Solutions to HW3
For affine geometries over the field F3 = {0, 1, 2}, every line (1-space) has 3 points; every
plane (2-space) has 9 points and 12 lines; every solid (3-space) has 27 points, 117 lines and
39 planes; and every 4-space has 81 points, 1080 lines, 1170 planes and 120 solids. All these
values were obtained in class, except for the last value, which is obtained in Exercise #1.
Here we reproduce the proof (from class) that every affine 3-space over the field
F3 = {0, 1, 2} (i.e. ‘solid’, which we denote here by S) has 117 lines and 39 planes. The
number of pairs (P, Q) of distinct points in S is clearly 27·26 = 702; on the other hand
since P and Q determine a unique line `, we may first choose ` in n ways (where n is the
number of lines in S) and then choose distinct points P and Q on ` in 3·2 = 6 ways, for a
total of 6n choices of the pair (P, Q). Since the two answers must agree, 6n = 702 which
gives n = 117 lines in S.
Similarly, let m be the number of planes in S. We count the number of pairs (P, `)
in S where ` is a line and P is a point not on `. There are 117 choices of `, and then
27 − 3 = 24 choices remaining for P , giving 117·24 = 2808 choices for (P, `). On the other
hand, (P, `) determines a unique plane π in S. If we first choose this plane (in m different
ways) and then choose ` in π (12 choices) and a point P in π but not on ` (9 − 3 = 6
choices) then we have 12·6m = 72m choices for (P, `). Since the two answers must agree,
72m = 2808 which gives m = 39.
Alternatively, every plane in S ∼
= F33 is defined by a linear equation of the form
Ax + By + Cz = D. There are 3 choices for each coefficient A, B, C, D ∈ F3 ; however,
(A, B, C) cannot equal (0, 0, 0), so this gives 33 − 1 = 26 choices for (A, B, C). Since
there is no restriction on D ∈ F3 , we obtain 26·3 = 78 linear equations defining planes
in S. Finally, there are two choices of linear equation for each plane, since the equations
Ax + By + Cz = D and 2Ax + 2By + 2Cz = 2D define the same plane; so there are only
78/2 = 39 planes in S.
1. There are exactly 120 solids in 4-space F43 .
First Solution: Count the number of pairs (π, S) in 4-space F43 , where π is a plane
and S is a solid containing π. On the one hand there are 1170 choices for π (counted
previously). Given π, there are 4 choices of solid S containing π. (Each such solid S
is generated by π together with one of the 81 − 9 = 72 points outside π; but in each
case there are 27 − 9 = 18 choices of point giving the same solid, so only 72/18 = 4
solids S containing π.) Altogether this gives 1170·4 = 4680 such pairs (π, S).
On the other hand, if there are n solids S and 39 planes in each, we get 39n pairs
(π, S). Since the two answers agree, 39n = 4680 and n = 120 for the number of solids
S in 4-space.
Second Solution: Every solid in 4-space is defined by a linear equation of the form
Ax + By + Cz + Dw = E. There are 3 choices for each coefficient A, B, C, D, E ∈
F3 ; however, (A, B, C, D) cannot equal (0, 0, 0, 0), so there are only 34 − 1 = 80
choices for (A, B, C, D). Since the choice of E ∈ F3 is arbitrary, there are 80·3 = 240
choices of linear equation. Finally, every solid has two choices of linear equation, since
Ax + By + Cz + Dw = E and 2Ax + 2By + 2Cz + 2Dw = 2E define the same solid.
So the number of solids is only 240/2 = 120.
2. a: y = − 15 x + 1
b: y = − 19 x + 1
c: y = 3x + 6
d: y = − 32 x + 6
e: y = 72 x + 7
f : y = − 57 x + 7
21
,
X: − 25
16 16
108 77
Y : − 65 , 65
15 56
Z:
11 , 11
The line y =
133
103 x
+
343
103
passes through points X, Y, Z.
3. a: y = 3x + 1
b: y = 4x + 1
c: y = 4x + 3
d: y = 5x + 3
e: y = 3x + 4
f : y = 5x + 4
X: (5, 2)
Y : (3, 6)
Since lines d and f are parallel, the point Z cannot be defined in the affine plane. In
the projective plane, Z is the point ‘at infinity’ where all affine lines of slope 5 meet.
The line y = 5x + 5 passes through the points X, Y, Z. (The reason it passes through
the point Z in the projective plane, is that its slope is 5.)