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\begin{slide}
\begin{center}
\LARGE \textcolor{highlight}{\textcolor{highlight}{Classical Rings}\\ } 

\vspace{0.75in}
%\vfill

\normalsize

\begin{tabular}{rcl}
John Adkins & $\Leftrightarrow$ &
\texttt{adkinsjl1@appstate.edu}\\
Stephen Graham & $\Leftrightarrow$ &
\texttt{grahamsp@appstate.edu}\\
Michael Kelley & $\Leftrightarrow$ &
\texttt{kelleyma1@appstate.edu}\\
\end{tabular}

\vspace{0.25in}       
       
\textcolor{highlight}{\large Thursday, May $2^{\mathrm{nd}}$ 2013}  % at 5:30pm
\end{center}
\end{slide}

\begin{slide}
{\color{highlight}Definition:} A {\color{highlight}ring} is a non-empty set $R$ equipped with two binary operations denoted {\color{highlight}$a+b$} (addition) and {\color{highlight}$ab$} (multiplication) such that the following axioms hold\dots
{\small
\begin{description}\itemsep0pt
\item[{\color{highlight}\sc Associativity}] For all $a,b,c \in R$, {\color{highlight}$(a+b)+c=a+(b+c)$} 
\item[{\color{highlight}\sc Identity}] There is some $0 \in R$ such that {\color{highlight}$a+0=a=0+a$} for all $a \in R$ 
\item[{\color{highlight}\sc Inverses}] For each $a\in R$ there exists $-a \in R$ such that\\ \force \hfill \hfill {\color{highlight}$a+(-a)=0=(-a)+a$} 
\item[{\color{highlight}\sc Commutativity}] For all $a,b\in R$, {\color{highlight}$a+b=b+a$}
\item[{\color{highlight}\sc Associativity}] For all $a,b,c \in R$, {\color{highlight}$(ab)c=a(bc)$} 
\item[{\color{highlight}\sc Identity}] There is some $1 \in R$ such that {\color{highlight}$a1=a=1a$} for all $a \in R$ 
\item[{\color{highlight}\sc Distributive}] For all $a,b,c\in R$, {\color{highlight}$a(b+c)=ab+ac$} and {\color{highlight}$(a+b)c=ac+bc$} 
\end{description}
}

If in addition, 
{\small
\begin{description}\itemsep0pt
\item[{\color{highlight}\sc Commutativity}] For all $a,b\in R$, {\color{highlight}$ab=ba$}
\end{description}
}

\vspace{-0.25in}
then $R$ is called a {\color{highlight}commutative ring}.

\end{slide}

\begin{slide}
\begin{center} \Large {\color{highlight}Examples of Rings} \end{center}

\begin{itemize}
\item $ {\color{highlight}\mathbb{Z}_{12}} = \{ 0,1,2,\dots,11 \}$ where for example {\color{highlight}$8+7=15=3$} and {\color{highlight}$5 \cdot 4 = 20 = 8$}. In this ring we add, subtract, and multiply integers as usual keeping in mind the {\color{highlight}$12=0$} (just like a clock).

\item $ {\color{highlight}\mathbb{Z}} = \{ \dots,-2,-1,0,1,2,\dots \}$ is a commutative ring.

\item $ {\color{highlight}\mathbb{R}}$ (real numbers) is a commutative ring.

\item $ {\color{highlight}\displaystyle \mathbb{R}^{2\times 2}} = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} \;\Bigg|\; a,b,c,d \in \mathbb{R} \right\}$ is a {\color{highlight}non-commutative} ring.
\end{itemize}

\end{slide}

\begin{slide}
{\color{highlight}Definition:} An element $x$ in a ring $R$ is called a {\color{highlight}unit} if it has a multiplicative inverse. This means there is some ${\color{highlight}x^{-1}} \in R$ such that $xx^{-1}=1=x^{-1}x$.

{\color{highlight}Examples:}

\vspace{-0.5in}

\begin{itemize}
\item $1 \cdot 1 = 1$, $5 \cdot 5 = 25 =1$, $7 \cdot 7 = 49 = 1$, $11 \cdot 11= 121 = 1$ thus {\color{highlight}$1,5,7,$ and $11$} are units in $\mathbb{Z}_{12}$.

\item ${\color{highlight}-1,1} \in \mathbb{Z}$ are units. No other element is. For example, $2^{-1}$ should be $1/2$ (which is not an integer).
\item {\color{highlight}Every non-zero real} number is a unit in $\mathbb{R}$.

\item A matrix in $\mathbb{R}^{2\times 2}$ is a unit if and only if it has a {\color{highlight}non-zero determinant}.

\end{itemize}
\end{slide}

\begin{slide}
{\color{highlight}Definition:} Let $a$ and $b$ be  {\color{highlight}non-zero} elements of a ring $R$. If {\color{highlight}$ab=0$}, then $a$ and $b$ are {\color{highlight}zero divisors}.

{\color{highlight}Examples:}

\vspace{-0.5in}

\begin{itemize}
\item $2 \cdot 6 = 12 = 0$, $3 \cdot 4 = 12 =0$, $8 \cdot 3 = 24 = 0$, $9 \cdot 4 = 36 = 0$, $10 \cdot 6 = 60 = 0$ thus {\color{highlight}$2,3,4,6,8,9,$ and $10$} are zero divisors in $\mathbb{Z}_{12}$.

\item $\mathbb{Z}$ and $\mathbb{R}$ have {\color{highlight}no zero divisors}. If $ab=0$ then either $a=0$ or $b=0$ in $\mathbb{Z}$ and $\mathbb{R}$.

\item A non-zero matrix in $\mathbb{R}^{2\times 2}$ is a zero divisor if and only if its {\color{highlight}determinant is zero}. For example,
\[ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 0 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]
\end{itemize}

\end{slide}

\begin{slide}
{\color{highlight}Definition:} A ring $R$ is called {\color{highlight}classical} if every element of $R$ is either zero, a zero divisor, or a unit.

{\color{highlight}Examples:}

\vspace{-0.5in}

\begin{itemize}
\item {\color{highlight}$\mathbb{Z}_{12}$} is {\color{highlight}classical}. $\mathbb{Z}_{12} = \{ 0 \} \cup \{ 1,3,5,7,11 \} \cup \{ 2,3,4,6,8,9,10\}$

\item {\color{highlight}$\mathbb{Z}$} is {\color{highlight}not classical}. $\pm 2, \pm 3, \dots$ are non-zero, non-unit, non-zero divisors!

\item {\color{highlight}$\mathbb{R}$} is {\color{highlight}classical}. Recall, in $\mathbb{R}$ all non-zero elements are units.

\item {\color{highlight}$\mathbb{R}^{2 \times 2}$} is {\color{highlight}classical}. A non-zero matrix is either a zero divisor or a unit depending on whether its determinant is zero or not.
\end{itemize}
\end{slide}


\begin{slide}
\begin{center} \Large Our Question: \end{center}


\begin{center} \Large {\color{highlight}When is a ring classical?} \end{center}
\end{slide}

\begin{slide}

\fbox{{\color{highlight}Theorem:} All finite rings are {\color{highlight}classical}.}

%\vspace{-0.25in}

Sketch of Proof:

\vspace{-0.25in}

Let $R$ be a ring and for $0 \not= r\in R$ define $L_r:R \to R$ by {\color{highlight}$L_r(x)=rx$}. 

\vspace{-0.45in}

\begin{itemize} \setlength{\itemsep}{0pt}
\item $r$ is a {\color{highlight}unit} $\Longrightarrow$ $L_r$ is {\color{highlight}onto}.
\item $r$ is {\color{highlight}\it not a zero divisor} $\Longrightarrow$ $L_r$ is {\color{highlight}one-to-one}.
\item Maps from finite sets to themselves are\\ {\color{highlight}one-to-one} $\Longleftrightarrow$ {\color{highlight}onto}.
\end{itemize}

{\color{highlight}Examples:}

\vspace{-0.25in}

\begin{itemize} \setlength{\itemsep}{0pt}
\item {\color{highlight}$\mathbb{Z}_{12}$} or more generally $\mathbb{Z}_n$.
\item {\color{highlight}$(\mathbb{Z}_{12})^{2 \times 2}$} or more generally square matrices over any finite ring.
\end{itemize}

\end{slide}

\begin{slide}

\begin{center}
\fbox{{\color{highlight}Theorem:} \begin{minipage}[t]{6in} Let $R$ be a commutative classical ring, then {\color{highlight}$R^{n \times n}$ is classical}.  \end{minipage}}
\end{center}

%\vspace{-0.25in}

Sketch of Proof:

\vspace{-0.45in}

\begin{itemize} \setlength{\itemsep}{0pt}
\item {\color{highlight}McCoy's Theorem} implies that matrices are zero divisors if their  {\color{highlight}determinants are zero divisors}.
\item A formula which follows from Cramer's rule implies that matrices whose  {\color{highlight}determinants are units} are themselves invertible.
\item If the base ring is {\color{highlight}classical},  {\color{highlight}\underline{\color{regcolor}every determinant}} is either zero, a zero divisor, or a unit.
\end{itemize}

{\color{highlight}Examples:}

\vspace{-0.45in}

\begin{itemize} \setlength{\itemsep}{0pt}
\item {\color{highlight}$\mathbb{R}^{2 \times 2}$} or more generally square matrices over any field (such as $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$).
\item {\color{highlight}$\mathbb{Z}^{2 \times 2}$} is {\color{highlight}not} classical since $\mathbb{Z}$ isn't.
\end{itemize}

\end{slide}


\begin{slide}

{\color{highlight}Other Results:}
\begin{itemize} \setlength{\itemsep}{0pt}
\item Every {\color{highlight}commutative} ring can be embedded in a classical ring ({\color{highlight} its total ring of quotients}). For example: $\mathbb{Z}$ can be completed to $\mathbb{Q}$.
\item {\color{highlight}Proper quotients of principal ideal domains} are classical.
\item {\color{highlight}Artinian} rings are classical.
\item Rings of {\color{highlight}Krull dimension 0} are classical.
\end{itemize}

{\color{highlight}Open Question:}\\
Artinian and Krull dimension 0 are sort of ``finiteness'' conditions which imply being classical. In general, {\color{highlight}what kind of finiteness condition is equivalent to being classical?}

\end{slide}

\begin{slide}

\begin{center}
{\color{highlight}\Large\sc $\mathcal{A}$ny $\mathcal{Q}$uestions?}
\end{center}

\end{slide}



\end{document}
