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The question I am dealing with is: Let F be a field, $f(x)\in F[x]$ be irreducible and let $N/F$ be normal field extension. Let

$$f(x) = g_1(x) \cdot \dots \cdot g_r (x)$$

be the factorization of $f(x)$ into irreducibles in $N[x]$. Show that $G_{N/F}$ acts transitively on the $g_i$'s.

It doesn't really spezify what kind of field F is. So I don't know if I can assume $N/F$ is Galois. Or if I can assume that $f$ is separable. Because if I cant assume those things than I don't know how to do it. Please help! Thanks

Thomas
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    I think $G_{N/F}$ acts on the $g_i$'s by acting on their coefficients. That action is like a permutation in the index set ${1,2,\dots, r}$. The question is, why is this action transitive? – Thomas Jan 25 '13 at 18:21
  • Is it easier if we can assume that f is separable? – Thomas Jan 25 '13 at 19:32
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    Consider the splitting field of $f$: $L/N$. Then $G_{L/F}$ acts transitively on the roots of $f$. If you want to find some $\sigma\in G_{N/F}$ that takes $g_i$ to $g_j$, consider some $\overline{\sigma}\in G_{L/F}$ that takes a root of $g_i$ to a root of $g_j$. The restriction of this element is your candidate. – Matt Jan 25 '13 at 20:33

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