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Let $F$ be a field of characteristic $p>0$, and let $a \in F - F^{p}$. Show that $x^{p} - a$ is irreducible over $F$.

I didn't get a good idea to solve that question. But I'm not looking for an answer, I would just like a hint. Thanks for the advance.

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  • $\begingroup$ If $\alpha$ is a root for $x^p - a$ in any splitting field, then $(x - \alpha)^p = x^p - a$. Suppose that $x^p -a$ is reducible in $F$; namely $x^p -a = f(x)g(x)$. Using unique factorization, how does $g$ (or $h$) look like? $\endgroup$ – user512346 May 17 '18 at 23:46
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Let $\bar F$ be the algebraic closure of $F$, $X^p-a$ has a root $b$ in $\bar F$ $b$ is not in $F$ since $a$ is not in $F^p$, $(X-b)^p=X^p-b^p=X^p-a$. Suppose that $X^p-a=f(X)g(X)=(X-b)^p, n,m>1$, we deduce that $f(X)=(X-b)^n, g(X)=(X-b)^m$ with $n+m=p$. The coefficient of $X^{n-1}$ in $(X-b)^n$ is $nb$, we deduce that $nb\in F$, since $gcd(n,p)=1$, there exists integer $u,v$ $un+vp=1$, $unb=(1-vp)b=b\in F$. Contradiction.

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