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I need to determine the degree of the field extension $\mathbb{Q}(\sqrt{(2+\sqrt{2})(3+\sqrt{3}}))/\mathbb{Q}$. I've determined that the minimal polynomial of $\sqrt{(2+\sqrt{2})(3+\sqrt{3}})$ is $$f(x) =x^8-24x^6+144x^4-288x^2+144$$ and that this field extension is the splitting field for this polynomial. Since the extension is primitive, I drew the conclusion that the field extension had $\deg f =8$.

But when computing the Galois group, we find that the number of automorphisms is greater than 8. Since the polynomial is even, and that $\partial f$ does not divide $f$, we have the roots $\pm \alpha, \pm \beta, \pm \gamma$ and $\pm \mu$. There are obviously more than 8 ways to permute these roots.

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Your minimal polynomial argument is sufficient to show that the degree of the extension is 8. However, unless you have been given reason to believe that the extension is Galois, the Galois group of a splitting field of $f$ might have more than 8 elements.

In this case the extension is Galois, and there are fewer valid automorphisms than immediately apparent, due to hidden algebraic relations. For instance, if $\alpha = \sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ and $\beta = \sqrt{(2-\sqrt{2})(3-\sqrt{3})}$, then we have $\alpha\beta = 2\sqrt{3}$. This tells us that $\alpha$ and $\beta$ can't be sent to arbitrary roots of $f$, in particular $\sigma(\alpha)\sigma(\beta) = \pm 2\sqrt{3}$. Going further, you can write $2\sqrt{3}$ as a polynomial in $\alpha$, giving an expression for $\beta$ in terms of $\alpha$ with rational coefficients. Thus, $\sigma(\beta)$ is uniquely determined by $\sigma(\alpha)$. Similar arguments show that $\sigma(\alpha)$ uniquely determines $\sigma$.

To do this explicitly, you might wish to save effort. For instance, consider the intermediate field $\Bbb Q(\sqrt{2}, \sqrt{3})$. Over this field $f$ factors into irreducibles as $(x^2-\alpha^2)(x^2-\beta^2)(x^2-\gamma^2)(x^2-\delta^2)$. The nontrivial automorphism $\sigma$ fixing this field must therefore satisfy $$\sigma(\alpha) = -\alpha, \sigma(\beta) = -\beta, \sigma(\gamma) = -\gamma, \sigma(\delta) = -\delta.$$ Note that the previous paragraph's argument implies that no nontrivial automorphism fixes any root of $f$, since all of these roots generate the whole field.

Next, we wish to examine the element $\psi$ satisfying $\psi(\alpha) = \beta$. We expand the polynomial $$(x^2-\alpha^2)(x^2-\beta^2) = x^4 -(12+2\sqrt{6}) + 12.$$ This tells us that the four Galois automorphisms fixing $\Bbb Q(\sqrt{6})$ send $\alpha$ to $\pm \alpha, \pm \beta$. Thus, $\psi$ fixes $\sqrt{6}$ and in turn we know $\psi(\sqrt{2}) = -\sqrt{2}, \psi(\sqrt{3}) = -\sqrt{3}$, since $\psi$ can't fix all of $\Bbb Q(\sqrt{2}, \sqrt{3})$. Thus, we infer from $\beta\psi(\beta) = \psi(\alpha\beta) = \psi(2\sqrt{3}) = -2\sqrt{3})$ that $\psi(\beta) = -\alpha$. To determine the action on the other roots, we fix $\gamma = \sqrt{(2-\sqrt{2})(3+\sqrt{3})}$ and $\delta = \sqrt{(2+\sqrt{2})(3-\sqrt{3})}$. Now we have the relations $\alpha/\gamma = \delta/\beta = 1+\sqrt{2}$. This lets us compute $$\frac{\beta}{\psi(\gamma)} = \frac{\psi(\delta)}{-\alpha} = 1-\sqrt{2}$$ Solving these equations gives $\psi(\gamma) = -\delta, \psi(\delta) = \gamma$.

Rolf Hoyer
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  • The extension is in fact Galois, we know that it is a normal extension. If $\alpha$ and $\beta$ can't be sent to arbitrary roots, how do we determine the restriction? –  May 13 '16 at 05:02
  • It's possible to do this by brute force. Brute force is to write $\beta = p(\alpha)$, for $p\in \Bbb Q[x]$, and then you can compute $\sigma(\beta) = p(\sigma(\alpha))$, for instance. This seems like the algebra becomes quite ugly to find the 8 permutations corresponding to the eight automorphisms. Using algebraic tricks you can save some effort, I will give an example by amending my answer. – Rolf Hoyer May 13 '16 at 05:13
  • @Elliot yes, this argument is there to show that an arbitrary Galois automorphism is determined uniquely by what it does to $\alpha$. – Rolf Hoyer May 13 '16 at 05:37
  • Shouldn't the polynomial $(x-\alpha^2)(x-\beta^2)(x-\gamma^2)(x-\delta^2)$ be $(x^2-\alpha^2)(x^2-\beta^2)(x^2-\gamma^2)(x^2-\delta^2)$? –  May 13 '16 at 05:46
  • @Elliot Thanks, fixed – Rolf Hoyer May 13 '16 at 05:47
  • Why do we know that from the polynomial $x^4-(12+2\sqrt{6})+12$ that the four Galois automorphisms fixing $\mathbb{Q}(\sqrt{6})$ send $\alpha$ to $\pm \alpha, \pm \beta$? –  May 14 '16 at 06:48
  • This doesn't give us any restriction, if we look at $(x^2-\gamma^2)(x^2-\alpha^2)$, we get the polynomial $x^4+(12_2\sqrt{6})x^2+12$. Note that I'm taking $\gamma = \sqrt{(2-\sqrt{2})(3+\sqrt{3})}$. Therefore, the four automorphisms fixing $\mathbb{Q}(\sqrt{6})$ send $\alpha \to \pm \alpha, \pm \gamma$. –  May 14 '16 at 07:20
  • @Elliot, with that choice of $\gamma$ I'm getting $(x^2-\alpha^2)(x^2-\gamma^2) = x^4 - (12 + 4\sqrt{3})x^2 + 24 + 12\sqrt{3}$, instead. The four automorphisms fixing $\Bbb Q(\sqrt{3})$ should be the ones sending $\alpha$ to $\pm \alpha, \pm \gamma$. – Rolf Hoyer May 15 '16 at 04:32