The general equation of a quadric is given by:

- F = 0.
- C = 0.
- A = 0.
- 4A + B + 4C + 2D + 3E + F = 0.
- A + 4B + C + D + 4E + F = 0.

- B + 2D + 3E = 0.
- 4B + D + 4E = 0.

2. What type of affine conic (parabola, hyperbola or ellipse) does the projective conic whose equation is given by xy + xz + yz -z^{2} = 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 B^{2}-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 = µ^{2} and c = µ^{3}. One the one hand we have a*(b+c) = µ*(µ^{2} + µ^{3}) = µ*(µ) = µ^{3}(µ) = µ^{4} = 2. On the other hand we have a*b = µ*(µ^{2}) = µ(µ^{2}) = µ^{3}. While a*c = µ*(µ^{3}) = µ^{3}(µ^{3}) = µ^{6}. Thus, a*b + a*c = µ^{3} + µ^{6} = 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*(b^{3}c) = a^{3}(b^{3}c) [since a square cubed is a square and the product of a square and a non-square is a non-square] = (a^{3}b^{3})c = (ab)^{3}c = (a*b)^{3}c = (a*b)*c.

*Case III*: b is a non-square, c is a square.

a*(b*c) = a*(bc) = a^{3}(bc) [since the product of a square and a non-square is a non-square] = (a^{3}b)c = (a*b)c = (a*b)*c.

*Case IV*: b and c are non-squares.

a*(b*c) = a*(b^{3}c) = a(b^{3}c) [since b^{3} is a non-square and the product of two non-squares is a square] = (ab^{3})c = (a^{3}b)^{3}c [since a^{9} = a] = (a*b)^{3}c = (a*b)*c.

5. Show that in the affine plane defined over N(9) the lines with equations y = x*µ^{5} and y = x*µ + 1 intersect at the point P with coordinates (µ^{7}, µ^{2}).

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 = x^{3}µ^{5} and y = x^{3}µ + 1. Subtracting the two and using the left distributive law (valid since we are working in the field) we get, 0 = x^{3}(µ - µ^{5}) + 1 = x^{3}(µ + µ) + 1 = x^{3}µ^{5} + 1. Thus, -1 = 2 = µ^{4} = x^{3}µ^{5}. So, x^{3} = µ^{7}, and x = (x^{3})^{3} = (µ^{7})^{3} = µ^{5}. Now, substituting this in to either equation we can solve for y. y = (µ^{5})^{3}µ^{5} = µ^{4}. As a check, we calculate y from the other equation as well. y = (µ^{5})^{3}µ + 1 = 1 + 1 = 2 = µ^{4}. So the correct point of intersection is **(µ ^{5}, µ^{4})**.

6. Determine the equation of the line in the affine plane defined over N(9) joining the points P = (µ^{7}, µ^{2}) and Q = (µ^{3}, µ^{4}). [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: (µ^{2} - µ^{4})/(µ^{7} - µ^{3}) = (µ^{2} + 1)/(µ^{7} + µ^{7}) = µ^{3}/µ^{3} = 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 = µ^{2} - µ^{7} = µ^{2} + µ^{3} = µ or µ^{4} - µ^{3} = µ^{4} + µ^{7} = µ. Thus, the equation of the line is **y = x + µ**.