In his Class Field Theory notes, J. Milne show the following example.
I can prove that $M_n(k)$ is a simple $k$-algebra by showing that the ideal generated by any matrix is necessarily $M_n(k)$ itself, but I don't understand Milne's argument. If $J$ is a nonzero ideal, then surely it has to contain a minimal left-ideal and a minimal right-ideal. How can I conclude Milne's argument to prove that $J=M_n(k)$?
PS: GT 7.15a has no information about this.
To me it seems there is an enormous gap between the content discussed before the phrase “it follows that” and the assertion after it. All the discussion suggests is that the ideal must contain left and right ideals of those types, but it isn’t obvious why that means it would be the whole ring.
The closest thing I can see is this: use a nonzero element of the ideal, and right multiplication to get a nonzero element with exactly one nonzero column j. By the conversation, the ideal must contain $L(\{j\})$. Using more right multiplications, it must contain $L(\{i\})$ for all the other indices too. Therefore it contains their sum, which is the entire ring.
But all this hinges on explaining the line that has the word “arbitrary” on it. It apparently is supposed to convey that you can get any column vector by choosing an appropriate $A$, but this deserves more explanation than was given.