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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?

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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.

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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.

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  • $\begingroup$ 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

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