What is an easy way to count points and lines in a finite Projective Plane over finite field $\mathbb F_q$, accessible for student with minor knowledge of linear algebra (spaces sub-spaces, e.c.t).
- Points in Projective Plane over finite field $\mathbb F_q$ are $1$-dimensional sub-spaces of $\mathbb F_q^3$ .
- Lines are $2$-dimensional sub-spaces of $\mathbb F_q^3$ .
I want to show that there are $\frac{q^3-1}{q-1}=q^2+q+1$ points and the same number of lines. As well as there are $q+1$ points on each line and there are $q+1$ lines going through each point.
How can one show this in easiest way?
When you have evaluated one of the two numbers you conclude for the duality that occurs between points and hyperplanes (in this case lines) in the projective space. Now your task for example is to count one dimensional nonzero subspaces so you find a nonzero generator , so you have q^3-1 choices. Now there are q-1 (a am excluding the zero of the field) multiples of this vector that generate the same space so we divide by this number. We have concluded. If you want to avoid the duality argument you can approach in this way the problem ( using the duality argument without knowing it ) : a two dimensional vector space in F_q is the datum of a cartesian equation. The row vector that describe the equation modulo multiplication by an invertible element of F_q describes uniquely the space, you can conclude as in the previous point. For the second part you take a line an obtain with analogous steps q^2-1/q-1= q+1 (one dimensional spaces in a two dimensional subspace), then we conclude by duality.. Do you want still avoid this tool? :)