A question about finite fields

#field_theory

Background

Cyclic groups

For each positive integer $m$ let's write $Z_{m}$ for the (commutative) ring $Z / m Z$ . Here are some quick facts:

The ring $Z_{m}$ is a field when $m$ is prime, but not even a domain when $m$ is composite.
If we forget the multiplicative operation, we can view $Z_{m}$ as an abelian group under addition. There should be notation to distinguish whether we are viewing $Z_{m}$ as a ring or as a group (and there is in category theory), but most algebraists just say (or imply) the type of structure they're considering.
Up to (group) isomorphism, for each positive integer $m$ there is a unique cyclic group of order $m$ , namely $Z_{m}$ .

We also have very precise information about the entire lattice of subgroups of a finite cyclic group:

Subgroups of finite cyclic groups

Suppose $G$ is a finite cyclic group of order $m$ , i.e, $G ≅ Z_{m}$ . Then for every subgroup $H \leq G$ we have:

$H$ is cyclic; and
$| H |$ divides $| G |$ .

Conversely, for each divisor $d$ of $m$ there is a unique subgroup $H \leq G$ of order $d$ . This subgroup consists precisely of those elements $g \in G$ that have order dividing $d$ . (If we use multiplicative notation in $G$ , then these are the elements $g$ that satisfy $g^{d} = 1$ .)

Moreover, if $H, K \leq G$ are subgroups of orders $d$ and $e$ , respectively, then $H \leq K$ exactly when $d$ divides $e$ .

In other words, the subgroups of $Z_{m}$ look exactly like $Z_{d}$ for the divisors $d$ of $n$ , and $Z_{d} \leq Z_{e}$ exactly when $d$ divides $e$ .

Finite fields

For each finite field $F$ , we recall the following basic facts:

The characteristic of $F$ is a positive prime integer $p$ .
There is a (unique, injective) field morphism $Z_{p} \to F$ , which allows us to consider $F$ as a field extension of the field $Z_{p}$ . In particular, this allows us to consider $F$ as a $Z_{p}$ -vector space. Since $F$ is finite (as a set), it must also be finite-dimensional as a $Z_{p}$ -vector space.
If $n$ is the dimension of $F$ as a $Z_{p}$ -vector space, then $| F | = p^{n}$ .
Every element $α \in F$ satisfies $α^{p^{n}} = α$ , which in turn allows us to prove that $F$ is a splitting extension over $Z_{p}$ for the polynomial $f (x) = x^{p^{n}} - x$ . Since splitting extensions are unique (up to unique isomorphism), this proves that $F$ is the unique field of order $p^{n}$ (up to isomorphism).
If $F \to E$ is a field morphism between finite fields, then $F$ and $E$ have the same characteristic (say $p$ ) and the dimension of $F$ as a $Z_{p}$ -vector space divides the dimension of $E$ as a $Z_{p}$ -vector space. In other words, we have $| F | = p^{n}$ and $| E | = p^{m}$ with $n \leq m$ .

Quick summary of the above facts: Every finite field $F$ has order $p^{n}$ for some prime $p$ and positive integer $n$ , and up to isomorphism there is exactly one field of each such order.

The big result we need from field theory

We need the following foundational result from field theory, which is sneakier to prove than it might seem:

Groups of units of fields

Suppose $F$ is any field and $F^{\times}$ is its group of (multiplicative) units. Then every finite subgroup of $F^{\times}$ is cyclic.

Corollary

Suppose $F$ is a finite field of order $p^{n}$ . Then $F^{\times} ≅ Z_{p^{n} - 1}$ .

So, if we're wondering about the subgroups of $F^{\times}$ for a finite field, we just need to understand the subgroups of the cyclic group $Z_{p^{n} - 1}$ . Fortunately, the result above tells us the entire lattice structure for the subgroups of $Z_{p^{n} - 1}$ : there is a unique subgroup $H$ of order $d$ for each divisor $d$ of $p^{n} - 1$ , and for two such subgroups $H, K \leq F^{\times}$ we have $H \leq K$ exactly when $| H |$ divides $| K |$ .

The subgroup of $k^{th}$ powers of $F^{\times}$

Now we wish to understand the subgroup $K \leq F^{\times}$ consisting of elements that are $k^{th}$ powers of elements in $F^{\times}$ . This subgroup is exactly the image of the group morphism $ϕ : F^{\times} \to F^{\times}$ defined by $ϕ (α) = α^{k}$ . What is the size of this subgroup? One way to answer this question is to use the First Isomorphism Theorem for groups, which tells us that

F^{\times} / \ker (ϕ) ≅ im (ϕ) .

In other words,

| K | = | im (ϕ) | = [F^{\times} : \ker (ϕ)] = \frac{| F^{\times} |}{| \ker (ϕ) |} = \frac{p^{n} - 1}{| \ker (ϕ) |} .

Can we compute $\ker (ϕ) ?$ This is the subgroup of $F^{\times}$ consisting of those elements $α$ that satisfy $α^{k} = 1$ , i.e., they are the elements of the field $F$ that are the roots of the polynomial $f (x) = x^{k} - 1$ . Or, strictly from the point of view of group theory, they are the elements in the group $F^{\times}$ of order dividing $k$ .

From the description of such $α$ as the roots of $f (x) = x^{k} - 1$ , you might at first think there are exactly $k$ such elements. But remember that we're working in positive characteristic, so we can easily be duped by repeated roots. For example, if $p = 2$ and $k = 4$ , then we have $f (x) = x^{4} - 1 = (x - 1)^{4}$ , which only has one root!

We can make things easier on ourself by imposing some conditions on $k$ . For example, if we restrict ourselves to those $k$ that divide $p^{n} - 1$ , then by our fact about finite cyclic groups above, the set of elements of order dividing $k$ is exactly the subgroup of $F^{\times}$ of order $k$ . In other words, in this case $| \ker (ϕ) | = k$ and hence $| K | = \frac{p^{n} - 1}{k}$ .

Now we can answer the following question:

A question about

k^{th}

powers in finite fields

Suppose $F$ is a finite field of order $p^{n}$ and $k$ is a positive integer that divides $p^{n} - 1$ . Let $H = Z_{p}^{\times}$ and let $K \leq F^{\times}$ be the subgroup of $k^{th}$ powers in $F^{\times}$ . When do we have $H \leq K$ ?

In this case, we have $| H | = p - 1$ and (based on our work above) $| K | = \frac{p^{n} - 1}{k}$ . Using our basic theory about subgroups of the finite cyclic group $F^{\times}$ , we can answer the question: we have $H \leq K$ exactly when $p - 1$ divides $\frac{p^{n} - 1}{k}$ .

If desired, we can rewrite the divisibility condition equivalently as: $H \leq K$ exactly when $k$ divides $\frac{p^{n} - 1}{p - 1}$ .