# 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 $\operatorname{min}(\omega\sqrt{2},\mathbb{Q})$?

• Do you mean $\Large\sqrt[3]{2}$, instead of $\Large\sqrt{2}$? – Zev Chonoles Aug 9 '15 at 20:00
• no it's $\sqrt2$ – Legolas Aug 9 '15 at 20:02
• Then your statement that $[\mathbb{Q}(\omega,\sqrt{2}):\mathbb{Q}]=6$ is incorrect. The correct value is $4$. – Zev Chonoles Aug 9 '15 at 20:02
• i will check it again – Legolas Aug 9 '15 at 20:03
• Note that the polynomial $x^3-1$ is not irreducible. – ajotatxe Aug 9 '15 at 20:08

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: $$\frac{1}{2}(\omega\sqrt{2})^3=\sqrt{2}$$ $$-\frac{1}{2}\left((\omega\sqrt{2})^2+2\right) = -\frac{1}{2}\left((-2\omega-2)+2\right)=\omega$$

• the answer for $\operatorname{Gal}(K/\mathbb{Q})$ is that has no elements of order four? – Legolas Aug 10 '15 at 8:29
• No, this is not the answer. But once you prove that in this group there are no elements of the order four, the answer would be easy to get. – Tzara_T'hong Aug 10 '15 at 8:34

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.

• i have only to find $a,b$? – Legolas Aug 11 '15 at 15:24
• No, I told you what the $a$ and the $b$ had to be. – Lubin Aug 11 '15 at 22:33
• I am just wondering for the same question that i asked,how i can find all the fields with $Q \subseteq L \subseteq K$ where $K=Q(ω,\sqrt2)$ and $ω$ a primitive third root of unity – Legolas Aug 13 '15 at 14:10
• To do that is much easier once you realize that $\Bbb Q(\omega)=\Bbb Q(\sqrt{-3}\,)$. – Lubin Aug 13 '15 at 14:13