# $j$-invariants for elliptic curves over $\mathbb{F}_p$

I'm reading an article about elliptic curve volcanos. I know how to compute the $$j$$-invariant given a curve in Weierstrass form, but i don't have any idea on how to compute every possible $$j$$-invariant possible for curves defined over $$\mathbb{F}_p$$, other than brute forcing every Weierstrass form curve.

In the paper the number of $$j$$-invariants is finite and every one of them is smaller than $$p$$.

How were those computed?

• What do the edges in the graphs mean? Is it the existence of an isogeny? Please clarify.
– Ted
May 26 '20 at 0:06
• The existence of an isogeny of degree $7$.
– José
May 26 '20 at 0:09

A curve defined over a given field $$K$$ the $$j$$-invariant of an elliptic curve is an element of that field. Therefore for a finite field of prime order the $$j$$-invariant can be represented by a number less than $$p$$.

As for which $$j$$-invariants are possible, they all are! The curve $$y^2 + xy = x^3 - \frac{36}{j_0 - 1728} x - \frac{1}{j_0 - 1728}$$

is well known and you can calculate its $$j$$-invariant to be $$j_0$$, the only edge case is $$j_0 = 1728$$ where the above formula breaks down, nevertheless, an elliptic curve of $$j$$-invariant 1728 are given by $$y^2 = x^3 - x$$ for $$p\ne 2$$.

In particular I assume that in the diagram it is not implied that those are all cordilliera, just some examples of some.

• For say $p\nmid 36$ there is no need of the $xy$ term May 26 '20 at 0:37
• @ruens, what do you mean? Sure you can clear it if you like, but the RHS will be marginally more complicated. May 26 '20 at 0:43
• So are we talking about any complex $j$-invariant? Starting building a volcano from an integer $j$-invariant all other $j$-invariants on that volcano will also be integers?
– José
May 26 '20 at 0:43
• The $j$-invariant is an element of whatever field the curve is defined over, so in this example they are all over finite fields. They look like integers because every element of a finite field of prime order can be represented as an integer, but really they are elements of $\mathbf F_p$. May 26 '20 at 0:45
• It won't be more complicated, with $a=b$ then $(4a^3-27b^2)/a^3=4-27/a$ May 26 '20 at 0:47

You may also want to look into modular polynomials. These are polynomials $$\Phi_\ell(X, Y) \in \mathbb{Z}[X,Y]$$ for primes $$\ell$$ such that the roots of $$\Phi_\ell(X, j_0)$$ over $$\mathbb{F}_p$$ are the $$j$$-invariants of curves over $$\mathbb{F}_p$$ that are $$\ell$$-isogenous to the curve $$j_0$$. The construction of these polynomials is complicated, but you can access them in Sage (from polmodular in Pari/GP). The degree of $$\Phi_\ell(X, j_0)$$ is $$\ell + 1$$ so, depending on the size of $$p$$, it can be much faster to just factor $$\Phi_\ell(X, j_0)$$ over $$\mathbb{F}_p$$ than to enumerate all the $$j$$-invariants and check them individually. You can use Vélu's formulae to write down the explicit isogeny if you want to see that (available as ellisogeny in Pari/GP, not sure for Sage).

• Thank you. I did know about modular polynomials, but had no idea they were available on Sage.
– José
May 26 '20 at 22:03