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?
Best Answer
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.