end{figure}
Note that the first two examples in Figure~\ref{fig:hasse} depict the same poset,
which illustrates that Hasse diagrams are not uniquely determined. Also, note that the
poset represented in the first two diagrams is a lattice. In contrast, the
poset depicted in the third diagram is not a lattice since neither $a \meet b$
nor $c \join d$ is defined -- the sets $\{a, b\}$ and $\{c, d\}$
have upper and lower bounds, but $\{a, b\}$ has no \emph{greatest} lower
bound, and $\{c, d\}$ has no \emph{least} upper bound.
% \begin{examples}
% Figures~\ref{fig:basiclattices} and~\ref{fig:m3n5} present the
% Hasse diagrams of a few small lattices, some of which are given names the
% meaning of which will become apparent later.
\begin{figure}[centering]
\caption{Hasse diagrams of some small lattices.}
\label{fig:basiclattices}
\begin{center}
\begin{tikzpicture}[scale=0.8]
\draw (0,-1) node {$\mathbf{1}$};
\node (0) at (0,0) [fill,circle,inner sep=1.2pt] {};
\draw (2,-1) node {$\mathbf{2}$};
\node (2) at (2,0) [fill,circle,inner sep=1.2pt] {};
\node (21) at (2,1) [fill,circle,inner sep=1.2pt] {};
\draw[semithick] (2) to (21);
\draw (4,-1) node {$\mathbf{3}$};
\node (3) at (4,0) [fill,circle,inner sep=1.2pt] {};
\node (31) at (4,1) [fill,circle,inner sep=1.2pt] {};
\node (32) at (4,2) [fill,circle,inner sep=1.2pt] {};
\draw[semithick] (3) to (31) to (32);
\draw (6,-1) node {$\mathbf{4}$};
\node (4) at (6,0) [fill,circle,inner sep=1.2pt] {};
\node (41) at (6,1) [fill,circle,inner sep=1.2pt] {};
\node (42) at (6,2) [fill,circle,inner sep=1.2pt] {};
\node (43) at (6,3) [fill,circle,inner sep=1.2pt] {};
\draw[semithick] (4) to (41) to (42) to (43);
\draw (9,-1) node {$\mathbf{2}\times\mathbf{2}$};
\node (2x2) at (9,0) [fill,circle,inner sep=1.2pt] {};
\node (2x21) at (8,1) [fill,circle,inner sep=1.2pt] {};
\node (2x22) at (10,1) [fill,circle,inner sep=1.2pt] {};
\node (2x23) at (9,2) [fill,circle,inner sep=1.2pt] {};
\draw[semithick] (2x2) to (2x21) to (2x23) to (2x22) to (2x2);
\draw (12,-1) node {$\mathbf{5}$};
\node (5) at (12,0) [fill,circle,inner sep=1.2pt] {};
\node (51) at (12,1) [fill,circle,inner sep=1.2pt] {};
\node (52) at (12,2) [fill,circle,inner sep=1.2pt] {};
\node (53) at (12,3) [fill,circle,inner sep=1.2pt] {};
\node (54) at (12,4) [fill,circle,inner sep=1.2pt] {};
\draw[semithick] (5) to (51) to (52) to (53) to (54);
\node (2x2t) at (15,0) [fill,circle,inner sep=1.2pt] {};
\node (2x2t1) at (14,1) [fill,circle,inner sep=1.2pt] {};
\node (2x2t2) at (16,1) [fill,circle,inner sep=1.2pt] {};
\node (2x2t3) at (15,2) [fill,circle,inner sep=1.2pt] {};
\node (2x2t4) at (15,3) [fill,circle,inner sep=1.2pt] {};
\draw[semithick] (2x2t) to (2x2t1) to (2x2t3) to (2x2t2) to (2x2t);
\draw[semithick] (2x2t3) to (2x2t4);
\node (t2x2) at (18,0) [fill,circle,inner sep=1.2pt] {};
\node (t2x21) at (18,1) [fill,circle,inner sep=1.2pt] {};
\node (t2x22) at (17,2) [fill,circle,inner sep=1.2pt] {};
\node (t2x23) at (19,2) [fill,circle,inner sep=1.2pt] {};
\node (t2x24) at (18,3) [fill,circle,inner sep=1.2pt] {};
\draw[semithick] (t2x2) to (t2x21) to (t2x22) to (t2x24) to (t2x23) to (t2x21);
\end{tikzpicture}
\end{center}
\end{figure}
\begin{figure}[centering]
\caption{Hasse diagrams of two important lattices.}
\label{fig:m3n5}
\begin{center}
\begin{tikzpicture}[scale=0.8]
\draw (0,-1) node {$M_3$};
\node (m3) at (0,0) [fill,circle,inner sep=1.2pt] {};
\node (m31) at (0,1.5) [fill,circle,inner sep=1.2pt] {};
\node (m32) at (-1.5,1.5) [fill,circle,inner sep=1.2pt] {};
\node (m33) at (1.5,1.5) [fill,circle,inner sep=1.2pt] {};
\node (m34) at (0,3) [fill,circle,inner sep=1.2pt] {};
\draw [semithick] (m3) to (m31) to (m34) to (m33) to (m3) to (m32) to (m34);
\draw (6,-1) node {$N_5$};
\node (n5) at (6,0) [fill,circle,inner sep=1.2pt] {};
\node (n51) at (5,1) [fill,circle,inner sep=1.2pt] {};
\node (n52) at (7,1.5) [fill,circle,inner sep=1.2pt] {};
\node (n53) at (5,2) [fill,circle,inner sep=1.2pt] {};
\node (n54) at (6,3) [fill,circle,inner sep=1.2pt] {};
\draw [semithick] (n5) to (n51) to (n53) to (n54) to (n52) to (n5);
\end{tikzpicture}
\end{center}
\end{figure}
%\end{examples}
\newpage
\part{Rings, Modules and Linear Algebra}
\newpage
\section{Rings}
A \defn{ring} is an algebra $\alg R = \algebra{R; +, -, \cdot, 0, 1}$
($-$ stands for unary negation) such that
$\algebra{R; +, -, 0}$ is an abelian group and
$\algebra{R; \cdot, 1}$ is a monoid (semigroup with $1$)
and
\begin{align*}
(a+b)c &= ac + bc \qquad a(b+c) = ab + ac \qquad \text{(distributive
laws)}\\
1x &= x1 = x.
\end{align*}
Since rings are defined by equational axioms, they are closed under
homomorphic images, subalgebras, and direct products. Note that if
$\alg S$ is a subalgebra (that is, subring) of $\alg R$, which is
denoted by $\alg S \le \alg R$, then $1 \in S$.
\begin{exercise} Show that for each $a$ in a ring $\alg R$ the map
$x \mapsto ax$ is endomorphism of the group $\algebra{R; +, -, 0}$.
\end{exercise}
A \defn{commutative ring} is one satisfying $xy = yx$. A ring
$\alg R$ is an
\defn{integral domain} (or just \defn{domain}) if
$a\ne 0$ and $b\ne 0$ imply $ab \ne 0$; that is,
$\algebra{R - \{0\}; \cdot, 1}$ is a monoid.
(Some books require that $\alg R$ be commutative.)
Note subrings of domains are domains but the class of all domains is
not closed under homomorphic images or direct products.
\todo{add pages R2--6 of the notes}
\subsection{The ring $\op M_n(\alg R)$}
We assume $\alg R$ is commutative. The \defn{determinant} of
$A \in \op M_n(\alg R)$ is defined as
\[
\det A = \sum_{\sigma\in S_n} \epsilon(\sigma)\prod_{i=1}^n a_{i,\sigma(i)}
\]
The set $\{a_{1,\sigma(1)}, \ldots, a_{n,\sigma(n)}\}$ is a
\defn{diagonal} or
\defn{generalized diagonal}.
$\{a_{1,1}\ldots, a_{n,n}$ is called the
\defn{main diagonal} and
$\{a_{1,n}\ldots, a_{n,1}$ is called the
\defn{sinister diagonal}.
The reader can show that if
$A$ has two equal rows then each generalized diagonal occurs
twice in the definition of $\det(A)$, once positive and once
negative, so $\det A = 0$. (If the characteristic of $\alg R$
is 2 then we get 2 times the diagonal product, so the determinant
is still 0.) Recall that
\[
\det(AB) = \det(A) \det(B).
\]
Let $A(i\mid j) \in \op M_{n-1}(\alg R)$ obtained from $A$ by
deleting the $i^\text{th}$ row and
$j^\text{th}$ column.
Recall the Laplace expansion theorem from linear algebra:
\begin{theorem}[Laplace Expansion Theorem]
\[
\det A = \sum_{k=1}^n (-1)^{i+k} a_{i,k} \det(A(i \mid k))
\]
\end{theorem}
\begin{corollary}
If $i \ne j$ then
\[
\sum_{k=1}^n (-1)^{j+k} a_{i,k} \det(A(j \mid k)) = 0.
\]
\end{corollary}
\begin{proof}
Let $B$ be the matrix which is the same as $A$ except the
$j^\text{th}$ row of $B$ equals the $i^\text{th}$ row of $A$.
Since the $i^\text{th}$ and $j^\text{th}$ rows of $B$ are the same,
$\det B = 0$. Expanding around the $j^\text{th}$ row with the
Laplace Expansion Theorem gives
\begin{align*}
0 = \det B &= \sum_{k=1}^n (-1)^{j+k} b_{j,k} \det(B(j \mid k))\\
&= \sum_{k=1}^n (-1)^{j+k} b_{i,k} \det(B(j \mid k))\\
&= \sum_{k=1}^n (-1)^{j+k} a_{i,k} \det(A(j \mid k)).
\end{align*}
\end{proof}
The \defn{adjoint} of a matrix $A$, denoted $\op{adj} A$,
is defined as
\[
(\op{adj} A)_{ij} = (-1)^{i+j} \det(A(j\mid i)).
\]
By the Laplace Expansion Theorem and its corollary we have
\[
A(\op{adj} A) = (\op{adj} A)A = (\det A)I_n
\]
Using this we get the next theorem.
\begin{theorem}
Let $\alg R$ be a commutative ring and let $A \in \op M_n(\alg R)$.
Then $A$ is invertible if and only if $\det A$ is invertible in~$\alg R$.
In particular, if $\alg R$ is a field, then $A$ in ivertible if and
only if $\det A \ne 0$.
\end{theorem}
\newpage
\subsection{Factorization in Rings}
In this sections $R$ denotes a commutative ring with $1$.
For $a$, $b\in R$, we say $a$ \defn{divides} $b$ if there exists
$x\in R$ such that $ax = b$. This is denotes $a \mid b$.
We say $a$ and $b$ are \defn{associates} if
$a \mid b$ and $b \mid a$.
An element $u$ is a \defn{unit} if there exists a $v\in R$ with
$uv = vu = 1$.
\begin{theorem}[Theorem 4 in class]\label{classthm4}
$R$ commutative with $1$, $a$, $b$, $u\in R$, then
\begin{enumerate}
\setlength{\itemsep}{0.0ex plus0.1ex}
\item $a \mid b$ iff $(b) \subseteq (a)$.
\item $a$ and $b$ are associates iff $(a) = (b)$.
\item $u$ is a unit iff $ u \mid r$ for all $r \in R$.
\item $u$ is a unit iff $(u) = R$.
\item The relation ``$a$ is an associate of $b$" is an equivalence
relation.
\item If $a = bu$, where $u$ is a unit, then $a$ and $b$ are
associates. If $R$ is an integral domain the converse is true.
\end{enumerate}
\end{theorem}
\begin{proof}
Exercise.
\end{proof}
An ideal $P$ is \defn{prime} in $R$ if $P \ne R$ and if $A$ and
$B$ are ideals then
\[
AB \subseteq P \implies A \subseteq P\text{ or } B \subseteq P
\]
\begin{theorem}[Theorem 5 in class]\label{thm:primeideal}
$R$ commutative with $1$. An ideal $P$ of $R$ is prime iff
\begin{equation}
\forall a, b \in R, \, ab \in P \implies a \in P\text{ or }
b \in P \tag{$*$}
\end{equation}
\end{theorem}
\begin{proof}
($\Leftarrow$)
Assume ($*$) and suppose $A$ and $B$ are ideal
with $AB \subseteq P$ and $A \not\subseteq P$. Let $a\in A - P$.
Note
$\forall b\in B, ab \in AB \subseteq P$ so by ($*$) $b\in P$.
So $B\subseteq P$.
($\Rightarrow$) Assume $P$ is a prime ideal.
Note $(a)(b) \subseteq (ab)$. So if $a$, $b\in P$ then $(ab)
\subseteq P$ and hence $(a)(b) \subseteq P$. So
$(a) \subseteq P$ or
$(b) \subseteq P$. Thus $a\in P$ or $b \in P$, so ($*$) holds.
\end{proof}
\begin{corollary}
Maximal ideals are prime.
\end{corollary}
\begin{proof}
Let $M$ be a maximal ideal and suppose $ab \in M$ with neither
$a \in M$ nor $b \in M$. Then, since $M$ is maximal,
$(a) + M = (b) + M = R$. So
\[
R = R^2 = ((a) + M)((b) + M) \subseteq (a)(b) + M = M.
\]
This contradiction together with
Theorem~\ref{thm:primeideal} shows that $M$ is prime.
\end{proof}
\begin{definition}
$R$ is a \defn{principal ideal domain}, or \defn{PID},
if it is an integral domain and all ideals are principal.
\end{definition}
\begin{definition}
$c \in R$ is \defn{irreducible} if
\begin{enumerate}
\setlength{\itemsep}{0.0ex plus0.1ex}
\item $c$ is a nonunit and not $0$, and
\item $c = ab$ implies $a$ or $b$ is a unit.
\end{enumerate}
$p \in R$ is \defn{prime} if
\begin{enumerate}
\setlength{\itemsep}{0.0ex plus0.1ex}
\item $p$ is a nonunit and not $0$, and
\item $p \mid ab$ implies $p \mid a$ or $p \mid b$.
\end{enumerate}
\end{definition}
\begin{theorem}[Theorem 6 in class]\label{classthm6}
Let $R$ be an integral domain,
and let $c$, $p\in R$. %$c \in R$, $p \ne 0 \ne c$.
\begin{enumerate}
\setlength{\itemsep}{0.0ex plus0.1ex}
\item $p$ is a prime iff $(p)$ is a nonzero prime ideal.
\item $c$ is irreducible iff $(c)$ is a maximal in the set of all
proper, principal ideals of $R$.
\item Every prime is irreducible.
\item If $R$ is a PID then $p$ is prime iff it is irreducible.
\item Associates of prime [irreducible] elements are prime
[irreducible].
\item The only divisors of an irreducible element are associates
and units.
\end{enumerate}
\end{theorem}
\begin{proof}
Coming.
\end{proof}
\begin{definition}
An integral domain is a \defn{unique factorization domain},
or \defn{UFD} if
\begin{enumerate}
\setlength{\itemsep}{0.0ex plus0.1ex}
\item If $a \in R$, $a \ne 0$, and $a$ is not a unit then
$a = c_1c_2 \cdots c_n$, where the $c_i$'s are irreducible.
\item If $a = c_1c_2 \cdots c_n$ and $a = d_1d_2 \cdots d_m$,
$c_i$, $d_j$ irredicible, then $n = m$ and there is a
$\sigma \in S_n$ such that $c_i$ and $d_{\sigma(i)}$ are
associates.
\end{enumerate}
\end{definition}
\begin{lemma}
If $R$ is a PID then $R$ satisfies the ACC (the ascending chain
condition); that is
if $I_i$ an ideal and
\[
I_1 \subseteq I_2 \subseteq \cdots
\]
then, for some $n$ and all $k \ge 0$, $I_{n+k} = I_n$.
\end{lemma}
\begin{proof}
Laters.
\end{proof}
\begin{theorem}[Theorem 7 in class]
If $R$ is a PID then it is a UFD.
\end{theorem}
\begin{proof}
Suppose $a\in R$ is not 0, is not a unit and cannot be written as a
product of irreducibles. Since $a$ is not a unit $(a) \ne R$ and so
$(a)$ is contained contained in a maximal ideal which has the form
$(c)$ since $R$ is a PID.
Note, by Theorem~\ref{classthm6}, $c$ is irreducible.
$(a) \subseteq (c)$ so $c \mid a$. Thus
$a = bc$. Now $b$ is not a unit; otherwise $a$ would be irreducible
by part 5 of Theorem~\ref{classthm6}. $b$ is not a product of
irreducibles, since otherwise $a = bc$ would also be a product of
irreducibles. Since $a = bc$, $(a) \subseteq (b)$. If
$(a) = (b)$ then by Theorem~\ref{classthm4} $a$ and $b$ are
associates and also by Theorem~\ref{classthm4} $a = bu$ for some
unit~$u$. By cancellation $u = c$, a contradiction since $c$ is not
a unit. Hence
\[
(a) \subsetneq (b)
\]
Now $b$ has the same properties as $a$ and so $(b)$ is also
properly contained another principal ideal $(d)$. Continuing in
this way we get an infinite ascending chain of ideals,
contradicting the lemma above. Hence every nonzero, nonunit element
of $R$ can be written as a product of primes.
For uniqueness suppose $c_1 c_2 \cdots c_n = d_1 d_2 \cdots d_m$
with each $c_i$ and $d_j$ irreducible. By Theorem~\ref{classthm6}
each of these elements is prime. So $c_1 \mid d_j$ for some~$j$. By
Theorem~\ref{classthm6}, $c_1$ and $d_j$ are associates. So
$c_1 = ud_j$. Using cancellation we get
\[
u c_2 \cdots c_n = d_1 \cdots d_{j-1} d_{j+1} \cdots d_m
\]
The proof can now be completed using induction. We leave the
details to the student.
\end{proof}
We proved that in a PID, an element is prime iff it is irreducible.
Now we extend this to UFD's.
\begin{theorem}
In a UFD $R$, a nonzero element is prime iff it is irreducible.
\end{theorem}
\begin{proof}
By Theorem~\ref{classthm6}(3), if an element is prime in $R$ it is
irreducible. So suppose $p \in R$ is irreducible and suppose
$p \mid ab$. So $pc = ab$ for some $c\in R$. If we write
$a= a_1a_2\cdots a_n$, and $b = b_1 b_2 \cdots b_m$ as a product of
irreducibles
and use the uniqueness of the
factorization and that $p$ is irreducible, we see that $p$ must be
an associate of one of the irreducibles in $a$ or one in $b$. WLOG
we may assume that $p$ and $a_1$ are associates so
$a_1 = up$ for some unit $u$. Hence $a = (up)a_2\cdots a_n$ so $p \mid a$.
\end{proof}
\subsection{Rings of Frations}
Coming.
\subsection{Euclidean Domain and the Eucidean Algorithm}
Coming.
\subsection{Polynomial Rings, Gauss' Lemma}
A \defn{greatest common divisor} or \defn{gcd} of elements $a$ and $b$
is an element $c$ such that $c \mid a$ and $c \mid b$ and if $d$ is
any element such that $d \mid a$ and $d \mid b$ then $d \mid c$.
The gcd will not in general be unique but any two are associates.
In a UFD we can find $\gcd(a,b)$ by letting $p_1,\ldots,p_n$ be the
distinct primes that occur in either $a$ or $b$. Then,
$a = u \prod_{i=0}^n p_i^{r_i}$ and
$b = v \prod_{i=0}^n p_i^{s_i}$, $u$ and $v$ units, $r_i$, $s_i \ge 0$.
Then if $t_i = \min(r_i,s_i)$, $\prod p_i^{t_i}$ is a gcd of $a$
and~$b$.
If $\gcd(a,b) = 1$ then we say
$a$ and $b$ are \defn{relatively prime}.
$\gcd(a_0,\ldots, a_n)$ is defined in an obvious way.
Let $R$ be a UFD and let $f = \sum_{i=0}^n a_i x^i \in R[x]$. The
\defn{content}, denoted $\op C(f)$, of $f$ is
$\gcd(a_0,\ldots,a_n)$. Again this is unique only up to
associates. If $\op C(f)$ is a unit (which is the same as saying
$\op C(f) = 1$) then $f$ is called \defn{primitive}.
\begin{exercise}
Show that if $R$ is a UFD and $a \in R$ then $\op C(af) = a\op
C(f)$.
\end{exercise}
\begin{lemma}[Gauss' Lemma]
If $R$ is a UFD and $f$, $g\in R[x]$ then
$\op C(fg) = \op C(f) \op C(g)$. In particular the product of
primitive polynomials is primitve.
\end{lemma}
\begin{proof}
Let $a = \op C(f)$ and $b = \op C(g)$. Then
$f = af_1$ and $g = b g_1$, for some $f_1$ and $g_1 \in R[x]$
which are primitive. Note $fg = abf_1g_1$. By the exercise it is
enough to show $f_1g_1$ is primitive. Let
$f_1 = \sum_{i=0}^n a_i x^i$ and
$g_1 = \sum_{i=0}^m b_i x^i$, then
$f_1g_1 = \sum _{i=0}^{m+n} c_i x^i$, where
$c_k = \sum_{i+j = k} a_i b_j$. Suppose $f_1g_1$ is not primitive,
then there is an irreducible element $p$ such that $p \mid c_i$ for
all~$i$. Since $f_1$ is primitive, there is a smallest $s$ such
that $p \nmid a_s$ and a smallest $t$ such that
$p \nmid b_t$. Now
$p \mid c_{s+t} = a_0b_{s+t} + \cdots + a_sb_t + \cdots + a_{s+t}b_0$.
This implies $p \mid a_sb_t$ (since it divides all other terms on
both sides). But since $p$ is irreducible and hence
prime, this implies $p \mid a_s$ or $p \mid b_t$, a contradiction.
\end{proof}
\begin{lemma}
Let $R$ be a UFD with quotient field $F$ and let $f$ and $g$ be
primitive polynomials in $R[x]$. Then $f$ and $g$ are associates in
$R[x]$ iff they are associates in $F[x]$.
\end{lemma}
\begin{proof}
Let $f$ and $g$ be associates in $F[x]$. Since $F[x]$ is an
integral domain, Theorem~\ref{classthm4}(6) there is a unit $u\in F[x]$
such that $f = ug$.
Units in $F[x]$ are just the nonzero elements of $F$. Hence
$u = b/c$ for some $b$, $c\in R$, $c \ne 0$. So
$cf = bg$. Since $C(f)$ and $C(g)$ are units in $R$,
\[
c \approx cC(f) \approx C(cf) \approx C(bg) \approx bC(g) \approx b
\]
where $\approx$ is the equivalence relation ``is an associate of."
Hence
$b = vc$ for some unit $v \in R$ and so $cf = bg = vcg$. Since $c
\ne 0$, $f = vg$, showing $f$ and $g$ are associates in $R[x]$.
The converse is easy.
\end{proof}
\begin{theorem}[This is also called Gauss' Lemma]
Let $R$ be a UFD with quotient field $F$ and let $f$ be a
primitive polynomial of positive degree in $R[x]$.
Them $f$ is irreducible in $R[x]$ iff it is irreducible in~$F[x]$.
\end{theorem}
\begin{proof}
Suppose $f$ is irreducible in $R[x]$ but $f = gh$ in $F[x]$ with
the degrees of $g$ and $h$ positive. Then
\[
g = \sum_{i=0}^n (a_i/b_i)x^i \text{ and }
h = \sum_{i=0}^m (c_i/d_i)x^i
\]
with $a_i, b_i, c_j, d_j \in R$ and $b_i \ne 0 \ne d_j$.
Let $b = b_0b_1\cdots b_n$ and for each $i$ let $b_i^* = b/b_i$.
Of course $b_i^*$ is in $R$. Let
$g_1 = \sum_{i=0}^n a_i b_i^*x^i \in R[x]$.
Then $g_1 = ag_2$ with $a = C(g_1)$ and $g_2$ is primitive in
$R[x]$. Now
\[
g = (1/b)g_1 = (a/b) g_2
\]
and $\deg g = \deg g_2$.
Similarly $h = (c/d)h_2$ with $c$, $d\in R$, $h_2$ primitive in
$R[x]$ and $\deg h = \deg h_2$. Hence $f = gh = (a/b)(c/d)g_2h_2$,
so $bdf = ac g_2h_2$. Since $f$ is primitive and $g_2h_2$ is
primitive by Gauss' Lemma,
\[
bd \approx bdC(f)
\approx C(bdf)
\approx C(acg_2h_2)
\approx acC(g_2h_2)
\approx ac.
\]
So $bd$ and $ac$ are associates in $R[x]$ which implies $f$ and
$g_2h_2$ are associates in $R[x]$. So $f$ is reducible in $R[x]$,
a contradiction. So $f$ is irreducible in $F[x]$.
Conversely suppose $f$ is irreduible in $F[x]$ but $f = gh$ in
$R[x]$. This is still a factorization in $F[x]$ so either $g$ or
$h$ is a unit in $F[x]$. The units of $F[x]$ are just the nonzero
constant elements of $F$ so one, say $g$, is constant (and in $R$).
Now $C(f) = gC(h)$. Since $f$ is primitive, $g$ must be a unit
in $R$ and so in $R[x]$. Therefore $f$ is irreducible in $R[x]$.
\end{proof}
\begin{theorem}
$R$ is a UFD iff $R[x]$ is.
\end{theorem}
\begin{proof}
Assume $R$ is a UFD. We will first show that every $f\in R[x]$ can be
factored into irreduibles and then prove the uniqueness.
As usual let $f = cf_1$, $c = C(f)$, $f_1$ primitive. Since $R$ is
a UFD, $c$ can be uniquely factored uniquely into irreducibles, so we only
need to factor $f_1$. So switching notation we assume the $f$ is a
primitive. We may also assume $\deg f > 0$.
Now $F[x]$ is a UFD (since it is a Euclidean domain) and so
$f$ can be factored into irreducibles in $F[x]$.
Using the proof of the second version of Gauss's
Lemma, $f$ can be factored in $R[x]$ and the factors are multiples
by elements in $F$ of the factors in $F[x]$. Since the content of
$f$ is 1, the content of each of these factors must be 1. So by the
last theorem each is irreducible in $R[x]$.
For the uniquess suppose $f = g_1\cdots g_r = h_1\cdots h_s$ are two
factorizations of $f$ into irreducibles. Since the content of $f$
is 1, the content of each $g_i$ and $h_j$ is 1. So they are
irreducible in $F[x]$ by Gauss. Since $F[x]$ is a UFD, $r=s$ and,
after renumbering, $g_i$ and $h_i$ are associates in $F[x]$, so
$g_i = (a/b)h_i$, for some nonzero $a$ and $b \in R$. So
$b g_i = a h_i$. The content of the left side is $b$ and of the
right side is $a$. Hence $a = ub$, for a unit $u\in R$. Thus
$g_i = u h_i$ so $g_i$ and $h_i$ are associates in $R[x]$,
completing the proof.
\end{proof}
\begin{corollary}
If $F$ is a field then $F[x_1,\ldots,x_n]$ is a UFD.
\end{corollary}
\begin{corollary}
If $F$ is a field then $F[x_1, x_2, \ldots]$ is a UFD.
\end{corollary}
\newpage
\subsection{Irreducibility Tests}
\begin{theorem}[Eisenstein criterion]
Let $R$ be a UFD with quotient field $F$. Let
\[
f(x) = a_n x^n + \cdots + a_1 x + a_0
\]
be in $R[x]$. If there exists a prime $p$ of
$R$ such that $p$ divides all the coeffients except the highest,
and $p^2 \nmid a_0$, that is,
\[
\text{$p \mid a_i$ $,i = 0,\ldots,a_{n-1},\qquad p \nmid
a_n,\qquad p^2 \nmid a_0$},
\]
then $f$ is irreducible in $F[x]$.
If $f$ is primitive it is also irreducible in $R[x]$.
\end{theorem}
\begin{proof}
$f = C(f)f_1$ for some primitive polynomial $f_1$. So $f$ and
$f_1$ are associates in $F[x]$ since $C(f)$ is a unit of $F$. This implies
$f$ is irreducible in $F[x]$ iff $f_1$ is. So by the second version of Gauss'
Lemma, it suffices to show $f_1$ is irreducible in $R[x]$.
Suppose $f_1 = gh$, $g$, $h \in R[x]$, where
\begin{align*}
g &= b_rx^r + \cdots + b_0, \text{ $\deg g = r \ge1$}\\
h &= c_sx^s + \cdots + c_0, \text{ $\deg h = s \ge1$}.
\end{align*}
Since $p \nmid a_n$, $p \nmid C(f)$, so if we write
$f_1 = \sum_{i=0}^n a_i^* x^i$, then the $a_i^*$ have the same
divibility properties with respect to $p$ as the $a_i$'s. Since
$p \mid a^*_0 = b_0c_0$, it divides one of them; say $b_0$. Since
$p^2 \nmid a_0^*$, $p \nmid c_0$. $p$ cannot divide every $b_i$
since otherwise $p$ would divide $g$ and hence $f_1 = gh$. Let
$k$ be the least integer such that $b_k$ is not divisible by $p$.
So $p \mid b_i$, $i
< k$ and $p\nmid b_k$. Since $r < n$, $k < n$. Now
\[
a_k^* = b_0c_k + b_1c_{k-1} + \cdots b_{k-1}c_1 + b_kc_0.
\]
This implies $p \mid b_kc_0$, a contradiction.
\end{proof}
\begin{exers}~
\begin{exercises}
\prob
Show that $2x^5 - 6x^3 + 9 x^2 - 15$ is irreducible in both
$\mathbb{Z}[x]$ and $\mathbb{Q}[x]$.
\prob
Let $f = y^3 + x^2y^2 + x^3 y + x \in R[x,y]$, where $R$ is a UFD.
Show that $f$ is irreducible in $R[x,y]$. (Hint: view
$R[x,y]$ as $(R[x])[y]$ and note that $x$ is irreducible (and hence
prime) in $R[x]$.)
\prob
Let $p$ be a prime in $\mathbb{Z}$ and
$f = \frac{x^p - 1}{x-1} =
1 + x + x^2 + \cdots + x^{p-1}$. Show
that $f$ is irreducible in $\mathbb{Z}[x]$.
(Hint: show that $g(x) = f(x+1)$ is irreducible and use this to
show $f$ is irreducible.)
\end{exercises}
\end{exers}
\newpage
\section{Modules}
\subsection{Basics}
Let $R$ be a ring. An $R$-\defn{module} is an algebra $M$ with
operations $+$, $-$, $0$, such that $\algebra {A; +, -, 0}$ is an
abelian group and for each $r\in R$ there is a unary operation
$a \mapsto ra$ such that for all $r$, $s\in R$ and $a$, $b\in M$,
\begin{align*}
r(a+b)a &= ra + rb \\
(r+s)a &= ra + sa \\
(rs)a &= r(sa)
\end{align*}
If $1 \in R$ and $1a = a$ then $M$ is a
\defn{unitary}[module!unitary] $R$-module.
The student can show that $0a = 0$ (note here the first $0$ is the
zero of $R$ and the second is the zero of $M$. Also $r(-a) = -ra$.
The notation ${}_RM$ means $M$ is a left $R$ module.
Right $R$-modules are defined in the obvious way and denote
$M_R$.
Unless otherwise stated, we will assume $R$ has a $1$ and modules
are unitary.
\begin{examples}~
\begin{enumerate}[(a)]
\setlength{\itemsep}{0.0ex plus0.1ex}
\item Each abelian group is a $\mathbb Z$ module (in an obvious way).
\item A vector space $V$ over a field $F$ is an $F$-module.
\item Column vectors of length $n$ with entries in $R$ are a module over
$M_n(R)$, the ring of $n \times n$ matrices.
\item $R$ itself is both a left and a right $R$-module, ${}_RR$ and
$R_R$. $I \subseteq R$ is a left idea if and only if it is a
submodule of ${}_RR$.
\item Let $F$ be a field, $V$ a vector space over $F$ and $T : V
\to V$ a linear transformation on $V$. Then $V$ is a $F[x]$-modules
under the action $f(x)v = f(T)v$.
\end{enumerate}
\end{examples}
Just as for abelian groups, the congruence associated with a module
homomorphism $\varphi : A \to C$ is determined by $f^{-1}(0)$,
which is a submodule (as you can easily check). Conversely if $B$
is a sumbmodule of $A$, we can form the quotient module
$A/B$ in the usual way. The usual isomorphism theorems hold;
see Dummit and Foote~\cite{DummitFoote:2004}, section~10.2.
Free modules can be defined in the usual way:
If $M$ is an $R$ module and $X \subseteq M$, then $M$ is
\defn{freely generated} by $X$ (or $M$ is \defn{free over $X$}) if
$M$ is generated by $X$ and if $\varphi : X \to N$ ($N$ another
$R$-module) then $\varphi$ can be extended to a homomorphism of
$\bar{\varphi} : M \to N$.
As in groups, we call a module $M$ \defn{cyclic}[module!cyclic] if it is
one-generated; that is, there is an element $a\in M$ such that
the smallest submodule containing $a$, which is
\[
Ra := \{ra : r \in R\}
\]
is $M$.
One easily verifies that the map $r \mapsto ra$ is an $R$-module
homomorphism of ${}_R R$ onto $Ra = M$. The kernel of this map is
called the \defn{annihilator} of $a$ and is denoted $\op{ann} a$;
thus $\op{ann} a := \{r \in R : ra = 0\}$ and so
\[
Ra \iso R/\op{ann} a
\]
The above homomorphism also shows that
${}_R R$ is the free $R$-module on the single generator~$1$.
In fact free $R$-modules are particularly simple to describe: they
are the direct sums of copies of ${}_R R$, as you will get to show
in the exercises.
\begin{exers}~
\begin{exercises}
\prob If $X$ and $Y$ are sets of the same cardinality (that is,
there is a one-to-one function from $X$ onto $Y$) then any free
$R$-module over $X$ is isomorphic to any free $R$-module over $Y$.
\prob
Let $R^n$ be $n$-tuples of elements of $R$, as usual. This is an
$R$-module in the obvious way:
$r(x_1,\ldots, x_n) = (r x_1,\ldots, r x_n)$ (dah).
This module is usually denoted ${}_R R^n$. Let
$e_i = (0,\ldots,0,1,0,\ldots 0)$, $1$ in the $i^{\text{th}}$
position and $X = \{e_1,\ldots,e_n\}$. Then ${}_R R^n$ is a free
$R$-module over $X$. In particular, every vector space over a
field $F$ is a free $F$-module. The free abelian group on $n$
generators is isomorphic to $\mathbb{Z}^n$.
\prob
Now consider ${}_R R^\omega$. Let $e_i =
(0,\ldots,0,1,0,\ldots)$,
$1$ in the $i^{\text{th}}$ position and $X = \{e_1,e_2, \ldots\}$.
The submodule generated by $X$ consists of all $\omega$-tuples
which are $0$ in all but finitely many coordinates (this is called
the \defn{direct sum}). Show this
submodule is free over~$X$.
\prob
Let the $e_1, e_2, \ldots,e_n$'s be the elements of ${}_R R^n$
given in the exercises above.
Show $\sum_{i=1}^n x_i e_i = \sum_{i=1}^n y_i e_i$
implies $x_i = y_i$ for all~$i$. Generators of a module that
have these properties are called a \defn{base}.
Show that if $M$ has a basis with $n$ elements, then
$M \iso {}_R R^n$ and hence is a free $R$-module.
\prob
Suppose $M_i$, $i = 1, \ldots, n$ are submodules of a
module $M$ such that the
submodule genereated by the $M_i$'s is all of $M$, that is,
$M = M_1 + \cdots + M_n$. Also assume
\[
M_i \cap (M_1 + \cdots + M_{i-1} + M_{i+1} + \cdots + M_n) = 0
\]
for $i = 1, \ldots, n$. Show that $M$ is isomorphic to the direct
product (or direct sum, they're the same for finite products) of
the $M_i$'s.
\end{exercises}
\end{exers}
Every vector space over a field $F$ has a
basis and so is a free $F$-module. But in general not all modules
are free. Also there are rings $R$ such that
${}_R R^n \iso {}_R R^m$ for distinct $n$ and $m$; see exercise 27
of section 10.3 of Dummit and Foote~\cite{DummitFoote:2004}.
But for commutative rings this is not possible:
\begin{theorem}
If $R$ is commutative, then ${}_R R^n \iso {}_R R^m$ implies
$n = m$.
\end{theorem}
\begin{proof}
See Chapter 3 of Jacobson~\cite{Jacobson:1985} for the proof of
this and the next result.
\end{proof}
\begin{corollary}
If $R$ is commutative and $M$ is a module with bases
$e_1,\ldots,e_n$ and
$f_1,\ldots,f_m$ (so $M$ is free by the exercises), then
$n = m$. Moreover if
\[
f_k = \sum_{i=1}^n a_{ki} e_i
\]
then $A = (a_{ij}) \in \op{GL}_n(R)$. Coversely if $e_1,\ldots,e_n$
is a basis and the $f_k$'s are defined as above for
$A \in \op{GL}_n(R)$, then $f_1,\ldots,f_n$ is a basis.
\end{corollary}
\subsection{Finitely Generated Modules over a PID}
\begin{theorem} [Dedekind]\label{thm:dedekind}
Let $R$ be a PID. Every submodule of a free $R$-module is free. If
$M$ is free of rank~$n$, then every submodule is free of rank at
most~$n$.
\end{theorem}
\begin{proof}
Jacobson~\cite{Jacobson:1985} gives an elementary, computational
proof of this in the finite rank case.
Hungerford~\cite{Hungerford:1974} proves the full
theorem.
Here is Jacobson's proof: let $M$ be a free $R$-module with basis
$e_1,\ldots,e_n$ and let $K$ be a submodule of $M$. We induct
on~$n$. The case $n=0$ is clear. Let $N \le M$ be the submodule
generated by $e_2,\ldots,e_n$. Of course $N$ is free of rank
$n-1$. If $K \le N$ induction applies. So we assume $K \not\le N$.
Let
\[
I = \{b \in R : be_1 + y \in K, \text{ for some $y\in N$}\}
\]
$I$ is ideal as one easily checks. It is nonzero since $K \not\leq N$.
Let $I = (d)$. $d\in I$ implies there is a $y_1\in N$ such that
$f_1 = de_1 + y_1 \in K$.
Let $L = K \cap N$. By induction $L$ is free of rank at most $n-1$.
So let $f_2, \ldots, f_m$ be a basis with $m-1 \le n-1$ (so $m \le
n$). We will show that $f_1, \ldots,f_m$ is a basis of $K$ which
will prove the theorem. Let $x \in K$. Then expressing $x$ in terms
of $e_1,\ldots,e_n$ we see that
$x = be_1 + y$ for some $b\in I = (d)$ and $y\in N$. So $b = c_1d$
and so
\[
x - c_1f_1 = c_1d e_1 + y - c_1(de_1 + y_1) = y - c_1y_1 \in L =
K\cap N
\]
Hence $x-c_1f_1 = \sum_2^m c_jf_j$ and so $x = \sum_1^m c_jf_j$.
This shows the $f_j$'s span $K$.
Now suppose $\sum_1^m c_jf_j = 0$. Then $c_1de_1 + c_1y_1 +
\sum_2^m c_jf_j = 0$.
Since $y_1$ and $f_j$, $j = 2,\ldots,m$ are in $N$ which is spanned
by $e_2,\ldots,e_n$, we get
\[
c_1de_1 + \sum_2^n b_ke_k = 0
\]
for some $b_k$'s in $R$. Hence $c_1d = 0$ and, since $d\ne 0$,
$c_1 = 0$. So $\sum_2^m c_jf_j = 0$. Since $f_2,\ldots,f_m$ is a
basis of $L$, we have that $c_j = 0$ for every $j$, proving
that $f_1, \ldots,f_m$ is a basis of~$K$.
\end{proof}
\begin{exers}~
\begin{exercises}\label{exer:submod}
\prob
Show that if $A$ is an $n$-generated module over a PID
and $B$ is a submodule of $A$, then $B$ can be generated by a set
with at most $n$ elements.
\end{exercises}
\end{exers}
Let $R$ be a PID and let $A$ and $B \in \mats_{m,n}(R)$. $A$ and
$B$ are \defn{equivalent} if there are invertible matrices $P \in
\mats_m(R)$ and $Q \in \mats_n(R)$ such that $B = PAQ$.
\begin{theorem}\label{smithnormalform}
If $A \in \mats_{m,n}(R)$, $R$ a PID, the $A$ is equvalent to a
diagonal matrix
\[
\op{diag} (d_1, d_2, \ldots, d_r, 0, \ldots, 0).
\]
where $d_i \ne 0$ and $d_i \mid d_j$ if $i \le j$.
The $d_i$'s are unique up to associates.
\end{theorem}
This diagonal matrix is called the \defn{Smith normal form} of $A$.
The $d_i$'s are called the \defn{invariant factors} of $A$. In
class we will
follow the proof given in Jacobson~\cite{Jacobson:1985}.
The uniqueness in the above theorem follows from the relation
between the invariant factors and the determinental divisors, which
we now define.
For $A \in \mats_{m,n}(R)$ we define the \defn{rank} of $A$ to be
the greatest $r$ such that $A$ has an $r \times r$ submatrix with
nonzero determinant. $\Delta_i$, the $i^{\text{th}}$
\defn{determinental divisor} of $A$ is the gcd of the determinants
of the $i \times i$ submatrices of $A$, $i = 1, \ldots, r$. The
uniqueness is shown by proving the following.
\begin{equation}\label{eq:Deltas}
\Delta_i = d_1 \cdots d_i, \quad i = 1, \ldots, r.
\end{equation}
Defining $\Delta_0 = 1$, this can be writen as
\begin{equation}\label{eq:dees}
d_i = \Delta_i/\Delta_{i-1}, \quad i = 1, \ldots, r,
\end{equation}
Note the Laplace expansion theorem implies $\Delta_{i-1} \mid
\Delta_i$.
\begin{theorem} [Cauchy-Binet]
Let $R$ be a commutative ring and let
$A \in \mats_{m,n}(R)$ and $B\in \mats_{n,k}(R)$,
$\alpha = (\alpha_1,\ldots,\alpha_r)$ be a sequence with
$1 \le \alpha_1 < \cdots < \alpha_r \le m$,
$\beta = (\beta_1,\ldots,\beta_r)$ be a sequence with
$1 \le \beta_1 < \cdots < \beta_r \le k$. For any matrix
$C\in\mats_{n,k}(R)$, let $C[\alpha,\beta]$ be the $r \times r$
submatrix of $C$ lying at the intersection of the $\alpha$ rows and
$\beta$ columns. Then
\[
\det AB[\alpha,\beta]
= \sum_\gamma \det A[\alpha,\gamma] \det B[\gamma,\beta,]
\]
where the sum is over $\gamma = (\gamma_1,\ldots,\gamma_r)$ with
$1 \le \gamma_1 < \cdots < \gamma_r \le n$.
\end{theorem}
\begin{corollary}
Up to associates, equivalent matrices have the same determinental
divisors.
\end{corollary}
\begin{proof}
If $d$ is the $r^{\text{th}}$ determinental divisor of $A$, the
Cauchy-Binet Theorem shows that $d$ divides all determinates of
$r \times r$ submatrices $PAQ$. Hence $d$ divides the
$r^{\text{th}}$ determinental divisor of $PAQ$. Since equivalence is
a symmetric relation, this shows equivalent matrices have the same
determinental divisors, up to associates.
\end{proof}
Note that \eqref{eq:Deltas} and \eqref{eq:dees} follow from this
corollary and from Theorem~\ref{smithnormalform}. And of course the
uniqueness of the invariant factors (the $d_i$'s), and that
equivalent matrices have the same invariant factors, up to
associates, follow from \eqref{eq:dees}. So we only need to prove
the first part of
Theorem~\ref{smithnormalform}, which we will do in class.
As an application of Smith normal form consider a system of linear
equations over $\mathbb Z$:
\begin{align*}
a_{11}x_1 + \cdots &+ a_{1n}x_n = b_1\\
&\;\;\vdots\\
a_{m1}x_1 + \cdots &+ a_{mn}x_n = b_m
\end{align*}
In matrix form $A\mathbf x = \mathbf b$. Let $D = PAQ$ be the Smith
normal form of $A$ and $\mathbf y = Q^{-1}\mathbf x$ and $\mathbf c
= P\mathbf b$.
Then, if $A\mathbf x = \mathbf b$,
\[
D\mathbf y = PAQ\mathbf y = PAQQ^{-1}\mathbf x = PA\mathbf x
= P\mathbf b = \mathbf c.
\]
Hence $A\mathbf x = \mathbf b$ holds if and only if
$D\mathbf y = \mathbf c$. So, if
$D = \op{diag} (d_1, d_2, \ldots, d_r, 0, \ldots, 0)$,
$D\mathbf y = \mathbf c$ has a solution if and only if
$d_i \mid c_i$, $i = 1,\ldots,r$ and $c_i = 0 $ for $i > r$.
If these conditions hold, the set of all solutions is
\[
\{(c_1/d_1,\ldots,c_r/d_r, k_{r+1},\ldots,k_n) :
k_j \in \mathbb Z\}.
\]
Multiplying these solutions by $Q$ we get all the solutions of
$A\mathbf x = \mathbf b$.
\medskip
Since $R$ is a PID each of the invariant factors can be written as
\[
d_i = p_1^{e_{i1}} \cdots p_m^{e_{im}}
\]
where $p_1, \ldots, p_m$ are distinct primes in $R$. Since $d_i
\mid
d_{i+1}$, $e_{ik} \le e_{i+1,k}$. A prime power $p_k^{e_{ik}}$ with
$e_{ik} > 0$ is called an \defn{elementary divisor} of $A$. The
\defn{list of elementary divisors} is the list of all of them
including repeats. For example, suppose $R = \mathbb Z$ and
$d_1 = 2^3\cdot 3$,
$d_2 = 2^4\cdot 3^3\cdot 5^2\cdot 7$, and
$d_3 = 2^4\cdot 3^3\cdot 5^5\cdot 7^2$. Then the list of elementary
divisors is
\[
2^3,2^4,2^4,3,3^3,3^3,5^2,5^5,7,7^2.
\]
The invariant factors can be recovered from the list of elementary
divisors and the rank of $A$. So if the rank of $B$ is four and it
has the above list of elementary divisors, then there will be 4
invariant factors. $d_4$ must be the product of all the highest
prime powers: $d_4 = 2^4\cdot 3^3\cdot 5^5\cdot 7^2$. $d_3$ is the
product of the highest remaining prime powers:
$d_3 = 2^4\cdot 3^3\cdot 5^2\cdot 7$.
And $d_2 = 2^3\cdot 3$. $d_1$ is the product of the rest. But
there aren't any so it is the product of the empty set:
$d_1 = 1$.
\begin{exercise}~
\begin{exercises}
\prob
\part
Let
\[
A = \begin{bmatrix}
2 & 1 & -3 & -1\\
1&-1&-3&1\\
4&-4&0&16
\end{bmatrix}
\]
Find the Smith Normal Form $D$ of $A$ and find
$P\in \op{GL}_3(\mathbb Z)$ and
$Q\in \op{GL}_4(\mathbb Z)$ such that
$D = PAQ$. \textbf{Hint:} Start with the threee matrices
$I_3$, $A$, and $I_4$. When you do a row operation to $A$ do the
same row operation to $I_3$.
When you do a column operation to $A$ do the same column
operation to $I_4$. $A$ will be converted to $D$ and at the same
time $I_3$ will be converted to $P$ and $I_4$ to $Q$.
\part
Using the method described above find the general solution to
$A\mathbf x = \mathbf b$, where
\[
\mathbf x = \begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}
\qquad\text{ and }\qquad
\mathbf b = \begin{bmatrix}
10\\
2\\
20
\end{bmatrix}
\]
\prob
Let $R$ be a Euclidean domain and let
$\op{SL}_n(R)$ be all $n \times n$ matrices over $R$ with
determinant~1. In this exercise you will show
that the group $\op{SL}_n(R)$ is generated by
the elementary matrices of the first kind (those of the form
$I_n + b E_{i,j}$, $i \ne j$).
First note that in putting a matrix into
Smith Normal Form the elementary matrices of the form
$D_i(u)$ (the third kind of elemenatry matrix) were
never used.
The elementary matrix corresponding
to interchanging two rows (or columns) has determinant~$-1$ but if
we multiply one of the rows by~$-1$, then the determinant is~$1$
and this elementary operation (interchanging two rows and
multiplying one of them by~$-1$) can be used (in the proof of the
Smith Normal Form) instead of just interchanging the rows. You can
take this as given---you do not need to write up proofs of them.
\part
Let $u$ be a unit is $R$. Show that
\[
\begin{bmatrix}
0&u\\
-u^{-1}&0
\end{bmatrix}
\]
can be written as a product of three elementary matrices of the
first kind, each of which has either $u$ or $-u^{-1}$ as its single
off-diagonal entry. Note the modified permutation matrix described
above is this matrix with $u = 1$.
\part
Show that if $u$ is a unit, then
\[
\begin{bmatrix}
u&0\\
0&u^{-1}
\end{bmatrix}
\]
is a product of two matrices from part~\textbf{a}.
\part
Let $A \in \op{SL}_n(R)$.
Using part~\textbf{a} and the comments above we see that $A$ can be
put into Smith Normal Form using elementary matrices of the first
kind. Now use the matrices of part~\textbf{b} to modify the Smith
Normal Form into the identity matrix and show this implies that $A$
is a product of elementary matrices of the first kind.
\end{exercises}
\end{exercise}
Let $M$ be an $R$-module, $N$ a submodule of $M$ and $a \in M$. The
\defn{annihilator} of~$a$ is $\op{ann} a = \op{ann}_R a
= \{ r\in R: ra = 0\}$; $\op{ann}_R N = \{r \in R : rN= 0\}$.
The \defn{torsion submodule} of $M$ is $\op{tor} M = \{m \in M :
\op{ann} a \ne 0\}$. The reader can show $\op{ann} a$ is a left ideal of $R$,
$\op{ann} N$ is a two-sided ideal of $R$ and $\op{tor} M$ is a submodule.
Also show that $R/\op{ann} a \iso Ra$.
$M$ is a \defn{torsion module}\indexit{module!torsion}
if $\op{tor} M = M$ and $M$ is
\defn{torsion free}[module!torsion free]\indexit{torsion free
module} if $\op{tor} M = 0$.
When $R$ is a PID, $\op{ann} a = (r)$, for some $r$ and we say
$a$ has \defn{order}~$r$. Of course any associate of~$r$ is also
``the" order of ~$a$. Note that in an abelian group, thought of as
a $\mathbb Z$-module, order agrees with the usual nottion except an
element of order~$n$ also has order~$-n$.
\begin{theorem}\label{torfreeisfree}
A finitely generated torsion free module over a PID is free.
\end{theorem}
We will prove this in class following
Hungerford~\cite{Hungerford:1974}.
\begin{theorem}
If $M$ is a finitely generated module over a PID $R$, then
$M = \op{tor} M \oplus F$, where $F$ is a finitely generated
free $R$-module.
\end{theorem}
\begin{proof}
Let $T = \op{tor} M$. Then $M/T$ is torsion free: if
$a + T \in M/T$ and $r(a+T) = T$ then $ra \in T$. But this implies
$sra = 0$ for some $s \ne 0$ in $R$. But then $a \in \op{tor} M = T$, and
so $a+T = T$. Of course $M/T$ is finitely generated since $M$ is.
By Theorem~\ref{torfreeisfree}, $M/T$ is free. Let $x_1, \ldots,
x_k$ be a free generating set, that is, a basis for $M/T$ and let
$\varphi : M \epi M/T$ be the natural homomorphism; that is
$a \mapsto a+T$.
Choose $y_i \in M$ with $\varphi(y_i) = x_i$.
Let $F$ be the submodule of $M$ gnerated by the $y_i$'s.
Let $\psi : M/T \to M$ be the homomorphism that extends
$\psi(x_i) = y_i$. Note $\varphi(\psi (x_i)) = \varphi(y_i) = x_i$
and so $\varphi \comp \psi$ is the identity homomorphism on $M/T$.
So $F \iso M/T$ and thus is free.
If $m \in M$ let $a = \psi(\varphi(m)) \in F$. Since
$\varphi \comp \psi$ is the identity,
$\varphi \comp \psi \comp \varphi = \varphi$.
Hence
\[
\varphi(m-a) = \varphi(m) - \varphi(a)
= \varphi\psi\varphi(m) - \varphi(a)
= \varphi(a) - \varphi(a) = 0
\]
So
$\varphi(m-a) = 0$, and so $m-a \in T$. Thus
\[
m = a + (m-a) \in F + T
\]
It is easy to verify that $F \cap T = 0$. Thus $M = F \oplus T$, as
desired.
\end{proof}
Let $M$ be a torsion module over a PID $R$. For $p$ a prime in $R$
let
\[
M(p) = \{a \in M : p^i a = 0, \text{ for some~$i$}\}.
\]
$M(p)$ is called the \defn{$p$-primary} component of~$M$.
\begin{theorem}\label{thm:primary}
Let $M$ be a torsion module over a PID $R$. Then
$M(p)$ is a submodule of $M$ and there is a set $S$ of primes in
$R$ such that
$M = \bigoplus_{p \in S} M(p)$.
If $M$ is finitely generated and $M(p) \ne 0$ for $p \in S$ then
$S$ is finite.
\end{theorem}
\begin{proof}
That $M(p)$ is a submodule is left as an exercise.
We can take $S$ to be an SDR (system of distict representatives) of
the set of all primes in $R$ under the equivalence relation
``$p$ is an associate of $q$".
Let $a \in M$ and let $r$ be the order of~$a$ so that
$\op{ann} a = (r)$. Since $R$ is s UFD, we can write
$r = u p_1^{n_1} \cdots p_k^{n_k}$ for distinct primes in $S$,
$n_i > 0$ and $u$ a unit.
Let $r_i = r/p_i^{n_i}$. Note $r_1, \ldots, r_k$ are relatively
prime. Since $R$ is a PID, this implies
$1 = s_1 r_1 + \cdots + s_kr_k$. Thus
\[
a = s_1 r_1 a + \cdots + s_kr_k a.
\]
Since $p_i^{n_i} s_i r_i a = s_i r a = 0$, $s_i r_i a \in M(p)$.
Hence $a \in \sum_{p\in S} M(p)$.
Suppose $a \in M(p) \cap \sum_{q\ne p} M(q)$. Then $p^m a = 0$ for
some $m \ge 0$ and $a = a_1 + \cdots + a_t$, where $a_i \in
M(p_i)$, $p_i$ primes distinct from each other and from~$p$. Let
$p_i^{m_i} a_i = 0$ and let $d = p_1^{m_1} \cdots p_t^{m_t}$
and note $da = 0$. Now $d$ and $p^m$ are relatively prime so there
exists $r$, $s\in R$ with $1 = rp^m + sd$. But then
$a = rp^ma + sda = 0$, showing $M(p) \cap \sum_{q\ne p} M(q) = 0$.
\end{proof}
\begin{theorem}\label{modulePIDthm}
Let $M$ be a finitely generated module over a PID $R$.
\begin{enumerate}
\setlength{\itemsep}{0.0ex plus0.1ex}
\item
$M$ is isomorphic to the direct sum of finitely many cyclic
modules:
\begin{align*}
M &= R^k \oplus Ra_1 \oplus \cdots \oplus Ra_m\\
&\iso R^k \oplus R/(d_1) \oplus \cdots \oplus R/(d_m)\\
&= R^k \oplus R/\op{ann} a_1 \oplus \cdots \oplus R/\op{ann} a_m
\end{align*}
where the $d_i$'s are nonzero, nonunit elements of $R$ and
\[
d_1 \mid d_2 \mid \cdots \mid d_m.
\]
\item
$M$ is isomorphic to the direct sum of finitely many primary cyclic
modules:
\[
M \iso R^k \oplus R/(p_1^{n_1}) \oplus \cdots \oplus R/(p_t^{n_t})\\
\]
where $p_1, \dots p_t$ are (not necessarily distinct) primes and
$n_i \ge 1$.
\end{enumerate}
Moreover, the decompositions in both parts are unique.
\end{theorem}
\begin{proof}
Let $x_1,\ldots,x_n$ generate $M$ and let $e_1,\ldots,e_n$ be the
natural basis of $R^n$. Let $\eta : R^n \to M$ be the homomorphism
defined by $\eta(e_i) = x_i$ and $K$ be the kernel of~$\eta$.
By Theorem~\ref{thm:dedekind} $K$ is free of rank $m$
and $m\le n$. Let $f_1,\ldots,f_m$ be a basis of $K$. Let
\[
f_\ell = \sum_{t=1}^n a_{\ell t}e_t
\]
and let $A = (a_{\ell t})$. Let
$D = PAQ^{-1}$ be the Smith normal form of $A$ and let
\[
e_i' = \sum_j q_{ij} e_j.
\]
Let $f_k' = \sum_\ell p_{k\ell} f_\ell$.
Writing $D = \op{diag} (d_1, d_2, \ldots, d_r, 0, \ldots, 0)$,
with $d_i \mid d_j$ for $i\le j$, and using $PA = DQ$ we calculate
\begin{align*}
f_k' &= \sum_{\ell=1}^m p_{k\ell} f_\ell\\
&= \sum_{\ell = 1}^m p_{k\ell} \sum_{i=1}^n a_{\ell i} e_i\\
&= \sum_{i = 1}^n \left(\sum_{\ell = 1}^m p_{k\ell} a_{\ell i}\right) e_i\\
&= \sum_{i = 1}^n (PA)_{ki} e_i\\
&= \sum_{i = 1}^n (DQ)_{ki} e_i\\
&= \sum_{i = 1}^n \sum_{t=1}^n d_{kt} q_{ti} e_i\\
&= \sum_{t = 1}^n d_{kt}\sum_{i=1}^n q_{ti} e_i\\
&= \sum_{t = 1}^n d_{kt} e_t'
= d_k e_k'
\end{align*}
So $f_k' = d_k e_k'$, $k = 1,\ldots,r$ and $f_k' = 0$, $k =
r+1,\dots,m$. Put $y_i = \sum_{j=1}^n q_{ij} x_j$. Then, since
$Q \in \op{GL}_n(R)$, $y_1,\ldots, y_n$ generate $M$ and
$\eta(e_i') = y_i$.
Now since $d_i e_i' = f_i' \in K$ we have $d_i y_i = 0$.
Suppose $\sum_{i=1}^n b_i y_i = 0$, $b_i \in R$. Then $\sum b_i e_i'
\in K$ which implies
\[
\sum b_ie_i' = \sum c_i f_i' = \sum c_i d_i e_i'.
\]
This gives $b_i=c_i d_i$ which implies
$b_i y_i = c_i d_i y_i = 0$.
These calculations prove $M = Ry_1 \oplus \cdots \oplus Ry_n$ and
$\op{ann}(y_i) = (d_i)$.
Since $f_k' = d_k e_k'$, if $d_k = 1$ then
$0 = \eta(f_k') = \eta(e_k') = y_k$. So if we assume $n$ is the
smallest size of a generating set of $M$, this, namely $d_k = 1$,
cannot happen. Of course $k_k =1$ gives a $0$ summand, which we
don't want.
This proves the first part.
The second part can be derived from the first using
Theorem~\ref{thm:primary} and Exercise~\ref{exer:submod}.
\end{proof}
The $k$ in this theorem is called the \defn{free rank} of $M$. The
$d_i$'s are called the \defn{invariant factors} of $M$ and the
$p_i^{n_i}$'s are the \defn{elementary divisors} of $M$. These
determine $M$ up to isomorphism:
\begin{theorem}
Let $M_1$ and $M_2$ be modules over a PID $R$.
\begin{enumerate}
\setlength{\itemsep}{0.0ex plus0.1ex}
\item
$M_1 \iso M_2$ iff they have the same free rank and invariant
factors (up to associates).
\item
$M_1 \iso M_2$ iff they have the same free rank and elementary
divisors (up to associates).
\end{enumerate}
\end{theorem}
In the case when $R=\mathbb Z$ Theorem~\ref{modulePIDthm}
specializes to fundamental theorem of abelian groups which says
every finitely generated abelian group is isomorphic to the direct
product of cyclic groups. The reader should write out the exact
formulation.
\subsubsection*{Linear Algebra}
Let $V$ be a vector space of dimension $n$ over a field $F$
and let $T : V \to V$ be a linear transformation of $V$.
$\gl \in F$ is an \emph{eigenvalue} of $T$ if there is a
$v \in V$, $v\ne 0$ with $Tv = \gl v$.
In this case $v$ is called an \emph{eigenvector}. Note eigenvectors
can never be $0$ but eigenvalues can. The \emph{eigenspace}
associated with the eigenvalue $\gl$ is
\[
\{v \in V : Tv = \gl v\},
\]
which is a subspace of $V$.
If $A \in \op{M}_{n,n}$ it can be viewed as a linear transformation
on the vector space $F^n$ so these concepts make sense for~$A$.
\begin{exers}~
\begin{exercises}
\prob
Show the following are equivalent for $\gl \in F$:
\begin{enumerate}
\item $\gl$ is a eigenvalue of $A$.
\item $\gl I - A$ is singular.
\item $\det(\gl I - A) = 0$.
\end{enumerate}
\end{exercises}
\end{exers}~
Viewing $\gl$ as an indeterminant, $c_A(\gl) = \det(\gl I - A) \in
F[\gl]$ is
a monic polynomial. It is known as the \emph{characteristic
polynomial} of $A$. The \emph{minimum polynomial} of $A$ is the
monic polynomial $m_A(\gl)$ of least degree such that
$m_A(A) = 0$.
Using the division algorithm one sees that if $f(A) = 0$, for some
$f(\gl) \in F[\gl]$, then $m_A(\gl) \mid f(\gl)$.
If $T$ is a linear transformation on $V$ and $u_1,\ldots,u_n$ is a
basis for $V$, then let $a_{ij}$ be defined by
\begin{equation}\label{eq:T}
T u_i = \sum_{j=1}^n a_{ij} u_j.
\end{equation}
In order to study $T$ (or $A$) we make the
vector space $V$ it acts on into a $F[\gl]$ module ${}_{F[\gl]}V$
by defining
\[
g(\gl) v = g(T) v
\]
for $g(\gl) \in F[\gl]$.
${}_{F[\gl]}V$ is a finitely generated module so we can decompose
it into cyclic modules as in Theorem~\ref{modulePIDthm}.
Looking at the proof of that theorem we first need a
generating set for the module. Clearly $u_1,\ldots,u_n$
is a generating set for $V$ as a
$F[\gl]$ module. Let $F[\gl]^n$ be the free $F[\gl]$ module with
basis $e_1,\ldots,e_n$. Let $\eta : F[\gl]^n \to V$ be given by
$\eta(e_i) = u_i$ and let $K = \ker \eta$. We need a basis for $K$.
\begin{lemma}
The elements
\[
f_i = \gl e_i - \sum_{j=1}^n a_{ij} e_j,\quad i = 1,\ldots,n
\]
form a basis of $K$.
\end{lemma}
\begin{proof}
$\eta(f_i) = \gl u_i - \sum a_{ij} u_j = 0$ by \eqref{eq:T} so
$f_i \in K$. Note
\[
\gl e_i = f_i + \sum_j a_{ij}e_j.
\]
Multiplying this by $\gl$ gives
\begin{align*}
\gl^2 e_i &= \gl f_i + \sum_j a_{ij}\gl e_j \\
&= \gl f_i + \sum_j a_{ij} (f_j + \sum_k a_{jk} e_k) \\
&= \gl f_i + \sum_j a_{ij} f_j + \sum_j\sum_k a_{jk} e_k) \\
&= \sum_j h_j(\gl) f_j + \sum_k b_k e_k,
\end{align*}
where $b_k \in F$.
What calculations like this show is that any element
$\sum_i g_i(\gl)e_i$ of ${}_{F[\gl]}V$ can be written in the form
$\sum_i h_i(\gl) f_i + \sum_i b_i e_i$ with $b_i\in F$.
If $\sum_i g_i(\gl)e_i \in K$ then applying $\eta$ gives that
$\sum_i b_i u_i = 0$. But the $u_i$'s are a basis so each $b_i = 0$
and so each element of $K$ can be expressed as $\sum_i h_i(\gl) f_i$.
Thus the $f_i$'s generate~$K$.
Now suppose that $\sum_i h_i(\gl) f_i = 0$. Then using the
definition of $f_i$ we have
\[
\sum_i h_i(\gl)\gl e_i = \sum_{i,j} h_i(\gl)a_{ij} e_j
= \sum_{i,j} h_j(\gl)a_{ji} e_i
\]
where we have simply interchanged the dummy variables $i$ and $j$
in the last sum. Since the $e_i$'s are a basis we have
\[
h_i(\gl)\gl = \sum_j h_j(\gl)a_{ji}.
\]
Assuming not all the $h_i(\gl)$ are 0, choose one of maximal degree
(and call it $h_i(\gl)$). By the above equation cannot hold because
the degree of the left side is strictly greater than the degree of
the right.
This contradictions shows $h_i(\gl) = 0$ for each~$i$ and hence
$f_1(\gl),\ldots,f_n(\gl)$ form a basis of~$K$.
\end{proof}
So we now have a basis of $K$ the next part of the proof found the
matrix relating the $f_i$'s and $e_i$'s and put that into Smith
normal form. Using the definition of $f_i$ it is easy to see the
matrix is $\gl I - A$. Let
$D = P(\gl I - A)Q^{-1}$ be the Smith normal
form and write $D =
\op{diag}(1,\ldots,1,d_1(\gl),\ldots,d_s(\gl))$.
We assume each nonzero $d_i(\gl)$ is monic.
Taking determinant we see
\[
d_1(\gl)d_2(\gl)\cdots d_s(\gl) = \det D
= \det(P) \det(\gl I - A) \det(Q^{-1})
= c_A(\gl)
\]
because $\det(P)$ and $\det(Q^{-1})$ are units (so in $F$) but all
the polynomials are monic so ${\det(P)\det(Q^{-1}) = 1}$.
This equation implies none of the $d_i(\gl)$'s is $0$. (This also
is equivalent to the fact the $V$ is a torsion $F[\gl]$ module,
which can be shown directly.)
By Theorem~\ref{modulePIDthm}
\[
{}_{F[\gl]}V \iso F[\gl]/(d_1(\gl)) \oplus \cdots
\oplus F[\gl]/(d_s(\gl)).
\]
Of course $d_s(\gl)$ generates the annihilator of
$F[\gl]/(d_s(\gl))$.
But since $d_1(\gl) \mid \cdots\mid d_s(\gl)$ everything in ${}_{F[\gl]}V$ is
annihilated by $d_s(\gl)$. This means $d_s(A)$ kills every vector
of $V$ and this implies $d_s(A) = 0$. From this we conclude that
the minimum polynomial of $A$ is $m_A(\gl) = d_s(\gl)$.
We say that $d_1(\gl), \ldots, d_s(\gl)$ are the invariant factors
of $A$ (even though they are really the invariant factors of
$\gl I - A$.
The next theorem gather together some easy but import consequences
of our theory. The proof is left to the reader.
\begin{theorem}\label{thm:minchar}
Let $A \in \op M_{n,n}(F)$ (or let $A$ be a linear transformation
of an $n$-dimensional vector space) with invariant factors
$d_1(\gl), \ldots, d_s(\gl)$.
\begin{enumerate}
\item $d_s(\gl) = m_A(\gl)$ is the minimum polynomial of $A$.
\item (Cayley-Hamilton) $A$ satisfies its characteristic
polynomial $c_A(\gl)$.
\item The eigenvalues of $A$ are precisely the roots of $m_a(\gl)$.
\item $m_A(\gl) \mid c_A(\gl)$ and $c_A(\gl) \mid m_A(\gl)^k$ for
some~$k$ (in fact $k = s$ works).
\end{enumerate}
\end{theorem}
\subsubsection*{The Campanion Matrix and Rational Canonical Form}
Let $f(x) = x^k + b_{k-1} x^{k-1} + \cdots + b_1x + b_0$ be a monic
polynomial in $F[x]$. The
\emph{companion matrix}\index{matrix!companion}
of $f(x)$ is
\[
\begin{bmatrix}
0&0&0&\cdots&0&-b_0\\
1&0&0&\cdots&0&-b_1\\
0&1&0&\cdots&0&-b_2\\
0&0&1&\cdots&0&-b_3\\
&&&\ddots&&\\
%0&0&0&\cdots&1&0&-b_{k-2}\\
0&0&0&\cdots&1&-b_{k-1}\\
\end{bmatrix}
\]
\begin{exers}~
\begin{exercises}
\prob
If $C$ is the companion matrix of $f(x)$ show that the Smith normal
form of $\gl I - C$ is $\op{diag}(1,1,\ldots,1,f(\gl))$
so the minimum polynomial and the characteristic polynomial of $C$
are both $f(\gl)$ and this is the only invariant factor.
There are two ways to do this problem. One is to use the method
presented in class (the method in Jacobson). The second is to
note
that if $e_i$, $i=1,\ldots,k$, are the unit vectors the
$Ce_i = e_{i+1}$, $i=1,\ldots,k-1$ and
$Ce_k = -b_0e_1 - b_1 e_2 - \cdots - b_{k-1} e_k$
and use this to show that $f(C)e_i = 0$ so that $f(C) = 0$
and that $C$ satisfies no polynomial of lower degree. Now use
Theorem~\ref{thm:minchar} to get the result.
\end{exercises}
\end{exers}~
Returning to our vector space $V$ and linear transformation $T$
and let $d_1(\gl), \ldots, d_s(\gl)$ be the
invariant factors of $T$.
We have by Theorem~\ref{modulePIDthm} that
\[
{}_{F[\gl]}V = F[\gl]/(d_1(\gl)) \oplus \cdots \oplus
F[\gl]/(d_s(\gl)).
\]
Suppose $d_i(\gl) = \gl^k + b_{k-1} \gl^{k-1} + \cdots + b_1\gl + b_0$.
Of course $F[\gl]/(d_i(\gl))$ is a $F[\gl]$ module and so it is an
$F$ module because $F$ is a subring of $F[\gl]$.
Let $V_i = F[\gl]/(d_i(\gl))$ be this subspace.
Note that $1$, $\gl$, $\gl^2, \ldots, \gl^{k-1}$ is a basis of
$V_i$ and the action of $\gl$ (and therefore $T$) sends
$1 \mapsto \gl$,
$\gl \mapsto \gl^2$,
$\gl^2 \mapsto \gl^3$, and
$\gl^{k-1} \mapsto \gl^k = -( b_{k-1} \gl^{k-1} + \cdots + b_1\gl +
b_0)$ since $(d_i(\gl)) = 0$ in $F[\gl]/(d_i(\gl))$.
If we let $T_i$ be the restriction of $T$ to $V_i$ we see that
under this basis the matrix of $T_i$ is companion matrix of
$d_i(\gl)$ displayed above. Since $V = V_1 \oplus \cdots \oplus V_s$
the union of the bases of the $V_i$'s is a basis for $V$ and the
matrix for $T$ relative to this basis is
\[
\begin{bmatrix}
C_1&&&\\
&C_2&&\\
&&\ddots&\\
&&&C_s
\end{bmatrix}
\]
where $C_i$ is the companion matrix of $T_i$. The matrix above is
know as the \emph{rational canonical form} or
\emph{Fr\"obenius normal form} of $T$.
Notice we have shown that the right choice of a basis for $V$, the
matrix for $T$ is its rational canonical form above.
Since a change of bases corresponds to similarity, we also have that
every matrix is similar to its rational canonical form.
\begin{theorem}\label{thm:similar}
Let $A$, $B\in \op M_n(F)$. Then the following are equivalent.
\begin{enumerate}
\item $A$ and $B$ are similar.
\item $\gl I - A$ and $\gl I - B$ are equivalent.
\item $\gl I - A$ and $\gl I - B$ have the same Smith normal form.
\item $A$ and $B$ have the same elementary divisors.
\item $A$ and $B$ have the same rational canonical form.
\end{enumerate}
\end{theorem}
\begin{exers}~
\begin{exercises}
\prob
Find the invariant factors and rational canonical form of
\[
A = \begin{bmatrix}
2&-2&14\\
0&3&-7\\
0&0&2
\end{bmatrix} \quad
B = \begin{bmatrix}
0&-4&85\\
1&4&-30\\
0&0&3
\end{bmatrix} \quad
C = \begin{bmatrix}
2&2&1\\
0&2&-1\\
0&0&3
\end{bmatrix} \quad
D=\begin{bmatrix}
1&2&-4&4\\
2&-1&4&-8\\
1&0&1&-2\\
0&1&-2&3
\end{bmatrix}
\]
\prob
Prove that two $3 \times 3$ matrices are similar if and only if
they have the same characteristic polynomial and same minimum
polynomial. Give an example showing this is false for $4\times 4$
matrices.
\prob
Suppose the characteristic polynomial of a linear transformation is
$x^2(x^2+1)^2$. Find all possibilities of the Smith normal form. (Of
course the Smith normal form of $A$ really means that of $\gl I - A$;
also it is enough to just make all possible lists of invariant
factors.)
\prob
Determine up to similarity all $A\in \op M_2(\mathbb Q)$ with $A^4 = I$.
\end{exercises}
\end{exers}~
\subsubsection*{Jordan Normal Form}
Assume that all the roots of the characteristic polynomial
$c_A(\gl)$ of $A$ (or of $T$) lie in $F$. Then $c_A(\gl)$ factors
into linear factors $\gl - r$. The elementary divisors are powers of
irreducible polynomials so have the form $(\gl - r)^e$. By part~2
of Theorem~\ref{modulePIDthm} can be decomposed as a direct sum of
submodules of the form $F[\gl]z \iso F[\gl]/\op{ann} z
\iso F[\gl]/((\gl - r)^e)$. This cyclic $F[\gl]$ module has an $F$
basis
\[
z, (\gl - r)z, (\gl - r)^2z, \ldots, (\gl - r)^{e-1}z
\]
We calculate
\begin{align*}
Tz &= \gl z = rz + (\gl -r)z\\
T(\gl - r) z &= \gl(\gl - r) z = r(\gl - r)z + (\gl - r)^2 z\\
&\,\,\;\vdots\\
T(\gl - r)^{e-2} z &= r(\gl - r)^{e-2}z + (\gl - r)^{e-1} z\\
T(\gl - r)^{e-1} z &= r(\gl - r)^{e-1}z
\end{align*}
So the matrix for $T$ using this basis is
\[
\begin{bmatrix}
r&0&0&\cdots&0&0\\
1&r&0&\cdots&0&0\\
0&1&r&\cdots&0&0\\
&&&\ddots\\
0&0&0&&r&0\\
0&0&0&\cdots&1&r\\
\end{bmatrix}
\]
This is called a \emph{Jordan block}. The
\emph{Jordan normal form} of $T$ is the matrix diagonal sum of
Jordan blocks for each elementary divisor $(\gl - r_i)^{e_i}$.
Many books write the Jordan form with the $1$'s in the diagonal
above the main diagonal rather than the diagonal below. The
following theorem justifies this.
\begin{theorem}
Let $A \in \op M_n(F)$. Then $A$ and $A^T$ are similar.
\end{theorem}
\begin{proof}
Since the determinant of a square matrix equals the determinate of
its tramspose, the $A$ and $A^T$ have the same deterimental
divisors and hence the same invariant factors. So by
Theorem~\ref{thm:similar} they are similar.
\end{proof}
\begin{exers}~
\begin{exercises}
\prob
Show that $A \in \op M_n(F)$ is similar to a diagonal matrix
(possibley in a field extension of $F$) if and only if
the minimum polynomial $m_A(\gl)$ has distinct roots.
\end{exercises}
\end{exers}~
\subsection{Tensor Products}
We will follow Dummit and Foote---they have a good explanation and
lots of examples. Here we will just repeat some of the important
definitions and results.
Let $M_R$ be a right $R$ module and ${}_R N$ be a left $R$ module.
Then $M \otimes N = M \otimes_R N$, the
\defn{tensor product} of $M$ and $N$, is an
abelian group (that is a $\mathbb Z$-module) obtained as follows.
First take the free $\mathbb Z$-module on the free generating set
$M \times N$. In this reguard, we are thinking of $(m,n)$ as a
formal symbol; in particular $(m,n) + (m',n') \ne (m + m',n+n')$.
Take the subgroup of this free abelian group generated by the
elements
\begin{align*}
&(m+m',n) - (m,n) - (m',n)\\
&(m,n+ n') - (m,n) - (m,n')\\
&(mr,n) - (m,rn)
\end{align*}
Then $M \otimes_R N$ is the quotient of this free group modulo
this subgroup. Let $m \otimes n$ be the coset of $(m,n)$. This is
called a \defn{simple tensor}[tensor!simple]. The elements of
$M \otimes_R N$ are finite sums of simple tensors and are called
tensors. Note tensors satisfy
\begin{align*}
(m + m') \otimes n &= m \otimes n + m' \otimes n\\
m \otimes (n + n') &= m \otimes n + m \otimes n'\\
mr \otimes n &= m \otimes rn\
\end{align*}
If $N$ and $M$ are as above and $L$ is an abelian group. A map
$\varphi : M \times N \to L$ is \defn{middle linear} with respect
to $R$ or $R$-\defn{balanced}[balanced map] if it is linear in each
of its argments (this is called \defn{mulitlinear}[multilinear
map]) and $\varphi(m,rn) = \varphi(mr,n)$. The natural map
$M \times N \to M \otimes_R N$ is balanced and universal for this
concept; that is, any balanced map can be factored through this.
\begin{theorem}\label{thm:balanced}
Let $M_R$ and ${}_R N$ be $R$ modules. Let $\iota : M \times N \to
M \otimes_R N$ be the natural map.
If $\varphi: M\times N \to L$ is an $R$-balanced map into an
abelian group $L$, then there is a group
homomorphism $\bar \varphi : M \otimes_R N \to L$ such that
the following diagram commutes:
\begin{center}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3em,column sep=3em]
{ M \times N & M \otimes_R N \\
& L \\ };
\path[->,font=\small]
(m-1-1) edge node[above] {$\iota$} (m-1-2)
edge node[below] {$\varphi$} (m-2-2);
\path[->, font=\small]
(m-1-2) edge node[right] {$ \bar\varphi $} (m-2-2);
\end{tikzpicture}
\end{center}
\end{theorem}
It is important to remember that $m \otimes n = m' \otimes n'$ (or
even $m \otimes n = m \otimes n'$) does not imply $m=m'$ or $n = n'$.
Every element of $M \otimes_R N$ is a sum of elements of the form
$m \otimes n$ so these elements generate $M \otimes_R N$, but they
are not a basis. This means that not every map from the set of
simple tensors into an abelian group $L$ can be extended to a
homomorphism. For example, if
\begin{equation}\label{tensor}
f: M_R \to M'_R \text{ and } g: {}_R N \to {}_R N'
\end{equation}
are homomorphisms, does the map
$m \otimes n \mapsto f(m) \otimes g(n)$ extend to a homomorphism
from $M \otimes_R N$ to $M' \otimes_R N'$? The answer in this case
is yes, but requires a proof, which we leave for the student. This
important homomorphism is denoted $f \otimes g$, so that
$(f \otimes g)(m \otimes n) = f(m) \otimes g(n)$.
If $R$ and $S$ are rings, then $M$ is a $(S,R)$-\defn{bimodule} is
$M$ is a left $S$-module and also a right $R$-module and, in
addition
\[
(sm)r = s(mr).
\]
This is denoted ${}_S M_R$.
If $M$ is an $(S,R)$-bimodule and $N$ is a left $R$-module, then
$M \otimes_R N$ is a left $S$ module in a natural way. Namely,
$s(m\otimes n) = sm \otimes n$.
We showed in class how Theorem~\ref{thm:balanced} can be used to
show this works.
\begin{theorem}
Suppose $f$ and $g$ are homomorphism as in \eqref{tensor}. Then
there is a homomorphism $f \otimes g : M \otimes_R N
\to M' \otimes_R N'$ such that $(f\otimes g) (m \otimes n) = f(m)
\otimes g(n)$. If ${}_S M_R$ and ${}_S M'_R$ are bimodules and $f$
is a bimodule homomorphism, then $f \otimes g$ is an $S$ module
homomorphism.
\end{theorem}
\begin{proof}
Use Theorem~\ref{thm:balanced}.
\end{proof}
Another use of Theorem~\ref{thm:balanced} is to show that tensor
products are associated; see Theorem~14 of Dummit and Foote. Also
tensor products distribute over direct sums. One consequence of
this is that if $R$ is a subring of $S$, then, viewing ${}_S S_R$
as a bimodule,
\[
S \otimes_R R^n \iso S^n
\]
(This uses that $S \otimes_R R \iso S$ under the map $s \otimes r
\mapsto sr$, which is another application of
Theorem~\ref{thm:balanced}.
When $R$ is commutative every left module is a right module, and
vice versa. In this case if $M_i$ are $R$-modules then
\[
M_1 \otimes_R \cdots \otimes _R M_k
\]
makes sense and is an $R$-module.
The fact that the tensor product distributes over direct sums
implies that if $V$ and $U$ are vector spaces over a field $F$,
of dimensions $m$ and $n$ respecitvely, then $V \otimes_F U$ is a
vector space over $F$ of dimension $nm$. In summary,
$F^m \otimes_F F^n \iso F^{mn}$.
\subsubsection{Algebraic Integers}
A complex number is an \defn{algebraic number} if it is a root of a
polynomial with coefficients in $\mathbb Q$ (or equivalently
$\mathbb Z$). It is an \defn{algebraic integer} if it is root of a
monic polynomial over $\mathbb Z$.
\begin{theorem}\label{thm:algInt}
$\alpha$ is an algebraic integer if and only if it is eigenvalue of a
matrix $A \in \op M_n(\mathbb Z)$.
\end{theorem}
\begin{theorem}
The set of all algebraic integers form a ring.
\end{theorem}
\begin{proof}
Suppose $\alpha$ and $\beta$ are algebraic integers. Then there are
matrices $A \in\op M_n(\mathbb Z)$ and $B \in\op M_m(\mathbb Z)$
and vectors $v \in \mathbb C^n$ and $u \in \mathbb C^m$ and such that
\[
Av = \alpha v \qquad Bu = \beta u
\]
Now $A\otimes B$ is a homomorphism (linear transformation) of
$\mathbb C^n \otimes \mathbb C^m$ to itself. If $e_1, \ldots, e_n$
is the standard basis of $\mathbb C^n$ and
$e'_1, \ldots, e'_m$
are the standard bases of $\mathbb C^n$ and $\mathbb C^m$,
respectively, then $e_i \otimes e'_j$ is a basis of
$\mathbb C^n \otimes \mathbb C^m$. Under this basis, ordered
lexicographically, the matrix
corresponding to $A \otimes B$ is the \defn{Kronecker product} of
$A$ and $B$, which is also denoted $A \otimes B$. This consists of
$n^2$ blocks, each of size $m \times m$, of the form
$a_{ij}B$. This is a straightforward calculation given in
Proposition~16 of Section~11.2 of Dummit and Foote.
In particular $A \otimes B$ has integer entries and so its
eigenvalues are algebraic integers by Theorem~\ref{thm:algInt}.
Now we calculate
\[
(A \otimes B)(v \otimes u) = Av \otimes Bu = \alpha v \otimes \beta
v = \alpha \beta (v \otimes u)
\]
showing that $v \otimes u$ is an eigenvector of $A \otimes B$ with
eigenvalue $\alpha\beta$. Thus algebraic integers are closed under
multiplication. To see that they are closed under addition, we
calculate
\[
(I_n \otimes B + A \otimes I_m)(v \otimes u)
= v \otimes \beta u + \alpha v \otimes u
= (\alpha + \beta) (v \otimes u)
\]
Thus $\alpha + \beta$ is an eigenvalue of the integer matrix
$I_n \otimes B + A \otimes I_m$ and so an algebraic integer.
\end{proof}
\begin{exers}~
\begin{exercises}
\prob
\part Show that if $\alpha$ is a nonzero algebraic number, then
$1/\alpha$ is also an algebraic number. Hint: suppose $f(\alpha) = 0$
where $f(x) \in \mathbb Q[x]$. Start by dividing the equation
$f(\alpha) = 0$ by $\alpha^k$, where $k$ is the degree of~$f$.
\part Show the if $\alpha$ is an algebra number then
$\alpha = \beta/n$ is also an algebraic number for all
$n \ne 0$ in $\mathbb Z$.
\part Show the $\alpha$ is an algebra number iff
$\alpha = \beta/n$ for some algebraic integer and
$n \in \mathbb Z$.
\part Show that the algebraic numbers form a field.
\end{exercises}
\end{exers}
\newpage
\subsection{Projective, Injective and Flat Modules; Exact Sequences}
This topic is covered well by Dummit and Foote. You should read
their Section 10.5. Hungerford also does a good job. Here we just
present some highlights and exercises.
A sequence of homomorphisms
\[
\cdots \to X_{n-1} \xrightarrow{\alpha} X_n
\xrightarrow{\beta} X_{n+1} \to \cdots
\]
is said to be \defn{exact}[exact sequence] if the image of $\alpha$
equals the kernel of~$\beta$ at each $X_n$ which has something to
its left and its right.
An exact sequence of the form
\begin{equation}\label{eq:short}
0 \to A \xrightarrow{\alpha} B \xrightarrow{\beta} C \to 0
\end{equation}
is called a \defn{short exact sequence}. Note this is equivalent to
saying $\alpha$ is injective and $\beta$ is surjective and that
$\alpha(A) = \ker \beta$.
We use $A \mono B$ to indicate a monomorphism and
$B \epi C$ to indicate an epimorphism.
Let $\mathcal V$ be a variety of algebras in the general sense.
An algebra $\alg P$
is said to be \defn{projective}[projective algebra] if
for each $\alg A$ and $\alg B \in \mathcal V$, epimorphism
$f : \alg A \epi \alg B$ and homomorphism $h : \alg P \to \alg B$,
there is a homomorphism $g : \alg P \to \alg A$ with
$h = fg$. Pictorially
\begin{center}
\begin{tikzpicture}
\matrix (m) [matrix of math nodes, row sep=3em,column sep=3em]
{ & \alg P \\
\alg A & \alg B \\ };
\path[->>,font=\small]
(m-2-1) edge node[above] {$f$} (m-2-2);
%edge node[below] {$\varphi$} (m-2-2);
\path[->,font=\small] (m-1-2) edge node[right] {$h$} (m-2-2);
\path[->, dashed, font=\small]
(m-1-2) edge node[above] {$g$} (m-2-1);
\end{tikzpicture}
\end{center}
A homomorphism $\rho$ from an algebra $\alg A$ to itself is called a
\defn{retraction} on $\alg A$ if $\rho^2 = \rho$.
We say that $\alg B$ is a \defn{retract} of $\alg A$ if $B = \rho(A)$
for some retraction $\rho$ on~$\alg A$.
\begin{exers}~
\begin{exercises}
\prob
If $\rho$ is a retraction of $\alg A$ onto $\alg B$, then
$\rho|_B$ ($\rho$ restricted to $B$) is the identity on $\alg B$.
\prob
Prove the following are equivalent for an algebra $\alg P$
in a variety $\mathcal V$:
\begin{enumerate}[(a)]
\setlength{\itemsep}{0.0ex plus0.1ex}
\item $\alg P$ is projective.
\item If $f : \alg A \epi \alg P$ is a epimorphism then there is
a homomorphism $g : \alg P \to \alg A$ so that $fg(x) = x$. Note
this forces $g$ to be a monomorphism.
\item $\alg P$ is isomorphic to a retract of a free algebra in $\mathcal V$.
\end{enumerate}
\prob
Let $R$ be a ring and
assume now that $\mathcal V$ is the variety of all $R$-modules.
\part
Show that if $A$ and $B$ are $R$-modules and if $\rho : A \to B$ is
a retraction, then $A$ has a submodule $C$ such that
$A = B \oplus C$ and $\rho (b,c) = b$.
\part
Use this to show that an $R$-module is projective iff it is a
direct summand of a free $R$-module.
\prob
Let $R$ be a PID. Use Theorem~\ref{thm:dedekind} to show that an
$R$ module is projective iff it is free. (We only proved
Theorem~\ref{thm:dedekind} in the finitely generated case, but you
can use it anyway.)
\prob
Let $F$ be the two-element field and let $R = F \times F$ be the
direct product. Let $P = \{(x,0) : x \in F \}$. Show that $P$ is
projective but not free.
\end{exercises}
\end{exers}
Note $\Hom_R(A,B)$ is an abelian group is an obvious way.
Let $D$ be another $R$-module.
If $\alpha : A \to B$ is a homomorphism, let
$\alpha' : \Hom_R(D,A) \to \Hom_R(D,B)$ be given by
$f \mapsto f' = \alpha \comp f$, for $f\in \Hom_R(D,A)$. This is a
homomorphism of abelian groups. Now suppose that \eqref{eq:short}
is a short exact sequence. Then the following sequence is exact:
\[
0 \to \Hom_R(D,A) \xrightarrow{\alpha'}
\Hom_R(D,B) \xrightarrow{\beta'} \Hom_R(D,C)
\]
Note the last $\to 0$ part is missing since $\beta'$ need not be
onto. But if $D$ is projective it is easy to see that it is. The
converse is also true so this is another characterization of
projective modules.
What does ``tensoring with $D$'' do to short exact sequences, where
now $D$ is a right $R$-module. Again assume that \eqref{eq:short}
is a short exact sequence. Then
\[
D \otimes_R A
\xrightarrow{1 \otimes \alpha}
D \otimes_R B
\xrightarrow{1 \otimes \beta}
D \otimes_R C
\to 0
\]
is exact. But this time we have a problem at the left end:
$1 \otimes \alpha$ is not necessarily injective. If it is injective
for all injective maps $\alpha : A \to B$ for all $A$ and $B$,
then $D$ is said to be a \defn{flat module}.
\textbf{Warning:} the element notation $a\otimes b$ can be misleading since
it does not indicate the ring $R$. For example,
$\mathbb Z/2\mathbb Z \otimes_{\mathbb Z} \mathbb Z
\iso \mathbb Z/2\mathbb Z$, but
$\mathbb Z/2\mathbb Z \otimes_{\mathbb Z} \mathbb Q = 0$, as we showed.
So $1\otimes 1 = 0$ in
$\mathbb Z/2\mathbb Z \otimes_{\mathbb Z} \mathbb Q$ but
$1\otimes 1 \ne 0$ in
$\mathbb Z/2\mathbb Z \otimes_{\mathbb Z} \mathbb Z$.
\begin{exers}~
\begin{exercises}
\prob
In this problem you will show
$\mathbb Q \otimes_{\mathbb Z} \mathbb Q \iso \mathbb Q$ as
$\mathbb Z$-modules.
\part
Show that
$\varphi : \mathbb Q \otimes_{\mathbb Z} \mathbb Q \to \mathbb Q$
given by $\varphi(r \otimes s) = rs$ for $r$ and $s \in \mathbb Q$
is a homomorphism. To do this you need to find the appropriate
middle linear map.
\part
Show that if $r$ and $s \in \mathbb Q$ then
$r \otimes s = 1 \otimes rs$. This is a little harder that it
looks: if we were working over $\mathbb Q \otimes_{\mathbb Q} \mathbb Q$
then we could just bring the $r$ to the other side. But in
$\mathbb Q \otimes_{\mathbb Z} \mathbb Q$ we can only move integers over.
Nevertheless a trick similar to the one I used in class showing
$\mathbb Z/2 \mathbb Z \otimes_{\mathbb Z} \mathbb Q = 0$ works.
\part Show that $r \mapsto 1 \otimes r$ is a homomorphism and is
the inverse of~$\varphi$.
(By the way, this same argument shows that if $K$ is the field of
fractions of an integral domain $R$, the
$K \otimes_R K \iso K$. On the other hand
$\mathbb C \otimes_{\mathbb R} \mathbb C \not\iso \mathbb C$. You
do not need to prove either of these.)
\prob
Suppose $R$ is commutative and $I$ and $J$ are ideals of $R$. Show
that
\[
R/I \otimes_R R/J \iso R/(I\join J).
\]
($I \join J$ is the ideal
generated by $I$ and $J$. It is often written $I + J$.)
For the map $R/I \otimes_R R/J \to R/(I\join J)$ you need to use the
usual trick of making a middle linear map
$R/I \times_R R/J \to R/(I\join J)$. For the other direction map
$R \to R/I \otimes_R R/J$ by $r \mapsto r(\bar1\otimes \bar{\bar 1})$,
where $\bar 1 = 1 + I\in R/I$ and $\bar{\bar 1} = 1 + J\in R/J$,
and show that $I \join J$ is contained in the kernel.
\prob
Using the previous problem (and the fundamental theorem of abelian
groups and that tensor products distribute over direct sum;
Theorem~14 of Dummit and Foote) describe $A \otimes_{\mathbb Z} B$,
where $A$ and $B$ are finite abelian groups.
This may be a little vague so, if you prefer, you can find
$A \otimes_{\mathbb Z} B$, where
$A= \mathbb Z / 4 \mathbb Z \oplus \mathbb Z / 16 \mathbb Z$ and
$B= \mathbb Z / 8 \mathbb Z \oplus \mathbb Z / 27 \mathbb Z$.
\prob
Let $B_1$ and $B_2$ be submodules of the left $R$-module $A$. Let
$D$ be a flat right $R$-module. Show
\begin{align*}
D \otimes_R (B_1 \join B_2) &=
(D \otimes_R B_1) \join (D \otimes_R B_2)\\
D \otimes_R (B_1 \cap B_2) &=
(D \otimes_R B_1) \cap (D \otimes_R B_2)
\end{align*}
proving the map $B \mapsto D \otimes_R B$ is a lattice homomorphism of
$\mathbf{Sub} (A) \to \mathbf{Sub} (D \otimes_R A)$.
Hint: the hardest part is proving the inclusion
$(D \otimes_R B_1) \cap (D \otimes_R B_2)
\subseteq D \otimes_R (B_1 \cap B_2)$.
To see this first note
$B_1/(B_1 \cap B_2) \iso (B_1 \join B_2)/B_2$. Hence we have the
short exact sequence
\[
0 \to B_1 \cap B_2 \to B_1 \xrightarrow{\varphi} (B_1 \join B_2)/B_2 \to 0
\]
and hence
\[
0 \to D \otimes_R (B_1 \cap B_2) \to D\otimes _R B_1
\xrightarrow{1\otimes\varphi}
D\otimes (B_1 \join B_2)/B_2 \to 0
\]
is exact so $\ker 1\otimes \varphi = D \otimes_R (B_1 \cap B_2)$.
Use this to show
$(D \otimes_R B_1) \cap (D \otimes_R B_2)
\subseteq D \otimes_R (B_1 \cap B_2)$.
% Now show ($D \otimes_R B_1) \cap (D \otimes_R B_2) \subseteq \ker
% 1\otimes \varphi$.
% Not sure this is really true.
%
% \prob
% Let $D_i$, $i\in I$, be a chain of submodules of a right $R$-module $D$.
% (This means for each $i$ and $j \in I$, either
% $D_i \subseteq D_j$ or $D_j \subseteq D_i$.)
% Show that if each $D_i$ is flat and $D=\bigcup D_i$ then $D$ is
% flat.
% \prob
% Show that $\mathbb Q$ is flat as a $\mathbb Z$-module. (You can
% use the previous problem.)
\end{exercises}
\end{exers}
\begin{theorem}
Let $_R A$ be a left $R$-module and let $D_R$ be a flat right $R$-module.
Then the map $\mathbf{Sub} (_R A) \to \mathbf{Sub} (D \otimes_R A)$
given by $B \mapsto D \otimes_R B$ is a lattice homomorphism.
\end{theorem}
\begin{proof}
First note that if $B_1 \le B_2$ are submodules of $A$, then
\[
0 \to B_1 \to B_2 \to B_2/B_1 \to 0
\]
is a short exact sequence. Since $D$ is flat this implies that
$D\otimes B_1$ is embedded into $D \otimes B_2$. It follows that
for arbitrary $B_1$, $B_2\le A$
\[
(D \otimes_R B_1) \join (D \otimes_R B_2) \le
D \otimes_R (B_1 \join B_2).
\]
If $b \in B_1 \join B_2$ the $b = b_1 + b_2$ for some $b_i \in B_i$.
But then, if $d \in D$, $d \otimes b = d \otimes b_1 + d \otimes b_2$,
proving the other inclusion.
As above
\[
D \otimes_R (B_1 \cap B_2) \le
(D \otimes_R B_1) \cap (D \otimes_R B_2).
\]
To see that $(D \otimes_R B_1) \cap (D \otimes_R B_2)
\subseteq D \otimes_R (B_1 \cap B_2)$ note
$B_1/(B_1 \cap B_2) \iso (B_1 \join B_2)/B_2$, by the Second Isomorphism
Theorem. Hence we have the
short exact sequence
\[
0 \to B_1 \cap B_2 \to B_1 \xrightarrow{\varphi} (B_1 \join B_2)/B_2 \to 0
\]
and hence
\[
0 \to D \otimes_R (B_1 \cap B_2) \to D\otimes _R B_1
\xrightarrow{1\otimes\varphi}
D\otimes (B_1 \join B_2)/B_2 \to 0
\]
is exact so $\ker 1\otimes \varphi = D \otimes_R (B_1 \cap B_2)$.
Let $\psi$ be the natural map from $B_1 \join B_2$ onto $(B_1 \join B_2)/B_2$
and note $\varphi$ is the restriction of $\psi$ to $B_1$.
Now let $\sum d_i\otimes c_i$ with $c_i\in B_2$ be an element of
$D \otimes_R B_2$. If this element is also in $D \otimes_R B_1$ then,
since $\psi(c_i) = 0$, we can calculate
\[
(1\otimes \varphi)(\sum d_i\otimes c_i) = (1\otimes \psi)(\sum d_i\otimes c_i)
= \sum d_i \otimes \psi(c_i) = 0.
\]
This shows $(D \otimes_R B_1) \cap (D \otimes_R B_2)
\subseteq \ker 1\otimes \varphi =
D \otimes_R (B_1 \cap B_2)$, completing the proof.
\end{proof}
\newpage
\part{Fields}
\newpage
\section{Basics}
Since we view $1$ as a fundamental (nullary) operation of a ring,
every ring has a unique smallest subring, the subring generated by~$1$.
This subring is called the \defn{prime subring}. Note the prime subring
is isomorphic to $\mathbb Z/n \mathbb Z$, for some $n > 0$, or to
$\mathbb Z$. In the former case we say the ring $R$
has \defn{characteristic} $n$; in the latter case $R$ is said to
have characteristic~$0$. We denote this as $\ch{R}$.
Also, since $\mathbb Z/n \mathbb Z$ is not an integral domain unless
$n$ is a prime, the characteristic of a field is either a prime or~$0$.
If $F$ is a subfield of a field $K$, then $K$ is a vector space over
$F$. The dimension is denoted $[K : F] = \dim_F(K)$.
If $f(x) \in F[x]$, $f$ may not have any roots in $F$; for example,
$x^2 - 2 \in \mathbb Q[x]$. But $x^2-2$ does have a root in
$\mathbb R$. This will be one of the primary foci of our study of fields.
\newpage
\appendix
\begin{center}
APPENDIX
\end{center}
\section{Prerequisites}
This section briefly lists some prerequisites from set theory needed in order to
read the main text. It consists primarily of paraphrased excerpts from the excellent
introductory textbook by Enderton, ``Elements of Set Theory''~\cite{Enderton:1977}.
\subsection{Relations}
\label{sec:relations}
Probably the reader already has an idea of what is meant by an
``ordered pair,'' $\$. It consists of two elements (or sets) $x$ and $y$, given in
a particular order. How to make this notion mathematically precise is not quite
so obvious. According to~\cite{Enderton:1977}, in 1921 Kazimierz
Kuratowski gave us the definition in general use today: given two
sets $x$ and $y$, the \defn{ordered pair} $\$ is defined to be the set
$\{\{x\}, \{x, y\}\}$.
It is not too hard to prove that this definition captures our
intuitive idea of ordered pair -- namely, $\$ uniquely determines both
what $x$ and $y$ are, and the order in which they appear. Indeed, it is a
theorem (Theorem 3A of~\cite{Enderton:1977}) that $\ = \$ iff $u =
x$ and $v = y$.
A \defn{relation} is a set of ordered pairs. Thus, if $X$ is a set, a
relation $R$ on $X$ is simply a subset of the Cartesian product; that is,
\[
R \subseteq X\times X := \{\ : x_1, x_2 \in X\}.
\]
For a relation $R$, we sometimes write $x \rR y$ in place of $\ \in R$.
For example, in the case of the ordering relation $<$ on the set \R\ of real numbers,
$<$ is defined to be the set
$\{\ \in \R \times \R : \text{$x$ is less than $y$}\}$,
and the notation ``$x < y$'' is preferred to ``$\ \in <$." See
Enderton~\cite{Enderton:1977} for more details.
For a relation $R$, we define the \defn{domain} of $R$ ($\dom R$), the
\defn{range} of $R$ ($\ran R$), and the \defn{field} of $R$ ($\fld R$) by
\begin{align*}
x \in \dom R \; &\Leftrightarrow \; \exists y \; \ \in R,\\
x \in\ran R \; &\Leftrightarrow \; \exists t \; \ \in R,\\
\fld R &= \dom R \cup \ran R.
\end{align*}
% \item {\bf $n$-ary relations.}
% We can extend the ideas behind ordered pairs to the case of ordered
% triples and, more generally, to ordered $n$-tuples. For triples we define
% \[
% \ = \<\, z\>.
% \]
% Similarly we can form ordered quadruples:
% \[
% \ = \<\, x_4\>
% = \<\<\,x_3\>,x_4\>.
% \]
% Clearly we could continue in this way to define ordered quintuples or ordered
% $n$-tuples for any particular $n$. It is convenient for reasons of uniformity to
% define also the 1-tuple $\ = x$.
% We define an $n$-ary relation on $A$ to be a set of ordered $n$-tuples with all
% components in $A$. Thus a binary (2-ary) relation on $A$ is just a subset of
% $A \times A$. And a ternary (3-ary) relation on $A$ is a subset of
% $(A \times A) \times A$.
% There is, however, a terminological quirk here. If $n > 1$, then any $n$-ary
% relation on $A$ is actually a relation. But a unary (1-ary) relation on $A$ is just
% a subset of $A$; thus it may not be a relation at all.
A relation $R$ on a set $A$ is called \defn{reflexive} iff $x \rR x$ for all $x
\in A$; \defn{symmetric} iff whenever $x\rR y$ then also
$y\rR x$; \defn{transitive} iff whenever $x\rR y$ and $y\rR z$,
then also $x\rR z$. A relation is an \defn{equivalence relation}
iff it is a binary relation that is reflexive, symmetric, and transitive. Given
a set $A$, we denote the set of all equivalence relations on $A$ by $\Eq(A)$.
\subsection{Functions}
\label{sec:functions}
A \defn{function} (or mapping) is a relation $F$ such that for each $x$ in $\dom F$ there
is only one $y$ such that $x \,F\, y$.
The following operations are most commonly applied to functions, are
sometimes applied to relations, but can actually be defined for arbitrary
sets $A$, $F$, and $G$.
\begin{enumerate}[(a)]
\setlength{\itemsep}{0.0ex plus0.1ex}
\item The \defn{inverse} of $F$ is the set
\[
F^{-1}=\{\ \mid v \,F \,u\}=\{\ \mid \ \in F \}.
\]
\item The \defn{composition} of $F$ and $G$ is the set
\[
F \circ G = \{\ \mid \exists t \,(u \,G \,t \;\&\; t\, F\, v)\}
= \{\ \mid \exists t \;(\\in G\; \& \;\ \in F)\}.
\]
\item The \defn{restriction} of $F$ to $A$ is the set
\[
F \res A = \{\ \mid u \, F \,v\; \& \; u \in A\}
= \{\ \mid \\in F \; \& \; u \in A\}.
\]
\item The \defn{image} of $A$ under $F$ is the set
\[
F\lb A\rb = \ran(F \res A)= \{v \mid (\exists u \in A)\; \ \in F\}.
\]
\end{enumerate}
$F\lb A\rb$ can be characterized more simply when $F$ is a function and
$A\subseteq \dom F$; in this case
\[
F\lb A\rb = \{F(u) \mid u \in A\}.
\]
In each case we can easily apply a subset axiom to establish the
existence of the desired set. Specifically,
\[
F^{-1} \subseteq \ran F \times \dom F, \quad
F \circ G \subseteq \dom G \times \ran F, \quad
F \res A \subseteq F, \quad
F\lb A\rb \subseteq \ran F.
\]
(A more detailed justification of the definition of
$F^{-1}$ would go as follows: By a subset axiom
there is a set $B$ such that for any $x$,
\[
x\in B \quad \Leftrightarrow \quad x \in \ran F \times \dom F \; \& \;
\exists u \; \exists v\; (x = \ \; \&\; \ \in F).
\]
It then follows that
\[
x\in B \quad \Leftrightarrow \quad \exists u \; \exists v\; (x = \ \; \&\; \ \in F).
\]
This unique set $B$ we denote by $F^{-1}$.)\\[8pt]
\begin{example}
Let
\[
F = \{ \<\emptyset, a\>, \<\{\emptyset\}, b\> \}.
\]
Observe that $F$ is a function. We have $F^{-1} = \{ \, \ \}$.
Thus, $F^{-1}$ is a function iff $a \neq b$. The restriction of $F$ to $\emptyset$ is $\emptyset$, but
$F \res \{\emptyset\} = \{\<0, a\>\}$.
Consequently, $F\lb \{\emptyset \}\rb = \{a\}$, in contrast to the fact that $F(\{\es\}) = b$.
\end{example}
\begin{theorem}
Assume that $F: A\rightarrow B$, and that $A$ is nonempty.
\begin{enumerate}[(a)]
\setlength{\itemsep}{0.0ex plus0.1ex}
\item There exists a function $G: B \rightarrow A$ (a ``left inverse'') such that $G \circ F$
is the identity function $\id_{A}$ on $A$ iff $F$ is one-to-one.
\item There exists a function $H: B \rightarrow A$ (a ``right inverse'') such that $F \circ H$
is the identity function $\id_{B}$ on $B$ iff $F$ maps $A$ \emph{onto} $B$.
\end{enumerate}
\end{theorem}
\noindent
{\bf Axiom of Choice 1.} For any relation $R$ there is a function
$H \subseteq R$ with $\dom H = \dom R$.
With this axiom we can prove the sufficiency direction of part (b) of the Theorem above:
take $H$ to be a function with $H \subseteq F^{-1}$ and $\dom H = \dom F^{-1} = B$. Then
$H$ does what we want: Given any $y \in B$, we have $\ \in F^{-1}$ hence
$\ \in F$, and so $F(H(y)) = y$.
\qed
\bibliographystyle{plain}
\bibliography{notes}
\printindex
\end{document}