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I want to prove that $K:=\mathbb{Q}(\sqrt2,\sqrt3,\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\mathbb{Q}$.

Let $\alpha:=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$. After some computations I was able to show that $[K:\mathbb{Q}(\sqrt2,\sqrt3)]=2$, thus proving that $[K:\mathbb{Q}]=8$.

After some more computations I was able to find that the polynomial $f=144-288 x^2+144 x^4-24 x^6+x^8$ has $\alpha$ as a root, but I don't know if this is useful.

My idea was:

1) To show that $K=\mathbb{Q}(\alpha)$, and thus $f$ would be the irreducible polynomial of $\alpha$ over $\mathbb{Q}$,

2) To show that $\pm \sqrt{(2\pm \sqrt{2})(3\pm \sqrt{3})}\in \mathbb{Q}(\alpha)$,

3) To show that these are all the roots of $f$.

This would prove that $K$ is a splitting field for $f$ over $\mathbb{Q}$, thus proving that $K/\mathbb{Q}$ is Galois.

I couldn't do part 1). Assuming part 1, part 2) is easy as can be seen multiplying those elements (they live on $\mathbb{Q}(\sqrt2,\sqrt3)$.) Part 3) is a straightforward, albeit tedious, computation.

I would appreciate any solution to this problem, but especially one along the lines I was trying to follow (should part 1) be true, of course).

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    I am working on this question. May I please ask how to show that all the roots are in $\Bbb Q(\alpha)$? How may I see that the roots can be obtained by multiplying elements in $\Bbb Q(\sqrt{2},\sqrt{3})$? I would really appreciate if get replied.2017-05-27

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To show that $K=\mathbb{Q}(\alpha)$ you can use the following: $\begin{array}{l} (1)\hspace{3pt}\alpha^2= (2+\sqrt{2})(3+\sqrt{3})=6+3\sqrt2+2\sqrt3+\sqrt6\in \mathbb{Q}(\alpha)\Rightarrow 3\sqrt2+2\sqrt3+\sqrt6\in \mathbb{Q}(\alpha)\\ (2)\hspace{3pt}(\alpha^2-6)^2=18+12+6+12\sqrt6+6\sqrt{12}+4\sqrt{18}=36+12(\sqrt6+\sqrt3+\sqrt2)\in\mathbb{Q}(\alpha)\\ (3)\hspace{3pt}\Rightarrow \sqrt6+\sqrt3+\sqrt2=[(\alpha^2-6)^2-36]/12\in\mathbb{Q}(\alpha) \end{array}$ From $(1)$ and $(3)$ we have $2\sqrt2+\sqrt3\in \mathbb{Q}(\alpha)$. Hence $(2\sqrt2+\sqrt3)^2=8+4\sqrt6+3=11+4\sqrt6\in \mathbb{Q}(\alpha)$, Which implies that $\sqrt6\in \mathbb{Q}(\alpha)$. Together with $(3)$ we get $\sqrt3+\sqrt2\in \mathbb{Q}(\alpha)$. And, finally, using $2\sqrt2+\sqrt3\in \mathbb{Q}(\alpha)$ we get $\sqrt2,\sqrt3\in \mathbb{Q}(\alpha)$.
So $\sqrt2,\sqrt3,\alpha\in \mathbb{Q}(\alpha)$
Can you see that this implies $K=\mathbb{Q}(\alpha)$?

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    Very nice and simple argument :-)2012-11-20