Showing that $x^p - a$ either splits or is irreducible for characteristic $p$ (prime) in a field F. Let $F$ be a field of characteristic $p$ and let $f (x) = x^p- a \in F[x]$.
Show that $f (x)$ is irreducible over $F$ or $f (x)$ splits in $F$.
If I assume that $f(x)$ doesn't split, then $f(x)$ cannot be written as a product of linear factors.  But I can't find any way to progress towards showing that $f(x)=g(x)h(x)$ where one of them is a unit.
Assuming it's reducible and showing that it splits:
If I let $f(x) = p(x)q(x)$ where neither $p(x)$ or $q(x)$ are units, I know I need to show that $f(x)$ can be written as a product of only linear factors, but I can't see how I would show this either.
 A: Let $L/F$ be a field extension which contains a root of $X^p-a,$ call it $\xi \in L.$
Since char$(F)=p$ we have char$(L)=p$ and so $(X-\xi)^p=X^p-\xi^p=X^p-a$ in $L[X].$
So if $X^p-a$ is reducible in $F[X]$, we must have that $(X-\xi)^r \in F[X]$ for some $1\leq r < p.$
$\Big[$Proof: Suppose that $X^p-a=g(X)h(X)$ where $g,h \in F[X]$ are both non-constant.
Since $F\subseteq L$, we can view this equation in $L[X].$ In which case, we have
$$(X-\xi)^p=g(X)h(X).$$
By uniqueness of factorization in $L[X],$ it follows that 
$$g(X)=(X-\xi)^r \text{ and } h(X)=(X-\xi)^{p-r} \text{ for some } 1\leq r <p.$$
Since $g(X) \in F[X],$ it thus follows that $(X-\xi)^r \in F[X].$ $\Big]$
Expanding out, we get $X^r-r\xi X^{r-1}+\ldots+(-\xi)^r \in F[X].$
In particular, $r\xi \in F$ and as $r\neq 0$ in $F$ (because $1\leq r <p$), we conclude that $\xi \in F.$
Hence $X^p-a$ splits completely in $F[X]$ as $(X-\xi)^p.$
This proves your claim :)
Appendix
Let $K$ be any field of characteristic $p.$
Claim: For any $x,y \in K,$ we have $(x+y)^p=x^p+y^p.$
Proof: Follows from binomial expansion and the fact that $\binom{p}{r}$ is divisible by $p$ for all $1\leq r <p.$
A: Hint:The Frobenius is an automorphism, so there exists $u\in F, u^p=a, (X-u)^p=X^p-a$.
