I know that $K = \mathbb{Q}(\sqrt{2},\sqrt{3})$ is the splitting field of $f(x)=(x^2-2)(x^2-3)$ over $\mathbb{Q}$ and so the Galois group $\Gamma(K:\mathbb{Q})$ permutes the roots $\{\sqrt{2},-\sqrt{2},\sqrt{3},-\sqrt{3}\}$ of $f$.

As elements of the Galois group must leave the set of roots of the irreducible factors (over $\mathbb{Q}$) of $f$ invariant, I can calculate that (in cyclic notation) $$\tag{1}\Gamma(K:\mathbb{Q}) = \{Id, \ (\sqrt{2},-\sqrt{2}), \ (\sqrt{3},-\sqrt{3}), \ (\sqrt{2},-\sqrt{2})(\sqrt{3},-\sqrt{3})\}$$

But, for $\alpha := \sqrt{2} + \sqrt{3}$, I can verify that also $K = \mathbb{Q}(\alpha)$ and in fact $K$ is also the splitting field of $g(x) = (x^2-5)^2 - 24$ (which is the minimal polynomial of $\alpha$). For $\beta := \sqrt{2} - \sqrt{3}$, the roots of $g$ are $$R = \{\alpha,-\alpha,\beta,-\beta\}$$ So using $(1)$ I can translate $\Gamma(K:\mathbb{Q})$ into permutations of elements of $R$ to find (again in cyclic notation) $$\tag{2}\Gamma(K:\mathbb{Q}) = \{Id, \ (\alpha, -\beta)(-\alpha,\beta), \ (\alpha,\beta)(-\alpha,-\beta), \ (\alpha,-\alpha)(\beta,-\beta) \}$$

So my question is, if I was just given $g$ and was told $K = \mathbb{Q}(\alpha)$ is the splitting field of $g$, how would I calculate $\Gamma(K,\mathbb{Q})$ in terms of $R$ as in $(2)$? How could I deduce that say $(\alpha,\beta) \notin \Gamma(K:\mathbb{Q})$? As $g$ is irreducible my first thought would be that the roots of $g$ could be permuted in whatever way I like and hence $\Gamma(K:\mathbb{Q}) \cong S_4$ which is clearly not true. Where am I going wrong?


If $\sigma$ is any ${\mathbb Q}$-homomorphism, we have $\sigma(-x)=-\sigma(x)$ for any $x$. In particular, if $\sigma(\alpha)=\beta$ then $\sigma(-\alpha)=-\beta$ also. This shows that $(\alpha,\beta) \notin \Gamma(K:\mathbb{Q})$.


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