A. Beutelspacher & U. Rosenbaum, Projective Geometry

Chapter 2: pg. 89 problem 5

Let P1 = <u>, P2 = <v> and P3 = <w>. Since these points are not collinear, they form a basis of the projective plane. The fourth point, P4, can be expressed as a linear combination of these three points, and, by using the replacement trick, we can write P4 = <u+v+w>. Now we have,
Q1 := P1P4P2P3
= <u, u+v+w> <v,w>
= <v+w>
Similarly, we obtain Q2 = <u+w> and Q3 = <u+v>. Now the line Q1Q2 = <v + w, u + w>.

If F has characteristic 2, then since (v+w) + (u+w) = u + v + 2w = u + v we have Q3 collinear with Q1 and Q2 [Note that the computation only uses commutativity of addition and the fact that 2 = 0 in a division ring of characteristic 2]. On the other hand, suppose that Q3 is on the line Q1Q2. Then there exist a,b in F so that u + v = a(v+w) + b(u+w) which can be rewritten as 0 = (b-1)u + (a-1)v + (a+b)w. Since u,v and w are a basis, all the coefficients must be 0. Thus, a = 1, b = 1 and a + b = 1 + 1 = 0 and the division ring F has characteristic 2.