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
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:)