## Lecture Notes 21: A Finite Non-Desarguesian Plane

We shall describe a non-Desarguesian projective plane of order 9. In order to do so we will need to describe two algebraic systems, the finite field of order 9 and a "nearfield" of order 9. The constructions we use are specializations of more general constructions to the case of order 9.

### GF(9)

Since 9 = 32, the prime field must be GF(3) whose elements we will represent by 0,1 and 2, and where addition and multiplication are done modulo 3. We seek an extension of degree 2 over the prime field, so our first task is to find a monic irreducible polynomial of degree 2 in GF(3)[x]. For large field this can be a difficult assignment, but for small fields it is not too bad. While there are some theorems that may help, the brute force procedure is effective if the prime field is small. We can in fact easily list all of the monic quadratics in this ring, they are:
• x2
• x2 + 1
• x2 + 2
• x2 + x
• x2 + x + 1
• x2 + x + 2
• x2 + 2x
• x2 + 2x + 1
• x2 + 2x + 2
Now the problem is to find the irreducible ones in this list. Clearly, any polynomial without a constant term is factorable (x is a factor), so the first, fourth and seventh can immediately be crossed out. For the remaining six polynomials, we may opt for one of two procedures. We could take each in turn and substitute all the field elements for x, if none of these substitutions evaluates to zero, the polynomial is irreducible (i.e., it has no root in the field). So, for example, substituting in x2 + 2 gives the values 02 + 2 = 2, 12 + 2 = 0 and 22 + 2 = 0, thus x2 + 2 factors, in fact x2 + 2 = (x + 1)(x + 2). On the other hand, the same procedure for x2 + 1 gives 02 + 1 = 1, 12 + 1 = 2 and 22 + 1 = 2 and so x2 + 1 is irreducible. The second possible procedure is to take all the linear factors (in this case, because we want quadratic products) and multiply them in all possible pairs to get a list of all the factorable quadratics, removing these from our list leaves all the irreducible quadratics. So,
• (x + 1)(x + 1) = x2 + 2x + 1
• (x + 1)(x + 2) = x2 + 2
• (x + 2)(x + 2) = x2 + x + 1
implying that
x2 + 1 , x2 + x + 2 and x2 + 2x + 2 are the only irreducible monic quadratic polynomials in GF(3)[x]. We could now choose any one of these letting þ be a zero of the chosen polynomial and write out the elements of GF(9) in its vector form representation using the basis {1, þ}. This however does not give us the most useful representation of the field. Rather, we will use the fact that the multiplicative group of the field is cyclic, so if we can find a primitive element (i.e., a generator of the cyclic group) we will have a handy representation of the elements. Now the primitive elements are to be found among the roots of the irreducible polynomials (they cannot be elements of the prime field). The cyclic group we are after has order 8, so not every root need be primitive. For example, letting þ be a root of x2 + 1, i.e., þ2 + 1 = 0, so þ2 = 2, we can write out the powers of þ.
þ1 = þ , þ2 = 2, þ3 = 2þ, þ4 = 2þ(þ) = 2þ2 = 2(2) = 1
and so þ has order 4 and does not generate the cyclic group of order 8, i.e., þ is not a primitive element. On the other hand, consider µ a root of the polynomial x2 + x + 2, so that µ2 + µ + 2 = 0 or µ2 = 2µ + 1. Now the powers of µ give us:
• µ1 = µ
• µ2 = 2µ + 1
• µ3 = µ(2µ + 1) = 2µ2 + µ = 2(2µ + 1) + µ = 2µ + 2
• µ4 = 2µ2 + 2µ = µ + 2 + 2µ = 2
• µ5 =
• µ6 = 2µ2 = µ + 2
• µ7 = µ2 + 2µ = 2µ + 1 + 2µ = µ + 1
• µ8 = µ2 + µ = 2µ + 1 + µ = 1
So µ is a primitive element and so we have represented the elements of GF(9) as the 8 powers of µ together with 0. Notice also that the bolded terms on the right are all the possible terms that can be written as linear combinations of the basis {1,µ} over GF(3). When working with finite fields it is convenient to have both of the above representations, since the terms on the left are easy to multiply and the terms on the right are easy to add. So for instance, if we wanted to calculate (2µ+2)3 + µ + 2, we would do so in this way, (2µ+2)3 = (µ3)3 = µ9 = µ and so (2µ+2)3 + µ + 2 = µ + µ + 2 = 2µ + 2 = µ3.

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)

The nearfield of order 9, denoted by N(9), is defined on the same set of elements as the field GF(9) and addition is the same in both systems. Multiplication in N(9) however is defined by (* denotes multiplication in N(9) and juxtaposition denotes multiplication in GF(9)):
x*y = xy if y is a square in GF(9) or x3y if y is a non-square.
(N(9)\{0},*) is not a group, but it does have many group properties. For instance, it satisfies the associative law and has an identity element (1). One can solve linear equations in this system in the following way:
Suppose that a is not zero. To solve the equation x*a = b for fixed a and b we consider two cases. If a is a square, then x*a = xa = b and x = ba-1 (where the calculation on the right is done in GF(9)). On the other hand, if a is a non-square then x*a = x3a = b, so x3 = ba-1 and x = (ba-1)3. In either case the solution is unique.
To solve the equation a*y = b, first consider the case that the solution y is a square. We would then have a*y = ay = b and so y = a-1b. If on the other hand y is a non-square, then a*y = a3y = b, and y = a-3b = a-2(a-1b). Notice that this solution is a non-square if and only if a-1b is a non-square. To summarize, the solution of a*y = b is y = a-1b if a-1b is a square, and y = a-3b if a-1b is a non-square. Again, the solution y is unique.

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.

### A non-Desarguesian Affine Plane of order 9

The points of this plane are the pairs (x,y) with x and y in N(9) (or GF(9) since it has the same set of elements). There are two kinds of lines, the vertical lines {(x,y)| x = k, for a fixed k in N(9)} and the non-vertical lines {(x,y)| y = x*m + b, for fixed m and b in N(9)}. Notice that the vertical lines and the non-vertical lines with m a square, are also lines in the Desarguesian affine plane of order 9. For convenience we will call these Desarguesian lines type I lines, and the non-vertical lines with non-square m, type II lines.

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.