Properties of f(x) when $Gal(K/Q) ≃ S4.

Question

Suppose that we have f ∈ $\Bbb Q[X]$ with $\deg(f) = 4$ and that for the splitting field $K$ of $f$ over $\Bbb Q$ we have that $\text{Gal}(K/Q) ≃ S_4$.

Show the following:

(a) $f$ is irreducible over Q and has four distinct roots $\Bbb α_1, α_2, α_3, α_4 $ in $K$.

(b) The element $β = \Bbb α_1α_2 + α_3α_4$ lies in $K \setminus \Bbb Q$.

(c) $K/\Bbb Q(β)$ is a Galois extension and $\text{Gal}(K/\Bbb Q(β)) ≃ D_4$.

This is the question I am working with at the moment. I am teaching myself advanced mathematics and Galois is a relatively new topic to me so my approach to this questions isn't as accurate as I would like it to be. I am confident with part (a) however, for part (b) and (c), what is the best approach to make? I would really appreciate your answers as well as any help or explanation you can offer. I really want to understand your methods ands reasoning. I am struggling to study this topic myself.

Thank you.

Answer 1

For (b), just use the fact that $K$ is a field. For (c), you can do this by "brute force". Suppose $\sigma \in \text{Gal}(K/\mathbb{Q})$ is such that $\sigma(\beta) = \beta$, then consider the cases below:

• If $\sigma(1) = 1$, then $\sigma(2) = 2$. Hence, $\sigma \in \{e, (34)\}$.
• If $\sigma(1) = 2$, then $\sigma(2) = 1$, whence $\sigma \in \{(12), (12)(34)\}$.
• If $\sigma(1) =3$, then $\sigma(2) = 4$ and $\sigma(3)\in \{1,2\}$. Hence $\sigma \in \{(13)(24), (1324)\} $
• If $\sigma(1) = 4$, then $\sigma(2) = 3$ and $\sigma(4)\in \{1,2\}$, whence $\sigma \in \{(14)(23), (1423)\}$.

These 8 options exhaust all elements of $\text{Gal}(K/\mathbb{Q})$ that fix $\beta$, and so $\text{Gal}(K/\mathbb{Q}(\beta))$ is the union of all these 4 sets which is $D_4$ (any subgroup of $S_4$ of order 8 must be isomorphic to $D_4$ because it is a 2-Sylow subgroup).