Finding galois group of Fq(x^(1/(q-1))) over Fq(x)

[oblixps]

i am trying to find $$G(F_{q}(x^{\frac{1}{q - 1}}/F_{q}(x))$$ where q is the power of some prime.

i know that $$F_{q}(x^{\frac{1}{q - 1}})$$ is an extension of $$F_{q}(x)$$ so i need to find the irreducible polynomial of $$x^{\frac{1}{q - 1}}$$ over $$F_{q}(x)$$.

i found this to be $$t^{q - 1} - x$$ which is irreducible over $$F_{q}[x]$$ by Eisenstein's criterion. i know that every automorphism in the galois group must map roots of polynomials to roots of the same polynomial but i am having trouble finding the roots of $$t^{q - 1} - x$$. besides $$x^{\frac{1}{q - 1}}$$, I am not sure what other roots it could have. can someone give me some hints on this?

 

[Deveno]

suppose a is a non-zero element of F_q. since the non-zero elements of F_q form a finite cyclic group, we have:

a^q-1 = 1, for all such a. thus the other q-2 roots are of the form ax^1/(q-1) for a in F_q- {0,1}.

this shows that F_q(x^1/(q-1)) is galois over F_q(x).

consequently, if b is a generator of F_q*, then any automorphism of F_q(x^1/(q-1)) that fixes F_q(x) sends x^1/(q-1) to b^kx^1/(q-1) for some k = 1,2...q-1.

on the other hand, [F_q(x^1/(q-1)) :F_q(x)] = q-1, so these q-1 automorphisms must be all of Gal(F_q(x^1/(q-1))/F_​q(x)).

since the automorphism x^1/(q-1) → bx^1/(q-1)​ has order q-1, it appears we have a cyclic group of order q-1.

 

[oblixps]

thank you! that was a very descriptive answer.