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 $F.$ 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 $F.$
Please help me. I'm absolutely clueless.
 A: Theorem (Capelli)
Let $p$ be a prime and $F$ be a field of any characteristic.
If the polynomial $x^p-a\in F[x]$ has no root in $F$, then it is irreducible.
Proof when $F$ is of characteristic $p$ ( the case actually asked by the OP)
Let $\alpha$ be a root of $x^p-a$ in a splitting field $K$ of that polynomial, so that $x^p-a=(x-\alpha)^p\in K[x]$,  and let $f(x)=Irr(\alpha, F,x)$ be the (irreducible !) minimal polynomial of $\alpha$ over $F$.
Then $f(x)$ divides $x^p-a$ (in $F[x]$ or $K[x]$) and so is of the form $(x-\alpha)^r$ with $1\leq r\leq p$, as is any divisor of $x^p-a=(x-\alpha)^p$ .
But since $f(x)\in F[x]$, we have $f(x)=(x-\alpha)^r=x^r-r\alpha x^{r-1}+\cdots \in F[x]$ so that $r\alpha\in F$ which [since $1\leq r\leq p$ and $\alpha\notin F$] is only possible if $r=p$.
But this means that $f(x)=(x-\alpha)^r=(x-\alpha)^p=x^p-a$ and so $x^p-a$ is indeed irreducible since $f(x)$ , being a minimal polynomial, is irreducible.  
Proof (much more difficult) when $F$ is not of characteristic $p$
Lang, Algebra Chapter VI, §9, Theorem 9.1  .
A: The following should be considered a long comment. I will answer your question in case $F$ is a finite field. Using Don Antonio's answer, it suffices to show that $x^p-a$ is reducible implies that it has a root in $F$. Consider $\phi:F\rightarrow F$ that sends $x$ to $x^p$. One verifies easily that  $f$ is a ring homomorphism. One also verifies easily that $Ker(f)$ is trivial, thus $f$ is injective. Since our field is finite, therefore $f$ is surjective. Thus, there exists $w\in F$ such that $w^p=a$.
Note: The above argument shows that $x^p-a$ always has root (hence always reducible) in case $F$ is finite and $Char(F)=p$ 
A: Let $E$ be the splitting field of $f(x)$ over $F$
and let $r$ be a root of $f(x)$ in $E$.
Then $f(r)=r^p-a=0$ and $r^p=a$ and $f(x)=x^p-a=x^p-r^p=(x-r)^p$ by $char F=p$ and Freshman's Dream.
Case I: If $r\in F$,
then $f(x)=(x-r)^p$ splits in $F[x]$.
Case II: If $r\notin F$.
Since $F[x]$ is a UFD,
suppose that $f(x)=g_1(x)g_2(x)\cdots g_n(x)$,
where $g_i(x)$ is irreducible and monic in $F[x]$ for each $i=1, 2, ..., n$.
(Note that $f(x)$ is monic.)
In addition,
$\deg{g_i(x)}\geq 2$ for each $i=1, 2, ..., n$.
For if $\deg{g_i(x)}=1$,
then $g_i(x)=(x-s)\in F[x]$
and $s\in F$ is a root of $f(x)$.
Which is the Case I.
Then we have $f(x)=g_1(x)g_2(x)\cdots g_n(x)=(x-r)^p$.
For each $i=1, 2, ..., n$,
since $\deg{g_i(x)}\geq 2$,
it follows that $g_i(x)$ has multiple root in $E$.
Recall that $g_i(x)$ is irreducible over $F$.
Which implies that $g_i(x)=h_i(x^p)$ for some $h_i(x)\in F[x]$.
(Here I use a theorem.)
Then $f(x)=x^p-a=h_1(x^p)h_2(x^p)\cdots h_n(x^p)$.
Compare the degree of this equation,
it must be $\deg{h_1(x)}=1$ and $h_2(x), h_3(x), ..., h_n(x)$ all are constant.
It follows that $g_2(x)=g_3(x)=\cdots =g_n(x)=1$ and $g_1(x)=f(x)$.
Therefore, $f(x)=g_1(x)$ is irreducible over $F$.
This question appears at the Gallian's Contemporary Abstract Algebra 8/e (Exercise 20.15).
A: Hint:
Suppose theres a root $\;\omega\in F\;$ of $\;f(x)\;$, so $\;\omega^p=a\;$,  but then
$$ x^p-a=(x-\omega)^p =\;\ldots$$
