Algebraic Geometry for Non-Experts
We investigate equations of the form x3 + y2 + bx + a = 0, where a and b are numbers. The solution space of an equation on this form is a curve in the xy-plane. See "What is Algebraic Geometry?" This curve has a certain shape, and we may wonder if other values of a and b give curves of the same shape. Instead of just trying arbitrary a's and b's and comparing the resulting curves, we use a theorem of algebraic geometry which says that the a and b values that give similar curves are a's and b's which themselves lie on a specific curve in the ab-plane.
Given an equation, for example x3 - y2 = 0, we can plot all solutions to this equation to get the curve in Figure 1. The solutions (0,0), (1,1), and (1,-1) are marked as examples of points on the curve. All subsequent plots will be done to the same scale, but the numbers will be left out.
Figure 1: The algebraic curve of f(x) = x3 - y2
Remark. The reader who is familiar with complex numbers will see that we get complex solutions for negative x-values. Due to the obvious difficulty of drawing complex numbers we restrict ourselves to drawing only real solutions.
The most important task in algebraic geometry is classification. That is, given a set of curves we want to group similar curves. We interpret "similar" to mean curves of the same shape. The Weierstrass family are curves given by equations of the form x3 - y2 + bx + a = 0, where a and b can vary. For example a = 0 and b = 0 gives x3 - y2 = 0 as in Figure 1. Let us choose some different a's and b's and see what the corresponding curves look like.
Figure 1 to Figure 4 show curves of different shape in the Weierstrass family. Given a and b, we want to be able to determine what kind of curve they will produce. To do this we must introduce a new curve, the discriminant, which is given by an equation of a and b.
Figure 2: a = sqrt(4/27), b = -1 Figure 3: a = b = -1 Figure 4: a = 0, b = 1
Figure 5: The discriminant
The discriminant is defined as the solution to 27a2 + 4b3 = 0. See Figure 5. The points in the figure are the a and b values that give the equations defining the curves in figure 1, 2, 3, and 4. For example, the point (1,1) (a = 1 and b = 1) in Figure 5 gives the equation x3 + y2 + 1x + 1 = 0. We now have a curve which points correspond to curves in the Weierstrass family. If a curve crosses itself (Figure 2) or has an abrupt change (the (0,0) point in Figure 1 we say that the curve is singular. The curves in Figure 3 and 4 are non-singular. As these examples indicate, it is possible to show that curves with (a,b)- values above the discriminant are of the form of Figure 3. Curves with (a,b) values under the discriminant are of the form of Figure 4 and curves with (a,b) values on the discriminant are of the form of Figure 2 apart from (0,0) which is Figure 1.
Further, it is possible to show that two non-singular curves defined by (a1,b1) and (a2,b2) are similar if and only if j(a1,b1) = j(a2,b2), the "j-function" being defined as
j(a,b) = b3 / (27a2 + 4b3).
As an arbitrary example, let us now see which a's and b's give j(a,b) = 1/8. The algebraic curve corresponding to
b3 / (27a2 + 4b3) - 1/8 = 0.
The two arbitrary points on the curve in Figure 6 give the curves in Figure 7. In fact, it can be shown that every set of similar curves in the Weierstrass family are given by points (a,b) lying on a curve, like the one in Figure 6. In this way, all similar curves in the Weierstrass family are classified.
Figure 6: Two arbitrary points
Figure 7: The two curves that corresponds to the two points in Figure 6