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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
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Let $\mathfrak{m}$ be an ideal in $M_{n \times n}(R)$ .
Claim 1 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 $\mathfrak{m}=M_{n \times n}(\mathfrak{i})$ .
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 $\mathfrak{m} \subseteq M_{n \times n}(\mathfrak{i})$ . 
A consequence of this is: If $F$ is a field, then $M_{n \times n}(F)$ is simple.
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