MatrixIdealsOverCommutativeRings 2003-10-10 14:23:00 2005-03-25 23:32:59 Topic % 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} 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 (two-sided) ideals of $R$ and the (two-sided) 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 \[\mathfrak{i}=\{x \in R \mid x\quad\mbox{is an entry of } A \in \mathfrak{m}\}\] 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.