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\title{Fields and Vector Spaces}
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\maketitle

A {\it Field} $F$ is a set equipped with binary operations
$+$ and $\times$ satisfying various axioms. We usually
suppress $\times$ and denote it by juxtaposition.
We use brackets to indicate the priority of operations
in an expression, and we omit them according to the usual
conventions. The field axioms are as follows.
\begin{enumerate}
\item $+$ is {\it closed} so $a + b \in F$ whenever $a,b \in F$.
\item $+$ is {\it associative} so
$(a+b) + c = a + (b+c)$ whenever $a,b,c \in F$.
\item $+$ has a distinguished identity element $0$ which acts as a
two-sided identity, so
\[ 0 + a = a = a + 0 \hbox{ whenever }a \in F.\]
\item $+$ admits of inverses, so for each $a \in F$ there is a unique
element $-a \in F$  such that 
\[ a+ (-a) = 0 = (-a) + a.\]
\item $+$ is commutative so $a + b = b + a$ whenever $a,b \in F$.
\item $\times$ is {\it closed} so $ab = ba$ whenever $a,b \in F$.
\item $\times$ is {\it associative} so
$(ab)c = a (bc)$ whenever $a,b,c \in F$.
\item $\times$ has a distinguished identity element $1$ which acts as a
two-sided identity, so
\[ 1a = a = a1 \hbox{ whenever }a \in F.\]
\item Non-zero elements have multiplicative inverses. To be explicit:
if $a \in F$ and $a \not = 0$, then there is a unique
$a^{-1} \in F$ such that $aa^{-1} = a^{-1}a = 1.$
\item Multiplication distributes over addition, so
\[ a(b+c) = ab + ac \hbox{ and } (b+c)a = ba + ca\] whenever
$a,b,c \in F$.
\item $0 \not = 1.$
\end{enumerate}

\newpage

A {\it vector space V} over a field $F$ is a set $V$ equipped
with a binary operation $+$, and a multiplication $\cdot :  
F \times V \rightarrow
V$
(often denoted by juxtaposition) such that teh following axioms
are satisfied.
\begin{enumerate}
\item $+$ is {\it closed} so $u + v \in V$ whenever $u,v \in V$.
\item $+$ is {\it associative} so
$(u+v) + w = u + (v+w)$ whenever $u,v,w \in V$.
\item $+$ has a distinguished identity element ${\mathbf 0}$ which acts as a
two-sided identity, so
\[ {\mathbf 0} + u = u = u + {\mathbf 0} \hbox{ whenever }u \in V.\]
\item $+$ admits of inverses, so for each $u \in V$ there is a unique
element $-u \in F$  such that
\[ u+ (-u) = {\mathbf 0} = (-u) + u.\]
\item $+$ is commutative so $u + v = v + u$ whenever $u,v \in V$.
\item $\cdot$ distributes over addition in $V$ so
\[ a \cdot (u + v) = a \cdot u + a \cdot v \mbox{ whenever }
a \in F,\ u,v \in V.\]
\item $\cdot$ distributes over addition in $F$ so
\[ (a + b) \cdot u  = a \cdot u + b \cdot u \mbox{ whenever }
a, b \in F,\ u \in V.\]
\item $\cdot$ enjoys a form of associativity whereby
\[ (ab)\cdot u = a \cdot (b \cdot u)\]
whenever $a,b \in F$ and $u \in V$. Here the product $ab$
is a product in $F$.
\item $1 \cdot u = u$ whenever $u \in V$.
\end{enumerate} 
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