Here $\mathbb{F}\_{q}$ means a finite field with $q = p^m$ elements where $p$ is the characteristic of the field in question and $\overline{\mathbb{F}\_{q}}$ means its algebraic closure.


I am studying an article about the existence of a normal basis with a primitive element for finite extensions of $\mathbb{F}\_{q}$. And there is this new definition for me, that I haven't found anywhere:


Let $\alpha \in \overline{\mathbb{F}\_{q}}$, and let $\sigma$ the Frobenius Authomorphism ($\sigma(\alpha) = \alpha^q$). One can turn $\overline{\mathbb{F}\_{q}}$ into a $\mathbb{F}\_{q}[X]$-module as follows:


If $f \in \mathbb{F}\_{q}[X]$, write $f = \sum\_{i=0}^{k} a\_{i}X^{i}$ and then define:


$f \circ \alpha := \sum\_{i=0}^{k} a\_{i}\sigma^{i}(\alpha)$


Which is essentially $f(\sigma)(\alpha)$. 


We then have that


$\alpha \in \mathbb{F}\_{q^n} \Leftrightarrow \alpha^{q^n} = \alpha \Leftrightarrow \sigma^{n}(\alpha) = \alpha \Leftrightarrow (X^n-1)\circ\alpha =0$,


and then the Annihilator of $\alpha$ over $\mathbb{F}\_{q}[X]$ is non-trivial for every $\alpha$. Then one can define the, let's call here additive order ($Ord(\alpha)$) of $\alpha$ by the monic polynomial generating the Annihilator of $\alpha$. It's easy to see that $\alpha \in \mathbb{F}\_{q^n} \Leftrightarrow Ord(\alpha)|X^n-1$. 


The point is, in the article, it is said that:


If $\alpha \in \mathbb{F}\_{q^n}$ has $Ord(\alpha) = g$, then $\alpha = h\circ\beta$, for some $\beta \in \mathbb{F}\_{q^n}$ and $h = \dfrac{X^n-1}{g}$.


The point is, I just cannot prove this (it seems trivial since every source I look into just states this and does not prove).


Well, if the characteristic of $\mathbb{F}\_{q^n}$ does not divide $n$, then it is clear for me, since $X^n-1$ is separable over $\mathbb{F}\_{q}[X]$ and then $(g,h) = 1$, therefore there exists $s,t \in \mathbb{F}\_{q}[X]$ such that


$sg + th = 1$, which implies $(sg + th)\circ \alpha = 1\circ \alpha$ and $ (sg)\circ \alpha + (th)\circ\alpha = (th)\circ\alpha = h\circ(t\circ\alpha) = \alpha$, defining $\beta = t \circ \alpha$ we get the result. $(sg)\circ \alpha = 0$ because $g$ is the order of $\alpha$ therefore $g\circ\alpha = 0$, and I used the fact that we have a module structure over $\mathbb{F}\_{q}[X]$.


Now, if $n$ divide the characteristic of $\mathbb{F}\_{q^n}$, I cannot prove by this way. Well, I think there is a simpler argument such that one doesn't need to split the proof between two cases. Can anyone help me with the second case or can anyone give me a hint or a new argument that helps me to solve this problem?



Viewing $\overline{\mathbb{F}_{q}}$ as a $\mathbb{F}_{q}[X]-$module
By the normal basis theorem, there exists $a \in \mathbb F\_{q^n}$ such that $\{\sigma(a)\mid \sigma \in \operatorname{Gal}(\mathbb F\_{q^n}/\mathbb F\_q)\}$ spans $\mathbb F\_{q^n}$ as $\mathbb F\_q$-vector space. Since $\operatorname{Gal}(\mathbb F\_{q^n}/\mathbb F\_q)$ is generated by Frobenius, this means that $\mathbb F\_{q^n}$ is spanned by the Frobenius powers of $a$, i.e. is generated as $\mathbb F\_q[X]$-module by $a$.


Thus $\mathbb F\_{q^n} \cong \mathbb F\_q[X]/I$, where $I = (\operatorname{Ann}(a))$. Since $\mathbb F\_{q^n}$ is killed by $X^n - 1$, we get $(X^n - 1) \subseteq I$. But
$$|\mathbb F\_q[X]/(X^n-1)| = q^n = |\mathbb F\_{q^n}| = |\mathbb F\_q[X]/I|,$$
so the inclusion $(X^n - 1) \subseteq I$ has to be an equality. That is,
$$\mathbb F\_{q^n} \cong \mathbb F\_q[X]/(X^n-1),$$
as $\mathbb F\_q[X]$-modules. The result now follows from the following lemma.


**Lemma.** Let $R$ be a PID, let $M = R/(f)$ for some $f \in R\setminus\{0\}$, and let $m \in M$. If $\operatorname{Ann}(m) = (g)$, then there exists $h \in R$ such that $gh = f$, and there exists $n \in M$ such that $hn = m$.


*Proof.* The first statement follows from the fact that $(f) \subseteq (g)$ since $\operatorname{Ann}(M) \subseteq \operatorname{Ann}(m)$. For the second statement, note that the sequence
$$0 \to (h)/(f) \to R/(f) \to R/(h) \to 0$$
is exact. But multiplication by $g$ induces an isomorphism $R/(h) \to (g)/(f)$. Hence, the sequence above turns into
$$0 \to (h)/(f) \to R/(f) \stackrel{\cdot g}\longrightarrow (g)/(f) \to 0.$$
This proves the second statement: if $gm = 0$, then $m$ is divisible by $h$. $\square$


