Matrix rings over simple rings are simple

Remarks

Let $J$ be a two-sided ideal of $M_n(R)$ and let $I$ be the set of all entries of all matrices in $J$. With the matrix units $E_{kl}$, one has $E_{1i} A E_{j1} = a_{ij} E_{11}$, and multiplying by scalar matrices and matrix units moves an entry to any position. From this, $I$ is a two-sided ideal of $R$ and $J = M_n(I)$. For example, the ideals of $M_2(\Z)$ are the sets $M_2(m\Z)$ for $m \ge 0$. If $R$ is simple, the only choices are $I = 0$ and $I = R$, so $M_n(R)$ has only the ideals $0$ and $M_n(R)$.

Going deeper

For a student

The bijection $I \mapsto M_n(I)$ between ideals is one instance of Morita equivalence: $R$ and $M_n(R)$ have equivalent module categories and the same lattice of two-sided ideals. Combined with Wedderburn's theorem it shows that the simple Artinian rings are exactly the rings $M_n(D)$ with $D$ a division ring. Note that $M_n(R)$ has many one-sided ideals even when $R$ is a field, such as the matrices whose only nonzero column is the first one.