## Galois group of x^6-9x^6-9

f = x^6-9 = (x^3-3)(x^3+3) Let L_f be splitting field therefore L_f = \mathbb{Q}[\sqrt[3]{3},e^{\frac{2\pi i}{3}}], [L_f:\mathbb{Q}] = 9. Also Gal\space x^3±3/\mathbb{Q} = S_3 and Gal\space f/\mathbb{Q}-orbits is roots of x^3-3 and x^3+3 therefore, i think, that G = Gal\space f/\mathbb{Q} = S^3\times S^3. But |G| = 36 \ne [L_f:\mathbb{Q}]. What’s wrong? Answer A few issues: … Read more

## If every polynomial of odd degree is reducible in $\mathbb{F}$ then $[E:\mathbb{F}]$ is not odd

$\mathbb{F}$ is a field of characteristic zero, and every odd-degree polynomial over $F$ has a root in $\mathbb{F}$. now I have two questions:- If $E$ is a finite extension of $F$ show that $[E:\mathbb{F}]$ is not odd. If $E$ is a finite Galois extension of $\mathbb{F}$, Show that $[E:\mathbb{F}]$ is a power of two. My … Read more

## Show that every finite extension of FF is cyclic. [duplicate]

This question already has an answer here: Galois Groups of Finite Extensions of Fixed Fields (1 answer) Closed 6 years ago. Let k be a field, ka an algebraic closure, and σ be an automorphism of ka leaving k fixed. Let F be the fixed field of σ. Show that every finite extension of F … Read more

## Finding Galois extension isomorphic to Zn\mathbb{Z}_n

Exist a method to find an Galois extension E such that Gal(E/Q)≅Zn? Only for n=6 how can I do? Answer Find m such that Gal(Q(ζm)/Q)≅Z∗m has a quotient group isomorphic to Zn. You can do this by letting m be a prime congruent to 1modn. By Galois theory, the fixed field of the kernel of … Read more

## Finding Galois extension isomorphic to \mathbb{Z}_n\mathbb{Z}_n

Exist a method to find an Galois extension E such that Gal(E/\mathbb{Q})\cong\mathbb{Z}_n? Only for n=6 how can I do? Answer Find m such that \text{Gal}(\mathbb{Q}(\zeta_m)/\mathbb{Q}) \cong \mathbb{Z}_m^{\ast} has a quotient group isomorphic to \mathbb{Z}_n. You can do this by letting m be a prime congruent to 1 \bmod n. By Galois theory, the fixed field … Read more