## Let A be a simple R-algebra, then M_n(A) is simple

We let $A$ be a simple $R$-algebra. Then $M_n(A)$, the n by n matrix algebra over $A$ is simple. In particular if $D$ is a division algebra, then $M_n(D)$ is simple.

We will split our proof into 2 Lemmas.

Lemma 1: If $I\lhd A$, then $M_n(I)\lhd M_n(A)$.

Proof: Let $X=(x_{ij}), Y=(y_{ij})\in M_n(I), Z=(z_{ij})\in M_n(A)$.

$(X-Y)_{ij}=x_{ij}-y_{ij}\in I$. Therefore $X-Y\in M_n(I)$.

$(XY)_{ij}=\sum_{k=1}^n x_{ik}y_{kj}\in I$, thus $XY\in M_n(I)$. Similarly $YX\in M_n(I)$. Thus $M_n(I)$ is indeed an ideal of $M_n(A)$.

Lemma 2 (Tricky part, takes some time to digest): If $J\lhd M_n(A)$, then $J=M_n(I)$ for some $I\lhd A$.

A technical thing we need to know is the $E_{ij}$ matrix, which is a n by n matrix with 1 in the (i,j) entry, 0 elsewhere. A key property is that $E_{ij}E_{kl}=E_{il}$ if $j=k$, and 0 otherwise (zero matrix).

The key idea here is that $I$ can be taken to be $I=\{r\in A\mid rE_{11}\in J\}$. We can check that $I$ is indeed an ideal. We need to know that it is possible to premultiply and postmultiply a given matrix by “elementary matrices” such that any entry of the given matrix is “shifted” to the (1,1) entry. An example will be $\begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&0\\1&0\end{pmatrix}=\begin{pmatrix}d&0\\0&0\end{pmatrix}$.

Let us begin the proof to first show $J\subseteq M_n(I)$. Let $X=(x_{ij})\in J$. We write $X=\sum_{i,j}^n x_{ij}E_{ij}$. For any $1\leq p,q\leq n$,

\begin{aligned}E_{1p}XE_{q1}&=E_{1p}(\sum x_{ij}E_{ij})E_{q1}\\ &=(\sum x_{pj}E_{1j})E_{q1}\\ &=x_{pq}E_{11}\in J \end{aligned}

Therefore $x_{pq}\in I$. Therefore $X\in M_n(I)$. What we are actually doing here is first pick an arbitrary matrix $X$ in $J$. Then, we do the “shifting” process to show that any (p,q) entry in $X$ can be “shifted” to the (1,1) entry, thus it is inside our ideal $I$. Since any arbitrary (p,q) entry in $X$ is inside the ideal $I$, we conclude that $X$ is an element of $M_n(I)$.

Next part is to show $M_n(I)\subseteq J$. Let $X=(x_{ij})\in M_n(I)$. We have $x_{ij}E_{11}\in J$ for all $i,j$. We compute that

\begin{aligned}E_{i1}(x_{ij}E_{11})E_{1j}&=x_{ij}E_{i1}E_{1j}\\ &=x_{ij}E_{ij}\in J \end{aligned}

Therefore $X=\sum x_{ij}E_{ij}\in J$. We will illustrate what we are doing here by an example in the 2 by 2 case. Let us have $X=\begin{pmatrix}a&b\\c&d\end{pmatrix}$. Since $X$ is in $M_n(I)$, what we have, by definition of $I$, is that $\begin{pmatrix}a&0\\0&0\end{pmatrix}$$\begin{pmatrix}b&0\\0&0\end{pmatrix}$$\begin{pmatrix}c&0\\0&0\end{pmatrix}$$\begin{pmatrix}d&0\\0&0\end{pmatrix}$ are all in $J$. We then proceed to “shift” them all into their correct positions: $\begin{pmatrix}a&0\\0&0\end{pmatrix}$$\begin{pmatrix}0&b\\0&0\end{pmatrix}$$\begin{pmatrix}0&0\\c&0\end{pmatrix}$$\begin{pmatrix}0&0\\0&d\end{pmatrix}$, all of which are still in the ideal $J$. $X$ is the sum of all of them, thus also in $J$.

## Final Conclusion

Since $A$ is simple, its ideals are 0 and itself. Thus, the ideals of $M_n(A)$ will be also the 0 matrix and itself, and thus is simple. A division algebra $D$ is simple (Any ideal containing a nonzero element $x$, when multiplied by its inverse will give 1. Thus the ideal will contain 1 and thus the entire ring $D$), thus $M_n(D)$ is also simple.