Could you help me in finding the Galois group of $\mathbb{Q}\left(\sqrt{2}, \sqrt{3}, \sqrt{(2+\sqrt{2})(3+\sqrt{3}}\right)$ over $\mathbb{Q}$?

I can only say that $\mathbb{Q}(\sqrt{2}) /\mathbb{Q}$ and $\mathbb{Q}(\sqrt{3}) /\mathbb{Q}$ are two normal extentions and that $\mathbb{Q}\left(\sqrt{2}, \sqrt{3}, \sqrt{(2+\sqrt{2})(3+\sqrt{3})}\right)=\mathbb{Q}\left(\sqrt{2},\sqrt{(2+\sqrt{2})(3+\sqrt{3})}\right)$ is an eighth-degree extention over $\mathbb{Q}$.

thank you!

edit: Is the following a valid argument?

$\mathbb{E}=\mathbb{Q}\left(\sqrt{2}, \sqrt{3} \right) \subset \mathbb{Q}\left(\sqrt{2}, \sqrt{3}, \sqrt{(2+\sqrt{2})(3+\sqrt{3})}\right)=\mathbb{F}$ therefore $\mathbb{F}/\mathbb{E}$ is a second-degree galois extention where $\sigma: \sqrt{(2+\sqrt{2})(3+\sqrt{3})} \mapsto -\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ is the only automorphism in the galois group.

Therefore $Gal(\mathbb{F}/\mathbb{Q})=\langle \pi,\tau,\sigma\rangle$ where

$\pi:\sqrt{2}\mapsto -\sqrt{2}\\ \tau:\sqrt{3}\mapsto -\sqrt{3}\\ \sigma: \sqrt{(2+\sqrt{2})(3+\sqrt{3})} \mapsto -\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$

We can easily verify that $\pi (\tau ( \sigma)) \neq \sigma( \pi (\tau))$ but $\pi(\tau)=\tau( \pi)$ Therefore $\langle \pi,\tau \rangle$ is the center of the galois group which is not commutative and then $Gal (\mathbb{F}/\mathbb{Q})=\mathbb{Z}_2\times \mathbb{Z}_2 \rtimes \mathbb{Z}_2$

  • $\begingroup$ Do you know whether your field is a Galois extension of $\Bbb{Q}$? $\endgroup$ – Jyrki Lahtonen Jun 3 '14 at 17:54
  • 2
    $\begingroup$ I haven't checked it either, but I suspect that the relations $(2+\sqrt2)(2-\sqrt2)=2$ and $(3+\sqrt3)(3-\sqrt3)=6$ come in handy because both these products have square roots in $\Bbb{Q}(\sqrt2,\sqrt3)$. $\endgroup$ – Jyrki Lahtonen Jun 3 '14 at 18:23
  • $\begingroup$ it's a problem from a past exam at my university, it doesn't ask you to prove that's a galois extention so i guess you can take that as given $\endgroup$ – giulio Jun 3 '14 at 21:06
  • 1
    $\begingroup$ what is $\pi(\sqrt{(2+\sqrt 2)(3+\sqrt 3)})$ ? $\endgroup$ – mercio Jun 4 '14 at 8:29

Let $\alpha = \sqrt{(2+\sqrt 2)(3+\sqrt 3)}$ and $K = \Bbb Q(\sqrt 2, \sqrt 3, \alpha)$. An automorphism $\sigma$ of $K$ over $L$ is determined by $\sigma(\sqrt 2),\sigma(\sqrt 3)$ and $\sigma(\alpha)$. Since $\sigma$ is an automorphism, those values have to satisfy the same algebraic relations of the original values, so $\sigma(\sqrt 2)^2 = 2, \sigma(\sqrt 3)^2 = 3$, and $\sigma(\alpha)^2 = (2+\sigma(\sqrt 2))(3+\sigma(\sqrt 3))$. In fact, $K = \Bbb Q(\alpha)$ so the image of $\alpha$ determines the images of $\sqrt 2$ and $\sqrt 3$.

So the first thing is to do is to check how many conjugates (over $\Bbb Q$) of $\alpha$ are in $K$. Obviously, we have $\alpha, - \alpha \in K$. Let $\beta = \sqrt{(2-\sqrt 2)(3+\sqrt 3)}$. Then $\beta\alpha = \sqrt 2(3+\sqrt 3) \in K$, so that $\beta, - \beta \in K$. Similarly, if $\gamma = \sqrt{(2+\sqrt 2)(3- \sqrt 3)}$ and $\delta = \sqrt{(2-\sqrt 2)(3-\sqrt 3)}$, we have $\gamma\alpha = (2+\sqrt 2)\sqrt 6$ and $\delta\alpha = 2\sqrt 3$.

Therefore, all $8$ conjugates of $\alpha$ are in $K$, so that $K$ is Galois over $\Bbb Q$.

Let $\pi$ be the automorphism sending $(\alpha,\sqrt 2,\sqrt 3)$ to $(\beta, - \sqrt 2, \sqrt 3)$. We have $\pi(\pi(\alpha)) = \pi(\beta) = \pi(\sqrt 2(3+\sqrt 3)/\alpha) = -\sqrt 2(3+\sqrt 3)/\beta = - \alpha$. Hence $\pi^2(\alpha,\sqrt 2,\sqrt 3) = (- \alpha,\sqrt 2,\sqrt 3)$, then $\pi^3(\alpha,\sqrt 2,\sqrt 3) = (- \beta, - \sqrt 2, \sqrt 3)$ and $\pi^4$ is the identity.

Let $\tau$ be the automorphism sending $(\alpha,\sqrt 2,\sqrt 3)$ to $(\gamma, \sqrt 2, -\sqrt 3)$. We have $\tau(\tau(\alpha)) = \tau(\gamma) = \tau((2+\sqrt 2)\sqrt 6/\alpha) = - \alpha$, so $\tau^2 = \pi^2$.
$\tau(\pi(\alpha)) = \tau(\beta) = \tau(\sqrt 2(3+\sqrt 3)/\alpha) = \sqrt 2(3-\sqrt 3)/\gamma = (\sqrt 2(3-\sqrt 3)/(2+\sqrt 2)\sqrt 6)\alpha = (6\sqrt 2/\alpha^2\sqrt 6)\alpha =2\sqrt 3/\alpha = \delta$
$\pi(\tau(\alpha)) = \pi(\gamma) = \pi((2+\sqrt 2)\sqrt 6/\alpha) = -(2-\sqrt 2)\sqrt 6/\beta = (-(2-\sqrt 2)\sqrt 6/\sqrt2(3+\sqrt 3))\alpha = (-2\sqrt 6/\alpha^2\sqrt 2)\alpha = -2\sqrt 3/\alpha = -\delta = \tau(\pi ^3(\alpha))$

So we have $\pi^4 = 1, \tau^2 = \pi^2, \pi \tau = \tau \pi^3$. $\tau^2 = \pi^2 = (\tau\pi)^2 = \tau\pi(\tau\pi)$. The Galois group is isomorphic to the quaternion group.


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.