Alternative axiomatization II

Task number: 3896

Let \( (X, \mathcal P) \) be a set system and for \( n \in \mathbb N \), \( n \geq 2 \), holds that:

  1. \(|X|=|\mathcal P|=n^2+n+1\),
  2. \(\forall P \in \mathcal P: |P|=n+1\),
  3. \(\forall P,Q\in\mathcal P: P \ne Q \Rightarrow |P \cap Q| \leq 1\).

Is then \( (X, \mathcal P) \) a finite projective plane of order \( n \)?

  • Hint

    Compare the number of all pairs of points with the number of pairs that define a line to prove that every two points uniquely identify a line.

  • Solution

    It follows from axiom (2) and the condition \( n \ge 2 \) that each line has at least three points. We choose the lines \(P,Q \) arbitrarily, then due to (3) \(P\) contains points \(a, b\) not lying on \(Q\) and the line \(Q\) has points \(c, d\) not lying on \(P\). The four points \(C = \{a, b, c, d \} \) are in the general position. If some line \(R\) satisfies \(|R \cap C| \ge 3 \), then either \(| R \cap P | = 2 \) or \( |R \cap Q | = 2 \), which would be in conflict with the axiom (3).

    On \(X\) we can create \( \binom{|X|}{2} = \binom{n^2+n+1}{2}\) pairs of points (1). Each line \( P \) is determined by \(\binom{|P|}{2}=\binom{n+1}{2}\) pairs of points (2).

    Due to (3) one pair can determine at most one line. To get \(n^2+n+1\) lines (1) we need at least \((n^2+n+1)\binom{n+1}{2}\) point pairs.

    However, since \((n^2+n+1)\binom{n+1}{2}=\binom{n^2+n+1}{2}\), there is no pair of points left that would not determine a line. Thus, we proved the next axiom of the projective plane, namely that each pair of points uniquely determines a line.

    Before proving the last axiom, we estimate the number of lines passing through any point \(a\). In addition to \(a\), on each line passing through \(a\) there are \( n \) points (2) from the remaining \(n^2+n\) (1), and these points must be different (3). Thus, the point \( and \) may lie on at most \(\frac{n^2+n}{n}=n+1\) lines.

    However, since \(n^2+n+1>n+1\), there exists a line \(P: a\notin P \) for the point \(a\). Each point of \(P\) determines, due to the just proved axiom, one line passing through \(a\), i.e. \(a\) belongs to at least \(n+1\) lines.

    (The existence of \(P\) alternatively: choose two different lines \(Q, R\) passing through \(a\). If \(a\) belongs only to one line, choose \(P\) from the remaining \(n^2+n\) lines arbitrarily. Else, on the lines \( Q, R \) select the points \(b, c\). These exist because \( n \ge 2 \) and (2). In the previous paragraph we have showed that the line \( P \) determined by \(b, c \) exists. It does not contan \(a\) due to (3).)

    Now let us similarly prove the axiom of the intersection of lines. According to (3) it suffices to show that every two lines intersect. There are \(\binom{n^2+n+1}{2} \) pairs of lines. Each single point corresponds to the intersection of at least \(\binom{n+1}{2}\) pairs of lines and since the total number of points is \(n^2+n+1 \), we find that no two lines remain that would not intersect.

Difficulty level: Hard task
Proving or derivation task
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