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    Divisibility proofs

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    Hello, I need a detailed solution to the attached problem:

    Let A be a finite dimensional F-algebra, a is in A a unit of finite order m.....

    Hint for part (a):

    p | x^k - 1 <=> x^k = 1 in F[x]/p <=> m|k since m = ord(x) in F[x]/p

    © BrainMass Inc. brainmass.com May 20, 2020, 5:37 pm ad1c9bdddf
    https://brainmass.com/math/basic-algebra/divisibility-proofs-minimal-polynomial-233949

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    Proof:
    Let's check the condition.
    Since a has order m in the group of units of A, then a^m = 1, then a^m - 1 = 0.
    So a is the root of the polynomial g(x) = x^m - 1.
    Since p(x) is the minimal polynomial of a, then p(a) = 0.
    From these two conditions, because p(x) is minimal, we must have p(x)|g(x), which
    means that p(x) ...

    Solution Summary

    This provides an example of divisibility proofs with a minimal polynomial.

    $2.19

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