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.

  • 4
    $\begingroup$ The map you have defined is the identity map. $\endgroup$ – Derek Holt Oct 8 at 15:41
  • $\begingroup$ 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. $\endgroup$ – beginarray Oct 8 at 16:03
  • $\begingroup$ 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 $x \mapsto x^{p^{e-1}}$. $\endgroup$ – Derek Holt Oct 8 at 16:39
  • $\begingroup$ I see. Thank you very much. $\endgroup$ – beginarray Oct 8 at 19:05
  • $\begingroup$ @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. $\endgroup$ – beginarray Oct 8 at 19:20

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