How to prove this well-known fact?

Question

Prove that

any polynomial of degree $n$ on $\operatorname{GF}(p)$ divides $X^{p^n-1}-1$?

for the case of $\operatorname{GF}(2)$ any polynomial of degree $n$ divides $X^{2^n-1}+1$

Answer 1

Let $f\in \mathbb F_p[X]$ irreducible of degree $n$ and $\alpha$ be a root of $f$. Then $[\mathbb F_p(\alpha) : \mathbb F_p] = \deg(f) = n$, so $\alpha\in\mathbb F_p(\alpha) = \mathbb F_{p^n}$. Since the unit group of finite fields is cyclic, $a^{p^n - 1} = 1$, so $a$ is a root of $X^{p^n - 1} - 1$.

Answer 2

Hints:

$$\begin{align*}(1)&\;\;[\Bbb F_{p^n}:\Bbb F_p]=n\;\;\text{and $\;\Bbb F_{p^n}\;$ is the splitting field of}\; \;x^{p^n}-x\in\Bbb F_p[x]\\{}\\ (2)&\;\;\text{For any irreducible}\;\;f(x)\in\Bbb F_p[x]\;,\;\deg f=n\;,\;\;\text{we have that}\;\;f(\alpha)=0\implies \alpha\in\Bbb F_{p^n}\end{align*}$$

Answer 3

Let $f$ be an irreducible polynomial of degre $n$ on $F=\mathbb G\mathbb F(p)$. Then $f$ has a root in the extension field $F[X]/(f)\cong \mathbb G\mathbb F(p^n)$. Let $\alpha$ be a generator of $\mathbb G\mathbb F(p^n)^\times$. Then $\alpha$ has order $p^n-1$, so is a root of $X^{p^n-1}-1$, i.e. $f$ and $X^{p^n-1}-1$ have a common factor. As $f$ is irreducible, $f\mid X^{p^n-1}-1$.

If $f$ is not irreducible, the claim need not be true. For example, take $p=3,n=2$: $(X-1)^2$ does not divide $X^8-1$.