# The Galois Group of $x^4 - 5x^2 + 6$ [duplicate]

As the title suggests, I'm asked to describe the Galois group of the polynomial $x^4 - 5x^2 + 6 \in \mathbb{Q}[x]$ over $\mathbb{Q}$.

I am pretty certain I have 95% of the problem completed. I'm just having difficulty fully putting together the last step (I think it may boil down to a lack of knowledge in the "types" of groups).

The roots of this polynomial are $\pm \sqrt{2}$ and $\pm \sqrt{3}$. The automorphisms are thus: $$f_1(a + b\sqrt{2}) = a - b\sqrt{2} \\ f_2(a - b\sqrt{3}) = a - b\sqrt{3}$$

This resembles... something like... a group $\{e, \sigma_1, \sigma_2\}$ that has two coordinates? Something like $\sigma_1 \times \sigma_2$ where $\sigma_1$ cycles between itself and its negative and $\sigma_2$ cycles between itself and its negative.

I feel like there's just one little thing I'm missing.

(Edit: I think I've got it. This group is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$, right?)

## marked as duplicate by Watson, Alex M., Rohan, Claude Leibovici, Arnaud D.Dec 26 '16 at 14:26

Yes. You have one automorphism of order 2 sending $\sqrt{2}\mapsto -\sqrt{2}$ (fixing the rest), and another doing the same for $\sqrt{3}$ (fixing the rest).
These are two distinct elements of order 2, and your galois group is order 4, so it must be $(\mathbb{Z}/2\mathbb{Z})^2$