Elliptic curves - connected or not
The graphs of some elliptic curves have one component like this one, the graph of y2 = x3 + 1.
Others have two components like this one, the graph of y2 = x3 – 2x + 1.
It turns out to be easy to tell these two cases apart. If the elliptic curve is defined by the equation y2 = x3 + ax + b and we write D = -4 a3 - 27 b2, then if D > 0 then the graph of the elliptic curve has two components and if D < 0 then the graph has one component. Here’s why.
D is the discriminant of the cubic f(x) = x3 + ax + b. So if we write
x3 + ax + b = (x – x1)(x – x2)(x – x3)
then we have that
D = (x1 – x2)2(x1 – x3)2(x2 – x3)2
Whether the graph of y2 = f(x) has one or two components is determined by how many real roots f(x) has. If there is a single real root then the graph crosses the x-axis once and the graph has a single component, like we get with y2 = x3 + 1. If there are three real roots then the graph then the graph crosses the x-axis three times and the graph has two disconnected components, like we get with y2 = x3 – 2x + 1.
If f(x) has three real roots then D is clearly positive. What about the case of a single real root and a complex conjugate pair of roots?
Let’s write the roots of f(x) as
x1 = a + bi
x2 = a - bi
x3 = c
where a, b and c are real. We can then write
D = (x1 – x2)2(x1 – x3)2(x2 – x3)2
= ((a + bi)– (a – bi))2(a + bi – c)2(a - bi – c)2
= (2bi)2((a - c) + bi)2((a – c) – bi)2
= -4b2((a - c)2 + b2)
which is always negative, just like we wanted.
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