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.

Reference: Associative Algebras (Graduate Texts in Mathematics)

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.

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