There is an integer $l$ such that $f_l = x^{2^l}+1$ is reducible in $\mathbb{F}_p[x]$ for any prime $p$

Question

So, the proposition is in the title, and the hint given is: show that $(U(\mathbb{Z}/2^n\mathbb{Z}),\cdot)$ is not a cyclic group if $n\ge 3$.

I proved the hint, noting the subgroups generated by $-1$ and $2^{n-1}+1$ are distinct and have both order $2$, which can't happen in a cyclic group.

Then I have the following solution that my past self wrote:

Suppose for contradiction that for any $l$ there is some prime $p$ such that $f_l$ is irreducible over $\mathbb{F}_p$, then $\large\frac{\mathbb{F}_p[x]}{(f_l)}$ contains all the roots, i.e. $\alpha,\alpha^p,\dots,\alpha^{\large p^{2^l-1}}$. Then $\lvert U(\mathbb{Z}/2^{l+1}\mathbb{Z})\rvert = 2^l$ and it is forced to be cyclic, which is a contradiction by the hint.

Now, I agree with the fact about the roots, because of the Frobenius automorphism, but I just really don't see why $U(\mathbb{Z}/2^{l+1}\mathbb{Z})$ would be forced to be cyclic...

Thanks for any help :)

Answer 1

The group of invertible elements of a finite field is always a cyclic group.

Finite subgroups of the multiplicative group of a field are cyclic