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Let $G$ be a finite group of automorphisms acting on a field $L$, with fixed field $K$.

My notes say "Let $ \alpha \in L $. Consider the set $ \{\sigma(\alpha)| \sigma \in G\}$ and suppose its distinct elements are $\alpha = \alpha_1, \alpha_2, ... \alpha_r $. Let $g = \Pi (X-\alpha_i).$ Then $g$ is invariant under $G$, since its linear factors are just permuted by elements of $G$"

I don't understand that last sentence. Why must an element of $G$ permute the linear factors? Why must an $\alpha_i$ be mapped to an $\alpha_j$, and not some other element in $L$?


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up vote 2 down vote accepted

Since $\alpha_i$ is in the orbit of $\alpha$ under $G$, there exists some $\sigma_i \in G$ such that $\sigma_i(\alpha) = \alpha_i$. Now suppose that $\sigma \in G$. Then $$\sigma(\alpha_i) = \sigma(\sigma_i(\alpha)) = (\sigma \circ \sigma_i)(\alpha)$$ Since $G$ is a group of automorphisms, $\sigma \circ \sigma_i \in G$. Thus $\sigma(\alpha_i)$ must be in the orbit of $\alpha$ under $G$, and thus must be $\alpha_j$ for some $j$.

Now looking at the behavior of $g$ under $G$, for any $\sigma \in G$, we have $$\sigma(g(X)) = \sigma \left (\prod(X-\alpha_i) \right) = \prod \sigma(X-\alpha_i) = \prod (X-\sigma(\alpha_i))$$ Since $\sigma$ is an automorphism, $\sigma(\alpha_i) \neq \sigma(\alpha_j)$ if $i \neq j$. But for any root $\alpha_i$, $$g(\sigma(\alpha_i)) = \sigma(g(\alpha_i)) = \sigma(0) = 0$$ So the image of $\alpha_i$ under $\sigma$ must be another root of $g$. This implies that the action of $G$ on $g$ permutes the roots, and thus doesn't change $g$.

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Both good and clear. I would recommend to OP to work as many examples as possible; but you need an extension of relatively high degree. Maybe the seventh roots of unity, over the rationals, a cyclic sextic extension. – Lubin Dec 7 '11 at 6:24

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