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Let $p$ be a prime number. Prove that for any field $k$ and any $a\in k$, the polynomial $f(x)=x^p-a$ is either irreducible or has a root.

I think if $\operatorname{Char}k=0$ then $f$ is an irreducible polynomial and if $\operatorname{Char}k=p$ for $p$ a prime number then $f=(x-a)^p$ so $f$ has a root.

This problem is in Galois Theory, by Miles Reid.

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  • $\begingroup$ If the field is $Z_p$, then $f(a)=0$ by Fermat's Little Theorem. $\endgroup$ – Joel Nov 4 '14 at 17:28
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    $\begingroup$ Your first assertion is false; if $k=\mathbb{R}$ or $k=\mathbb{C}$, then $f$ is not irreducible for most values of $p$, by the fundamental theorem of algebra. $\endgroup$ – mdp Nov 4 '14 at 17:36
  • $\begingroup$ Your argument for characteristic $p$ is only almost correct. The frobenius isomorphism $\phi\colon x\mapsto x^p$ need not leave $a$ untouched. However, we do find that $\phi^{-1}(a)$ is a root of $f$. - Also note that the characteristic may be nonzero and stil $\ne p$. $\endgroup$ – Hagen von Eitzen Nov 4 '14 at 18:48