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An exercise asks to describe (i.e. basically tell what it is isomorphic to, rather than listing the automorphisms explicitly) the Galois group of $\mathbb{Q}(\sqrt{2},\sqrt{3})$ and suggests computing its degree over $\mathbb{Q}$. I already know it's $4$ and it is easy to prove. But I don't understand why this is needed, as we saw in class that the Galois group associated with an irreducible polynomial acts transitively on its roots and this is enough to conclude.

I would solve the exercise the following way:

$\mathbb{Q}(\sqrt{2},\sqrt{3})$ is the splitting field of the polynomial $(X^2-2)(X^2-3)$, each factor of degree $2$ being irreducible over $\mathbb{Q}$ using Eisenstein's criterion with $p=2, 3.$ Since an automorphism preserves each of these two polynomials, any permutations of the roots of $(X^2-2)(X^2-3)$ is a product of a permutation of the roots of $(X^2-2)$ and a permutation of the roots of $(X^2-3)$, and they are disjoint. (If they were not disjoint, i.e. if a root of a factor were to be sent to the root of the other, the polynomial factor wouldn't be preserved, since the two factors have different roots...).

Therefore, the Galois group is a subgroup of $S_2\times S_2$. However, since each of the polynomial factor is irreducible, its Galois group's action on its roots must be transitive, so the Galois group is actually isomorphic to $G_1\times G_2$ where $G_1$, $G_2$ are transitive groups of $S_2$, which is $S_2$ itself. Therefore, the Galois group is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. $\square$

I think the same method works for $\mathbb{Q}(\sqrt{p_1},...,\sqrt{p_n})$ where the $p_i$ are distinct primes. A corollary would be that these square roots are linearly independent over $\mathbb{Q}$.

Is there anything wrong with my proof? I think this is interesting because if this works then it allows the description of the Galois group of such extensions without having to bother showing linear independence of the square roots (i.e. computing the degree of the extension, which can be troublesome)

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  • $\begingroup$ @TobErnack I see, thank you! I implicitly assumed independence, resulting in the direct product. I think you should post this as an answer $\endgroup$ – Evariste Dec 6 '18 at 11:50
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(Moved from comments)

You need to know that $\sqrt{2}$ and $\sqrt{3}$ are linearly independent, otherwise you could have something like $\sqrt{3} = a\sqrt{2} + b$ for some $a, b \in \mathbb{Q}$. Then you would not be able to choose an automorphism that sends $\sqrt{2}$ to its conjugate while fixing $\sqrt{3}$. So it would turn out that the Galois group is just $S_2$.

For example the argument you gave would carry out just as well for the polynomial $(X^2 - 2)(X^2 - 18)$ (both factors are irreducible by the rational roots test). But the Galois group here is actually just $S_2$, and this is because $\sqrt{18} = 3\sqrt{2}$ so they are linearly dependent. The splitting field is $\mathbb{Q}(\sqrt{2}, \sqrt{18}) = \mathbb{Q}(\sqrt{2})$, so the degree of the extension is only $2$.

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