 Simple spectrum gives simultaneous diagonalization

Suppose $D$ has $n$ distinct eigenvalues over a field and $C$ commutes with $D$. Each eigenspace of $D$ is one-dimensional. Since commuting maps preserve eigenspaces, $C$ acts as a scalar on each of them. The eigenvectors of $D$ give a common eigenbasis and hence simultaneous diagonalization. For real matrices, the result gives a real eigenbasis only if the eigenvalues of $D$ are real; otherwise it applies over $\mathbb{C}$.