Answer to Question #202348 in Real Analysis for Arun

Question

Prove that any n-th root of unity is a primitive d-th for a uniqued/n ?

Answer 1

Let F

F be a field and let n∈N

 and let ε∈F

 (F is an algebraic clausure of F) be a 

n_th root of unity (

ε is a root of X

n−1∈F[X])

$X_n−1∈F[X]$ . Can I state that there exists d∈N

d∈N, with d|n, such that ε is a d

d_th primitive root of unity? (

ε is d-th primitive root of unity if ε

ε is d-th root of unity and ε generates all the elements in the group of d

d-th roots of unity, which is a subgroup in $F _{0}$ . I am going to denote by

W_k the group of k_th roots of unity and W∗k the elements in 

W_k which are primitive.

I have not found any counter example. For instance, if we take 

F=C, then 

W₄={1,i,−1,i}. In this group, the only elements which do not belong to W∗₃

 are 1

1 and −1

−1; however, −

−1∈W₄∗ and 1∈W

∗

1∈W₁∗, so no counterexample can be found here. On the other hand, if we take 

F as the finite field of 

9 elements, 

F=GF(9) (whose operations between the elements can be found

here: http://www.cs.miami.edu/home/burt/learning/Csc609.032/notes/gf9example.html),

we can see that

W₄={1,2,x,2x}. Here, 

W₄∗={x,2x}, but 

2∈W₂∗ and 

1∈W₁∗. Finally and it is my last example, if we take

F=GF(4), we have that 

X₄−1=(X−1)4 and 

X₂−1=(X−1)2, so 1

1 is a primitive element, as 

W₄=W₂={1}.

My attempt for the proof: let d

d the order for ε

ε as element in W_n

W_n (each element in a group generates a cyclic group), which is a cyclic and finite group. Then 

ε_d=1 and ⟨ε⟩={1,ε,…,εd−1}. Apparently, we are done because 

ε_d=1, which shows that 

ε∈Wd, but (εj)d=1j=1 for all j=0,1,…,d−1

j=0,1,…,d−1, which shows that 

⟨ε⟩⊂W_d and 

|W_d|≤d, which shows that 

⟨ε⟩=W_d.

The main problem which I have found in the proof is that I cannot state that the group 

Wn has n elements (see my last example), it happens in finite groups or in groups which characteristic is not zero. Generally, I can state that 

m≤d (it is an equality in C

C por example), where 

m=|Wd| but I cannot state that m=d

m=d.

hence proved.