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.

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.