# j-invariant of an elliptic curve

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.

• Maybe this link from Math SE helps? – AleksanderRas Nov 23 '18 at 16:18
• @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. – Youssef El Housni Nov 28 '18 at 19:23