# What is the inverse of this Frobenius endomorphism?

Let $$p$$ be a prime. The Frobenius map: $$x \mapsto x^3$$ is bijective from $$\mathbb{F}_p \longrightarrow \mathbb{F}_p$$. I'm trying to write its inverse map: $$x \mapsto x^{?}$$. I suppose the ($$?$$) must be a power of $$p$$. Is it something like $$x \mapsto x^{{p}^{p-1}}$$ ?

Any help would be greatly appreciated.

• The map you have defined is the identity map. – Derek Holt Oct 8 at 15:41
• Yes, I know that every $a \in \mathbb{F}_p$ satisfies $a^p=a$. I just thought it was possible to express the inverse map with a power of $p$. Forget about it, my question makes no sense...ps: thank you for answering. – beginarray Oct 8 at 16:03
• The Frobenius map $x \mapsto x^p$ is more interesting when applied to the finite field ${\mathbb F}_{p^e}$ for $e \ge 1$. The automorphism has order $e$, and the inverse map is $x \mapsto x^{p^{e-1}}$. – Derek Holt Oct 8 at 16:39
• I see. Thank you very much. – beginarray Oct 8 at 19:05
• @Derek Holt; just a brief explanation. I was trying to prove that, for $p \equiv 2 \pmod 3$, the number of projective solutions modulo $p$ of the Fermat equation $x^3+y^3+z^3=0$ is $p+1$. We can use the fact that $x \mapsto x^3$ is a bijection from $\mathbb{F}_p$ to itself. – beginarray Oct 8 at 19:20