PlanetMath: ideals in matrix algebras

Let

be a ring with 1. Consider the ring $M_{n \times n}(R)$ of $n \times n$ -matrices with entries taken from

It will be shown that there exists a one-to-one correspondence between the (two-sided) ideals of

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

and 0 in all other places. It can be easily checked that

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

and

respectively, say $x=a_{ij}$ and $y=b_{kl}$ . Then the matrix $A \cdot E_{jl} +E_{ik}\cdot B \in \mathfrak{m}$ has

at position

, 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

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

thus, $rx, xr \in \mathfrak{i}$ . This shows that $\mathfrak{i}$ is an ideal in

. 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

$\displaystyle A=\sum\limits_{1 \le i,j \le n} a_{ij}E_{ij}, a_{ij} \in \mathfrak{i}$

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