| Version current |
Version 6 |
| Let $R$ be a ring with 1. Consider the ring $M_{n \times n}(R)$ of $n \times n$-matrices with entries taken from $R$. |
Let $R$ be a ring with 1. Consider the ring $M_{n \times n}(R)$ of $n \times n$-matrices with entries taken from $R$. |
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| It will be shown that there exists a one-to-one |
It will be shown that there exists a one-to-one |
| correspondence between the (two-sided) ideals of $R$ and the |
correspondence between the (two-sided) ideals of $R$ and the |
| (two-sided) ideals of $M_{n \times n}(R)$. |
(two-sided) ideals of $M_{n \times n}(R)$. |
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| For $1 \le i,j \le n$, let $E_{ij}$ denote the $n |
For $1 \le i,j \le n$, let $E_{ij}$ denote the $n |
| \times n$-matrix having entry 1 at position |
\times n$-matrix having entry 1 at position |
| $(i,j)$ and 0 in all other places. It can be |
$(i,j)$ and 0 in all other places. It can be |
| easily checked that |
easily checked that |
| \begin{equation} |
\begin{equation} |
|
E_{ij} \cdot E_{kl}=\left\{
|
E_{ij} \cdot E_{kl}=\left(
|
| \begin{array}{lllll} |
\begin{array}{lllll} |
| 0 & \mbox{iff}& k \ne j \\ |
0 & \mbox{iff}& k \ne j \\ |
| E_{il} & \mbox{otherwise.} |
E_{il} & \mbox{otherwise.} |
| \end{array}\right. |
\end{array}\right. |
| \end{equation} |
\end{equation} |
| Let $\mathfrak{m}$ be an ideal in $M_{n \times |
Let $\mathfrak{m}$ be an ideal in $M_{n \times |
| n}(R)$. |
n}(R)$. |
| \begin{claim} |
\begin{claim} |
| The set $\mathfrak{i}\subseteq R$ given by |
The set $\mathfrak{i}\subseteq R$ given by |
|
\[\mathfrak{i}=\{x \in R \mid x\quad\mbox{is an entry
|
\[\mathfrak{i}=\{x \in R \mid x \mbox{is an entry
|
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of } A \ A \in \mathfrak{m}\}\]
|
of } A \mid A \in \mathfrak{m}\}\]
|
| is an ideal in $R$, and |
is an ideal in $R$, and |
| $\mathfrak{m}=M_{n \times n}(\mathfrak{i})$. |
$\mathfrak{m}=M_{n \times n}(\mathfrak{i})$. |
| \end{claim} |
\end{claim} |
| \begin{proof} |
\begin{proof} |
| $\mathfrak{i} \ne \emptyset$ since $0 \in |
$\mathfrak{i} \ne \emptyset$ since $0 \in |
|
\mathfrak{i}$. Now let $A=(a_{ij})$ and $B=(b_{ij})$
|
\mathfrak{i}$. Now let $A=(a_{ij}$ and $B=(b_{ij}$
|
| be matrices in $\mathfrak{m}$, and $x,y \in R$ be |
be matrices in $\mathfrak{m}$, and $x,y \in R$ be |
|
entries of $A$ and $B$ respectively, say
|
entries of $A$ and $B$ respectively-- say
|
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$x=a_{ij}$ and $y=b_{kl}$. Then the matrix $A \cdot
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$x=a_{ij}$ and $y=b{kl}$. Then the matrix $A \cdot
|
| E_{jl} +E_{ik}\cdot B \in \mathfrak{m}$ has $x+y$ |
E_{jl} +E_{ik}\cdot B \in \mathfrak{m}$ has $x+y$ |
| at position $(i,l)$, and it follows: If $x,y \in |
at position $(i,l)$, and it follows: If $x,y \in |
| \mathfrak{i}$, then $x+y \in \mathfrak{i}$. Since |
\mathfrak{i}$, then $x+y \in \mathfrak{i}$. Since |
| $\mathfrak{i}$ is an ideal in $M_{n \times n}(R)$ |
$\mathfrak{i}$ 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 |
it contains, in particular, the matrices $D_r \cdot A$ and $A \cdot D_r$, where |
| \begin{equation*} |
\begin{equation*} |
| D_r :=\sum_{i=1}^n r\cdot E_{ii}, r \in R. |
D_r :=\sum_{i=1}^n r\cdot E_{ii}, r \in R. |
| \end{equation*} |
\end{equation*} |
| thus, $rx, xr \in \mathfrak{i}$. This shows |
thus, $rx, xr \in \mathfrak{i}$. This shows |
| that $\mathfrak{i}$ is an ideal in $R$. |
that $\mathfrak{i}$ is an ideal in $R$. |
| Furthermore, $M_{n \times n}(\mathfrak{i}) |
Furthermore, $M_{n \times n}(\mathfrak{i}) |
| \subseteq \mathfrak{m}$. |
\subseteq \mathfrak{m}$. |
|
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| By construction, any matrix $A \in \mathfrak{m}$ |
By construction, any matrix $A \in \mathfrak{m}$ |
| has entries in $\mathfrak{i}$, so we have |
has entries in $\mathfrak{i}$, so we have |
| \begin{equation*} |
\begin{equation*} |
| A=\sum\limits_{1 \le i,j \le n} a_{ij}E_{ij}, |
A=\sum\limits_{1 \le i,j \le n} a_{ij}E_{ij}, |
| a_{ij} \in \mathfrak{i} |
a_{ij} \in \mathfrak{i} |
| \end{equation*} |
\end{equation*} |
| so $A \in m_{n \times n}(\mathfrak{i})$. Therefore |
so $A \in m_{n \times n}(\mathfrak{i})$. Therefore |
| $\mathfrak{m} \subseteq M_{n \times n}(\mathfrak{i})$. |
$\mathfrak{m} \subseteq M_{n \times |
|
n}(\mathfrak{i}$. |
| \end{proof} |
\end{proof} |
| A consequence of this is: If $F$ is a field, then $M_{n \times n}(F)$ is simple. |
A consequence of this is: If $F$ is a field, then $M_{n \times n}(F)$ is simple. |
