## Homework Assignment 5

1. Determine the equation of a quadric which passes through the 5 points in PG(2,5) given by (0,0,1), (0,1,0), (1,0,0), (2,3,1), and (1,4,1).

The general equation of a quadric is given by:

Q(x,y,z) = Ax2 + Bxy + Cy2 + Dxz + Eyz + Fz2 = 0.
Substituting the five points into this general equation gives 5 equations:
• F = 0.
• C = 0.
• A = 0.
• 4A + B + 4C + 2D + 3E + F = 0.
• A + 4B + C + D + 4E + F = 0.
Using the first three equations to simplify the last two, we get
• B + 2D + 3E = 0.
• 4B + D + 4E = 0.
Adding these equations (and recalling that 5 = 0), gives 3D + 2E = 0. Thus, 3D = -2E = 3E, so D = E. Substituting this into the first equation gives B + 5D = B = 0. The general equation is thus, Dxz + Dyz = 0. Since D = 0 would give a degenerate quadric, we can assume that D is not zero and divide by it to obtain: xz + yz = 0.

2. What type of affine conic (parabola, hyperbola or ellipse) does the projective conic whose equation is given by xy + xz + yz -z2 = 0 in PG(2,5) become when the line z = 0 is removed to form an affine plane?

Since the characteristic of the field GF(5) is odd, we may examine the discriminant of the quadratic equation to answer this. The discriminant is B2-4AC, where A,B and C are the coefficients found in the general form of the equation given above. In this case, A = C = 0 and B = 1. So, the discriminant is 1. Since 1 is a square, the affine conic will be an hyperbola.

3. Show by example that the nearfield N(9) does not satisfy the left distributive law (i.e., a*(b+c) = a*b + a*c).

There are many examples, in fact any choice of b and c which are non-zero, one a square and the other a non-square will work for any a different from 0 or 1. So, in particular, let a = µ, b = µ2 and c = µ3. One the one hand we have a*(b+c) = µ*(µ2 + µ3) = µ*(µ) = µ3(µ) = µ4 = 2. On the other hand we have a*b = µ*(µ2) = µ(µ2) = µ3. While a*c = µ*(µ3) = µ33) = µ6. Thus, a*b + a*c = µ3 + µ6 = 2µ + 2 + µ + 2 = 2+ 2 = 1.

4. Prove that the associative law (a*(b*c) = (a*b)*c) holds for all non-zero elements of N(9). [Hint: There are 4 cases to consider, depending on whether b and c are squares or non-squares.]

Case I: b and c are both squares.
a*(b*c) = a*(bc) = a(bc) [since the product of two squares is a square] = (ab)c = (a*b)c = (a*b)*c.
Case II: b is a square, c is a non-square.
a*(b*c) = a*(b3c) = a3(b3c) [since a square cubed is a square and the product of a square and a non-square is a non-square] = (a3b3)c = (ab)3c = (a*b)3c = (a*b)*c.
Case III: b is a non-square, c is a square.
a*(b*c) = a*(bc) = a3(bc) [since the product of a square and a non-square is a non-square] = (a3b)c = (a*b)c = (a*b)*c.
Case IV: b and c are non-squares.
a*(b*c) = a*(b3c) = a(b3c) [since b3 is a non-square and the product of two non-squares is a square] = (ab3)c = (a3b)3c [since a9 = a] = (a*b)3c = (a*b)*c.

5. Show that in the affine plane defined over N(9) the lines with equations y = x*µ5 and y = x*µ + 1 intersect at the point P with coordinates (µ7, µ2).

Oops. I goofed and gave the wrong point of intersection. We will calculate the correct one.

Since the "slopes" are both non-squares, we rewrite the equations as equations over the field. That is, y = x3µ5 and y = x3µ + 1. Subtracting the two and using the left distributive law (valid since we are working in the field) we get, 0 = x3(µ - µ5) + 1 = x3(µ + µ) + 1 = x3µ5 + 1. Thus, -1 = 2 = µ4 = x3µ5. So, x3 = µ7, and x = (x3)3 = (µ7)3 = µ5. Now, substituting this in to either equation we can solve for y. y = (µ5)3µ5 = µ4. As a check, we calculate y from the other equation as well. y = (µ5)3µ + 1 = 1 + 1 = 2 = µ4. So the correct point of intersection is 5, µ4).

6. Determine the equation of the line in the affine plane defined over N(9) joining the points P = (µ7, µ2) and Q = (µ3, µ4). [Hint: Determine the slope of the line PQ in the Desarguesian affine plane to decide whether this is a type I or a type II line.]

The slope of PQ calculated in the field is: (µ2 - µ4)/(µ7 - µ3) = (µ2 + 1)/(µ7 + µ7) = µ33 = 1. As this is a square, the line PQ is a type I non-vertical line, so it has an equation of the form y = x*1 + k = x + k. We can find k by substituting either point into this equation. k = y - x = µ2 - µ7 = µ2 + µ3 = µ or µ4 - µ3 = µ4 + µ7 = µ. Thus, the equation of the line is y = x + µ.