Welcome to our community

Be a part of something great, join today!

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


May 20, 2012
i am trying to find [tex] G(F_{q}(x^{\frac{1}{q - 1}}/F_{q}(x)) [/tex] where q is the power of some prime.

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

i found this to be [tex] t^{q - 1} - x [/tex] which is irreducible over [tex] F_{q}[x] [/tex] 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 [tex] t^{q - 1} - x [/tex]. besides [tex] x^{\frac{1}{q - 1}} [/tex], im not sure what other roots it could have. can someone give me some hints on this?


Well-known member
MHB Math Scholar
Feb 15, 2012
suppose a is a non-zero element of Fq. since the non-zero elements of Fq form a finite cyclic group, we have:

aq-1 = 1, for all such a. thus the other q-2 roots are of the form ax1/(q-1) for a in Fq- {0,1}.

this shows that Fq(x1/(q-1)) is galois over Fq(x).

consequently, if b is a generator of Fq*, then any automorphism of Fq(x1/(q-1)) that fixes Fq(x) sends x1/(q-1) to bkx1/(q-1) for some k = 1,2....q-1.

on the other hand, [Fq(x1/(q-1)) :Fq(x)] = q-1, so these q-1 automorphisms must be all of Gal(Fq(x1/(q-1))/F​q(x)).

since the automorphism x1/(q-1) → bx1/(q-1)​ has order q-1, it appears we have a cyclic group of order q-1.
Last edited:


May 20, 2012
thank you! that was a very descriptive answer.