Question

. Let Π be a finite incidence geometry. Prove that, if every line in Π has exactly n points and every point in Π lies on exactly n + 1 lines, then Π is an affine plane. Come up with a similar criterion for finite geometries satisfying (EP) (those geometries are called projective planes).

Answer 1

Let Π be a finite incidence geometry. Prove that, if every line in Π has exactly n points and every point in Π lies on exactly n + 1 lines, then Π is an affine plane. Come up with a similar criterion for finite geometries satisfying (EP) (those geometries are called projective planes).

Proof

From the finite Projective Plane Geometry, we have 4 Rules that are below

Rule1. For any two distinct points, there is exactly one line incident with both points.

Rule2. For any two distinct lines, there is at least one point incident with both lines.

Rule3. Every line has at least three points incident with it.

Rule4. There exist at least four distinct points of which no three are collinear.

By the definition of a projective plane of order n, there exists a line l with exactly n + 1 points incident to it, call them P1 , P2 ,…..,Pn+1.

By Rule4, there is a point Q not incident with l. Thus by Rule1, there exist lines QP1 , QP2 ,…….,QPn+1. We need to show the lines are distinct and that there are no other lines through Q.

Suppose QPi = QPj for some i ≠ j.

Then by Rule1, Q, Pi , and Pj would be on line l = Pi Pj , but this contradicts that Q is not on line l. Hence, the n + 1 lines QP1 , QP2 ,…… ,QPn+1 are distinct.

Now let m be a line incident to Q. By Rule2, lines l and m are incident with a point R. Since P1 , P2 , ….. ,Pn+1 are the only points incident to l, R = Pi for some i ϵ {1, …. ,n + 1}. Hence, m = QR = QPi which is one of the n + 1 lines through Q. Therefore, the point Q is incident with exactly n + 1 lines.

Henced proved that, if every line in Π has exactly n points and every point in Π lies on exactly n + 1 lines.