# Isogenous elliptic curves over finite fields have the same number of points

I'm stuck in this question, it is the first part of exercise 5.4 from Silverman - The arithmetic of elliptic curves.

Let $C,D$ be two isogenous elliptic curves over a finite field $\mathbb{F}_q$. Then $$\#C(\mathbb{F}_q)=\#D(\mathbb{F}_q)$$

Any idea would be appreciated.

I also wonder if the following is true. Suppose $C,D$ are 2-isogenous curves over $\mathbb{Q}$, and for any $p$ prime that does not divide the discriminant, the reduction of these curves modulo $p$ are such that 4 divides their orders. Is it true that the reduced curves are also 2-isogenous?

## 2 Answers

In the spirit of chapter 5 of Silverman: use that $f:C\to D$ to be an isogeny defined over $\mathbb F_q$ means that $f \circ \phi_C = \phi_D \circ f$, where $\phi_C$ and $\phi_D$ are the Frobenius morphisms on $C$ and $D$ respectively. Then $$f \circ ( 1_C - \phi_C) = (1_D - \phi_D) \circ f.$$ Take the degree of both sides, and use the fact that $\deg u\circ v = \deg u \cdot \deg v$, and $\deg u\not= 0$ if $u$ is an isogeny. Now, use that $E(\mathbb F_q) = \ker (1 -\phi)$, for any elliptic curve $E$ over $\mathbb F_q$, and that $1-\phi$ is separable.

To answer your second question - I think that you are asking whether the isogeny $f$ over the rationals extends to one (call it $f$ again) over the open set $S$ of $\mathop{\rm Spec} \mathbb Z$ where the two curves have good reduction?

According to lemma 6.2.1 of S's "Advanced Topics in the Arithmetic of Elliptic Curves," a rational map from a smooth scheme to a proper scheme over a dedekind domain only fails to be defined on a set of at worst (at least) codimension 2, "so $f$ extends," and does so uniquely, as implicit in the definitions is 'separated.'

For the extended $f$ to be a group homomorphism one needs that $f$ commute with addition; but that's a Zariski closed condition which holds generically over $S$, so it must hold identically over $S$ (the separated condition). The degree of $f$ doesn't change - use the above to extend the dual isogeny $\check f$, and the relation $f \circ \check f = [m]$, where $m$ is the degree of $f$.

I hope I haven't screwed this up! Even if I haven't, I am sure there are better arguments.

Yes, they have the same number of points. They have the same characteristic polynomial of Frobenius acting on the Tate module, hence the same number of points over $\mathbb{F}_p$.