Exercise 5.8.3 (Number of points of $\mathbb{P}_2(\mathbb{Z}_p)$)

Proof. By definition, ${\mathbb{P}}_2({\mathbb{Z}}_p)$ is the quotient $({\mathbb{Z}}_p^3\setminus \{(0,0,0)\})∕ℛ$, where $ℛ$ is the equivalence relation defined by

Then the class of every triple $(x,y,z)\in ({\mathbb{Z}}_p^3\setminus \{(0,0,0)\})$ is the projective point

Therefore each class has $p-1$ elements, and the number of elements of ${\mathbb{Z}}_p^3\setminus \{(0,0,0)\})$ is

(Alternatively, if we choose some projective line in the projective plane ${\mathbb{P}}_2({\mathbb{Z}}_p)$, the complementary of this line has a structure of affine plane, with $p^2$ elements, and the line has $p+1$ element, therefore the projective plane has $p^2+p+1$ elements.)