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Let $F$ be a field of characteristic 0, $p$ an odd prime and $a\in F$ not a p-th power. Then $x^p-a$ is irreducible over $F$. To prove this we assume the contrary, put $\alpha\not\in F$ a root of some irreducible factor of $x^p - a$ of degree $m$ and take the norm: $$ N^{F(\alpha)}_F(a) = a^m = N^{F(\alpha)}_F(\alpha^p) = (N^{F(\alpha)}_F(\alpha))^p = \gamma^{p},\quad \gamma\in F. $$ Since $(m,p) = 1$, we have $1 = xm + yp$ and $\gamma^{xp} = a\cdot a^{-yp}$, and it follows that $a$ is a p-th power (of $\gamma^x a^y$) in $F$.

This is part (a) of problem VII.3.3 of Lang's Undergraduate Algebra. The task of part (b) is to prove that if $a\in F$ is not a p-th power, then for any positive integer $r$ the polynomial $x^{p^r} - a$ is irreducible. I would like to extend this method, but it runs into a problem of taking roots: just as before, we can write $$ N^{F(\alpha)}_F(a) = a^m = N^{F(\alpha)}_F(\alpha^{p^r}) = (N^{F(\alpha)}_F(\alpha))^{p^r} = \gamma^{p^r},\quad \gamma\in F. $$

Now, if $m$ and $p$ are relatively prime, the same argument goes through well enough. However, if $p|m$, we have $m = p^k\cdot n$ where $k<r$ and $(n,p) = 1$, from which we get $$ \left(a^n\right)^{p^k} = \left(\gamma^{p^{r-k}}\right)^{p^k}. $$ This gives $$ a^n = \gamma^{p^r-k}\zeta^d, $$ where $\zeta$ is a primitive $p^k$-th rood of unity, $\zeta^d\in F$ and by inductive hypothesis we see that $a^n\zeta^{-d}$ is a p-th power in $F$ (because $\gamma$ is a root of $x^{p^{r-k}} - a^n\zeta^{-d}$). That is $$ a^n = \delta^p\zeta^d,\quad\delta\in F. $$ If $p$ does not divide $d$, then I don't see a way of getting rid of the root of unity. Can this argument be finished?

Thank you!

P.S. Lang does give a hint, but I don't get it.

share|cite|improve this question
What is the hint given by Lang? – user38268 Jan 11 '13 at 13:09
"Use induction. Distinguish the cases whether a root $\alpha$ is a p-th power in $F(\alpha)$ or not, and take the norm from $F(\alpha)$ to $F$." – Artem Jan 14 '13 at 11:22

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