Finding Galois group of $K=\Bbb{Q}(\omega,\sqrt2)$, showing that $K=\Bbb{Q}(\omega\sqrt2)$, and finding $\operatorname{min}(\omega\sqrt2,\Bbb{Q})$ Let $\omega$ be a primitive third root of unity, and $K=\mathbb{Q}(\omega,\sqrt{2})$.
I found that the degree of $[K:\mathbb{Q}]=6$. How can I find the Galois group $\operatorname{Gal}(K/\mathbb{Q})$?
After this, could anyone give me any hints for showing that $K=\mathbb{Q}(\omega\sqrt{2})$ and finding $\operatorname{min}(\omega\sqrt{2},\mathbb{Q})$?
 A: It is straightforward to check, that an element $k\in K$ can be represented us (one should not forget that $\omega^2 = -1-\omega$)
$$k = q_1\cdot 1+q_2\cdot\sqrt{2} +q_3\cdot \omega+q_4\cdot \omega\sqrt{2}$$
and that representation is unique. It means that $[K:Q]=4$.
There are only two groups of order four ($\mathbb{Z}_2\oplus \mathbb{Z}_2$ and $\mathbb{Z}_4$).
The next step is to show, that $\operatorname{Gal}(K/\mathbb{Q})$ has no elements of order four.
Hint. If $\alpha \in \operatorname{Gal}(K/\mathbb{Q})$ and $x_0$ is a root of $x^2-2$, then $\alpha(x_0)$ is also a root.
To show that $K=\Bbb{Q}(\omega\sqrt2)$ let us notice the following:
$$\frac{1}{2}(\omega\sqrt{2})^3=\sqrt{2}$$
$$-\frac{1}{2}\left((\omega\sqrt{2})^2+2\right) = -\frac{1}{2}\left((-2\omega-2)+2\right)=\omega$$
A: The response of @Andrey says it all, except how to get the minimal polynomial for $\omega\sqrt2$ over $\Bbb Q$.
This is easy enough: the conjugates of $\omega$ are $\{\omega,\omega^2\}$, and the conjugates of $\sqrt2$ are $\pm\sqrt2$. So you take the product of the four linear polynomials $X-ab$, where $a$ is a conjugate of $\omega$ and $b$ is a conjugate of $\sqrt2$. This is a fun thing to do, and I leave it to you.
