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'ideals in matrix algebras'
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| Title of object: |
ideals in matrix algebras |
| Canonical Name: |
MatrixIdealsOverCommutativeRings |
| Type: |
Topic |
| Created on: |
2003-10-10 14:23:00 |
| Modified on: |
2003-10-20 10:25:35 |
| Classification: |
msc:15A30 |
Revision comment (for changes between this and next version):
| Changes for correction #5745 ('typos and other improvements'). |
Preamble:
% this is the default PlanetMath preamble. as your knowledge
% of TeX increases, you will probably want to edit this, but
% it should be fine as is for beginners.
% almost certainly you want these
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
% used for TeXing text within eps files
%\usepackage{psfrag}
% need this for including graphics (\includegraphics)
%\usepackage{graphicx}
% for neatly defining theorems and propositions
\usepackage{amsthm}
% making logically defined graphics
%\usepackage{xypic}
% there are many more packages, add them here as you need them
% define commands here
\newtheorem*{claim}{Claim} |
Content:
Let $R$ be a ring with 1. Consider the ring $M_{n \times n}(R)$ of $n \times n$-matrices with entries taken from $R$.
It will be shown that there exists a one-to-one
correspondence between the ideals of $R$ and the
ideals of $M_{n \times n}(R)$.
For $1 \le i,j \le n$, let $E_{ij}$ denote the $n
\times n$-matrix having entry 1 at position
$(i,j)$ and 0 in all other places. It can be
easily checked that
\begin{equation}
E_{ij} \cdot E_{kl}=\left(
\begin{array}{lllll}
0 & \mbox{iff}& k \ne j \\
E_{il} & \mbox{otherwise.}
\end{array}\right.
\end{equation}
Let $\mathfrak{m}$ be an ideal in $M_{n \times
n}(R)$.
\begin{claim}
The set $\mathfrak{i}\subseteq R$ given by
\[\mathfrak{i}=\{x \in R \mid x \mbox{is an entry
of } A \mid A \in \mathfrak{m}\}\]
is an ideal in $R$, and
$\mathfrak{m}=M_{n \times n}(\mathfrak{i})$.
\end{claim}
\begin{proof}
$\mathfrak{i} \ne \emptyset$ since $0 \in
\mathfrak{i}$. Now let $A=(a_{ij}$ and $B=(b_{ij}$
be matrices in $\mathfrak{m}$, and $x,y \in R$ be
entries of $A$ and $B$ respectively-- say
$x=a_{ij}$ and $y=b{kl}$. Then the matrix $A \cdot
E_{jl} +E_{ik}\cdot B \in \mathfrak{m}$ has $x+y$
at position $(i,l)$, and it follows: If $x,y \in
\mathfrak{i}$, then $x+y \in \mathfrak{i}$. Since
$\mathfrak{i}$ is an ideal in $M_{n \times n}(R)$
it contains, in particular, the matrices $D_r \cdot A$ and $A \cdot D_r$, where
\begin{equation*}
D_r :=\sum_{i=1}^n r\cdot E_{ii}, r \in R.
\end{equation*}
thus, $rx, xr \in \mathfrak{i}$. This shows
that $\mathfrak{i}$ is an ideal in $R$.
Furthermore, $M_{n \times n}(\mathfrak{i})
\subseteq \mathfrak{m}$.
By construction, any matrix $A \in \mathfrak{m}$
has entries in $\mathfrak{i}$, so we have
\begin{equation*}
A=\sum\limits_{1 \le i,j \le n} a_{ij}E_{ij},
a_{ij} \in \mathfrak{i}
\end{equation*}
so $A \in m_{n \times n}(\mathfrak{i})$. Therefore
$\mathfrak{m} \subseteq M_{n \times
n}(\mathfrak{i}$.
\end{proof}
A consequence of this is: If $F$ is a field, then $M_{n \times n}(F)$ is simple. |
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