Suppose $K$ is a finite extension of $\mathbb{Q}$. Suppose there is some complex number in $K$. Is it necessary that $\tau$, complex conjugation, is a member of $\text{Aut}(K/\mathbb{Q})$?

I feel like this should be true, but consider $\mathbb{Q}(\sqrt{2} + i\sqrt{3})$. I'm not even sure if $\sqrt{2} - i\sqrt{3}$ is in this field.

  • $\begingroup$ Sure it is. $(\sqrt{2}+i\sqrt{3})^{-1}=\frac{1}{5}(\sqrt{2}-i\sqrt{3})$. $\endgroup$ – David Hill Apr 22 '16 at 17:15
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    $\begingroup$ Ah, sure, I suppose in general $(a+ib)^{-1} = \frac{1}{a^2+b^2}(a - ib)$, but $a^2+b^2$ may not necessarily be rational or even in the field itself... $\endgroup$ – cemulate Apr 22 '16 at 17:24
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    $\begingroup$ It's true if $K$ is Galois. In this case complex conjugation acting on $\mathbb{C}$ fixes the image of any embedding $K \to \mathbb{C}$. $\endgroup$ – Qiaochu Yuan Apr 22 '16 at 17:42
  • $\begingroup$ @QiaochuYuan I would like to show the claim is true if $K$ is Galois, but I'm not quite seeing how from your comment. Could you expand your argument a bit more or maybe even post an answer? Thanks! $\endgroup$ – cemulate Apr 22 '16 at 21:40

Not necessarily. Consider the extension of $\mathbb{Q}$ by $\alpha = \sqrt[3]{2}\cdot \zeta $, where $\zeta$ is a primitive cube root of unity. Then $\alpha$ is a non-real cube root of $2$.

Let us show that $\mathbb{Q}(\alpha)$ is not stable under complex conjugation. Indeed, we have that $$ \overline{\sqrt[3]{2}\cdot \zeta} = \sqrt[3]{2}\cdot \zeta^2, $$ so if $\mathbb{Q}(\alpha)$ were stable under conjugation, we would have that all three cube roots of $2$ $$\sqrt[3]{2}\cdot \zeta, \sqrt[3]{2}\cdot \zeta^2, \sqrt[3]{2} = -\sqrt[3]{2}\cdot \zeta-\sqrt[3]{2}\cdot \zeta^2$$ would be in $\mathbb{Q}(\alpha)$, and $\mathbb{Q}(\alpha)$ would be a Galois extension of $\mathbb{Q}$. However, we know that it isn't.

Hence $\overline{\alpha}$ is not in $\mathbb{Q}(\alpha)$.

Added: Note, as Qiaochu Yuan has remarked, that the claim is true if $K$ is a Galois extension of $\mathbb{Q}$. One way to see this is to view $K$ as the splitting field of a polynomial $F$ with coefficients in $\mathbb{Q}$. Since the conjugate of a root of $F$ is still a root of $F$, we get that $K$ is stable under conjugation, and since $K$ contains (by hypothesis) a non-real element, we get that conjugation is a non-trivial automorphism of $K$.


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