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.

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    $\begingroup$ I haven't carefully read through your work past the first two paragraphs, but one problem that I see is that there's a subtle issue with your using Eisenstein over $\mathbb{Q}(i)$. In particular, let $P = (1+i)\mathbb{Z}[i]$; then $(1+i)(1-i)=2$, and $(1-i) = -i(1+i)$, so $1-i$ and $1+i$ are associates, so $P = (1-i)\mathbb{Z}[i]$, and therefore $P^{2} = 2\mathbb{Z}[i]$. In particular, $2$ is in $P^{2}$, so Eisenstein at the prime ideal $P$ does not apply. $\endgroup$ – Alex Wertheim Jan 1 '18 at 23:51
  • $\begingroup$ @AlexWertheim I see thank you for pointing this out. In that case, could I claim it is irreducible over complex rationals by saying that if it weren't, it must have two linear and one quadratic factors, as the roots come in complex conjugate pairs. Then the quadratic factor must contain only real coefficients. This then contradicts the irreducibility of $f$ in $\mathbb Q$, since the two linear factors corresponding to the two complex roots must have a product in $\mathbb Q[x]$? $\endgroup$ – user366818 Jan 2 '18 at 0:26
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    $\begingroup$ An observation: Your analysis of the Galois group over $\Bbb{Q}$ reveals that the only quadratic subfield of $L$ must be $M=\Bbb{Q}(\sqrt{-4864})=\Bbb{Q}(\sqrt{-19})$. Because $i\notin M$ we can conclude that $i\notin L$. Therefore $[L(i):L]=2$. This might generalize to the quintic case, but I haven't checked. $\endgroup$ – Jyrki Lahtonen Jan 2 '18 at 8:57
  • $\begingroup$ @JyrkiLahtonen, thank you. For the quintic case I did the same thing that I did in the quartic case to conclude that $|L(i):L| =2$ by showing that the discriminant is not a square and so the Galois group cannot have order $60$ $\endgroup$ – user366818 Jan 2 '18 at 10:15

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