I posted the following in another thread and got some clarification. But i need more help. Hence I'm reposting the same problem.
There is the following problem from my text book. I just wanna know if I understand the solution correctly.
For which of the following values of n does the finite field F5n with 5n elements contain a non trivial 93 rd root of unity? 1. 92 2. 30 3. 15 4. 6
The theorem I used was as follows
"Let F be a field and H , the group of nth roots of unity in F, be a subgroup of the multiplicative group FX. Then H is cyclic of some order m such that m divides n. If, in addition, F is finite with order q, |H| = (n, q-1)"
First up, the multiplicative group FX has |F| -1 elements. Since H is a subgroup of FX , o(H) should divide |F| -1. Since H is cyclic, there is a generator whose order is the same as the order of H. So if 'a' is a generator of H, o(a) should divide |F| -1.
Since H is a group of the nth roots of unity in F, the nth power of every element is 1, the identity element with respect to H and FX . So, a93 =1. This means 93 should divide |F| -1.
To sum it up, 93 should divide 5n - 1.
Have I been right so far?
Can someone help me find the right option from here on?
well, not quite.
suppose (for example) that F contains all 3rd roots of unity, but no primitive 9th root of unity. well all 3rd roots of unity are also 9th roots of unity, if:
a3 = 1 (mod q-1) then:
a6 = (a3)3 = 13 = 1 (mod q-1).
but the subgroup of 9-th roots of unity in F doesn't have order 9, but order 3.
for example F25 has a cyclic multiplicative group of order 24, which therefore has a cyclic subgroup of order 3, but no cyclic subgroup of order 9 since 9 does not divide 24
(if the generator of F* is b, then b8 is a cube root of 1, so b8 is a 9-th root of 1 as well:
(b8)9 = b72 = (b24)3 = 13 = 1).
so all you can say is 93 and 5n-1 have a common factor > 1. this common factor may be 3,31 or 93.
if F5n is supposed to contain ALL 93-rd roots of unity, THEN 93 would have to divide 5n-1.
Can someone help me with the next step towards the solution?