Let $f(x) = x^4 - 4x + 2$.
I am asked to find its Galois groups over $\mathbb Q $ and $\mathbb Q(i)$. I believe I have done this, but have arrived at a result that I was not expecting, so I wanted to ask if there was anything wrong with what I tried. Here are my thoughts:
Using primes $2$ and $1+i$ over $\mathbb Q$ and $\mathbb Q(i)$ respectively, we see by Eisenstein's Criterion that $f(x)$ is irreducible over $\mathbb Q$ and $\mathbb Q(i)$. This tells us that $Gal_{\mathbb Q}(f), \; Gal_{\mathbb Q(i)}(f)$ are both transitive subgroups of $S_4$
The discriminant of $f$ is $-4864$ which is not a square in $\mathbb Q$ or $\mathbb Q(i)$, thus the Galois groups over these fields are not in $A_4$. This tells us that $Gal_{\mathbb Q}(f), Gal_{\mathbb Q(i)}(f) \in \{C_4, D_8, S_4 \}$
Now we note that $f$ has a two real and two complex conjugate roots. This means that over $\mathbb Q$, complex conjugation is the equivalent of a transposition. However, $C_4,D_8$ both only have the identity, $4$-cycles, or double transpositions. Therefore, $Gal_{\mathbb Q}(f) \cong S_4$
Now suppose the splitting field over $f$ over $\mathbb Q$ is $L$. Then the splitting field over $\mathbb Q(i)$ is $L(i)$. We also note that $|L:\mathbb Q| = 24,\; |\mathbb Q(i):\mathbb Q|= 2$, so if we let $|L(i):L|=k,$ then by the Tower Law: $|L(i): \mathbb Q(i)| = 12k$.
We also notice that $k \leq 2$, so if we assume that $k=1$, then we see $|L(i):\mathbb Q(i)| = 12 \Rightarrow Gal_{\mathbb Q(i)}(f) \cong A_4$, a contradiction.
Therefore, we see that $k=2, \;|L(i):\mathbb Q(i)| = 24, \;Gal_{\mathbb Q(i)}(f) \cong S_4$
Put simply I was expecting this to be an example of a polynomial with two different Galois groups over these two fields, so I was surprised when I concluded that they were both $S_4$. If in fact the two Galois Groups are really the same, I would like to ask if the polynomial $x^5 -4x + 2$ also has that Galois group $S_5$ over $\mathbb Q$ and $\mathbb Q(i)$.
Thank you in advance for any help you may be able to offer.