Why is $x^p - \alpha$ irreducible in $F[x]$?

Question

Let $F$ be a field of odd prime characteristic.

We have $\alpha \in F$ such that $x^p - \alpha$ has no root in $F$.

How can I show that $x^p - \alpha$ is irreducible in $F[x]$?

Answer 1

If $p$ is the prime characteristic of $F$, then we have the equality $(a+b)^p = a^p + b^p$.

So especially if $x^p-\alpha$ has one root $\sqrt[p]{\alpha}$, then $x^p-\alpha = (x-\sqrt[p]{\alpha})^p$. So we know (we can calculate in a field where we have adjoined $\sqrt[p]{\alpha}$) if $x^p-a = P \cdot Q \in F[x]$ irreducible, then $P = (x-\sqrt[p]{\alpha})^{k},Q = (x-\sqrt[p]{\alpha})^{p-k}$ for $1\leq k < p$.

By distributive law $P = x^k - k\sqrt[p]{\alpha} x^{k-1} + \dots \in F[x]$, so $F$ contains $\sqrt[p]{\alpha}$, a contradiction.

Answer 2

Whether $p$ is equal to the characteristic or not, in both cases $f=X^p-\alpha$ factors as $f=\prod_{i=1}^p (X-\zeta_i \alpha^{1/p})$ where the $\zeta_i$ are $p$-th roots of unity (when $p$ is the characteristic, we have $\zeta_i=1$ for all $i$, otherwise the $\zeta_i$ are all distinct, but this won't even matter for what follows). Hence if $g$ is a monic irreducible polynomial dividing $f$, then its constant term must be $g(0)=\zeta\alpha^{d/p}$, where $d = \deg g$ and where $\zeta$ is some $p$-th root of unity.

Now, if $0 < d < p$, then there exist integers $a,b$ with $ad+bp=1$, and therefore $(g(0)^a\alpha^b)^p=g(0)^{ap} \alpha^{bp}=\zeta^{ap}\alpha^{ad+bp}=\alpha$, hence the element $g(0)^a\alpha^b$ contained in the base field $F$ is a root of $f$. Hence if $f$ is reducible, then it has a root in $F$.