MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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 $P(X)\in F[X]$?


share|cite|improve this question
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
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.