Finite affine and projective incidence

Try

The four binary coordinate pairs (0,0), (1,0), (0,1), (1,1) form an affine plane over the field F₂. Arithmetic is modulo two: 1+1=0. Its lines are not the entire strokes of a Euclidean drawing; they are specified sets of binary points. Each line contains exactly two points.

This example is useful because every pair and intersection can be checked exhaustively. It teaches incidence without pretending to reproduce length, angle or global order. A valid finite incidence plane is not automatically a model of every geometric axiom discussed later.

Predict

Predict how many lines are needed if each of the six unordered pairs of four points determines a different line. The answer is six. Predict whether the two diagonals of a drawn square must intersect in this model. Their Euclidean crossing is not one of the four binary points, so the model may call them parallel.

The picture’s extra ink is not part of the point set. This is an early warning against proving incidence by looking at a screen.

Experiment

List all lines and check each pair occurs once.

Equation in F₂	Points
x=0	(0,0), (0,1)
x=1	(1,0), (1,1)
y=0	(0,0), (1,0)
y=1	(0,1), (1,1)
x+y=0	(0,0), (1,1)
x+y=1	(0,1), (1,0)

Choose one point and one line not containing it. Of the three lines through the point, exactly one is disjoint from the given line. Verify this directly, then repeat for a different point. The small model satisfies affine parallel uniqueness, even though it cannot satisfy the coming order extension axioms.

Understand

For any field F, define projective points as one-dimensional subspaces of F³ and projective lines as two-dimensional subspaces. Incidence means containment. Distinct projective points have independent representatives p and q. Their span is two-dimensional and is the unique plane containing both: any other such plane contains the span and has the same dimension.

Two distinct two-dimensional subspaces U,V meet in dimension one. Indeed dim(U∩V)=dim U+dim V−dim(U+V)≥1. An intersection of dimension two would imply U=V. Thus every two distinct projective lines meet in exactly one projective point.

Over a field with q elements there are q³−1 nonzero triples. Each one-dimensional subspace contains q−1 nonzero representatives, so the number of projective points is q²+q+1. Each line similarly contains (q²−1)/(q−1)=q+1 points. For q=2 this is the seven-point projective plane.

Apply

Explain why counting vectors without dividing by scalar multiples gives the wrong projective count. The coordinates (1,1,0) and (2,2,0) over F₃ describe the same one-dimensional subspace. They are two representatives of one geometric point.

For optional depth, count projective lines by nonzero linear functionals up to nonzero scaling. A functional’s kernel is a two-dimensional subspace, and two nonzero functionals with the same kernel differ by scale. This produces the same q²+q+1 count and anticipates point–line duality.

Do not add an arbitrary “between” relation and assume a finite plane becomes an ordered plane. Repeated extension along a strictly ordered line will force infinitely many points.

Continue

Write a model verification that states the point set, line set and incidence relation before proving any axiom. Include the dimension argument, not just the numerical count. Continue to order and Pasch to identify the additional assumptions responsible for segments, sides and crossings.