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Let $E$ be a field and let $F$ be a finite Galois extension of $E$.

Let $h(x)$ be an irreducible monic polynomial in $E[x]$, and $h_{1}(X),h_{2}(X)$ be two irreducible monic polynomials in $F[X]$, both of which divide $h(x)$.

I want to show that exists an automorphism $\theta$ of $F[X]$ such that $\theta$ leaves all elements in $E[X]$ fixed and furthermore $\theta (h_{1})=h_{2}$.

case one is h1,h2 are both linear polynomial, then its done. Suppose not, case two, I have to show that they have the same order, this is where I stuck

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@ YACP , this is not a homework, I am just doing problems preparing for the exam. for this problem, case one is $h_{1},h_{2}$ are both linear polynomial, then its done. Suppose not, case two, I have to show that they have the same order, this is where I stuck –  user53800 Dec 29 '12 at 22:21

1 Answer 1

up vote 3 down vote accepted

Let $G$ be the Galois group of $F/E$. Let $P = (h(X))$ be the ideal of $E[X]$. Let $Q_i = (h_i(X))$ be the ideal of $F[X]$ for $i = 1, 2$. Since $P \subset Q_1 \cap E[X]$ and $P$ is a maximal ideal, $P = Q_1 \cap E[X]$. Let $g = \prod_{\sigma\in G} \sigma(h_1)$. Then $g \in Q_1 \cap E[X] = P$. Hence $g \in Q_2$, i.e. $g$ is divisible by $h_2$. Hence there exists $\sigma \in G$ such that $\sigma(h_1) = h_2$.

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Thanks for your help, but I have a question, how do you conclude that $g\in Q_{1}\cap E[X]$, of course g is in $Q_{1}$, but why it is also in E[X]? –  user53800 Dec 30 '12 at 14:45
    
@user53800 Because it is invariant under the action of $G$. –  Makoto Kato Dec 30 '12 at 17:30
    
Got it, thanks a lot. –  user53800 Dec 30 '12 at 17:39

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