Can there exist a finite extension $K$ where $K$ is Galois over $\mathbb Q(i)$ but $K$ is not Galois over $\mathbb Q?$

Question

Can there exist a finite extension $K$ where $K$ is Galois over $\mathbb{Q}(i)$ but $K$ is not Galois over $\mathbb{Q}$?

I am trying to come up with a specific example to show it is possible. My attempt: Let $\alpha^2 = 1 + i$, and let $K = \mathbb{Q}(i,\alpha)$. Then $K$ has degree 2 over $\mathbb{Q}(i)$, so it is Galois. But the minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $x^4-2x^2+2$, which has roots $\pm \alpha = \pm\sqrt{1+i}, \pm\sqrt{1-i}$. So all that's left to do is to show that $\sqrt{1-i} \notin \mathbb{Q}(i,\sqrt{1+i})$, and then $K$ cannot be the splitting field of a polynomial over $\mathbb{Q}$. Except I'm not sure how to do this... I can say $\sqrt{1+i}\sqrt{1-i} = \sqrt{2}$, but how do I show that $\sqrt{2}$ is not in $\mathbb{Q}(i,\sqrt{1+i})$? And is this sufficient?

Answer 1

This ought to do it, if you are familiar with Kummer Theory. The quadratic extensions of any field $F$ of characteristic zero are in one-to-one correspondence with the nontrivial elements of $F^\times/(F^\times)^2$. You don’t actually need to know Kummer, just mimic the proof that I’m sure you know that the only quadratic extensions of $\Bbb Q$ are $\Bbb Q(\sqrt d\,)$ for $d$ a square-free integer unequal to $1$.

All we need to do is show that, for $F=\Bbb Q(i)$, $F\left(\sqrt{1+i}\,\right)=K_1$ is different from $F\left(\sqrt{1-i}\,\right)=K_2$, by showing that $\frac{1+i}{1-i}=i$ is not a square in $F$. But you know that to be the case.

Answer 2

Your idea works. Assume towards a contradiction that $\sqrt{1-i} \in \mathbb{Q}(\sqrt{1+i})$

We look at roots of unity. Observe that $$\sqrt{2} \cdot \zeta_8 = 1 + i \ \ \ \ \Longrightarrow \ \ \ \ \ \sqrt{1+i} = \sqrt[4]{2} \cdot \zeta_{16}$$

Similarly, $\sqrt{1-i} = \sqrt[4]{2} \cdot \zeta_{16}^{-1}$. Therefore, $\mathbb{Q}(\sqrt{1+i})$ contains $\sqrt[4]{2} ( \zeta_{16} + \zeta_{16}^{-1})$. We will solve for this explicitly.

Recall that $ \zeta_{16} + \zeta_{16}^{-1} = 2\text{cos}\bigg( \frac{2\pi}{16}\bigg) $.

\begin{align} & \text{cos}^2\bigg( \frac{2\pi}{16}\bigg) = \frac{1}{2\sqrt{2}} + \frac{1}{2} \ \ \ \ \ \ \text{(Double Angle Formula for cos($x$) )} \\ \Longrightarrow \ & \text{cos}\bigg( \frac{2\pi}{16}\bigg) = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} \ \ \ \ \ \ \ (\text{Taking square roots + algebra}) \\ \Longrightarrow \ & \sqrt[4]{2} \cdot 2\text{cos}\bigg( \frac{2\pi}{16}\bigg) = \sqrt{2 + 2\sqrt{2}} \ \ \ \ (\text{More algebra}) \end{align}

$\sqrt{2 + 2\sqrt{2}}$ is of degree $4$ over $\mathbb{Q}$. But $\mathbb{Q}(\sqrt{1+i})$ is also of degree $4$ over $\mathbb{Q}$, so we have $\mathbb{Q}(\sqrt{2+2\sqrt{2}}\ ) = \mathbb{Q}(\sqrt{1+i}) $. This is obviously not true as one of these fields contains nonreal complex numbers and the other does not.

Answer 3

This is a slight improvement of Dionel's answer.

Assume for a contradiction that $\sqrt{1-i}\in\mathbb Q(\sqrt{1+i})$, and let $\alpha=\sqrt{1-i}+\sqrt{1+i}$. By hypothesis, $\mathbb Q(\alpha)\subseteq\mathbb Q(\sqrt{1+i})$.

We have $\alpha^2=(\sqrt{1-i}+\sqrt{1+i})^2=2+2\sqrt2$, so we know $\mathbb Q(\alpha)/\mathbb Q$ is of degree $4$. Thus, we must have $\mathbb Q(\alpha)=\mathbb Q(\sqrt{1+i})$. This cannot be true, since $i\in\mathbb Q(\sqrt{1+i})$ but $\mathbb Q(\alpha)\subseteq\mathbb R$.