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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?

José
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2 Answers2

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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.

Alex J Best
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  • For say $p\nmid 36$ there is no need of the $xy$ term – reuns May 26 '20 at 00:37
  • @ruens, what do you mean? Sure you can clear it if you like, but the RHS will be marginally more complicated. – Alex J Best May 26 '20 at 00:43
  • 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 00:43
  • 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 $\mathbf F_p$. – Alex J Best May 26 '20 at 00:45
  • It won't be more complicated, with $a=b$ then $(4a^3-27b^2)/a^3=4-27/a$ – reuns May 26 '20 at 00:47
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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).

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