2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ Note that the polynomial $x^3-1$ is not irreducible. $\endgroup$ – ajotatxe Aug 9 '15 at 20:08
3
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.