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Let $k\subset K$ be a Galois extension, i.e. $$K=\langle{\xi}_{1}, {\xi}_{2},\ldots, {\xi}_{n}\rangle,$$ $$k\supset K=\{a_0+a_1{\xi}_{1}+\ldots+a_n{\xi}_{n}|a_0,\ldots,a_n\in k\}.$$ Is it true that $$\forall i~~\exists j:\sigma({\xi}_{i})={\xi}_{j},$$ where $\sigma$ is any element of $\operatorname{Gal}{(K,k)}$. In other words, is it true that elements of $\operatorname{Gal}{(K,k)}$ relocate generators of $K$?


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2 Answers 2

up vote 7 down vote accepted

I believe you are asking whether the Galois group of a finite Galois extension $L/K$ must permute a given $K$-basis $\xi_1,\ldots,\xi_n$ of $L$. The answer is no: for instance take $K = \mathbb{Q}$, $L = \mathbb{Q}(\sqrt{2})$ and the basis $\xi_1 = 1$, $\xi_2 = \sqrt{2}$.

However the Normal Basis Theorem says that you can always find some $K$-basis of $L$ which the Galois group acts on by permutations. In the above example, taking e.g. $\xi_1 = 1 + \sqrt{2}$, $\xi_2 = 1 -\sqrt{2}$ will work.

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This is not true. Take $k=\mathbb{Q}$ and $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. Then $K/k$ is Galois, but there is no Galois element that will take $\sqrt{2}$ to $\sqrt{3}$.

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I mean that $\sigma({\xi}_{i})$ is also generator (but no linear combination of generators with сoefficients in $k$) –  Aspirin Sep 4 '11 at 15:33

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