# the proof of a proposition of galois theory

Let f(x) be a separable polynomial over the field K, with roots $r_1 , ... , r_n$ in it's splitting field F. Then f(x) is irreducible over K if and only if Gal(F/K) acts transitively on the roots of f(x).

How can I prove this theorem?

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Do you know the extension theorem for field embeddings? It will be very helpful I think. – Mike B Sep 28 '12 at 2:12

Let $G = Gal(F/K)$. Let $R = \{r_1,\dots,r_n\}$.

Suppose $f(X)$ is irreducible. Let $\pi\colon K[X] \rightarrow K[X]/(f(X))$ be the canonical homomorphism. There exists $K$-isomorphism $\psi_i\colon K[X]/(f(X)) \rightarrow K(r_i)$ such that $\psi_i(\pi(X)) = r_i$. Hence $\psi_i\circ \psi_1^{-1}\colon K(r_1) \rightarrow K(r_i)$ is a $K$-isomorphism. Hence there exists $\sigma \in G$ such that $\sigma|K(r_1) = \psi_i\circ \psi_1^{-1}$. Since $\sigma(r_1) = r_i$, $G$ acts transitively on $R$.

Conversely suppose $f(X)$ is not irreducible. There exists $g(X), h(X) \in K[X]$ such that $f(X) = g(X)h(X)$, deg $g(X) \ge 1$, deg $h(X) \ge 1$. Let $S$ be the set of the roots of $g(X)$ in $F$. Clearly $S \subset R$. Then $\sigma(s) \in S$ for $\sigma \in G, s \in S$. Since $S \neq R$, $G$ does not act on $R$ transitively.

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Hint: $f$ is irreducible if and only if all of $f$ roots are conjugate

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the definition that two elements are conjugate is that there exist an automorphism of $Gal(F/K)$ that takes one into the other? – Daniel Sep 28 '12 at 2:02
no, otherwise the hint would of just stated what you want to prove.two algebric elements are called conjugate if they have the same minimal polynomial – Belgi Sep 28 '12 at 2:06
that result only holds in Galois extensions right? – Daniel Sep 28 '12 at 2:11
Why would you think so ? – Belgi Sep 28 '12 at 2:18

Perhaps this will help you: show that if $\,\overline K\,$ is any algebraically closed field containing both $\,F\,,\,K\,$ and $\,\sigma:F\to\overline K\,$ is any monomorphism over $\,K\,$ (i.e., fixing $\,K\,$ pointwise) , then $\,\forall w\in F\,$ , both $\,w\,,\,\sigma w\,$ are roots of the very same irreducible polynomial in $\,F[x]\,$

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