Irreducible factors of $X^n -1$ in $\mathbb{F}_q[X]$

Question

I am stuck trying to prove the following corollary:

Let $f = X^n -1 \in \mathbb{F}_q[X]$ with $gcd(q,n)=1$. Let k = order of q mod n. Then the degree of every irreducible factor of $f$ divides k.

It is supposed to be a relatively straightforward consequence of the following fact:

Let $q$ be a prime power and $n$ such that $gcd(n,q)=1$. Let $e$ be the order of the congruence class of $q$ in $(\mathbb{Z}/ n \mathbb{Z})^*$. Then the splitting field K of $X^n-1 \in \mathbb{F}_q[X]$ is isomorphic to $\mathbb{F}_{q^e}$.

I unfortunately fail to see how I can apply the second statement to show the first.

Answer 1

If $f$ is an irreducible factor of $t^n-1$ over $\mathsf k=\mathbb F_q$, then if $K$ is the splitting field of $g(t)=t^n-1$, $f$ splits in $K$, and hence if we take the subfield generated by one of its roots, we get an intermediate field $E\cong \mathsf k[t]/\langle f\rangle$ and so

$$ e=[K:\mathsf k] = [K:E][E:\mathsf k] = [K:E]\deg(f), $$ so that $\deg(f)\mid e$.

It is also not too hard to derive the result from scratch:

If $K$ is the splitting field of $t^n-1$ then $K$ is a finite extension of $\mathsf k$ and so is $\mathbb F_{q^m}$ for some integer $m$. But now $\mathbb F_{q^m}^{\times}$ is a (finite subgroup of) the multiplicative group of a field, it is cyclic. But then since $t^n-1$ splits in $K$ it forms a subgroup of $K^\times$ of order $n$ and hence we must have $n\mid q^m-1$ so that $q^m\cong 1 \mod n$ so that $e\mid m$ where $e$ is the order of $q$ in $(\mathbb Z/n\mathbb Z)^\times$. But the same argument shows that if $e$ is the order of $q$ in $(\mathbb Z/n\mathbb Z)^\times$ then $t^n-1$ must split in $\mathbb F_{q^e}$ so that $K=\mathbb F_{q^e}$.

Proof that any finite subgroup of the multiplicative group of a field is cyclic:

Let $G_n$ be a subgroup of the multiplicative group of a field $K$ with $|G_n|=n$. Any subgroup $H$ of $G_n$ has, by Lagrange's theorem, order a divisor, $d$ say, of $n$, and its elements all have order dividing $d$, so that, since a polynomial has at most its degree many roots over a field, $H = \{x \in K: x^d=1\}=G_d$. It follows that $G_n$ has at most one subgroup $G_d$ of order $d$ for each divisor $d$ of $n$.

But if $\varphi$ denotes the Euler $\varphi$-function, then it is easy to see that if $G_n$ contains an element $\zeta$ of order $d$, then $\zeta\in G_d$ and $G_d$ is therefore cyclic, so that it contains exactly $\varphi(d)$ elements of order $d$. Thus if we set $c_d =1$ if $G_n$ contains an element of order $d$ and $c_d=0$ otherwise, it follows that $$ n=|G_n| = \sum_{d\mid n} c_d\varphi(d)\leq \sum_{d\mid n} \varphi(d) $$ But by, for example, considering the case $K=\mathbb C$ where $c_d =1$ for all $d \mid n$ we see that $\sum_{d\mid n} \varphi(d) =n$, and hence we must have $c_d=1$ for all $d \mid n$. Thus $G_n$ is a cyclic group as required.

Answer 2

I came up with a proof that uses less terminology but that may be different from what you brought up in your post [my eyes still glaze over at the words 'congruence class', I am sorry!]

Proof #1: Let $k$ be the smallest positive integer s.t. $q^k \equiv 1 \pmod n$, or equivalently, $n$ divides $q^k-1$. Let $f$ be an irreducible polynomial that divides $X^n-1$ in $\mathbb{F}_q[X]$ an let $\alpha$ be a root of $f$. Then the equation $\alpha^n = 1$ holds, and as $n$ divides $q^k-1$ it follows that $\alpha^{q^k-1} = 1$ and $\alpha^{q^k} = \alpha$. Now, let $d$ be the smallest positive integer such that the equation $\alpha^{q^d}=\alpha$ holds.

Then check that $\alpha = \alpha^{q^d} = (\alpha^{q^d})^2=(\alpha^{q^d})^3 = \cdots = (\alpha^{q^d})^{\ell} = \cdots$, and thus in particular $\alpha^{q^{\ell d}} = \alpha$ for all positive integers $\ell$. And so writing $k = \ell d +r$ [with $r \in \{0,\ldots, d-1\}$] plugging in the equation $\alpha^{q^k}=\alpha$ yields $$\alpha = \alpha^{q^k}$$ $$= \alpha^{q^{\ell d+ r}} = \alpha^{q^{\ell d}q^{r}}$$ $$= (\alpha^{q^{\ell d}})^{q^{r}} = \alpha^{q^r}.$$ Given that $d$ is the smallest positive integer such that $\alpha^{q^d}=\alpha$ and $\alpha^{q^{r}} = \alpha$ with $r \in \{0,1,\ldots, d-1\}$, it follows that $r$ must be $0$, and thus on the one hand, $d$ must divide $k$.

On the other hand, as $\alpha^{q^d}-\alpha = 0$ and $f$ is a irreducible polynomial in $\mathbb{F}_q[x]$, it follows that $f$ must divide $X^{q^d}-X$ [because they both share a common root namely $\alpha$]. However, from an elementary result in finite fields, we know the following: Let $f$ be an irreducible polynomial in $\mathbb{F}_q[x]$ that divides $x^{q^d}-x$. Then letting $c$ be the degree of $f$, the relation $c$ divides $d$ holds. So in particular, the degree $c$ of the irreducible polynomial $f \in \mathbb{F}_q[x]$ that has $\alpha$ as a root, must divide $d$.

So letting $\alpha$ be a root of $X^n-1$ in $\mathbb{F}_q[X]$, and letting $d$ be the smallest positive integer s.t. $\alpha^{q^d}-\alpha = 0$, we have established the following: On the one hand, $d$ must divide $k$. On the other hand, the degree $c$ of the irreducible polynomial $f \in \mathbb{F}_q[x]$ that has $\alpha$ as a root, divides $d$. So $c$ must divide $k$, and the result follows.

---

ETA: Here is another proof Proof #2, that is along the lines of what was outlined in your OP: Let $\mathbb{F}_{q^k}$ denote the field on $q^k$ elements. Note that $(\mathbb{F}_{q^k})^{\times}$ is a cyclic group on $q^k-1$ elements. As $n|q^k-1$ then it follows that there are exactly $n$ elements $\alpha \in (\mathbb{F}_{q^k})^{\times}$ s.t. $\alpha^n=1$, or equivalently, $X^n-1$ factors completely in $(\mathbb{F}_{q^k})^{\times}$. [Use the fact that a cyclic group $G$ on $N$ elements has exactly $n$ elements $g \in G$ satisfying $g^n = e$ iff $n|N$.] So now then let $f$ be an irreducible polynomial in $\mathbb{F}_q[X]$ that divides $X^n-1$. Then, as $X^n-1$ factors completely in $\mathbb{F}_{q^k}$ there is a element $\alpha \in (\mathbb{F}_{q^k})^{\times}$ s.t. $\alpha$ that satisfies $f(\alpha)=0$. Furthermore, as $f$ is irreducible in $\mathbb{F}_q[X]$ and $\alpha$ is in $\mathbb{F}_{q^k}$, it turns out that $\mathbb{F}_q(\alpha)$ is a subfield of $\mathbb{F}_{q^k}$ of cardinality $q^c$, where $c$ is the degree of $f$. We finish this proof that noting that a field of $q^k$ elements has a subfield of $q^c$ elements iff $c|k$, and so the result follows.