When is $x^{2^n} + 1$ Reducible in $\mathbf{F}_p$ For All Primes $p$

Question

Define $f_n = x^{2^n} + 1$.

Then we want to show that there is an integer $n$ such that $f_n$ is reducible in $\mathbb{F}_p[x]$ for all primes, $p$.

However, I want to do this using the hint

The group of units modulo $2^r$, $(\mathbb{Z}/2^r\mathbb{Z})^*$, is not cyclic for $r \ge 3 $.

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Indeed, there are ways to show this without the hint - one such method forms the basis of this answer.

However, I just can't seem to figure out how to use the hint! Hopefully using the hint will also shine some light on the question of for which $n$ does the above hold?

Answer 1

It's always reducible modulo $2$.

Modulo an odd prime $p$, its zeroes are the primitive $2^{n+1}$-th roots of unity. Let $\alpha$ be any of them. Then $\Bbb F_p(\alpha)=\Bbb F_{p^k}$ where $k$ is the least positive integer with $\alpha^{p^k}=\alpha$. That's equivalent to $p^k\equiv1\pmod{2^{m+1}}$. If $k<2^m$, then the degree of $\Bbb F_p(\alpha)$ over $\Bbb F_p$ is less than $2^m$, so that $x^{2^m}+1$ cannot be irreducible over $\Bbb F_p$.

Answer 2

Assume $f_n$ is irreducible in some $K=\mathbb{F}_p$, where $p$ is odd. Then let $\omega \in L=\mathbb{F}_q$ a root of $f_n$ in some field extension.

Then, since $L$ is a Galois extension of $K$, the Galois group of $L/K$ (a generator of which is the Frobenius automorphism) has to act transitively on the roots of $f_n$. In other words, every root of $f_n$ is $\omega^{p^k}$ for some $k$. Now, $f_n$ has $2^n$ distinct roots, which are the odd powers of $\omega$. Thus, every odd integer mist be congruent mod $2^{n+1}$ to some $p^k$. Thus $p$ is a generator of your non-cyclic group.

When $p=2$, then $f_n=(X+1)^{2^n}$.