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I picked a problem from Dummit and Foote's book that asked to find the Galois group of $x^5+x-1$. The polynomial can be factored as

$x^5+x-1=(x^2-x+1)(x^3+x^2-1)$

The polynomial of degree 3 has Galois group equal to $S_3$ (it is not hard to calculate using the discriminant). Also I saw it has a real root and two complex roots (I give this information if is helpful). The Galois group of the polynomial of degree 2 is $Z_2$ (also not hard to calculate).

The answer should be the direct product of this two groups but I need to prove the intersection of the two splitting fields is $Q$ and I have problem with that.

My idea is assume that the intersection is not $Q$. Since the intersection is contained in $K_2$ (the splitting field of polynomial of degree 2) then it has to be equal to $K_2$. Then $K_3$ contains $K_2$. If one the of roots of the polynomial of degree 3 is contained in $K_2$ then the field generated by that root is contained in $K_2$ but the polynomial $x^3+x-1$ is irreducible over $Q$ then the field generated by the root has degree 3 which not divide 2 (contradiction).

My problem is when the root is not in $K_2$. I don't know how to prove a contradiction because the field generated doesn't necessarily contain $K_2$.

Thank you for your help:)

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  • $\begingroup$ All the subfields of $K_3$ of degree 2 over $\mathbb{Q}$ correspond to subgroups of order $3$ in the galois group $S_3$. Maybe that helps. $\endgroup$ – Verdruss Dec 21 '17 at 13:42
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Note that $S_3$ has a unique index 2 subgroup isomorphic to $C_3$; the corresponding fixed field $F$ is therefore the only quadratic subfield inside $K_3$. Then the Galois group of the full polynomial is not the direct product $S_3 \times C_2$ (and therefore just $S_3$) if and only if $K_3 \cap K_2 \neq \mathbb{Q}$ if and only if $F=K_2$.

A different way of saying this is that the Galois group of the cubic over $K_2$ is only $C_3$. To check this, you can again just apply the discriminant trick (i.e. checking if $-31$ is a square), this time square meaning in $K_2= \mathbb{Q}(\sqrt{-3})$. This is false, hence the Galois group of the cubic over $K_2$ is still $S_3$ and hence the full Galois group is $S_3 \times C_2$.

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  • $\begingroup$ How you prove that K_2 cant be that unique field of degree 2 over Q? That field is the field that contain the automorphism that send the complex root to his conjugated $\endgroup$ – LadyGugu Dec 21 '17 at 22:04

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