# $\mathbb{Q}(\sqrt2,\sqrt3,\sqrt{(2+\sqrt{2})(3+\sqrt{3})})$ is Galois over $\mathbb{Q}$

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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Part 3 is not tedious at all, just apply the automorphism $\sqrt 2 \to -\sqrt 2$ to the polynomial evaluated at $\alpha$ to see that it remains zero. Repeat for $\sqrt 3$. – Phira Nov 19 '12 at 22:51
@Phira: oh, that's true, how silly of me. Thank you for your comment. – user46225 Nov 19 '12 at 23:07

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 $K=\mathbb{Q}(\alpha)$?
Thanks. I should be more patient when attempting these boring computations :P Yes, I see how it implies that $K=\mathbb{Q}(\alpha)$. – user46225 Nov 19 '12 at 23:04
+1 Just pointing out that (1) alone proves that $\mathbb{Q}(\sqrt2,\sqrt3)\subseteq \mathbb{Q}(\alpha)$. This is because the known Galois group of $\mathbb{Q}(\sqrt2,\sqrt3)$ over the rationals allows us to list all the proper subfields of $\mathbb{Q}(\sqrt2,\sqrt3)$. They are $\mathbb{Q}(\sqrt n)$ for $n=1,2,3,6$. Manifestly $\alpha^2$ does not belong to any of these, so we can already conclude that $\mathbb{Q}(\alpha^2)=\mathbb{Q}(\sqrt2,\sqrt3)$. – Jyrki Lahtonen Nov 20 '12 at 14:21