hollow matrix ringsHollowMatrixRings2007-12-19 12:07:022009-04-01 15:24:41Examplering\usepackage{latexsym}
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\section{Definition}
\begin{defn}
Suppose that $R\subseteq S$ are both rings. The \emph{hollow matrix ring} of $(R,S)$
is the ring of matrices:
\begin{equation*}
\begin{bmatrix} S & S \\ 0 & R\end{bmatrix} :=\left\{
\begin{bmatrix} s & t \\ 0 & r\end{bmatrix} : s,t\in S, r\in R\right\}.
\end{equation*}
\end{defn}
It is easy to check that this forms a ring under the usual matrix addition and
multiplication.
This definition is slightly simplified from the obvious higher dimensional examples
and the transpose of these matrices will also qualify as a hollow matrix ring.
The hollow matrix rings are highly counter-intuitive despite their simple definition.
In particular, they can be used to prove that in general a ring's left ideal
structure need not relate to its right ideal structure. We highlight a few
examples of this.
\section{Left/Right Artinian and Noetherian}
We specialize to an example with the fields $\mathbb{Q}$ and $\mathbb{R}$, though
the same argument can be made in much more general settings.
\begin{equation}
R := \begin{bmatrix} \mathbb{R} & \mathbb{R}\\ 0 & \mathbb{Q}\end{bmatrix}
= \left\{\begin{bmatrix} a & b \\ 0 & c\end{bmatrix} : a,b\in \mathbb{R}, c\in\mathbb{Q}\right\}.
\end{equation}
\begin{claim}
$R$ is left Artinian and left Noetherian.
\end{claim}
\begin{proof}
Let $I$ be a left ideal of $R$ and suppose
that $r:=\begin{bmatrix} x & y \\ 0 & z\end{bmatrix}\in I$ for some
$x,y\in\mathbb{R}$ and $z\in\mathbb{Q}$.
Suppose that $z\neq 0$. Hence, $s_q:=\begin{bmatrix} 0 & 0 \\ 0 & q/z\end{bmatrix}\in R$
for each $q\in\mathbb{Q}$
and so $s_qr=\begin{bmatrix} 0 & 0 \\ 0 & q\end{bmatrix}\in I$ for all $q\in\mathbb{Q}$. In particular,
$\begin{bmatrix} x & y \\ 0 & 0 \end{bmatrix}=r-s_{1}r\in I$. So in all cases it follows that
$\begin{bmatrix} x & y\\ 0 & 0 \end{bmatrix} \in I$. So now we take
$r=\begin{bmatrix} x & y \\ 0 & 0 \end{bmatrix}$ and assume that $I$ does not contain any $r$ with $z\neq 0$.
By observing matrix multiplication it follows that $I$ is now a left $\mathbb{R}$-vector space, and so any
chain of left $R$-modules is a chain of subspaces. As $\dim_{\mathbb{R}} I\leq 2$,
it follows that such chains are finite.
Hence, there can be no infinite descending chain of distinct left ideals
and so $R$ is left Artinian and Noetherian.
\end{proof}
\begin{claim}
$R$ is not right Artinian nor right Noetherian.
\end{claim}
\begin{proof}
Using $\pi$ (the usual $3.14\dots$), or any other transcendental number, we define
\begin{equation}
I_n := \begin{bmatrix} 0 & \mathbb{Q}[\pi;n] \\ 0 & 0 \end{bmatrix},
\end{equation}
where
\begin{equation}
\mathbb{Q}[\pi;n] := \{ q(\pi)\pi^{n} : q(x)\in\mathbb{Q}[x]\}.
\end{equation}
Since $\mathbb{Q}[\pi;n]$ properly contains $\mathbb{Q}[\pi;n+1]$ for all
$n\in\mathbb{Z}$, it follows that $\{I_n:n\in\mathbb{Z}\}$ is an infinite proper
ascending and descending chain of right ideals. Therefore, $R$ is neither right Artinian
nor right Noetherian.
\end{proof}
\begin{coro}
$R$ does not have a ring anti-isomorphism. Thus $R$ is not an
involutory ring.
\end{coro}
\begin{proof}
If $R$ is a ring with an anti-isomorphism, then the set of left ideals
is mapped to the set of right ideals, bijectively and order preserving.
This is not possible with $R$.
\end{proof}