We also note that for this field, cubing (raising to the 3rd power) is an automorphism. This can be seen because (x+y)3 = x3 + 3x2y + 3xy2 + y3 = x3 + y3, since 3 = 0 in this field, and (xy)3 = x3y3. Since an automorphism is a bijection, the cubing function has an inverse. Because x9 = x for all x in this field, the cubing map is its own inverse, i.e., taking a cube root is the same as cubing.
Given the representation of the field by powers of a primitive element it is easy to determine which elements are squares and non-squares. Namely, any element which can be represented by a primitive element raised to an even power is a square. 0 is also a square. (In general, this only makes sense if the characteristic of the field is odd, as it is in this case).
N(9) satisfies the right distributive law, but not the left distributive law (so it is known as a right nearfield). Consider (a + b)*c. If c is a square, (a+b)*c = (a+b)c = ac + bc = a*c + b*c. If c is a non-square, then (a+b)*c = (a+b)3c = (a3 + b3)c = a3c + b3c = a*c + b*c.
Proposition 21.1: Every two distinct points lie on a unique line.
Let P=(x1,y1) and Q=(x2,y2) be distinct points. If x1 = x2 then P and Q lie on the vertical line with equation x = x1. It is easy to check that these points can not be together on any non-vertical line. So we may assume that x1 -x2 is not zero. Now, consider the equations y1 = x1*m + b and y2 = x2*m +b. Subtracting the second from the first gives, y1 - y2 = x1*m - x2*m = (x1 - x2)*m, using the right distributive property. As we have seen, this equation has a unique solution for m, which is (x1-x2)-1(y1-y2) if this quantity is a square, or (x1-x2)-3(y1-y2) otherwise. In the first case, where m is a square, we can multiply the first equation by x2 and the second by x1 and then subtract to get x2y1 - x1y2 = x2b - x1b = (x2 - x1)b, so b = (x2-x1)-1(x2y1-x1y2). When m is a non-square, we have to modify this calculation by multiplying the first equation by x23 and the second by x13. We then obtain b = (x2-x1)-3(x23y1-x13y2). Thus, in either case, m and b are uniquely determined by the points P and Q.
Before proving our next geometric result, we need two algebraic lemmas concerning the field GF(9).
Lemma 21.2: For any element b in GF(9), b4 is 0, 1 or 2 depending on whether b is 0, a square or a non-square respectively.
Pf: In a field, a power of b can be 0 if and only if b = 0. We may therefore assume that b is not 0. Let µ be a primitive element of GF(9). Any non-zero element of GF(9) is represented as µx for some integer x. Since (µx)8 = (µ8)x = (1)x = 1, we see that every non-zero element of GF(9) raised to the 8th power is 1. Since (b4)2 = b8 = 1, we see that b4 is a root of the equation z2 - 1 = 0. As the solutions of this equation are 1 and -1 (=2), we have that b4 = 1 or 2 for any non-zero b. Now, let b = µx, so b4 = µ4x. If b4 = 1, then 4x must be a multiple of 8 and so, x must be even, i.e., b is a square. If b is a non-square, x is odd and so b4 can not be 1, so it must be 2.
Lemma 21.3: In GF(9), the equation x3 - bx - c = 0 has a unique solution (= -cb3 -c3) if b is a non-square.
Pf: We first check to see that -cb3 -c3 is a solution to this cubic equation. Indeed, (-cb3 -c3)3 -b(-cb3 -c3) - c = -c3b -c + cb4 +c3b -c = cb4 -2c = cb4 + c = c(b4+1) = 0. This computation used the fact that c9 = c, and that b4+1 = 0, since b is a non-square (Lemma 21.2).
Now, divide x3 - bx -c by the linear factor x - (-cb3 -c3) to get the quadratic x2 + (-cb3 -c3)x + (-cb3 -c3)2-b. Any additional roots of the cubic must be roots of this quadratic. However, the discriminant of this quadratic is (-cb3 -c3)2 - 4(-cb3 -c3) = b which is a non-square, so the quadratic has no roots in this field.
Proposition 21.4: Any two lines with different slopes (infinity for the vertical lines and m for the non-vertical lines) intersect in a unique point.
Pf: The statement is clearly true for any pair of type I lines since it is true in the appropriate Desarguesian affine plane. Consider two type II lines, y = x*m1 + k1 and y = x*m2 + k2, with m1 not equal to m2. These can be written as y = x3m1 + k1 and y = x3m2 + k2. By subtracting one from the other we obtain 0 = x3(m1 -m2) + k1-k2 (using left distributivity in GF(9)). From this we can solve for x = ((k2-k1)/(m1-m2))3. The value of y can now be determined from either of the GF(9) equations, and so the point of intersection is uniquely determined. We now consider the intersection of a type II line with a type I line. If the type I line is a vertical line, it is easy to see that the statement is true, so suppose that the type II line has equation y = x3m1 + k1 and the type I line has equation y = xm2 + k2. Subtracting gives the equation 0 = x3m1 - xm2 + (k1-k2). Since m1 is a non-square, and so, not zero, we can divide by it to obtain 0 = x3 -(m2/m1)x - (k2-k1)/m1. Since m2 is a square and m1 is a non-square, m2/m1 is a non-square and by Lemma 21.3, this cubic equation has a unique solution for x. We can find y from either of the original equations, and so, there is a unique point of intersection in this case as well.
Proposition 21.5: The lines with the same slope are parallel and every point is on precisely one line of each of these parallel classes.
Pf: The statement is clearly true for type I lines. Suppose that two type II lines with the same slope, i.e., with equations y = x*m + k1 and y = x*m + k2, meet at a point (a,b). Then, b = a*m + k1 and b = a*m+ k2. Subtracting gives 0 = k1 - k2, i.e., k1 = k2. Thus, two distinct lines with the same slope must be parallel. The point (a,b) is on the line y = x*m + k, with k = b - a*m for each parallel class with slope m.
Theorem 21.6: The plane constructed above is an affine plane of order 9.
Pf: Axiom A1 is Proposition 21.1. Axiom A2 (Playfair's axiom) follows immediately from Propositions 21.4 and 21.5. Each line has 9 points and there are 90 lines (54 type I and 36 type II).
Theorem 21.7: The affine plane constructed above is not Desarguesian.
Pf: Consider the triangles ABC and A'B'C' with coordinates A =(0,0), B = (1,0), C = (µ,1), A' = (0,1), B' = (1,1) and C' = (µ,0). Corresponding vertices lie on the parallel lines with equations x = 0, x = 1 and x = µ. The line AC is a type II line with equation y = x*µ5. The line A'C' is a type II line with equation y = x*µ + 1. The intersection of these lines is the point P with coordinates (µ7, µ2). The line BC is a type I line with equation y = x*µ2 + µ6, and B'C' is a type I line with equation y = x*µ6 + µ3. The intersection of these lines is the point Q with coordinates (µ3, µ4). Line AB has equation y = 0 and line A'B' has equation y = 1. Since these are parallel lines, if Desargues theorem were to hold the line PQ would have to be parallel to both and thus have an equation of the form y = k for some k. Since P and Q do not have the same second coordinate, the line PQ is not of this form.