<s>[INST] Find the Galois group of $x^4-5x^2+6$ over $\mathbb{Q}$ 
Find the Galois group of $x^4-5x^2+6$ over $\mathbb{Q}$

$$x^4-5x^2+6 = a^2-5a+6 = (a-3)(a-2) = (x^2-3)(x^2-2)$$
The roots are $\pm \sqrt{3}, \pm \sqrt{2}$, and the Galois group is made of the automorphisms that mixes the roots of $x^2-3$ and $x^2-2$. The possibilities of automorphisms:
$$e: \sqrt{2}\to\sqrt{2}, \sqrt{3}\to\sqrt{3}\\a: \sqrt{2}\to-\sqrt{2}, \sqrt{3}\to\sqrt{3}\\b: \sqrt{2}\to\sqrt{2}, \sqrt{3}\to-\sqrt{3}\\c: \sqrt{2}\to-\sqrt{2}, \sqrt{3}\to-\sqrt{3}$$
Is this the Galois group? 
 [/INST] The roots of $x^4-5x^2+6$ are $\pm\sqrt2$ and $\pm\sqrt3$ as you have pointed out.
Let $r$ be an automorphism of $\Bbb Q(\sqrt2,\sqrt3)$ that fixes $\Bbb Q$.
Then, $r(\sqrt2)^2 = r(2)=2$, so $r(\sqrt2)=\pm\sqrt2$.
Similarly, $r(\sqrt3)=\pm\sqrt3$.
Since an automorphism of a field is completely determiend by its generators, there are only $4$ possible automorphisms. There are three involutions, i.e. three elements with order $2$, so the group is $V_4$.
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