Exercise 5.8.6 (Order of the group $E_f(\mathbb{Z}_p)$)

Suppose that the polynomial x 3 + a x 2 + bx + c has no repeated root ( mod p ) , and put f ( x , y ) = y 2 ( x 3 + a x 2 + bx + c ) . Show that the group of points of the elliptic curve 𝒞 f ( p ) has order

| E f ( p ) | = p + 1 + x = 1 p ( f ( x ) p ) .

Note: Here f ( x ) is f ( x ) = x 3 + a x 2 + bx + c (note of R.G.).

Answers

Proof. Since E f ( p ) = 𝒞 f ( p ) { O } , where O is the point at infinity of the curve, we obtain

| E f ( p ) | = | 𝒞 f ( p ) | + 1 .

Put, for some fixed z p ,

N ( y 2 = z ) = Card { y p y 2 = z } .

We know that for all z p ,

N ( y 2 = z ) = 1 + ( z p ) .

(Consider the three cases z = 0 , z is a nonzero residue, and z is a nonresidue.)

For every fixed x 0 put A x 0 = { ( x 0 , y ) p y 2 = f ( x 0 ) } . Then A x 0 is in bijective correspondence with B x 0 = { y p y 2 = f ( x 0 ) } by ( x 0 , y ) y , and 𝒞 f ( p ) = x p A x (disjoint union). Hence we obtain

| 𝒞 f ( p ) | = x p N ( y 2 = f ( x ) ) = x p ( 1 + ( f ( x ) p ) ) = p + x p ( f ( x ) p ) .

Therefore

| E f ( p ) | = p + 1 + x p ( f ( x ) p ) .

Example: Consider the group of the curve y 2 = x 3 7 x + 7 on the field p , where p = 37409 (see p. 286).

sage: p = 37409
sage: is_prime(p)
True
sage: def f(x): return (x^3 - 7*x + 7) % p
sage: N = p + 1 + sum(kronecker(f(x),p) for x in range(p)); N
37620
sage: factor(N)
2^2 * 3^2 * 5 * 11 * 19

So | E f ( p ) | = 37620 = 2 2 3 2 5 11 19 .

This “lengthy calculation” takes 0.15 s on my modest computer.

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2025-07-01 10:02
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