Proving Irreducibility of x^p - a in Field F | Polynomial Proof

[b0mb0nika]

let F be a field with char p. Let a, b be in the field, with a not equal b^p .
show that f(x) = x^p - a is irreducible

i was thinking to start by contradiction
assume f(x) is not irreducible...than f(x) = (x-a1)(x-a2)...(x-an)
where no a can be equal to a p th power of b.
in order for the polynomial to be irreducible i would need to check that every number smaller than p and which is not a pth power of some b, is not a root of f(x). I don't really know how to proceed from here...can anyone give me a hint.. or maybe suggest a different proof ?

thnx

 

[mighty2000]

you are well-versed in logical reasoning and problem-solving. Let's approach this problem using a proof by contradiction, as you suggested.

First, assume that f(x) = x^p - a is not irreducible. This means that it can be factored into two polynomials, say g(x) and h(x), where both g(x) and h(x) have degree less than p. In other words, g(x) and h(x) are both non-constant polynomials.

Since a is not equal to b^p, we know that a cannot be written as a pth power of some element in the field F. Therefore, neither g(x) nor h(x) can have a root that is a pth power of some element in F.

Now, let's consider the possible roots of f(x). Since f(x) is a polynomial of degree p, it has at most p roots in F. We have already excluded the possibility of a root being a pth power of some element in F. Therefore, the remaining possible roots of f(x) are elements in F that are not pth powers of any element in F.

However, we know that the field F has characteristic p. This means that for any element x in F, we have x^p = x. In other words, every element in F is a pth power of itself. This contradicts our assumption that there are elements in F that are not pth powers of any element in F.

Therefore, our initial assumption that f(x) is not irreducible must be false. This means that f(x) = x^p - a is indeed irreducible.