About the Galois group of the polynomial $f(x)=x^4+ax^2+b$ over $\mathbb Q$

Question

I am trying to prove:

Let $K$ be the splitting field of $f(x)=x^4+ax^2+b$, whose roots are denoted by $\pm\alpha,\ \pm\beta$. Prove that $Gal(K/\mathbb Q)\cong V_4$ if and only if $\alpha\beta\in\mathbb Q$.

My question is: if we let $$\sigma:\alpha\mapsto-\alpha,\ \beta\mapsto\beta$$ $$\tau:\alpha\mapsto\alpha,\ \beta\mapsto-\beta$$ Then if both are well-defined automorphisms of $K$, the group $$\{1,\sigma,\tau,\sigma\tau\}$$ is isomorphic to $V_4$. However, $\sigma,\tau$ do not fix $\alpha\beta$, which means that this group must not be $Gal(K/\mathbb Q)$ (otherwise $\alpha\beta\notin\mathbb Q$, contradictory to what I am trying to prove) What is the reason?

p.s. I am aware that there questions asking about the same proposition. But they don't mention the problem that I am having here.

It seems to me that this is covered in this question : https://math.stackexchange.com/questions/649466/galois-group-of-a-biquadratic-quartic — Arnaud D., Oct 10 '19 at 13:57
Possible duplicate of Computing the Galois group of $x^4+ax^2+b \in \mathbb{Q}[x]$ — mrtaurho, Oct 10 '19 at 13:59
@ArnaudD. I don't see directly how. Is it possible that transitivity is the problem with the group that I constructed here? — trisct, Oct 10 '19 at 14:06
Yes. Basically, you're not looking at the correct automorphisms of $K$, because you don't have anything that takes $\alpha$ to $\pm\beta$. (I'm assuming your quartic polynomial is irreducible.) — Arnaud D., Oct 10 '19 at 14:14

score 0 · Answer 1 · answered Oct 10 '19 at 14:14

Your roots have two different values of $\alpha\beta$, depending the signs you choose. Call these possible values $+\gamma,-\gamma$. You thus have a set of two products $\{+\gamma,-\gamma\}$.

When you apply $\sigma$ or $\tau$, then true, you do not map either of the individual elements $+\gamma,-\gamma$ into itself. But you do map the set $\{+\gamma,-\gamma\}$ into itself. Thus the set is invariant, and that is all you really need to identify a symmetry element. The fact that you changed the order within the set does not change the set.

About the Galois group of the polynomial $f(x)=x^4+ax^2+b$ over $\mathbb Q$

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