Let $F$ be a field of characteristic $p > 0$ and $C$ be an element of $F$ that is not a $p$-power. For a positive integer $s$, show that $x^{p^s} − C$ is irreducible, and its splitting field is of the form $F(a)$ for some $a$ such that $a^{p^s} = C$.
For irreducibility, I thought to use the following theorem: Let $q > 1$ be a prime power, and let $f(x) \in F_q[x]$ be a non-constant
polynomial of degree $n > 1$. Then $f(x)$ is irreducible over $F_q$ if and only if for all $d$ such
that $1 \leq d \leq \frac{n}{2}$ we have $\operatorname{gcd}(f(x), x^{q^d} − x) = 1$. In this case, I'd have to show $\operatorname{gcd}(x^{p^s} − C, x^{p^{sd}} − x) = 1$, which at first glance seems immediate, but I'm afraid I can't think of a good reasoning to show it. Also, I cannot grasp why the splitting field is of this form, even though it also seems very intuitive.
Best Answer
Consider the following set:
$I=\{k\in\mathbb{Z}: c^k=b^{p^s} \ \textbf{for some} \ \ b\in F\}$
It's very easy to check that this is an ideal of $\mathbb{Z}$. It obviously contains $p^s$, and since every ideal of $\mathbb{Z}$ is principal we conclude that its generator is a power of $p$. Now let $a$ be an element in the algebraic closure such that $a^{p^s}=c$. Then:
$x^{p^s}-c=(x-a)^{p^s}$
This already tells us that the splitting field must be $F(a)$, there are no other roots. Now we have to show the polynomial is irreducible. Take an irreducible factor over $F$, it must be of the form $(x-a)^m$ for some $m$. Then $a^m\in F$, and note that:
$(a^m)^{p^s}=c^m$
This means $m\in I$. In particular, if $I=k\mathbb{Z}$ then $k\leq m$. We'll show that $k=m$. Indeed, take $b\in F$ such that $c^k=b^{p^s}$. Then $(a^k)^{p^s}=b^{p^s}$, and so $a^k=b\in F$. But remember, $k$ must be a power of $p$, so write $k=p^r$. Then $(x-a)^k=x^k-a^k\in F[x]$ divides $(x-a)^m$. Since we assumed $(x-a)^m$ is irreducible over $F$, it follows that $k=m$.
So now we have $m=p^r$. If we assume $r<s$ then this means $c=(a^{p^r})^{p^{s-r}}$, contradicting the assumption that $c$ is not a $p$-power in $F$. Thus we must have $r=s$. So we have shown that $(x-a)^{p^s}$ is the only irreducible factor of the polynomial, and so it is irreducible.