Bezout’s Theorem is an important theorem in algebraic geometry. It is far more general than what is presented here. It is, however, useful to look at this specific case of the theorem when we have two curves in a plane and want to know how they intersect.

**Definition.** The degree of a polynomial in two variables on the form *x*^{r}*y*^{s} + *x*^{r-1}*y*^{s-1} + … + a = 0 is max(r_{i} + s_{j}).

**Example.** *x*^{2}*y* + *y* = 0 has degree 3.

**Definition.** The degree of a curve is equal to the degree of the polynomial that defines the curve. Bezout’s Theorem. If we have two curves of degree r and s, they will meet in exactly r*s points in **C**^{2} – the plane of complex numbers.

**Example.** Let us consider the two curves given by the polynomial equations *x*^{2} – *y* = 0 and *x* + 2 – *y* = 0. The degrees of these polynomials are 2 and 1 respectively and hence so are the degrees of the corresponding curves. Bezout’s Theorem tells us that the curves intersect in exactly 2 * 1 = 2 points.

Let us first do this algebraically. To find how the two curves meet, we must set the two polynomials equal to each other and solve the resulting equation.

*x*^{2} – *y* = *x* + 2- *y*

*x*^{2} – *x* – 2 = 0

A second order equation on the form a*x*^{2} + b*x* + c = 0 has two solutions given by

x_{1} = (-b + sqrt(b^{2} – 4ac)) / 2a and x_{2} = (-b + sqrt(b^{2} – 4ac)) / 2a.

We get x_{1} = 1/2 + sqrt(1 – 4(-2))/2 = 1/2 + 3/2 = 2 and x_{2} = -1. We get the points of intersection by inserting these roots in one of the equations for the curves. For example the linear equation gives us y_{1} = x_{1} + 2 = 2+ 2 = 4 and y_{2} = x_{2} + 2 = -1 + 2 = 1.

We now plot the solutions to the two polynomial equations in Figure 1 and observe that the intersections are as predicted by our algebra and the number of intersections are as predicted by Bezout’s Theorem. The maximum values for both axes are 5.

Figure 1: The number of intersections between a quadratic and a straight line is 2