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

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

  • $\begingroup$ the answer for $\operatorname{Gal}(K/\mathbb{Q})$ is that has no elements of order four? $\endgroup$ – Legolas Aug 10 '15 at 8:29
  • $\begingroup$ 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. $\endgroup$ – 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.

  • $\begingroup$ i have only to find $a,b$? $\endgroup$ – Legolas Aug 11 '15 at 15:24
  • $\begingroup$ No, I told you what the $a$ and the $b$ had to be. $\endgroup$ – Lubin Aug 11 '15 at 22:33
  • $\begingroup$ 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 $\endgroup$ – Legolas Aug 13 '15 at 14:10
  • $\begingroup$ To do that is much easier once you realize that $\Bbb Q(\omega)=\Bbb Q(\sqrt{-3}\,)$. $\endgroup$ – Lubin Aug 13 '15 at 14:13

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