Given an elliptic curve $(E/\mathbb{K})$ where $char(\mathbb{K}) \ne 2,3$ defined by the Weierstrass equation $y^2=x^3+ax+b$. The $j$-invariant is $j=1728 \frac{4a^3}{4a^3+27b^2}$.

I want to understand very clearly how this j-invariant is constructed and especially from where does the 1728 come.

A rather simple but interesting explanation is, given $(E/\mathbb{K})$ in Legendre form $y^2=x(x-1)(x-\lambda)$, one looks for metric that is invariant with respect to $\lambda$ permutations; that is invariant w.r.t. $\lambda, \frac{1}{\lambda}, \lambda-1, \frac{1}{\lambda-1}, \frac{\lambda}{\lambda-1}, \frac{\lambda-1}{\lambda}$. One can try to multiply or sum these permutations but the result is not only invartiant w.r.t $\lambda$ but also w.r.t the elliptic curve. Finally trying to sum the squares results in the $j$-invariant but without the constant 1728.

So where does it come from? An answer with all the math details of the complex multiplication theory would be great.

  • 3
    $\begingroup$ Maybe this link from Math SE helps? $\endgroup$ – AleksanderRas Nov 23 '18 at 16:18
  • $\begingroup$ @AleksanderRas According to the link, the 1728 factor is there to "cancel out the coefficients that would make it otherwise impossible to define the j-invariant" but I don't get why it has to be exactly 1728. When converting a general Weierstrass equation into a short one, the coefficients 36 and 108 appear in the denominator in the second substitution which indeed cancel out in presence of 1728 ($1728=16 \times108$ and $1728=48 \times 36$) but it would have been the same with 108 as a factor in the j-invariant, which is the least common multiple. $\endgroup$ – Youssef El Housni Nov 28 '18 at 19:23

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