Question

Consider n-dimensional finite affine space: F(n,p) over the field with prime p-elements.

a. Show tha if l and l' are two lines in F(n,p) containing the origin 0, then either l intersection l' ={0} or l=l'

b. how many points lie on each line through the origin in F(n,p)?

c. derive a formula for L(n,p), the number of lines through the origin in F(n,p)

Answer 1

By definition, the n-dimensional finite affine space :F(n,p) = F_pⁿ = { (a₁,a₂,....a_n) : a_i F_p , i=1,2,...n }, where F_p is the field with prime p elements.
This forms a vector space over F_p of dimension n.

a.
Now, any line through the origin 0 is the subset { ka : k F_p } for some vector a F(n,p) \ {0}.
This is a vector subspace of F(n,p) of dimension 1, namely, the linear span of a.

Let l and l' be two lines in F(n,p) containing the origin 0. Then, l and l' are subspaces of F(n,p).
Now, l l' is again a subspace of F(n,p) and is a subspace of both l and l'.
Hence, dimension of l l' is less than or equal to the dimension of l or l' that is, 1.
Now, dim(l l') 1 implies that dim(l l') = 0 or 1.

If dim(l l') = 0, then l l' = {0}.

If dim(l l') = 1 = dim(l) = dim(l'), then l = l l' = l' as l l' is a subspace of l(and l') of the same dimension as l( and l'). Thus, in this case, l = l'.



b.
Suppose that L = { ka : k F_p } be a line through the origin in F(n,p), where a is a vector in F(n,p).
Since in a vector space, each singleton set {v} is linearly independent over the base field( where v is not 0), hence, for any two scalars k₁ and k₂, k₁ v = k₂ v iff k₁ = k₂.
Thus, | L | = | { ka : k F_p } | = | F_p | = p.

Thus, each line through the origin in F(n,p) contains p points.




c.
| F(n,p) | = | F_pⁿ | = pⁿ.
Any line through the origin in F(n,p) contains exactly p points and any two distinct lines through the origin in F(n,p) intersect at 0.
Now, if L₁ and L₂ are two distinct lines through the origin, then L₁ \ {0} and L₂ \ {0} are disjoint and each has p-1 points.
Also, F(n,p) \ {0} has pⁿ - 1 points.
Also, any point in F(n,p) \ {0} lies on a unique line through the origin.
Now,

  

where W is the set of all lines through the origin in F(n,p).
Thus, pⁿ-1 = | W | (p-1) = L(n,p) . (p-1)

Thus, the total number of lines through the origin in F(n,p) is:

L(n,p) = (pⁿ-1)/(p-1) = 1+p+p²+....p^n-1