ag.algebraic geometry – How do I take the n-th root of a linear operator? Answer

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ag.algebraic geometry – How do I take the n-th root of a linear operator?

Suppose I’ve a vector house $$V/overline{mathbb Q}$$ over some algebraically closed bailiwick and a semi-simple operator $$T$$ on it with eigenvalues $$alpha_1,dots,alpha_r$$ such that $$|alpha_i| = 1$$ underneath each complicated embedding.

Is there a pleasant (functorial?) route to outline a vector house $$V^{1/n}$$ of dimension $$n$$ with an operator $$T^{1/n}$$ with eigenvalues $$alpha_1^{1/n},dots,alpha_r^{1/n}$$ (the place we take all workable roots)?

(Motivation: I got here throughout Abelian varieties over finite fields the place the corresponding Frobenius operators have the above relation.)

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