Affine Plane

Definition

An affine plane is an incidence structure \(\sigma = (\mathscr{P}, \mathscr{L}, \mathrm{I})\) satisfying:

• A1: Two points determine exactly one line.
• A2: For line \(L\) and \(p \notin L\), exactly one line through \(p\) parallel to \(L\).
• A3: Every line has at least two points.
• A4: For every line, some point lies off it.
• A5: There exists at least one line.

Examples

The complete four-point with 4 points and 6 lines is the smallest affine plane. It has three pairs of parallel lines.

Young's configuration: 9 points, 12 lines, 4 sets of 3 parallel lines.

Theorems

An affine plane contains three non-collinear points.

Let \(\alpha = (\mathscr{P},\mathscr{L},\mathrm{I})\).

• By A5, there is a line \(L\).
• By A3, there are points \(p, q \in L\).
• By A4, there is a point \(r \notin L\).

Then, \(p, q, r\) are three non-collinear points. \(\blacksquare\)

An affine plane contains a four-point.

Let \(\alpha = (\mathscr{P},\mathscr{L},\mathrm{I})\) be an affine plane.

• Let \(L\) be a line (A5).
• Let \(a, b \in L\) (A3).
• Let \(c \notin L\) (A4).
• Then \(\exists!\) line \(M \parallel L\) through \(c\) (A2).
• \(M\) has another point \(d\) (A3).

So affine plane contains a four-point such as \(a,b,c,d\) . \(\blacksquare\)

Let \( \alpha=(\mathscr{P,L,I}) \) be an incidence structure satisfying

1. Two points determine a line
2. If \(L\) is a line and \(p\) be a point not on \(L\), there is exactly one line on \(p\) and parallel to \(L\)
3. There is a set of three non-collinear points

Then prove that \(\alpha \) is an affine plane.

1. (1) ⇒ This implies A1.
2. (2) ⇒ This implies A2.
3. (3) ⇒ This implies A5 (by 3, any two of three points define a line, by 1).
4. (3) ⇒ This implies A4 (third point lies off any line through the other two points).
5. A3 (≥2 points on each line):
   By (3), assume that \(p,q,r\) are three non-collinear points.
   Now suppose, \(L\) is arbitrary line, then by (A4), there is a point (at least one from \(p,q,r\) ) not on \(L\)
   say \(p \notin L\).
   Case 1
   If \(pq\) and \(pr\) both meet at \(L\), then \(L\) has two points.
   
   Case 2
   If one of the lines, say \(pq\) does not meet \(L\),then by (2)-\(pr\) must meet \(L\).
   Also, by (2), there exists a line \(M \in q\) such that \(M || pr\).
   Since \(pq\) does not meet \(L\), the line \(M\) must meet \(L\), otherwise it contradicts (2).
   
   Hence \(L\) has two points.
   This satisfies (case 1 and case 2) axiom A3 of affine plane ⇒ (A3)

Hence all axioms holds (A1 to A5)→ It shows that \( \alpha=(\mathscr{P,L,I}) \)ia an affine plane. \(\blacksquare\)

Note: The three conditions are equivalent to the five axioms. This simplifies verification.

Exercise

Solve the following problems.

1. Prove that if \(L\) and \(M\) are two lines in a plane, there is at most one point on both.
2. Concerning a complete four-point \( \alpha \), explain that:
   1. \( \alpha \) has exactly \(\cdots\) points.
   2. Each point on \( \alpha \) is on \(\cdots\) lines.
   3. Total number of lines in \( \alpha \) is \(\cdots\).
   4. There exists \(\cdots\) pencils of parallels in \( \alpha \).
3. Concerning a Young's configuration \( \alpha \), explain that:
   1. \( \alpha \) has exactly \(\cdots\) points.
   2. Each point on \( \alpha \) is on \(\cdots\) lines.
   3. Total number of lines in \( \alpha \) is \(\cdots\).
   4. There exists \(\cdots\) pencils of parallels in \( \alpha \).