Prove that if $ x^p −1 $ splits completely, then $x^p−a $ for $a\in F$ is either irreducible or splits I am stuck on proving  if $ x^p −1 $ splits completely, then $x^p−a $ for $a\in F$ is either irreducible or splits. I have searched the related questions: If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. 
Showing that $x^p - a$ either splits or is irreducible for characteristic $p$ (prime) in a field F.
But it seems that my question is a little different and so far, I cannot get the answer of it by thinking of methods in these questions. Could someone please help? Thanks a lot! 
I have not seen a question with the assumption "$x^p-1$ splits" in any related question, also the similar questions have not consider the case $char(F)=0$. So what is special here? May I please ask if with some modification we cam apply the proof in other questions to this question?
I think if I can show that $\sqrt[p]{\alpha}\in F$, then we are done, so  how can I do that  without the uasge of Galois theory?
 A: Suppose that $x^p-a$ is reducible over $F$. Then there is an irreducible factor $f(x)$ of $x^p-a$ with $\deg f<p$.
Let $\alpha$ be a root of $f$, and let $E=F(\alpha)$. Then $\alpha^p=a$ since $f(x)$ divides $x^p-a$. Hence if $d=\deg f$, then
$$ a^d=N_{E/F}(a)=N_{E/F}(\alpha^p)=N_{E/F}(\alpha)^p$$
Next, $d$ and $p$ are coprime since $d<p$, so there are integers $r$ and $s$ such that $rd+sp=1$. Therefore
$$a=a^{rd}a^{sp}=N_{E/F}(\alpha)^{rp}a^{sp}$$
so in fact $a$ is the $p$th power of an element of $F$, say $a=b^p$.
This means that $x^p-a$ has a root in $F$ (namely $b$), but by hypothesis $F$ contains all of the $p$th roots of unity (though note that if $\mathrm{char}(F)=p$ as in the linked questions then there is only one $p$th root of unity), so in fact $F$ contains all $p$th roots of $a$, so $x^p-a$ splits in $F$.
A: Here is a more elementary way to do the case where the characteristic of $F$ is $0$. 
Notation
Let $\alpha$ be a root of $X^p-a$, and let $\zeta\not=1$ be a root of $X^p-1$. Then the roots of $X^p-1$ are $\zeta^i$ for $i=0,\dots,p-1$; and the roots of $X^p-a$ are $\alpha\zeta^i$ for $i=0,\dots,p-1$.
A useful fact
I assume that you know the Eisentein criterion proof that $X^{p-1}+X^{p-2}+\dots+X+1$ is irreducible over $\mathbb{Q}$.
By the irreducibility of $X^{p-1}+X^{p-2}+\dots+X+1$  over $\mathbb{Q}$ we see that if $a_0+a_1\zeta+\dots+a_{p-1}\zeta^{p-1}=0$ with all $a_i\in\mathbb{Q}$, then for some $c$, $a_i=c$ for all $i$.
The result
Suppose $X^p-a$ has a non-constant factor $g(X)=X^r-cX^{r-1}+\dots$.
Then $c\in F$ is a sum of some of the roots of $X^p-a$; that is some
$$\alpha(\zeta^{r_1}+\zeta^{r_2}+\dots+\zeta^{r_k})\in F$$
so that either $\alpha$ (and so all roots of $X^p-a$) are in $F$, or else
$(\zeta^{r_1}+\zeta^{r_2}+\dots+\zeta^{r_k})=0$. 
By the Useful Fact this latter possibility can only happen when we have all the roots present, and then $g(X)=X^p-a$ after all.
A: Since this was already answered, here's a way to do it with just a bit of Galois theory. Let $\omega$ be a primitive $p$-th root of unity which is in $F$ since $x^p-1$ splits. Let $L$ be the splitting field of $x^p-a$.
Then if $\beta\in L$ is a root of $x^p-a$, the complete set of roots in $L$ must be $\{\beta,\beta\omega,\beta\omega^2,\dots,\beta\omega^{p-1}\}$. This implies that if $F$ contains a root of $x^p-a$, then it must contain them all.
Suppose $F$ does not contain a root of $x^p-a$. Then the minimal polynomial $m_{\beta,F}$ must have degree at least 2. So there is some other root $\beta\omega^j$ of $m_{\beta,F}$ in $L$ where $1\leq j<p$. Then there is an
$F$-homomorphism $\sigma:L\to L$ such that $\sigma(\beta)=\beta\omega^j$. But since $\sigma$ fixes elements of $F$ we have $\sigma(\beta\omega^j)=\sigma(\beta)\sigma(\omega^j)=\beta \omega^j \omega^j$. Since $j$ is coprime to $p$ it's easy to see that $\sigma$ has order $p$. But then $\sigma^k(\beta)$ is a root of $m_{\beta,F}$ for all $k\geq 1$, which implies that $m_{\beta,F}$ has $p$ distinct roots. Since $m_{\beta,F}$ is a divisor of $x^p-a$ it follows that $m_{\beta,F}=x^p-a$. Hence $x^p-a$ is irreducible in this case.
