Show that $\cdots\circ$ Fr$^{3!}\circ$Fr$^{2!}\circ$ Fr$^{1!} \in\operatorname{Gal}(\overline{\mathbb F_p}/\mathbb F_p)\setminus\langle$ Fr $\rangle$

Question

Let $\varphi=\cdots\circ \operatorname{Fr}^{3!}\circ \operatorname{Fr}^{2!}\circ \operatorname{Fr}^{1!}$, where $\operatorname{Fr}$ is the Frobenius endomorphism. Show that $\varphi \in \operatorname{Gal}(\overline{\mathbb F_p}/\mathbb F_p)\setminus\langle\operatorname{Fr}\rangle$.

First of all, I had to show that $\varphi$ is well-defined. For this argument I proved that for every $\mathbb F_p\subset\mathbb K \subset \overline{\mathbb F_p}$, if $[\mathbb K:\mathbb F_p]< \infty$, then there exists $N$ such that for every $n\geq N$, $\operatorname{Fr}^{n!}|_\mathbb K=\operatorname{Id}$. How do I continue from here?

Answer 1

You should make the following observations, and/or justify the claims:

• $\overline{\Bbb{F}_p}$ is the union of the fields $\Bbb{K}$ that you have considered. By your result this implies that the infinite composition is actually well defined. Furthermore, it also follows that it is an automorphism of $\overline{\Bbb{F}_p}$.
• But, the composition $\varphi$ is not in the group generated by $\mathbf{Fr}$. For otherwise we would have $\varphi=\mathbf{Fr}^n$ for some integer $n$. But, if $k$ is a positive integer such that $n\mid k!$ then $n\not\equiv n+k!\pmod{(k+1)!}$. This implies that the restriction of $\varphi$ to either $\Bbb{F}_{p^{k!}}$ or $\Bbb{F}_{p^{(k+1)!}}$ is not equal to $\mathbf{Fr}^n$.