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Let $E=\mathbb{Q}(\sqrt{2},\sqrt{3})$ and $$ L = E \left( \sqrt{ ( \sqrt{2}+2 ) ( \sqrt{3} + 3)} \right) \ . $$ I want to show that $L/\mathbb{Q}$ is a Galois extension with the Quaternion group as its Galois group.

I know $E/\mathbb{Q}$ is Galois and $L/E$ is also Galois, but it is not true in general that if $K_1/K_0$ is Galois and $K_2/K_1$ is Galois then $K_2/K_0$ is Galois (take $K_0 = \mathbb{Q}$, $K_1 = \mathbb{Q}(\sqrt{2})$ and $K_2 = \mathbb{Q}(\sqrt[4]{2})$ as a counter-example).

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3 Answers 3

A hands-on method is to check that all the conjugates of
$$\theta = \sqrt{(\sqrt{2}+2)(\sqrt{3}+3)} $$ are in $\mathbb Q[\theta]$. It is easy to check that the eight conjugates are $\theta$, $$\theta_1 = \sqrt{(-\sqrt{2}+2)(\sqrt{3}+3)}, \quad\theta_2 = \sqrt{(\sqrt{2}+2)(-\sqrt{3}+3)} , \quad\theta_3 = \sqrt{(-\sqrt{2}+2)(-\sqrt{3}+3)}$$ and their negatives.

Now $\sqrt{2}, \sqrt{3}$, and $\sqrt{6} \in \mathbb Q[\theta]$, for example $$ (\theta^2 - 6 -2\sqrt{3} )^2 = 24 + 12\sqrt{3} $$ and it follows that $\sqrt{3} \in \mathbb Q[\theta]$. But now $$ \theta\theta_3 = 2\sqrt{3} $$ so it follows $\theta_3 \in \mathbb Q$ and a similar argument shows that $\theta_1$ and $\theta_2$ are also in $\mathbb Q[\theta]$. So $\mathbb Q[\theta]$ is the splitting field of the minimal polynomial of $\theta$ and as a consequence is Galois over $\mathbb Q$.

Now you can check by hand that the automorfisms $\theta\to\pm\theta_i$ behave as expected but I suppose there are much better methods to do this.

To do it you should work out the expression of $\theta_i$ in terms of $\theta$, for example: $$\begin{align} \theta_1 &= \frac{-1}{24}(\theta^7 -20\theta^5 + 60\theta^3 + 24\theta) \\\\ \theta_2 &= \frac{1}{12}(\theta^5 -18\theta^3 + 36) \\\\ \theta_3 &= \frac{1}{12}(\theta^7 -22\theta^5 + 102\theta^3-120\theta) \end{align}$$ to work in the field $\mathbb Q[\theta]$ you have to compute with polynomials modulo the minimal polynomial of $\theta$ ($x^8 -24x^6 + 144x^4 -288x^2 +144$).I recommend you to use some computer algebra system to perform these computations.

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Let us check that $i = ( \theta \mapsto \theta_1 )$ satisfies $i^2 = -1$. Then $$ i^2 \left( \sqrt{ ( \sqrt{2} + 2 )( \sqrt{3} + 3) } \right) = i \left( \sqrt{ ( -\sqrt{2} + 2 )( \sqrt{3} + 3) } \right) = \sqrt{ ( \sqrt{2} + 2 )( \sqrt{3} + 3) } $$ so $i^2 = 1$ and not $-1$. Where is the mistake? –  LinAlgMan Dec 1 '12 at 12:00
    
In fact, one needs to define $i(\theta_1) = -\theta$ and $i(\theta_2)=\theta_3$ and $i(\theta_3) = -\theta_2$ in order for things work out. –  LinAlgMan Dec 1 '12 at 12:14
    
Note that by $$(\theta^2 - 6 - 2 \sqrt{3})^2 = 6 + \sqrt{2}$$ you already assumed $\sqrt{3}$ in $\mathbb{Q}(\theta)$. We need a more elementary way to prove $\sqrt{3} \in \mathbb{Q}(\theta)$. –  LinAlgMan Dec 1 '12 at 16:51
    
I worked it out, compute first $( \theta^2 - 6)^2$, then $(\theta^2 - 6 - 2 \sqrt{6} - 2 \sqrt{3} - 2 \sqrt{2})^2$ and then $( \theta^2 - 6 - 3 \sqrt{6} - 3 \sqrt{3} - 3 \sqrt{2} )^2$. –  LinAlgMan Dec 1 '12 at 17:48
    
I have corrected the right hand side of $(\theta^2-6-2\sqrt{3})^2$, sorry for that. I have also added some computations in the end to help you with the automorphisms. –  Esteban Crespi Dec 1 '12 at 22:19

The existing answer is very brute-force and not to my taste. I present an alternate method, essentially copy-pasted from another answer of mine.

Let $\sigma$ be the automorphism of $\mathbb Q(\sqrt 2, \sqrt 3)$ over $\mathbb Q$ sending $\sqrt 2$ to $-\sqrt 2$ and fixing $\sqrt 3$. Compute $\sigma(\alpha^2)/\alpha^2$. You will get $(2-\sqrt 2)/(2+\sqrt 2)$. This is $3-2\sqrt 2=(-\sqrt 2 + 1)^2$. So $\sigma(\alpha^2)=\alpha^2(-\sqrt 2 + 1)^2$. If $\alpha$ were in $\mathbb Q(\sqrt 2, \sqrt 3)$, then $(\sigma(\alpha))=\pm (-\sqrt 2+1)\alpha$ and $\sigma(\sigma(\alpha))=\alpha(1+\sqrt 2)(1-\sqrt 2)=-\alpha$, a contradiction as $\sigma$ has order $2$. This proves $\alpha$ has degree $2$ over $\mathbb Q(\sqrt 2, \sqrt 3)$.

We now see that $\sigma^4(\alpha)=\alpha$, so $\sigma$ generates a subgroup of order $4$ in the Galois group.

Let $\tau$ be the automorphism of $\mathbb Q(\sqrt 2, \sqrt 3)$ over $\mathbb Q$ sending $\sqrt 3$ to $-\sqrt 3$ and fixing $\sqrt 2$. You can do a similar computation with $\tau$. You will find they are both of order $4$ and anti-commute. The only group of order $8$ with such elements is the quaternions, as the only non-commuative groups of order $8$ are $Q$ and $D_8$, and $D_8$ does not have two anti-commuting elements of order $4$.

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I am clueless! Is this answer adapted from Prof. Milne's solutions here or Milne adapted your proof? :) –  kan Apr 20 '13 at 4:26
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@KannappanSampath I've seen the solution to a similar problem before in another book (not Milne's). –  Potato Apr 20 '13 at 4:46
    
Oh, OK. Thanks. –  kan Apr 20 '13 at 6:02

This example is motivated by Richard Dean's article: A Rational Polynomial whose Group is the Quaternions Author(s): Richard A. Dean Source: The American Mathematical Monthly, Vol. 88, No. 1 (Jan., 1981), pp. 42-45 http://www.jstor.org/stable/2320711.

The key point is that if $K/\mathbb{Q}$ is Galois, with Galois group $Q$, then it will have a field lattice reflecting the peculiar properties of $Q$. That is $Q$ has 3 distinct normal and cyclic subgroups of order 4 and a normal subgroup of order 2 (whose quotient is the Klein group $\mathbb{Z}/2 \times \mathbb{Z}/2$). No other group of order $8$ has such a normal subgroup structure. (Indeed, there are only 3 Abelian groups of order $8$ and $2$ non-Abelian groups, namely the dihedral group and $Q$, so you can check this by hand).

So from the fundamental theorem of Galois Theory (i.e. the Galois Correspondence), we are therefore looking for a biquadratic field $L = \mathbb{Q}(\sqrt{A}, \sqrt{B})$, with $AB$ non-square in $\mathbb{Q}$ (and $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ will of course suffice), which admits a quadratic extension $K/L$ so that the field extensions $K/\mathbb{Q}(\sqrt{A}),K/\mathbb{Q}(\sqrt{B})$ and $K/\mathbb{Q}(\sqrt{AB})$ are all cyclic of degree $4$.

Richard Dean does not look at the algebraic number you have defined, but rather he considers $\theta = \sqrt{ (2 + \sqrt{2})(3 + \sqrt{3})(6+ \sqrt{6})}$ which is clearly a quadratic extension of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$. It therefore remains to show that 1) $\theta$ defines cyclic degree $4$ extensions of $\mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{6})$ and 2) $\theta$, and its obvious conjugates, are the entire set of roots of a degree $8$ polynomial over $\mathbb{Q}$ and all the conjugates lie in $\mathbb{Q}(\theta)$. Part 1) is the involved bit, and requires a bit of argument. Part 2) is not so tricky - a consequence of part 1) is a degree 4 polynomial over $\mathbb{Q}(\sqrt{2})$ which multiplied by the conjugate gives a degree $8$ extension with the required property.

Milne has simplified this construction in his notes on Galois Theory http://www.jmilne.org/math/CourseNotes/FT.pdf (see pages 46 & 129). Indeed he shows property 1) but with the algebraic number $\alpha = \sqrt{(2 + \sqrt{2})(3 + \sqrt{3})}$. By considering the effect of the automorphism $\sigma$ which sends $\sqrt{2}$ to $-\sqrt{2}$ and fixes $\sqrt{3}$ on $\alpha^2$ (as in Potato's answer) he can show that $\sigma^2$ is not the identity. Now $\sigma$ is an element of the Galois group of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}(\sqrt{3})$ (you should be able to write down the $4$ conjugates of $\alpha$ over $\mathbb{Q}(\sqrt{3})$ and convince yourself they are all elements of $\mathbb{Q}(\sqrt{3},\alpha)$, so that $\mathbb{Q}(\alpha)$ really does define a Galois extension of $\mathbb{Q}(\sqrt{3})$). Therefore $\mathbb{Q}(\alpha)/\mathbb{Q}(\sqrt{3})$ is cyclic of degree $4$. A similar argument shows that it is cyclic of degree $4$ over $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{6})$.

To convince yourself of part 2) write down a degree $4$ polynomial satisfied by $\alpha$ over $\mathbb{Q}(\sqrt{3})$ and its conjugate polynomial obtained by sending $\sqrt{3}$ to $-\sqrt{3}$ - the product of these two polynomials will be defined over $\mathbb{Q}$ and have $\alpha$, and its eight conjugates, as roots - moreover using similar reasoning to the previous paragraph you can show they all lie in the same field $\mathbb{Q}(\alpha)$.

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