The j-invariant of an elliptic curve

If we have an elliptic curve

y2 = x3 + ax + b

then the j-invariant of the curve is given by

j(a,b)= 4a3 / (4a3 + 27b2)

or perhaps by some constant multiple of this. Sometimes there's an additional factor of 1728 thrown in there that makes some calculations come out cleaner, but it's not really necessary. Why does this definition make sense? Elliptic curves with the same j-invariant are isomorphic, or are really the same curve in disguise, and here's why.

Suppose that we have a lattice L in the complex plane with basis ω1 and ω2. When we multiply each of these basis elements by the same non-zero constant to get the basis cω1 and cω2 , we end up with essentially the same lattice, and the j-invariant captures that idea.

The Weierstrass invariants g2 and g3 for L are given by

g2 = 60 ∑ ω-4


g3 = 140  ∑ ω-6

where the sums are over all non-zero ω = nω1 + mω2.

If we multiply both ω1 and ω2 by a non-zero constant c we get the basis for a lattice L′ we get that

g2' = 60 ∑ (cω)-4 = c-4 60 ∑ ω-4 = c-4 g2


g3' = 140 ∑ (cω)-6 = c-6 140 ∑ ω-6 = c-6 g3

The value

j(g2,g3) = g23 / (g23 – 27 g32)

is invariant under this change of variables, so that

j(g2' ,g3' ) = j(g2,g3)

How is this reflected in an elliptic curve that's parameterized by the Weierstrass ℘-function that we get from these lattices?

If we have an elliptic curve

y2 = 4x3g2xg3


(y/2)2 = x3g2/4 xg3/4

and write this as

y2 = x3 + ax + b

then we have that g2 = -4a and g3 = -4b so that

j(g2,g3) = j(-4a,-4b)

= (-4a)3 / ((-4a)3 – 27(-4b)2))

= 4a3 / (4a3 + 27b2)

which is the form that's usually given for the j-invariant, at least up to a constant.

We can also see how the mapping of L to cL, which maps z to cz, is reflected in the coordinates of a point on an elliptic curve.

The x-coordinate of a point on an elliptic curve that's parameterized by the Weierstrass ℘-function is given by

L(z) = z-2 + ∑ ((z – ω)-2 – ω-2)

so that

cL(cz) = (cz)-2 + ∑ ((czcω)-2 – (cω)-2)

= c2 (z-2 + ∑ ((z – ω)-2 – ω-2))

= c2L(z)

or that x gets mapped to c2x.

Similarly, we have that

℘′L(z) = -2 ∑ (z – ω)-3

so that

℘′cL(cz) = c-3 ℘′L(z)

or that y gets mapped to c-3y.

Here's a table that summarizes the correspondences between (x,y) on E: y2 = x3 + ax + b and (x′,y′) on the isomorphic E′: (y′)2 = (x′)3 + a′x′ + b′:




a′ = c-4a


b′ = c-6b


x′ = c-2x


y′ = c-3y


j′ = j

Now suppose that we want to find an elliptic curve with a particular j-invariant, say J. If we have

a = b = (27/4) (J / (1 – J))

then we get that

j(a,b) = J

so the elliptic curve

y2 = x3 + ax + b

has that j-invariant.

Now, if I could just learn why a j-invariant is called a j-invariant I'd be happy. I've never been able to track down that particular bit of history.

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