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Hi,

Suppose that $E/F$ is a Galois extension. If $P(X)\in E[X]$ is a (EDIT: monic) polynomial of degree $n > 0$, such that $P(X)^n\in F[X]$, does it follow that $P(X)\in F[X]$?

Thanks

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Not if $n=0$ :P –  Kevin Buzzard Jun 9 '11 at 22:56
    
Actually, not for $n=2$ either. Try $E$ the complexes, $F$ the reals, and $P(X)=i(X^2+2X+3)$. –  Kevin Buzzard Jun 9 '11 at 23:00
1  
Seems to me counterexamples are thick on the ground. Let $a$ be any element of $F$ with no $n$th root in $F$, let $E$ be the splitting field of $x^n-a$, let $b$ be a zero of $x^n-a$ in $E$, let $Q$ be any polynomial of degree $n$ in $F[x]$, and let $P=bQ$. Is there some other condition you meant to add? If $P$ is monic and we're in characteristic zero then the conclusion does follow, and it isn't hard to prove. –  Gerry Myerson Jun 10 '11 at 1:23

1 Answer 1

up vote 3 down vote accepted

If $P$ is monic, the answer is Yes (in any characteristic, if the extension is separable), because a field automorphism over $F$ fixes $P^n$ and $(P^{\sigma})^n=P^n$ implies $P^{\sigma} = P$ for monic polynomials. In general, if $c$ is the leading coefficient of $P$, then $c^n \in F$ and $c^{-1}P\in F[X]$. Thus the counterexamples from the comments are the only ones.

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And of course if the extension is not separable you get things like $(x-\root p\of p)^p=x^p-p$. –  Gerry Myerson Jun 10 '11 at 23:13

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