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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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Because the endomorphism is annihilated by a polynomial with simple roots.

(The fact that the eigenvalues are roots of unity is clear. Indeed, the diagonal matrix has also finite order.)

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how do you verify your first claim? –  Matt Jan 3 '12 at 18:11
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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

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