Binomial of Mersenne prime power.

Question

Let $f(x)$ be irreducible over $\mathbb{Z}_2$ of degree $p$, where $p$ is prime. Let $2^p-1$ be a Mersenne prime. I have to show \begin{equation*} f(x) \mid x^{2^p-1}-1. \end{equation*}

I am really not sure where to go with this. So far, since $2^p-1$ is prime, I have that $x^{2^p-1}-1$, in $\mathbb{Z}_2$, splits into: $$x^{2^p-1}-1=(x-1)(x^{2^p-2}+x^{2^p-3}+\cdots+1)=(x+1)(x^{2^p-2}+x^{2^p-3}+\cdots+1)$$ i.e. it splits into $x-1$ and the respective cyclotomic polynomial.

But the cyclotomic polynomials are irreducible, so I don't know how I can show that $f(x)$ can divide this.

Answer 1

Let $f\in\mathbb F_2[X]$ be an irreducible polynomial of degree $n$. Let $q=2^n$. Then $f$ has a root $\alpha$ in $\mathbb F_2[X]/(f)\cong \mathbb F_q$. As $\mathbb F_q^\times$ is cyclic of order $q-1$, we have $\alpha^{q-1}=1$, i.e., $\alpha$ is also a root of $X^{q-1}-1$ and so $\gcd(f,X^{q-1}-1)\ne 1$. As $f$ is irreducible, we conclude $f\mid X^{q-1}-1$.

Your problem is a special case of this (namely when $n=p$ and $q-1=2^n-1$ are prime).