# How to prove that the Galois group of a normal extension transitively permutes the factors of an irreducible polynomial?

How to do the following problem?

Let $K$ be a normal extension of $F$, and let $f(x)\in F[x]$ be an irreducible polynomial over $F$. Let $g(x)$ and $p(x)$ be monic irreducible factors of $f(x)$ in $K[x]$. Prove that there is a $z\in\operatorname{Gal}(K/ F)$, with $z(g)=p$.

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I think this is contained in the book: Foundations Of Algebraic Geometry by A.Weil. If you are interested you can have a look into it. I will vouch for this after I have the access to that book again. – awllower Feb 4 '12 at 13:29

Let $N/F$ be a normal extension, let $X$ be an indeterminate, let $f\in F[X]$ be an irreducible polynomial, and let $g_1,g_2\in N[X]$ be irreducible monic factors of $f$.

We want to find an $F$-automorphism of $N$ which maps $g_1$ to $g_2$.

Let $N^a$ be an algebraic closure of $N$, and $\alpha_i$ a root of $g_i$ in $N^a$.

The minimal polynomial of $\alpha_i$ over $F$ being $f$, there is an $F$-automorphism $\sigma$ of $N^a$ mapping $\alpha_1$ to $\alpha_2$.

As $N/F$ is normal, we have $\sigma N=N$.

The minimal polynomial of $\alpha_i$ over $N$ being $g_i$, our automorphism $\sigma$ maps $g_1$ to $g_2$.

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An optimally elegant proof ! – Georges Elencwajg Feb 5 '12 at 10:10

I wrote up a solution to this problem a while back. You can find it at the bottom of this page (3.13).

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Under the hypotheses, $f$ splits completely over $\bf K$, so $g$ and $p$ are linear. Thus the problem reduces to showing that if $\alpha$ and $\beta$ are roots of $f$ in $\bf K$ then there's an element of the Galois group that takes $\alpha$ to $\beta$. The proof of that result is in every exposition of Galois Theory.

EDIT: as awllower indicates, I misread the problem, and thought it was assumed that $f$ has a root in $\bf K$.

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Why does $f$ split completely over $K$? It is supposed to be irreducible only, is it not? – awllower Feb 4 '12 at 13:31
If $f$ has one root in $K$, then it splits of course, but I see no reason to assume so... – awllower Feb 4 '12 at 13:33
Just consider $KL$, where $L$ is the splitting field of $f$. – Arturo Magidin Feb 4 '12 at 20:43