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What order is the field: $\mathbb{F}_p(\zeta_{p-1}^{1/n})$?

$\mathbb{F}_p^{\times}$ is size $p-1$ and cyclic (with a generator we shall call $\zeta_{p-1}$). Naively, it seems that by taking the $n^{th}$ root of $\zeta_{p-1}$ we get a group of units of size $C_{n(p-1)}$. This, of course, doesn't make sense. What should be true is that $C_{n(p-1)}$ is some subgroup of the group of units of the $\mathbb{F}_{p^r}$ we get.


What is the order of the field $\mathbb{F}_p[x]/(m_{\zeta_{p-1}^{1/n}})$ where $m_{\zeta_{p-1}^{1/n}}$ is the minimal polynomial of $\zeta_{p-1}^{1/n}$ (which, unless I'm mistaken, is the same as the splitting field of $x^n-\zeta_{p-1}$)? Does it depend on whether $p|n$ or not?

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Take p=2 as an example to see you have to make some adjustments in your bolded question. –  Jack Schmidt Dec 4 '10 at 18:38
Okay, I changed the question. –  Shannon Dec 4 '10 at 18:41
Cool. And yup, the field and the splitting field are the same. The answer in brief is just "find the smallest field with a subgroup of order n(p-1)", but hopefully the longer answer is helpful. –  Jack Schmidt Dec 4 '10 at 18:50
It is! I can't upvote because of my low reputation, but I picked your answer. –  Shannon Dec 4 '10 at 18:53
No worries. I always thought this was kind of a spooky result. "I want an element of order k in a mod p world, so I looked for the order p mod k." This gives you neat ways to write down sums of roots of unity too: sqrt(-3) = ζ3+1/ζ3, so to write down sqrt(-3) in a finite field of characteristic not 3, you find the order of p mod 3 (either 1 or 2) and use that to find a primitive 3rd root of unity say z, and then "sqrt(-3) mod p" is just equal to z+1/z. –  Jack Schmidt Dec 4 '10 at 18:58
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up vote 2 down vote accepted

Every nonzero element of a finite field is a root of unity. Roots of unity behave very nicely, since the other roots of unity of the same order are just powers of the first.

For instance i is a primitive 4th root of unity, and the other one, -i, is just a power i,-1,-i,1.

The roots of unity form these nice cyclic subgroups of the group of units of a field.

If you want one primitive k'th root of unity, you get all of them for free, so we are just interested in the smallest finite field of characteristic p with a primitive k'th root of unity.

Now you get into trouble if p divides k: a finite field of characteristic p has size p^m, so its group of units has order p^m-1, and so k has to divide p^m-1. Such numbers are coprime to p.

In your question, you'll want to assume n is coprime to p, otherwise you won't get very good roots. For instance, when p=2, the 8th roots of unity are exactly {1}; none of them are primitive.

Groups of units of finite fields are very well behaved: they are all cyclic. So to find a primitive k'th root of unity, you just want to find the smallest m such that k divides p^m-1:

The field obtained by adjoining a primitive k=n(p-1)st root of unity is the field of size pm, where m is the order of p in the group of units of the ring Z / k Z.

Note that k has to be coprime to p in order for a primitive k'th root of unity to exist. Just divide k by p until it is coprime if you want to adjoin the "closest to primitive" root.

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