I aim to prove that $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\Bbb Q$. To do this, I aim to show that it is the splitting field of the minimal polynomial of $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$.

Thee minimal polynomial is $144-288 x^2+144 x^4-24 x^6+x^8$. And all its roots are:$\pm \sqrt{(2\pm \sqrt{2})(3\pm \sqrt{3})}$

I do notice this post:$\mathbb{Q}(\sqrt2,\sqrt3,\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\mathbb{Q}$

But I cannot fully understand how I can do that. My questions are:

  1. Call $\sqrt{(2+ \sqrt{2})(3+ \sqrt{3})}=\alpha$. How can we show that $\pm \sqrt{(2\pm \sqrt{2})(3\pm \sqrt{3})}\in\Bbb Q(\alpha)$?

  2. There is a comment which mentioned a trick to check if we have found all the roots.

    Part 3 is not tedious at all, just apply the automorphism $\sqrt{2}\rightarrow -\sqrt{2}$ to the polynomial evaluated at $α$ to see that it remains zero. Repeat for $\sqrt{3}$.

I would like to learn about the trick mentioned in the comment but I cannot fully understand it. S ocould someone please show me how to do that?

And now I need to determine the Galois group of $\Bbb Q$. Is there any rather simple way to see what the Galois group is?

Thanks for any help!

EDIT: Now I had stucked on it for some time and I think I do need help...

We know that the automorphism is uniquely determined by where it sent the generater to. So we only need to determine its effect on $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$.

As $(\sigma(\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})})))^2=\sigma((\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3})}))^2)=\sigma(\mathbb{Q}((2+\sqrt{2})(3+\sqrt{3})))=6+2\sigma(\sqrt{3})+3\sigma(\sqrt{2})+\sigma(\sqrt{6})$. I need to determine where to send $\sqrt{2},\sqrt{3},\sqrt{6}$.

But I only know that I can send $\sqrt{2}\mapsto\pm\sqrt{2}$, same to $\sqrt{3},\sqrt{6}$. But there are then too many possiblities. So what should I do now to rule out some possiblities?

Now I realize that maybe I need not determine where to send $\sqrt{6}$ to because once we determine $\sqrt{2},\sqrt{3}$ we have already know that where to send $\sqrt{6}$...

Is that correct?


From the link you gave, we have $\mathbb Q(\alpha) = \mathbb Q(\alpha,\sqrt 2,\sqrt 3)$.

All the roots of the minimal polynomial of $\alpha$ over $\mathbb Q$ are of the form $$\pm\sqrt{(2\pm\sqrt2)(3\pm\sqrt 3)}$$ (which we will denote with $\{\pm\alpha,\pm\beta,\pm\gamma,\pm\delta\}$ and will soon prove). To answer 1., note that

$$\alpha\delta = 2\sqrt 3,\ \alpha\beta = (3+\sqrt3)\sqrt 2,\ \beta\gamma = 2\sqrt 3.$$

For the part 2., first notice that the minimal polynomial of $\alpha$ $$f(x) = x^8-24x^6+144x^4-288x^2+144$$ is of the form $g(x^2)$, where $g(x) = x^4-24x^3+144x^2-288x+144$ and thus all the roots of $f$ are $\pm\sqrt{x_0}$, where $x_0$ is a root of $g$. Thus, we know that $\alpha^2$ is root of $g$ and it suffices to show that the other roots of $g$ are $\beta^2,\gamma^2,\delta^2$. Note that $\alpha^2,\beta^2,\gamma^2,\delta^2$ are elements of $\mathbb Q(\sqrt 2,\sqrt 3)$, and are of the form $\varphi(\alpha^2)$ for automorphism $\varphi$ of $\mathbb Q(\sqrt 2,\sqrt 3)$ ($\varphi$ is determined by its action on $\sqrt 2$ and $\sqrt 3$, see the last paragraph). Since $\varphi$ is an automorphism, we have that $g(\varphi(\alpha^2)) = \varphi(g(\alpha^2)) = 0$, and thus $\varphi$ sends roots of $g$ to roots of $g$, so $\alpha^2,\beta^2,\gamma^2,\delta^2$ are roots of $g$ and these are all the roots since $\deg g = 4$.

Finally, to determine the Galois group, from $\mathbb Q(\alpha) = \mathbb Q(\alpha,\sqrt 2,\sqrt 3)$ we know that any automorhpism of $\mathbb Q(\alpha)$ will send $\sqrt 2$ to $\pm\sqrt 2$ and $\sqrt 3$ to $\pm\sqrt 3$. From your observation that for automorhpism $\sigma$ of $\mathbb Q(\alpha)$ we have \begin{align}\sigma(\alpha)^2=\sigma(\alpha^2) &= 6 + 2\sigma(\sqrt 3) + 3\sigma(\sqrt 2) + \sigma(\sqrt 6)\\ &= 6 + 2\sigma(\sqrt 3) + 3\sigma(\sqrt 2) + \sigma(\sqrt 2)\sigma(\sqrt 3)\\ &= \begin{cases} \alpha^2, & \sigma(\sqrt2)=\sqrt 2,\ \sigma(\sqrt 3)=\sqrt 3;\\ \beta^2, & \sigma(\sqrt2)=-\sqrt 2,\ \sigma(\sqrt 3)=\sqrt 3;\\ \gamma^2, & \sigma(\sqrt2)=\sqrt 2,\ \sigma(\sqrt 3)=-\sqrt 3;\\ \delta^2, & \sigma(\sqrt2)=-\sqrt 2,\ \sigma(\sqrt 3)=-\sqrt 3;\\ \end{cases}\end{align}

which implies that $$\sigma(\alpha) = \begin{cases} \pm\alpha, & \sigma(\sqrt2)=\sqrt 2,\ \sigma(\sqrt 3)=\sqrt 3;\\ \pm\beta, & \sigma(\sqrt2)=-\sqrt 2,\ \sigma(\sqrt 3)=\sqrt 3;\\ \pm\gamma, & \sigma(\sqrt2)=\sqrt 2,\ \sigma(\sqrt 3)=-\sqrt 3;\\ \pm\delta, & \sigma(\sqrt2)=-\sqrt 2,\ \sigma(\sqrt 3)=-\sqrt 3;\\ \end{cases}$$ we see that all the automorphisms are of the form $\pm\sigma_{\pm,\pm}$ where $\pm$'s in the index determine where $\sqrt 2$ and $\sqrt 3$ are sent.

Hopefully, from this you can work out the Galois group yourself.


Let $\alpha=\sqrt{(2+\sqrt2)(3+\sqrt3)}$ and $\beta=\sqrt{(2-\sqrt2)(3+\sqrt3)}$. One of the things you must show is that $\beta\in\Bbb Q(\alpha)$. I hope you can show $\sqrt2$, $\sqrt3\in\Bbb Q(\alpha)$. Now $\alpha\beta=\sqrt2(3+\sqrt3)\in\Bbb Q(\alpha)$. Therefore $\beta\in\Bbb Q(\alpha)$.

There are a couple more calculations of this ilk you need to finish off the proof.

  • $\begingroup$ Thanks! I get it. And I am just wondering if there is some rater simple way to determine the Galois group since the calculation is really tedious and I am thinking about determine the Galois group through rulling out all the other possiblility of what the Galois group would be. I guess the Galois group is quaternion group. $\endgroup$ – PropositionX May 27 '17 at 11:43

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.