I am looking for a detailed proof (or a reference containing a detailed proof) of the following fact:
I know that $\mathbb F_p$-isogenous elliptic curves over $\mathbb F_p$ have the same number of $\mathbb F_p$-points. Is the converse true?
Characterization of Isogenous Elliptic Curves over Finite Fields
algebraic-geometryelliptic-curvesnumber theory
Related Solutions
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.
Regarding Mindlack's answer, I'll point out that Serre's proof (if I remember correctly) uses the fact that there are only finitely many $K$-isomorphism classes of elliptic curves with good reduction outside of a given finite set of primes $S$, i.e., Shafarevich's theorem for elliptic curves, which was extended to abelian varieties by Faltings. And the fact about isogenies is also an immediate consequence of this fact, since $K$-isogenous curves have the same set of primes of bad reduction. The most direct way to prove the fact about primes of good reduction outside $S$ is to first expand $S$ so that the ring of $S$-integers is a PID and so that $S$ includes all primes over 2 and 3. Then every curve $E/K$ with good reduction outside $S$ has a model $y^2=x^3+Ax+B$ with $A,B\in R_S$ and with $4A^3+27B^2\in R_S^*$. Now using the fact that $R_S^*/{R_S^*}^6$ is finite, we end up needing to show that for a given $c\in R_S^*$, the equation $u^3+v^2=c$ has only finitely many solutions $u,v\in R_S$. So now we're reduced to the fact that an elliptic curve has only finitely many $R_S$-integral points. This is the generalized version of Siegel's theorem due (I think) to Mahler. It's proof depends on something like the Thue-Siegel-Roth theorem on diophantine approximation, although one doesn't need the full strength of Roth's result. However, one does need a result that I would classify as a deep theorem.
Best Answer
Yes, this is true but not obvious! It was proved by John Tate (1966) in his paper "Endomorphisms of Abelian Varieties over Finite Fields", p. 139.
You can find the proof also in the book draft here, Corollary 16.25. (One just needs to use that if two elliptic curves over a finite field have the same number of points, then they have the same zeta function; it follows from the rationality and the functional equation of the zeta function: the numerator is just a polynomial $P(t) = 1 + a_1 t + ... + a_{2g} t^{2g}$ of degree $2g$ where $g=1$ being the genus of the curve, and such that $a_{2g-i} = q^{g-i} a_i$ for $1 \leq i \leq g$).