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 dnoted 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 (); 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 example 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.

Illustration of Bezout's Theorem

The varieties illustrated are ellipses and thus are of degree 2. According to Bezout's Theorem the number of intersection points should be 2x2=4. In Figure 1 there are four intersection points. In Figure 2 the tangent intersection at has multiplicity two so there are again four intersection points. In Figure 3 the intersection at A involves not only a common point and tangent for the two ellipses but also a common curvature. Therefore the multiplicity of A is three and hence the number of intersections is again four. Figure 4 involves a different problem for Bezout's Theorem. There appear to be only twp point of intersection of multiplicity one each. The equations for these two circles are:

x²+y²=1
and
(x-1)²+y²=1
 

Their common points appear to be only (1/2, 3/2) and (1/2, -3/2). In homogeneous coordinates the equations of the circles are:

X²+Y²-Z² = 0
and
X² - 2XZ + Y² = 0
 

If the second equation is subtracted from the first the result is:

-Z²+2XZ = 0
 

so if Z is not zero then X=Z/2 and this substituted into the first equation yields the solution Y=3Z/2. Thus, in homogeneous coordinates the points (Z/2, 3Z/2, Z) and (Z/2, -3Z/2, Z) are points of intersection. But if Z=0 the two equations reduce to +X² + Y² = 0. In the real numbers this equation has only the solution X=0 and Y=0. In the complex number field there are also the solutions Y=iX and Y=-iX. Therefore there are two intersection points, (X, iX, 0) and (X, -iX, )) on the complex line at infinity. Hence in the complex projection plane the two circles have four intersection points as required by Bezout's Theorem.

Now consider the case of Figure 5 where the circles appear to have only one intersection point of multiplicity two. The equations for the circles are:

(x-1)²+y²=1 (x+1)²+y²=1
 

In homogeneous coordinates these become:

X² - 2XZ + Y² = 0
and
X² + 2XZ + Y² = 0
 

Subtracting the first equation from the second gives:

4XZ = 0
 

If Z is not equal to zero then X=0 and Y=0 and thus the intersection is the point (0,0,Z). But if Z=0 then X can be anything and Y must is iX or -iX. Therefore the points (X,iX,0) and (X,-iX,0) are intersection points. The point (0,0,Z) happens to be of multiplicity two so the number of intersection points is again four.

In Figure 6 there appear to be no intersection points of the circles, which have equations:

(x-2)²+y²=1 (x+2)²+y²=1
 

In homogeneous coordinates these become:

X² - 4XZ + 4Z² + Y² = Z²
and
X² + 4XZ + 4Z²+ Y² = Z²
 

Subtracting the first equation from the second gives:

8XZ = 0
 

If Z is not equal to zero then X=0 and Y=+3^1/2Z and Y=-3^1/2Z. Thus there are two intersection points (0,3^1/2Z,Z) and (0,-3^1/2Z,Z). But if Z=0 then X can be anything and Y must is iX or -iX. Therefore the points (X,iX,0) and (X,-iX,0) are also intersection points. The number of intersection points is again four.
 

There are of course many more possibilities than are covered in the six figures and many would appear to violate Bezout's Theorem, but it should be clear from the above illustrations that when intersections are considered in the complex projective plane Bezout's Theorem holds.