The projective line

The projective line will be a very important object for us in this project.

Let K be a field. A point of the projective line P^1 over K is by definition a 1-dimensional subspace of K^2. As in the case of the projective plane we use the notation (a : b) to denote the point corresponding to the line spanned by the column vector (a, b). Thus, whenever we write this we assume that a and b are not both zero. Moreover, the point (a : b) is the same as the point (a' : b') if and only if there exists a nonzero scalar λ ∈ K such that a = λ a' and b = λ b'.

Exercise 7: If K = Z/pZ how many points does P^1 have?

Suppose that we have an invertible 2×2 matrix A with coefficients in K. Then A induces a linear map K^2 —> K^2 given by x |—> Ax. In particular, 1-dimensional subspaces map to 1-dimensional subspaces, hence the points of P^1 get shuffled. In other words, A defines a morphism P^1 —> P^1. If the entries of A are (a_{ij}) with i, j ∈ {0, 1} then we can give a formula for this map, namely

(x_0 : x_1) |—→ (a_{00}x_0 + a_{01}x_1 : a_{10}x_0 + a_{11}x_1)

Note that since A is invertible the map has an inverse. Moreover, note that if we replace A by a nonzero scalar times A, then the morphism is unchanged.

We now define an automorphism of P^1 to be a morphism given by a matrix A as above.

Exercise 8: Show that if P, Q, R are three pairwise distinct points on P^1 then there exists a matrix A as above such that P, Q, R map to (1 : 0), (0 : 1), and (1 : 1).

Exercise 9: In case K = Z/pZ. Write a script that takes a triple of pairwise distinct points on P^1 and finds the matrix A as in the previous exercise.

Continue reading about conics. Go back the the start page.

projective_line.txt · Last modified: 2012/05/30 12:45 by johan
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