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\title{Classical Rings}

\author{William J. Cook} 

%\thanks{}

\maketitle 

\begin{abstract}
We investigate several criterion which lead to rings whose elements are all either zero, a zero divisor, or a unit.
In particular, finite rings, non-trivial quotients of principal ideal domains, and Artinian rings consist entirely of zero,
zero divisors, and units. 
\end{abstract}

\section{Introduction} 
A frequent mistake made by students first learning ring theory is that every non-zero element of a ring is either a zero divisor or a unit. We take up the question of when this does occur and make several observations about how rings with this property have some nice properties.

For what follows, let $R$ be a ring with $1$. We say $R$ is a {\em classical ring} if every non-zero element of $R$
is either a zero divisor or a unit. 

When $R$ is a commutative ring we can form its {\em total quotient ring}. This construction is essentially the same as that of forming the field of quotients of an integral domain (see \cite{df} section 7.5 for more details). 
Let $S = \{ r \in R \,|\, r \mbox{ is not zero or a zero divisor.} \}$.
$R \times S$ is then given the equivalence relation $(a,b) \sim (c,d)$ if and only if $ad=bc$. Let $S^{-1}R$ be the 
set of equivalence classes and denote the equivalence class of $(a,b)$ by $a/b$. In this case we can turn $Q(R) = S^{-1}R$
into a (commutative) ring by giving it the (well-defined) operations $a/b+c/d = (ad+bc)/ad$ and $(a/b)(c/d) = (ac)/(bd)$.
Moreover, $R$ can be identified with the subring of all elements of the form $r/1$. 

The total quotient ring, $Q(R)$, now contains an inverse for every non-zero element which is not a zero divisor. Thus
$Q(Q(R))=Q(R)$. Therefore, $R$ is a commutative classical ring if and only if $R=Q(R)$. 

\section{Finite Rings}

When student come to the false belief that all non-zero elements are either zero divisors or units, they are typically thinking
of finite rings. When $R$ is finite, this is actually true. An easy way to see this is to investigate left and right multiplication
operators.

Let $R$ be a ring with $1$ (not necessarily commutative). Let $r \in R$ and define
$L_r:R \rightarrow R$ and $R_r:R \rightarrow R$ by $L_r(x)=rx$ and $R_r(x)=xr$ for all $x \in R$.
Notice that the distributive laws tell us that $L_r(x+y)=r(x+y)=rx+ry=L_r(x)+L_r(y)$ and likewise $R_r(x+y) = R_r(x)+R_r(y)$ for all $x,y \in R$. So for all $r \in R$, the left and right multiplication operators $L_r$ and $R_r$ are {\em group} homomorphisms (when $R$ is thought of as a group under addition). 

\begin{prop} Let $r\in R$ and $r\not=0$. Then $r$ is a left zero divisor if and only if $L_r$ is not injective.
Likewise $r$ is a right zero divisor if and only if $R_r$ is not injective.
\end{prop}

\bproof 
$r$ is a left zero divisor $\Longleftrightarrow$ there exists $x \in R$, $x\not=0$ such that $rx=0$
$\Longleftrightarrow$ there exists $x \in R$, $x\not=0$ such that $L_r(x)=0$
$\Longleftrightarrow$ $L_r$ has a non-trivial kernel
$\Longleftrightarrow$ $L_r$ is not one-to-one.

The proof for $R_r$ is similar.
\eproof

\begin{prop} Let $r \in R$. Then $r$ is a left unit if and only if $L_r$ is surjective. Likewise, $r$ is a right unit if and only
if $R_r$ is surjective.
\end{prop}

\bproof
If $r$ is a left unit, then there exists some $x \in R$ such that $rx=1$. Let $y \in R$. 
$L_r(xy) = r(xy) = (rx)y = 1y =y$ and thus $L_r$ is onto. Conversely suppose $L_r$ is
onto. Then there exists some $x$ such that $L_r(x)=1$ and thus $rx=1$ so that $r$ is a left unit.

The proof for $R_r$ is similar.
\eproof

\begin{prop} Let $r \in R$ and $r \not=0$. $r$ cannot be both a left unit and a right zero divisor.
Likewise, $r$ cannot be both a right unit and a left zero divisor. \label{leftright}
\end{prop}

\bproof
Suppose that $r$ is a left unit. Then there exist $x \in R$ such that $rx=1$.
Suppose $yr=0$ then $0 = 0x = (yr)x = y(rx) = y1 = y$. Thus $r$ is not a right zero divisor.

The other statement is proven similarly.
\eproof

Now recall maps from a finite set to itself are either injective and surjective or neither injective nor surjective.
Since these properties are tied together, our propositions tells us that not being a zero divisor is the same as
being a unit (for non-zero elements). Also, recall that if $r$ is both a left and right unit, then $r$ is a unit:
$rx=1$ and $yr=1$ imply that $x = 1x = (yr)x = y(rx) = y1 = y$.

\begin{prop}  \label{finite_result}  % This is  a theorem label
Let $R$ be a finite ring. Left units are automatically right units. Similarly left zero divisors are automatically
right zero divisors. It then follows that, every non-zero element of $R$ is either a zero divisor or a unit.
In other words, all finite rings are classical rings. 
\end{prop}

\bproof Recall that since $R$ is finite, multiplication operators are injective if and only if they are surjective.
Let $r \in R$ and $r \not=0$. 

$r$ is not a left zero divisor $\Longleftrightarrow$ $L_r$ is injective
$\Longleftrightarrow$ $L_r$ is surjective
$\Longleftrightarrow$ $r$ is a left unit.
So all non-zero elements are either left zero divisors or left units.

Likewise, $r$ is not a right zero divisor $\Longleftrightarrow$ $r$ is a right unit and so
all non-zero elements are either right zero divisors or right units.

Suppose $r$ is a left unit. Then by Proposition \ref{leftright}, $r$ is not  a right zero divisor. Hence $r$ is a right unit. Therefore, $r$
is a unit. Likewise, if $r$ is a right unit, then $r$ is a unit. There are no one-sided units in finite rings.

On the other hand, if $r$ is a left zero divisor, then $r$ is not  a right unit  and so $r$ is a right zero divisor.
Likewise, if $r$ is a right zero divisor, then it is also a left zero divisor. All zero divisors are both left and right
zero divisors.
\eproof

% This is how you reference something
I would like to reference Proposition \ref{finite_result} for no apparent reason. If there are question marks, I should re-compile. \LaTeX takes about 2 passes before it resolves references and citations. 

\section{Quotients of Principal Ideal Domains}

Consider the ring of polynomials $\mathbb{F}[x]$ where $\mathbb{F}$ is a field. We know $\mathbb{F}[x]$ is a principal ideal domain (PID) and so each ideal can be generated by a single element. Let $I$ be a non-zero ideal and 
suppose $I=(f(x))$ where $0 \not= f(x) \in \mathbb{F}[x]$. It is easy to show that if the degree of $f(x)$ is $n$, then
the quotient ring $\mathbb{F}[x]/(f(x))$ is a vector space (over $\mathbb{F}$) of dimension $n$. It is not hard to show
that the left and right multiplication operators in this quotient ring are not just group homomorphisms (preserving addition)
but are in fact linear maps. If we recall that linear maps between finite dimensional vector spaces of the same dimension
are either both injective and surjective or neither, it follows by the same reasoning as that for finite rings, that $\mathbb{F}[x]/(f(x))$ is a classical ring. It turns out that this result holds for any PID. In fact, it holds for any Bezout ring.

Prime = Maximal

\section{Artinian Rings}

There is nothing to see here. To hear something please refer to \cite{beatles}. {\it Note:} Please refer to rooftop concert for ``Revolution.''

\begin{center}
\includegraphics[width=2in]{yosef.jpg}

Figure 123: {\sc Yosef} just thinks he's cool.
\end{center}


\begin{thebibliography}{RT-bib}
\bibitem[B]{beatles} McCartney et. all, {\em The Beatles White Album} 1968.
\bibitem[DF]{df} Dummit D.S., Foote R.M., Abstract Algebra, 3rd ed., John Wiley \& Sons, Hoboken, 2004.
\end{thebibliography}

\end{document}


