3
$\begingroup$

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 $p$. enter image description here

How were those computed?

$\endgroup$
2
  • $\begingroup$ What do the edges in the graphs mean? Is it the existence of an isogeny? Please clarify. $\endgroup$
    – Ted
    May 26 '20 at 0:06
  • $\begingroup$ The existence of an isogeny of degree $7$. $\endgroup$
    – José
    May 26 '20 at 0:09
1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ For say $p\nmid 36$ there is no need of the $xy$ term $\endgroup$
    – reuns
    May 26 '20 at 0:37
  • $\begingroup$ @ruens, what do you mean? Sure you can clear it if you like, but the RHS will be marginally more complicated. $\endgroup$ May 26 '20 at 0:43
  • $\begingroup$ 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? $\endgroup$
    – José
    May 26 '20 at 0:43
  • $\begingroup$ 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 $\mathbf F_p$. $\endgroup$ May 26 '20 at 0:45
  • $\begingroup$ It won't be more complicated, with $a=b$ then $(4a^3-27b^2)/a^3=4-27/a$ $\endgroup$
    – reuns
    May 26 '20 at 0:47
1
$\begingroup$

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).

$\endgroup$
1
  • $\begingroup$ Thank you. I did know about modular polynomials, but had no idea they were available on Sage. $\endgroup$
    – José
    May 26 '20 at 22:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.