Math 419/819  Homework Assignment 4  November 21, 2003

due date: November 28, 2003

1. Suppose that F is a homogeneous polynomial of degree d in X, Y, and Z. If we make the substitution Z = tX, we get a polynomial

F_{t} = ∑_{j=0}^{d} a_{j}(t)X^{j}Y^{d−j}

whose coefficients a_{j}(t) are polynomials in t. Explain why deg(a_{j}) ≤ j, and also explain which terms of F contribute to the top coefficient of t in a_{j}(t).

2. Suppose that we fix a value t_{0} of t. Explain how a solution (in X and Y) to the equation F_{t_{0}} = 0 gives you a point on F = 0 in ℙ^{2}. How to you find the Z coordinate of this point?

3. Suppose that G is a homogeneous polynomial of degree e in X, Y, Z, and

G_{t} = ∑_{j=0}^{e} b_{j}(t)X^{j}Y^{e−j}

the polynomial resulting from the substitution Z = tX. For a fixed value t_{0} of t, explain why finding a point p in the intersection {F = 0} ∩ {G = 0} in ℙ^{2} such that p is also on the line Z = t_{0}X is the same as finding a common solution (in X and Y) to the equations F_{t_{0}} = 0 and G_{t_{0}} = 0.

4. Suppose that no point of {F = 0} ∩ {G = 0} lies on the line X = 0. Explain why all points of the intersection have to lie on some line of the form Z = t_{0}X, for some t_{0}. Suppose further that no two points of the intersection {F = 0} ∩ {G = 0} lie on the same line of this form. Explain why points of the intersection are in one to one correspondence with the values of t such that the determinant

R(t) := | \begin{matrix}a_{d}(t) & a_{d−1}(t) & a_{d−2}(t) & ⋯ & a_{1}(t) & a_{0}(t) & 0 & 0 & ⋯ & 0 & 0 \\ 0 & a_{d}(t) & a_{d−1}(t) & ⋯ & a_{2}(t) & a_{1}(t) & a_{0}(t) & 0 & ⋯ & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 & ⋯ & a_{1}(t) & a_{0}(t) \\ b_{e}(t) & b_{e−1}(t) & b_{e−2}(t) & ⋯ & b_{0}(t) & 0 & 0 & 0 & ⋯ & 0 & 0 \\ 0 & b_{e}(t) & b_{e−1}(t) & ⋯ & b_{1}(t) & b_{0}(t) & 0 & 0 & ⋯ & 0 & 0 \\ ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ & ⋮ & ⋮ & ... & ⋮ & ⋮ \\ 0 & 0 & 0 & ⋯ & 0 & 0 & 0 & 0 & ⋯ & b_{1}(t) & b_{0}(t) \end{matrix} |

of the (d + e) × (d + e) matrix above is zero.

5. Using the bounds on deg(a_{j}) (and of course deg(b_{j})) from question 1, show that deg(R(t)) ≤ d · e. The formula

det(M) = ∑_{σ∈S_{n}} sgn(σ)M_{1σ(1)}M_{2σ(2)}M_{3σ(3)} ⋯ M_{n−1σ(n−1)}M_{nσ(n)}

for the determinant of an n × n matrix M might be useful. Here S_{n} is the permutation group on n symbols, and sgn(σ) of a permutation σ is ±1 depending on whether σ is an even or odd permutation. (Hint: Extend the bound on degree to the zero entries of the matrix in question 4 so as to give you a pattern which is easy to analyze. Since the entries are zero, any value of the “degree” is permitted for these entries, even negative ones – any piece of the formula involving these entries will be zero anyway. Use the pattern to compute the degree in t of any monomial in the sum above.)

6. We now want to show (under our assumptions about F and G) that R(t) must be of degree precisely d · e. This won’t be true for arbitrary F and G. For instance, for F = X^{2} + 2Y^{2} + 18Z^{2} + 12Y Z and G = X^{3} + Y^{3} + 27Z^{3}, we get the matrix

[ \begin{matrix}(18t^{2} + 1) & 12t & 2 & 0 & 0 \\ 0 & (18t^{2} + 1) & 12t & 2 & 0 \\ 0 & 0 & (18t^{2} + 1) & 12t & 2 \\ (27t^{3} + 1) & 0 & 0 & 1 & 0 \\ 0 & (27t^{3} + 1) & 0 & 0 & 1 \end{matrix} ]

with determinant 2196t^{4} + 54t^{2} + 72t + 9, of degree strictly smaller than 6.

Consider the matrix made up by replacing each entry in the resultant matrix (of the form a_{j}(t) or b_{j}(t)) by the coefficient of the top possible power of t in that spot (i.e., the coefficient of t^{j} in each case). In the example above, this would be the matrix

[ \begin{matrix}18 & 12 & 2 & 0 & 0 \\ 0 & 18 & 12 & 2 & 0 \\ 0 & 0 & 18 & 12 & 2 \\ 27 & 0 & 0 & 1 & 0 \\ 0 & 27 & 0 & 0 & 1 \end{matrix} ] .

Explain why (no matter what F and G are) that the coefficient of the d · e-th power of t in R(t) is the determinant of this matrix, so that deg(R(t)) = d · e if and only if this determinant is not zero.

7. Considering where the coefficient of t^{j} in a_{j}(t) comes from (problem 1) explain why the vanishing of this determinant is the same as saying that the polynomials F′ and G′ in Y and Z obtained from F and G by setting X = 0 have a common root.

8. Explain why this is equivalent to {F = 0} ∩ {G = 0} having a point of intersection on the line X = 0. Thus, under the assumptions of question 4, degree R(t) = d · e.