Alternative axiomatization I
Task number: 3893
Show that in the definition of the projective plane, the axiom of the existence of four points in the general position can be replaced by one of the following axioms, while the modified set will again define finite projective planes:
Variant
There are two different lines with at least three points, formally: \(\exists P,Q \in \mathcal P: P \not= Q \wedge |P| \geq 3 \wedge |Q| \geq 3\).
Hint
Draw pictures for the axioms.
Solution
If \((X,\mathcal P)\) is a finite projective plane according to the original definition, consider the set \(C=\{a, b, c, d \} \) of four points in the general position.
The line \(P\) connecting \(a\) and \(b\) with the line \(Q\) passing through \(c\) and \(d\) intersects at the third point outside \(C\), so these two lines satisfy the altered axiom.
Conversely, if \((X,\mathcal P)\) satisfies the altered set of axioms, consider the lines \(P\) and \(Q\). In addition to the common intersection, let \(P\) contain two other points \(a \) and \(b\), and \(Q\) again \(c\) and \(d\). Then \(C = \{a, b, c, d \}\) is the desired set of four points in the general position, and thus \( (X, \mathcal P) \) is a projective plane.
If a line \( P’\in \mathcal P \) intersects \( C \) in at least three points, then \( P \) or \( Q \) intersects in at least two points, which is forbidden by the axiom that two points uniquely define a line.
The remaining two axioms (that two points define a line and that two lines intersect at exactly one point) are in the original and altered definition and therefore do not need to be verified in any way.
Variant
\(X\) cannot be covered by two lines.
Solution
If \( (X, \mathcal P) \) is a finite projective plane according to the original definition, consider the set \( C = \{a, b, c, d \} \) four points in the general position.
If, for a contradiction \( X \) was covered by two lines \(P\) and \(Q\), they both have to cover exactly two points from \( C \). Without loss of generality, \(P\) will cover \(a\) and \(b\) while \( Q \) the remaining two points.
The line \(P’\) connecting \(a\) and \(c\) with the line \( Q’\) passing through \(b\) and \(d\) intersects at a point outside \( P \cup Q \), so these two lines do not cover \(X\), which is a contradiction. The amended axiom therefore follows from the original ones.
That the original set of axioms follows from the original can be proved, for example, by the inverse statement – i.e. from the non-existence of four points in the general position we can derive the coverage by two lines.
If \( (X, \mathcal P) \) does not have three points in the general position, all points lie on a single line. Let us therefore consider three different points \(a\), \(b\) and \(c\) in the general position, where \(a\) and \( b \) specify the line \(P\), and similarly the remaining two pairs \(a, c \) and \(b, c\) determine the lines \(Q\) and \(R\).
In each quadruple of type \(a, b, c, x \) it is possible to find a line with three points, i.e. the remaining points \(X \setminus \{a, b, c \} \) can be divided into three groups according to the lines \(P, Q\) and \(R\). If some two of these lines both have at least three points, e.g. \(P\) would have extra \(x\) and \(Q\) would have extra \(y\), we would get four points \(b, c , x, y\) in the general position. Thus, one of the three lines contains all the remaining points of \( X \setminus \{a, b, c \}\), i.e. it covers \(|X|-1\) points. Any line passing through it is sufficient to cover the only remaining point.
Alternatively, this implication can be proved directly. The set \(X\) has three points \( a, b, c \) in the general position, otherwise we would cover it with one line. Let \(P\) pass through \(a, b\), the line \(Q\) through \(a, c \) and the line \(R\) through \(b, c\).
Because \(P\) and \(Q\) do not cover \(X\), there is a point \(d\) outside \(P\cup Q\). If \(d\notin R \), we are done and \(C = \{a,b,c,d \}\) are the required four points. Otherwise, \(P\cup R \) also does not cover \(X\), so there is an uncovered point \(e\), and to avoid an analogy with the previous case, assume that \(e \in Q\). Then of course \(C = \{b, c, d, e \}\) are the required four points.
