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