# Divisibility proofs

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 ad1c9bdddfhttps://brainmass.com/math/basic-algebra/divisibility-proofs-minimal-polynomial-233949

#### Solution Preview

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