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If you consider any quadratic extension $K$ of $\mathbb{Q}$, it has to be fixed by complex conjugation, because from $[K : \mathbb{Q}] = 2$ we know $K | \mathbb{Q}$ has to be a normal extension and as $\mathbb{Q}$-Homomomorphism complex conjugation on $\mathbb{C}$ becomes a $K$-Automorphism, when restricted to $K$.

I'm now wondering, what happens if you take any arbitrary field $K \supseteq \mathbb{Q}$, which has not to be an algebraic extension of $\mathbb{Q}$. Is every quadratic extension of $K$ still fixed by complex conjugation?

At the moment I'm neither able to prove this nor to find an counter example.

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  • $\begingroup$ Do you intend that $K$ should be a subfield of $\Bbb R$ as well? If so, then the answer is yes: any non-real element of the quadratic extension of $K$ is a root of a quadratic polynomial with coefficients in $K$; the quadratic formula then shows that the two roots are complex conjugates of each other and their sum is in $K$, which implies that the complex conjugate is in the field generated by $K$ and the first root. $\endgroup$ – Greg Martin Aug 19 '15 at 22:48
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Edit: I've been unclear about the difference between being fixed pointwise by conjugation (i.e. for all $k \in K$, $\overline{k} = k$) and being closed under conjugation (i.e. $\overline{K} = K$). Hopefully it is now clarified.

I'm assuming that you're considering $\mathbb{Q} \subseteq K \subseteq \mathbb{C}$, since otherwise complex conjugation is meaningless.

Now, consider $L \supseteq K$, a quadratic extension. This extension is automatically Galois, generated by (using the quadratic formula) $\sqrt{D}$ for some $D \in K$. If $K$ is fixed by complex conjugation (i.e. if $K \subseteq \mathbb{R}$), then the complex conjugate of $L$ is an isomorphic field extension of $K$; this means that $\left(\overline{\sqrt{D}}\right)^2 = \overline{\sqrt{D}^2} = \overline{D} = D$, since $D \in K$ and complex conjugation fixes $K$. Since $\overline{L}$ is generated by $\overline{\sqrt{D}}$ and $\overline{\sqrt{D}}$ satisfies the same minimal polynomial over $K$ as does $\sqrt{D}$, it must be the case that $\overline{\sqrt{D}} \in L$ and thus $\overline{L} = L$, so $L$ is closed under conjugation.

Now, if $K$ is not fixed by complex conjugation, the picture is more complicated.

First of all, we see that even when $K$ is a finite but non-normal extension of $\mathbb{Q}$, $L$ may not be closed under conjugation. Note that in this example, $K$ is not closed under conjugation either.

Consider $K = \mathbb{Q}[\omega]$, where $\omega = e^{2\pi i/3}\sqrt[3]{2}$ is a non-real cube root of $2$. $K$ is not Galois over $\mathbb{Q}$, since we may embed $K$ into $\mathbb{C}$ by mapping $\omega$ to the real cube root of $2$, $\sqrt[3]{2}$, which is a subfield of $\mathbb{R}$ and thus does not contain the non-real cube roots of $2$. However, if we use the non-real embedding, $K$ is not closed under complex conjugation. Now consider $L = K[\sqrt{\omega}]$. If $L$ were closed under complex conjugation, in particular, $L$ would contain $\overline{K}$. Thus $L$ would contain $\overline{\omega} = e^{-2\pi i/3} \sqrt[3]{2}$, so it would also contain $\omega \overline{\omega} = \sqrt[3]{2}^2$ and $\frac{2}{\sqrt[3]{2}^2} = \sqrt[3]{2}$, so it would contain $e^{2\pi i/3}$. Now, the field $F$ generated over $\mathbb{Q}$ by $e^{2\pi i/3}$ and $\sqrt[3]{2}$ is the splitting field of $X^3 - 2$ over $\mathbb{Q}$. In particular, it is a degree $6$ Galois extension of $\mathbb{Q}$.

However, since $L$ is a degree $6$ extension of $\mathbb{Q}$ containing $F$, we must have that $L = F$. Since $L$ is generated over $K$ by $\sqrt{\omega}$ and $\omega$ satisfies $\omega^6 = 2$, we have that $\omega$ is a root of $X^6 - 2$ over $\mathbb{Q}$. Thus, since $F = L$ is normal over $\mathbb{Q}$, $F$ contains all $6$ roots of this (irreducible) polynomial, so in particular the real root $\alpha = \sqrt[6]{2}$. However, the field $E$ generated over $\mathbb{Q}$ by $\alpha$ is contained in $\mathbb{R}$ and of degree 6 over $\mathbb{Q}$. Since $E$ is a subfield of $F$ of degree $6$, we must have $E = F$, but this is a contradiction since $F$ contains the non-real cube roots of unity, and thus cannot be included into $\mathbb{R}$.

Finally, we will show that if $K$ is closed under conjugation, then $L$ is closed under conjugation as well if and only if $L$ is Galois over $K \cap \mathbb{R}$.

Lemma: Let $F$ be a finite normal extension of a field $E$ which is contained in $\mathbb{R}$. Then $K$ is closed under conjugation.

To see this, recall that by the primitive element theorem, $F = E(\alpha)$. Let $p$ be the minimal polynomial over $E$ of $\alpha$. Then, since $F$ is normal, $F$ contains all of the roots of $p$. We see that $\overline{\alpha}$ is a root of $p$ by using the hypothesis that $E \subseteq \mathbb{R}$ and thus $E$ is fixed by conjugation:

We have $$0 = p(\alpha) = a_0 + a_1 \alpha + \cdots + a_n\alpha^n$$ with $a_i \in \mathbb{Q}$ for all $i$. Then, conjugating both sides, we have $$0 = \overline{0} = \overline{a_0} + \overline{a_1} \overline{\alpha} + \cdots + \overline{a_n} \overline{\alpha}^n = a_0 + a_1\overline{\alpha} + \cdots + a_n \overline{\alpha}^n = p(\overline{\alpha})$$

Since $F$ is normal, therefore, $\overline{\alpha} \in F$ and thus $\overline{F} = F$.

Now, let $K \supseteq \mathbb{Q}$ be an extension which is closed under complex conjugation and let $L \supseteq K$ be a quadratic extension. By the above lemma, if $L$ is Galois over $K \cap \mathbb{R}$, then $L$ is closed under complex conjugation as well.

Conversely, assume $L$ is closed under complex conjugation. If $L \subseteq \mathbb{R}$, then $K \subseteq \mathbb{R}$ and $L$ is Galois over $K \cap \mathbb{R} = K$ since it is quadratic over $K$. Otherwise, since conjugation is an automorphism of order two of $L$ with fixed field $L \cap \mathbb{R}$, $L$ must be a quadratic extension of $L \cap \mathbb{R}$. If $K \subseteq \mathbb{R}$, then as above, we see that $L$ is Galois over $K \cap \mathbb{R}$. If $K$ is not contained in $\mathbb{R}$, then $K$ is quadratic over $K \cap \mathbb{R}$. If $L \cap \mathbb{R} = K \cap \mathbb{R}$, then since $L$ is a quadratic extension of $K \cap \mathbb{R}$, we have that it is Galois. Otherwise, $L$ is the compositum of the two Galois extensions $K/K \cap \mathbb{R}$ and $L \cap \mathbb{R}/K \cap \mathbb{R}$, and thus is Galois.

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  • $\begingroup$ The case I'm mainly interested in, is $K$ beeing not necessarily fixed, but closed under complex conjugation, i.e. mapped into itself by conjugation. Any suggestions on that one? $\endgroup$ – cosinus Aug 20 '15 at 0:01

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