ideals in matrix algebras

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

(1)

Let $𝔪$ be an ideal in $M_{n\times n}(R)$.

Claim.

The set $\mathrm{i}\mathrm{\subseteq }R$ given by

$$𝔦=\{x\in R\mid x\mathit{\hspace{1em}}\mathit{\text{is an entry of }}A\in 𝔪\}$$

is an ideal in $R$, and $\mathrm{m}\mathrm{=}{M}_{n\mathrm{\times }n}\mathit{}\mathrm{(}\mathrm{i}\mathrm{)}$.

Proof.

$𝔦\ne \mathrm{\varnothing }$ since $0\in 𝔦$. Now let $A=(a_{ij})$ and $B=(b_{ij})$ be matrices in $𝔪$, and $x,y\in R$ be entries of $A$ and $B$ respectively, say $x=a_{ij}$ and $y=b_{kl}$. Then the matrix $A\cdot E_{jl}+E_{ik}\cdot B\in 𝔪$ has $x+y$ at position $(i,l)$, and it follows: If $x,y\in 𝔦$, then $x+y\in 𝔦$. Since $𝔦$ 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

$$D_r:=\sum _{i=1}^{n}r\cdot E_{ii},r\in R.$$

thus, $rx,xr\in 𝔦$. This shows that $𝔦$ is an ideal in $R$. Furthermore, $M_{n\times n}(𝔦)\subseteq 𝔪$.

By construction, any matrix $A\in 𝔪$ has entries in $𝔦$, so we have

$$A=\sum _{1\le i,j\le n}{a}_{ij}{E}_{ij},a_{ij}\in 𝔦$$

so $A\in m_{n\times n}(𝔦)$. Therefore $𝔪\subseteq M_{n\times n}(𝔦)$. ∎

A consequence of this is: If $F$ is a field, then $M_{n\times n}(F)$ is simple.

Title	ideals in matrix algebras
Canonical name	IdealsInMatrixAlgebras
Date of creation	2013-03-22 13:59:28
Last modified on	2013-03-22 13:59:28
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	10
Author	mathcam (2727)
Entry type	Topic
Classification	msc 15A30