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let $f_1=x^4+14x-7 \in \mathbb{Q}[x]$ and $f_2=x^4-6x^2+4 \in \mathbb{Q}[x]$ be two irreducible polynomials. Now I've got determine $Gal(f_i(x)$) for $i=1,2$ and all intermediate fields $\mathbb{Q}\subsetneq K \subsetneq L$ in these two cases.

My attempt: I've looked up that $f_1$ has two real and two complex roots. $f_2$ has only 4 real roots. But I'm very lost on how to go on. For $f_2$ I've got the factorization: $(x- \sqrt (3-\sqrt5))(x+ \sqrt (3-\sqrt5))(x- \sqrt (3+\sqrt5))(x+ \sqrt (3+\sqrt5))$. How can I deduce the Galois group from there? Thanks for any help!

Thesinus
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  • It can only be $C_4,D_4,V_4,A_4,S_4$, see Conrad's article, or this site (for example here) how to find it. Compute $\Delta$. – Dietrich Burde Jun 08 '20 at 14:14
  • It is a difficult task to determine in general the galois group by hand. For the degree 4 polynomial there exist an auxiliary polynomial, the solving cubic that helps finding the group. Cubic+ discriminant of the cubic can find the solutions. – Fabrizio Jun 08 '20 at 14:15

1 Answers1

1

$f_1$ case

Denote by $\alpha$ one of $f_1$ real roots. As $f_1$ is irreducible over $\mathbb Q$, $[\mathbb Q(\alpha):\mathbb Q]=4$. If $\beta$ is one of the non real roots of $f_1$, $\beta \notin \mathbb Q(\alpha)$ and we can write in $\mathbb Q(\alpha)[x]$:

$$f_1(x) = (x- \alpha)(x - \tilde{\alpha})q_1(x)$$ where $\tilde{\alpha} \in \mathbb Q(\alpha)$ and $q_1$ is an irreducible polynomial of $\mathbb Q(\alpha)$. We have $[L:\mathbb Q] = [L:\mathbb Q(\alpha)][\mathbb Q(\alpha):\mathbb Q]=8$ and $\mathbb Q \subset \mathbb Q(\alpha)$ is not a Galois extension. According to the fundamental theorem of the Galois theory this implies that $\mathrm{Gal}(L/\mathbb Q(\alpha))$ is not a normal subgroup of $\mathrm{Gal}(L/\mathbb Q)$.

And finally that $\mathrm{Gal}(f_1) = D_8$, the dihedral group of order 8, which is the only group of order $8$ having non normal subgroups.

$f_2$ case

$\alpha = \sqrt{3 -\sqrt 5}$ is one of $f_2$ roots and the set of roots is $\{\alpha, -\alpha, 2/\alpha, -2/\alpha\}$. Therefore $$\vert \mathrm{Gal}(f_2) \vert = [\mathbb Q(\alpha):\mathbb Q]=4$$ and $\mathrm{Gal}(f_2) $ is either the cyclic group $\mathbb Z_4$ or the Klein-four group $\mathbb Z_2 \times \mathbb Z_2$ which are the only groups of order $4$. One can verify that $\mathrm{Gal}(f_2) $ has no element of order $4$.

Therefore it is equal to the Klein-four group.

mathcounterexamples.net
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