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 $p$th 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!