## Homework Assignment 6

### pg. 146 : 5

How many points are in the projective space PG(n,2)? How many lines?

Since the only non-zero scalar is 1, the number of points in this projective space is the number of non-zero vectors in the vector space of algebraic dimension n+1 over the field Z2. This is 2n+1-1. The lines are the subspaces of algebraic dimension 2. Again, since we don't have to worry about scalar multiples, we can calculate the number of these subspaces by first choosing two different points in all possible ways and then dividing this number by the number of points on a line (points in the subspace). Since there are only two scalars, there are exactly 4 linear combinations of two non-zero vectors, but one of these is the zero vector, so there are 3 points in every line. The number of lines is therefore (2n+1-1)(2n+1-2)/(2)(3) = (2n+1-1)(2n-1)/3.

### pg. 146 : 6

How many points and lines are in the projective space PG(3,p) where p is a prime number?

The vector space of algebraic dimension 4 over Zp has p4-1 non-zero vectors. However, there are p-1 non-zero scalars, so a point is a 1-dimensional (algebraic) subvector space with p-1 non-zero vectors in it. The total number of points is therefore p4-1/p-1 = p3 + p2 + p + 1. A line contains p+1 points. This can be seen since there are p2-1 linear combinations of two vectors which are not the zero vector. However, each of the p-1 scalar multiples of a vector appears as a linear combination, so the total number of points is p2-1/p-1 = p + 1. The total number of lines is thus (p3 + p2 + p + 1)(p3 + p2 + p)/2(p+1) = (p2+1)(p3 + p2 + p)/2, obtained by counting the number of pairs of distinct points and dividing by the number of points on a line.

### pg. 150 : 1

Show that, in any projective space, the intersection of a line and a plane not containing it is either empty or a point.

Let M be a plane and l a line not in M. If Ml is not empty, then it contains points of l. If there were at least two points in this intersection, then since M is a subspace (flat), the line l would be contained in M, contradicting the assumption.

### pg. 150 : 4a-c

a) Each line of a projective space is a subspace (flat).
All the points of a line are contained in the line.

b) Each plane of a projective space is a subspace.
As a plane is the intersection of all the subspaces containing it, this follows from c) below.

c) The intersection of an arbitrary (finite or infinite) collection of subspaces is a subspace.
Let M be an arbitrary collection of subspaces, and S the intersection of all the subspaces in M. If P and Q are two points in S, then P and Q are in every subspace of M. Since these are subspaces, all the points of the line determined by P an Q are in every subspace of M. Thus, all the points of PQ are in S and so, S is a subspace.

### pg. 154 : 1

In PG(3,R) let P = (0,0,0,1), A = (1,0,0,1), A' = (3,0,0,1), B = (0,1,0,1), B' = (0,4,0,1), C = (0,0,1,1), C' = (0,0,5,1). Show that these points satisfy the hypotheses for Desargues' Theorem, and find the points A", B" and C". Show that these points are collinear.
The point P is the intersection of the three planes whose equations are x = 0, y = 0 and z = 0. A and A' are in the planes y = 0 and z = 0, so they lie on the line which is the intersection of these two planes. B and B' are in the planes x = 0 and z = 0, so they lie on the line which is the intersection of these two planes. C and C' are in the planes x = 0 and y = 0, and lie on their line of intersection. These three lines all contain P so the hypotheses for Desargues' Theorem are satisfied. The points of AB (except B) have coordinates of the form (1,0,0,1) + a(0,1,0,1) = (1,a,0,a+1) and the points of A'B'(except B') have coordinates of the form (3,0,0,1) + b(0,4,0,1) = (3,4b,0,b+1). We must find a and b so that a vector of the first form is a scalar multiple of a vector of the second form. If we multiply the first form by the scalar 3, we are then looking for a and b so that (3,3a,0,3a+3) = (3,4b,0,b+1). This gives us two equations, 3a = 4b and 3a+3 = b+1 to solve for a and b. The solutions are a = -8/9 and b = -2/3 and so, the point of intersection is (1,-8/9,0,1/9) or in standard form (9,-8,0,1) = C". Repeating this method for the other two points, we get the points of AC of the form (1,0,a,a+1) and those of A'C' of the form (3,0,5b,b+1). Set (3,0,3a,3a+3) = (3,0,5b,b+1) to get equations 3a = 5b and 3a+3 =b+1, from which we obtain a = -5/6 and b = -1/2. Thus, B" = (1,0,-5/6,1/6) = (6,0,-5,1). Finally, BC has points of the form (0,1,a,a+1) and B'C' has points of the form (0,4,5b,b+1). Set (0,4,4a,4a+4) = (0,4,5b,b+1) to get equations, 4a = 5b and 4a+4 = b+1, with solutions a = -15/16 and b = -3/4. So, A" = (0,1,-15/16,1/16) = (0,16,-15,1). To show that these three points are collinear we will need to show that one of them is a linear combination of the other two (up to scalar multiples). So, consider a(9,-8,0,1) + b(6,0,-5,1) = (0,16,-15,1). This gives 4 equations, 9a + 6b = 0, -8a = 16, -5b = -15 and a + b = 1. The middle two equations give a = -2 and b = 3. These values also satisfy the other two equations, so we have found that A" is a linear combination of B" and C", i.e., A", B" and C" are collinear.

### pg. 154 : 2

Show that any point in PG(3,F) can be represented uniquely as (x1,x2,x3,1), (x1,x2,1,0),(x1,1,0,0) or (1,0,0,0).
Since the all zero vector is not a point, at least one coordinate must be non-zero. By dividing all coordinates by the rightmost non-zero coordinate we will obtain one of these forms.

### pg. 154 : 3

Extend the result in Exercise 2 to PG(n-1,F).
Any point in PG(n-1,F) can be represented uniquely as (x1,x2,x3,...,xn-1,1), (x1,x2,...,xn-2,1,0),...,(x1,1,0,...,0) or (1,0,...,0). Proof is the same.

### pg. 154 : 4a

Use the results in Exercises 2 and 3 to count the number of points in PG(n-1,F) where F = GF(pk).
For simplicity, let q = pk. Consider the result of Exercise 3. There are qn-1 points of the first type, qn-2 points of the second type, ..., q points of the next to last type and 1 point of the last type. So, the total number of points is qn-1 + qn-2 + qn-3 + ... + q2 + q + 1 = pk(n-1) + pk(n-2) + ... + pk + 1.