Let $\mathbb{F}_q$ be a finite field with $q$ elements and let $V \subset \mathbb{P}^n$ be a variety defined over $\mathbb{F}_q$.
The $q^{th}$ power map $\varphi: = [ X_0^q , . . . , X_n^q ]$ is a morphism $\varphi: V\to V$. It is called the Frobenius morphism.
I would like to prove that $V(\mathbb{F}_q)=\{ P \in V :\varphi(P)=P\}$.
I know absolute projective plane over $\mathbb{F}_q$ is characterized as fixed points of action by absolute Galois group over $\mathbb{F}_q$ (this proof needs Hilbert theorem 90).
Any other proof (not using above fact) is also appreciated.
Thank you in advance.