for any ring $A$ the matrix ring $M_n(A)$ is simple if and only if $A$ is simple

All ideals below are two-sided ideals. I also assume $A$ has a unit.

• If $I$ is an ideal of $A$, then $J=M_n(I)$ is an ideal of $M_n(A)$.
  This is easy to prove from the definition of matrix addition and multiplication.

• If $J$ is an ideal of $M_n(A)$, then $J=M_n(I)$ for some $I$ is an ideal of $A$.
  Take $I$ to be the set of all $a \in A$ such that there is $M\in J$ having $a$ in one entry. Then $I$ is an ideal of $A$ and $J=M_n(I)$ because you can pre- and post- multiply matrices in $J$ by elementary matrices $E_{ij}$ to move entries to any desired place. ($E_{ij}$ is the matrix will all zeros except for the entry $ij$ which is $1$.)