## Projective Geometry Homework Answer

A. Beutelspacher & U. Rosenbaum, *Projective Geometry*
**Chapter 2: pg. 89 problem 5**

Let P_{1} = <u>, P_{2} = <v> and P_{3} = <w>. Since these points are not collinear, they form a basis of the projective plane. The fourth point, P_{4}, can be expressed as a linear combination of these three points, and, by using the replacement trick, we can write P_{4} = <u+v+w>. Now we have,
Q_{1} := P_{1}P_{4}P_{2}P_{3}

= <u, u+v+w> <v,w>

= <v+w>
Similarly, we obtain Q_{2} = <u+w> and Q_{3} = <u+v>. Now the line Q_{1}Q_{2} = <v + w, u + w>.
If F has characteristic 2, then since (v+w) + (u+w) = u + v + 2w = u + v we have Q_{3} collinear with Q_{1} and Q_{2} [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 Q_{3} is on the line Q_{1}Q_{2}. 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.