TITLE: Why is the ring of matrices over a field simple?
QUESTION [50 upvotes]: Denote by $M_{n \times n}(k)$ the ring of $n$ by $n$ matrices with coefficients in the field $k$.  Then why does this ring not contain any two-sided ideal?  
Thanks for any clarification, and this is an exercise from the notes of Commutative Algebra by Pete L Clark, of which I thought as simple but I cannot figure it out now.

REPLY [44 votes]: A faster, and more general result, which Arturo hinted at, is obtained via following proposition from Grillet's Abstract Algebra, section "Semisimple Rings and Modules", page 360:

Consequence: if $R:=D$ is a division ring, then $M_n(D)$ is simple.
Proof: Suppose there existed an ideal of $M_n(D)$. By the proposition, it'd be of the form $M_n(I)$, for $I\unlhd D$, but division rings do not have any ideals (other than $0$ and $D$), so this is a contradiction. $\blacksquare$