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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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