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\newtheorem{ex}{\bf Example}
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\begin{document}

\title{Introduction to algebra --- handout 10}
\maketitle

\section{Rings and fields}

\begin{df}
Algebraic structure $\langle R, +,\cdot\rangle$ is a {\bf ring} if $+$
and $\cdot$ are two binary operations having the following properties:
\begin{itemize}
\item $\langle R, +\rangle$ is a commutative group.
\item The multiplication $\cdot$ is associative, i.e., $(a\cdot b)\cdot c= a\cdot (b\cdot c)$.
\item The multiplication $\cdot$ is left and right {\bf distributive} over the addition $+$, i.e., $c\cdot(a+b)=c\cdot a+c\cdot b$ and $(a+b)\cdot c=a\cdot c+b\cdot c$.  
\end{itemize}
\end{df}

\begin{df}
{\bf Unit rings} or {\bf Rings with identity} are rings having an identity element $1$ such that $1\cdot a=a\cdot 1=a$ for all $a\neq 0$.  
\end{df}


\begin{df}
A ring with identity is called a {\bf skew field} if every nonzero
element has an inverse in it.
\end{df}

\begin{df}
{\bf Commutative rings} are  rings in which $\cdot$ are commutative. 
\end{df}

\begin{df}
Commutative skew fields are called {\bf fields}. 
\end{df}

\begin{df}
$a\neq 0$ is called left {\bf zero divisor} if $ab=0$ for some
  $b\neq0$, in this case $b$ is called right zero divisor.
\end{df}


\begin{df}
A commutative ring with identity is called {\bf integral domain} if it contains no zero divisor.  
\end{df}

\begin{ex} ${}$
\begin{enumerate}
\item $\langle \Z,+,\cdot\rangle$, $\langle \Q,+,\cdot\rangle$, $\langle \R,+,\cdot\rangle$, $\langle \C,+,\cdot\rangle$. 
\item $\langle \Z_m,+,\cdot\rangle$.
\item $\Z[i]:=\{a+bi:a,b\in\Z\}$ Gauss integers. 
\item $\Q[i]:=\{p+qi:p,q\in\Q\}$ Gauss rationals.
\item $R[x]$ -- polynomials over a ring $R$. 
\item $R(x)$ -- rational functions over $R$.
\item $R^{n\times n}$ -- $n\times n$ matrices over ring $R$.  
\item $C(X)$ -- Continuous functions from $X$ to $\R$. 
\item $\Z\times \Z$, $\Q\times\Q\times\Q$, etc.. 
\end{enumerate}
\end{ex}

\begin{e}
Show that every finite integral domain is a field.
\end{e}
\begin{thm}
Every integral domain $R$ can be embedded into a field (called the {\bf field of fractions} for $R$).
\end{thm}

\begin{proof}
The construction of $\Q$ from $\Z$ works in the general case.
\end{proof}

\begin{ex} ${}$
\begin{itemize}
\item The field of fractions for $\Z$ is $\Q$.
\item The field of fractions for $\Z[i]$ is $\Q[i]$. 
\item Let $R$ be an integral domain, e.g., $\Z$, $\Q$, or $\R$. The
  field of fractions for $R[x]$ is $R(x)$.
\item The field of fractions for a field $F$ is $F$. 
\end{itemize}
\end{ex}





\end{document}