# Finite order endomorphisms

Why the finite order endomorphisms are diagonalizable over the complex numbers (or any algebraically closed field where the characteristic of the field does not divide the order of the endomorphism) with roots of unity on the diagonal?

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Dear @Matt: Let $a$ be the endomorphism and let $n$ be its order. Then $a$ is annihilated by $X^n-1$, whose roots are simple in view of your assumption that the characteristic doesn't divide $n$. – Pierre-Yves Gaillard Jan 3 '12 at 18:18