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I am having a difficult time trying to figure this out. I suspect it is true, but cannot really convince myself to it. Here is the setting and the question:

I have a Galois field extension of $\Q$, let's call it $K$. Also I know that the maximum real subfield of $K$, let's call it $L:=K\cap\mathbb R$, is not Galois over $\Q$ but with Galois closure $K$. Let $f$ be the minimal polynomial of a primitive element of $K$ over $\Q$ and denote $Z(f)$ the set of zeros of $f$ in $K$ (so all roots of $f$ are non-real and we can write $K=\Q(\alpha)$ for some $\alpha\in Z(f)$). I want to convince myself that there is a field automorphism $\sigma\in\mathrm{Gal}(K/\Q)$ and an element $z\in Z(f)$ such that its complex conjugate pairs are not mapped by sigma to another complex conjugate pair i.e. $$\{\sigma(z),\sigma(\bar z)\}\neq \{\sigma(z),\overline{\sigma(z)}\} $$

I feel this is true but cannot really show this.

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Well, what does it mean that $L$ is not Galois over $\mathbb{Q}$? It means that $Gal(K/L)$ is not a normal subgroup of $Gal(K/\mathbb{Q})$. But $Gal(K/L)$ just contains the identity and complex conjugation, and so this is a normal subgroup iff complex conjugation commutes with every element of $Gal(K/\mathbb{Q})$. So, there exists some $\sigma\in Gal(K/\mathbb{Q})$ that does not commute with complex conjugation. This means that $\sigma(\overline{\alpha})\neq\overline{\sigma(\alpha)}$, since $\alpha$ generates $K$ over $\mathbb{Q}$.

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  • $\begingroup$ I wish I could add two answer as answers. Thank you both (Eric and Jyrki). I just flipped a coin to know who to mark. $\endgroup$ – quantum Feb 22 '18 at 7:54
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Hint:

  • Let $\tau$ be the usual complex conjugation. If $\tau\sigma=\sigma\tau$ for all $\sigma\in Gal(K/\Bbb{Q})$, then the subgroup generated by $\tau$ is normal, and hence its fixed field is Galois over $\Bbb{Q}$.
  • Because $Z(f)$ consists of primitive elements $\sigma$ is fully determined if we know $\sigma(z)$ for some $z\in Z(f)$.
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