Projective Plane for F3

Question

I have to calculate the points of a projective Plane on $\mathbb{Z}_3^2$. I thing I understood the way how to do this for the Fano-Plane but I am not sure how to do this here because I have more than one nonzero scalar?

Answer 1

As in all projective planes over fields, the points are $[x_1,x_2,x_3]$, where $(x_1,x_2,x_3)\neq (0,0,0)$ and $[x_1,x_2,x_3]=[kx_1,kx_2,kx_3]$ for all $k\neq 0$.

In this case the points are $A=[0,0,1]$, $B=[0,1,0]$, $C=[1,0,0]$, $D=[0,1,1]$, $E=[1,0,1]$, $F=[1,1,0]$, $G=[0,1,2]$, $H=[1,0,2]$, $I=[1,2,0]$, $J=[1,1,1]$, $K=[1,1,2]$, $L=[1,2,1]$, $M=[2,1,1]$.

The interesting part is giving the lines. You have to select every two points, for example $A$ and $B$, and the other two points on their line are given by every possible linear combinations of $A$ and $B$. In this case they are $A+B$ and $A+2B$.

So the lines are $(A,B,D,G)$ $(A,C,E,H)$, $(A,F,J,K)$, $(A,I,L,M)$, $(B,C,F,I)$, $(B,E,J,L)$, $(B,H,K,M)$, $(C,D,J,M)$, $(C,G,K,L)$, $(D,E,K,I)$, $(D,F,L,H)$, $(E,F,M,G)$, $(G,H,J,I)$.

Answer 2

In general, you are just counting one dimensional subspaces of $F_q^3$.

There are $q^3-1$ nonzero vectors. These partition into groups of $q-1$ scalar multiples of each other. So you will always have $\frac{q^3-1}{q-1}$ points in the projective plane over $F_q$.

Answer 3

Assuming that by calculating points of projective plane on $\mathbb Z^2_3$ you mean finite projective plane of the order $9 = 3^2$, then as the first you should generate Galois field $GF(3^2)$. You can use Wikipedia article $GF(p^2)$ for an odd prime. After having multiplicative group of the Galois Field $GF(3^2)$, use it to rotate permutation sub-matrices $C_{ij}$ in incidence matrix in canonical form (See Paige L.J., Wexler Ch., A Canonical Form for Incidence Matrices of Projective Planes...., In Portugalie Mathematica, vol 12, fasc 3, 1953). This is incidence matrix of points and lines and it defines the projective plane.