We let be a simple -algebra. Then , the n by n matrix algebra over is simple. In particular if is a division algebra, then is simple.
We will split our proof into 2 Lemmas.
Lemma 1: If , then .
Proof: Let .
. Therefore .
, thus . Similarly . Thus is indeed an ideal of .
Lemma 2 (Tricky part, takes some time to digest): If , then for some .
A technical thing we need to know is the matrix, which is a n by n matrix with 1 in the (i,j) entry, 0 elsewhere. A key property is that if , and 0 otherwise (zero matrix).
The key idea here is that can be taken to be . We can check that 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 .
Let us begin the proof to first show . Let . We write . For any ,
Therefore . Therefore . What we are actually doing here is first pick an arbitrary matrix in . Then, we do the “shifting” process to show that any (p,q) entry in can be “shifted” to the (1,1) entry, thus it is inside our ideal . Since any arbitrary (p,q) entry in is inside the ideal , we conclude that is an element of .
Next part is to show . Let . We have for all . We compute that
Therefore . We will illustrate what we are doing here by an example in the 2 by 2 case. Let us have . Since is in , what we have, by definition of , is that , , , are all in . We then proceed to “shift” them all into their correct positions: , , , , all of which are still in the ideal . is the sum of all of them, thus also in .
Since is simple, its ideals are 0 and itself. Thus, the ideals of will be also the 0 matrix and itself, and thus is simple. A division algebra is simple (Any ideal containing a nonzero element , when multiplied by its inverse will give 1. Thus the ideal will contain 1 and thus the entire ring ), thus is also simple.