Elliptic curves – connected or not

The graphs of some elliptic curves have one component like this one, the graph of y2 = x3 + 1.

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Others have two components like this one, the graph of y2 = x3 – 2x + 1.

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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 = (xx1)(xx2)(xx3)

then we have that

D = (x1x2)2(x1x3)2(x2x3)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 = abi

x3 = c

where a, b and c are real. We can then write

D = (x1x2)2(x1x3)2(x2x3)2

= ((a + bi)– (a bi))2(a + bic)2(a bi – c)2

= (2bi)2((a c) + bi)2((ac) – bi)2

= -4b2((ac)2 + b2)

which is always negative, just like we wanted.

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