Why is x^p - a irreducible over a field of characteristic p?

[imurme8]

If K is a field of characteristic p, and there exists an element a \in K
which is not a pth power (i.e. the Frobenius endomorphism is not
surjective), then I am told we can show x^p - a is an irreducible polynomial
(and since it is not separable our field is imperfect). I see that
x^p - a has no roots in K, but how do we know that there does not exist
any factorization of x^p -a into factors of lesser degree?

 

[Petek]

Hint: Let f(x) = x^p - a, let F be a splitting field for f and let \alpha be a root of f(x) in F. Can you take it from there?

 

[imurme8]

OK, how about this? Over F we have \alpha as the only root of f(x) (with multiplicity p). Let f(x)=p_1(x)\dotsb p_n(x) be a factorization of f(x) in K[x] into monic irreducibles. Then each of these must be the minimal polynomial of \alpha over K. So they all must have the same degree. So the degree of p_1(x)=\dotsb=p_n(x) divides p, so it is either 1 or p. But it cannot be 1, so it must be p, so f(x) itself is irreducible.

Is that what you would suggest? I don't see an easier way to show (x-\alpha)^m\notin K[x] for all positive integers m<p.

 

[Petek]

Your argument looks OK. Here's what I had in mind: We have that f(x) = (x - \alpha)^p = x^p - \alpha^p and so a = \alpha^p. Suppose that f(x) = g(x)h(x) is a proper, non-trivial factorization of f(x) over K. Then, by unique factorization, g(x) = (x - \alpha)^s, for some 0 < s < p. The constant term of g(x) is \alpha^s, so \alpha^s \in K. Since gcd(s, p) = 1, there exist integers m and n such that sm + pn = 1. Therefore, \alpha = \alpha^{sm+pn} = (\alpha^s)^m (\alpha^p)^n is an element of K. But a = \alpha^p, contradicting the assumption that a is not a pth power in K. Therefore, no such factorization of f(x) exists.

 

[imurme8]

I like your way, very elegant. Thanks for the help!