Background in Algebraic Geometry

Algebraic geometry is concerned with sets of solutions to polynomical equations. Let F be a field and Fⁿ the set of ordered n-tuples from F. The set of polynomials in n variables; x₁, ..., x_n; is denoted as F[x₁, ..., x_n]. A polynomial f(x₁, ..., x_n) in F[x₁, ..., x_n] maps Fⁿ into F. The zeroes of a polynomial f are the points of Fⁿ that f maps into the additive identity of F, 0. If we have a mapping g, A→B then the elements of the field A which are mapped into the additive identity of the field B are called the kernel of the mapping.

A variety V is a set of points in Fⁿ such that there exists a set of polynomials {f₁, ..., f_m} in F[x₁, ..., x_n] where a point p belongs to V if and only if p is a common zero of {f₁, ..., f_m}; i.e., p is in the intersection of the kernels of {f₁, ..., f_m}. Thus a variety is a function of a set of polynomials in F[x₁, ..., x_n] and may be donoted as V(f₁, ..., f_m). The degree of a variety V is the maximum degree of the polynomials generating V.

The homogeneous coordinates for a point p = (x,y) in F² is the triple (X,Y,Z) such that x=XZ^-1 and y=YZ^-1. For the fields of real or complex numbers this would be denoted as x=X/Z and y=Y/Z.

Homogeneous coordinates of the form (X,Y,0), where X and Y are not both zero, represent ponts on the "horizon" or the "line at infinity." Each point in F² corresponds to a line through the origin in F³. The set of points in F³ with the point (0,0,0) deleted is called a projective space. A "point" in projective space is a line through the origin, but not including the origin.

As an illustration of an intersection "at the horizon" consider two parallel lines, L₁ and L₂, given by the equations:

y = a₁ + bx and y = a₂ + bx
 

where a₁ is not equal to a₂. In homogeneous coordinates these become:

Y/Z = a₁ + bX/Z
and
Y/Z = a₂ + bX/Z
or, equivalently
Y = a₁Z + bX
and
Y = a₂Z + bX.
 

If Z=0, then any point such that Y=bX satisfies both equations. Therefore any point in projective space of the form (X,bX,0) is on both L₁ and L₂ and is thus an intersection point of L₁ and L₂.

Now consider the general case for lines. Let L₁ and L₂, given by the equations:

y = a₁ + b₁x and y = a₂ + b₂x
 

In homogeneous coordinates these become:

Y/Z = a₁ + b₁X/Z
and
Y/Z = a₂ + b₂X/Z
or, equivalently
Y = a₁Z + b₁X
and
Y = a₂Z + b₂X
or a₁Z = b₁X − Y
and
a₂Z = b₂X − Y.
 

This latter set of equations is a system of two equations in two unknowns, X and Y, with Z as a parameter. If Z=0 then the two equations have a nontrivial solution if and only if the determinant of the coefficient matrix is zero. That determinnant is given by

− b₁ − (−b₂) = b₂−b₁
which is zero if and only if
b₁ = b₂

If a₁=a₂ then the lines are coincident. If a₁≠a₂ then the lines are parallel and have an intersection at infinity, as indicated previously.

If Z≠0 then the system of equations will have a solution if and only if the determinant of the coefficient matrix is not equal to zero. Thus if b₁≠b₂ then there will be a solution. Let X* and Y* be the solution for Z=1. Then the general solution will be (x*Z,Y*Z,Z). Therefor if b₁=b₂ there is either a coincidence of lines or one intersection (at infinity). If b₁≠b₂ there is one intersection (not at infinity).

The Intersection of a Degree 2 Variety with a Degree 1 Variety

The equations are

a₁ + b₁x + c_1,1x² + c_1,2xy + c_2,2y² + d₁y = 0
and
a₂ + b₂x + d₂y = 0

When (X/Z) and (Y/Z) are substituted for x and y and the results multiplied by the appropriate power of Z the equations become

aZ² + b₁XZ + c_1,1X² + c_1,2XY + c_2,2Y² + d₁YZ = 0
and
a₂Z + b₂X + d₂Y = 0

For the case of Z=0 these reduce to

c_1,1X² + c_1,2XY + c_2,2Y² = 0
and
b₂X + d₂Y = 0

This only has the trivial solution X=0 and Y=0 unless all of the c_i,j are equal to zero. If Z=0, then any point such that Y=bX satisfies both equations. Therefore any point in projective space of the form (X,bX,0) is on both L₁ and L₂ and is thus an intersection point of L₁ and L₂.