I am looking for a full proof (references containing a full proof are more than welcome as well) of the following fact:
If $E$, $E'$ are $\mathbb Q$-isogenous elliptic curves over $\mathbb Q$, are their reductions mod $p$ $\mathbb F_p$-isogenous if $p$ is a prime of good reduction for the curves? The answer here was not quite complete; it did not explain why the reduced Tate modules were isomorphic. In addition, the last comment in that post says that one can ignore Tate modules and just reduce the isogeny mod $p$ to get the result, but I do not see why this is true.
Best Answer
Let $f : E \to E'$ be an isogeny over $\mathbb{Q}$ and let $p$ be a prime of good reduction for $E, E'$.
One way to see that the reductions (of global minimal integral Weierstrass models) $\tilde{E}, \tilde{E'}$ are isogenous over $\mathbb F_p$ is to use Tate modules (which would allow to treat the case of general abelian varieties over number fields, using Tate's isogeny theorem). The linked answer explains that for $\ell \neq p$, the map $V_{\ell}(f) : V_{\ell}(E) \to V_{\ell}(E')$ is a isomorphism of $G_{\mathbb{Q}}$-representations, with inverse $\frac{1}{\deg(f)} V_{\ell}(f^\vee)$.
If we denote by $r : E \to \tilde{E}$ the reduction mod $p$ map, then statement VII.3.1.b) in Silverman's book "Arithmetic of Elliptic curves" implies that the reduction map induces isomorphisms $E[m] \cong \tilde E[m]$ for all $m$ coprime to $p$, and thus we get an isomorphism $V_{\ell} r : V_{\ell} E \to V_{\ell}(\tilde E)$, and the same holds for $V_{\ell} r' : V_{\ell} E' \to V_{\ell}(\tilde E')$.
We may thus define an map $\overline{V_{\ell} f} : V_{\ell}(\tilde E) \to V_{\ell}(\tilde E')$ so that the following diagram commutes:
Because all other 3 maps are isomorphisms, we see that $\overline{V_{\ell} f}$ is an isomorphism at least of $\mathbb Z_\ell$-modules. To conclude we need to see that $\overline{V_{\ell} f}$ is an isomorphism at least of $G_{\mathbb{F}_p}$-representations. If we denote by $D_p \cong \mathrm{Gal}(\overline{\mathbb{Q}_p} / \mathbb{Q}_p) \hookrightarrow G_{\mathbb{Q}}$ the decomposition subgroup at $p$, we have a surjective morphism $\pi : D_p \twoheadrightarrow G_{\mathbb{F}_p}$.
Then the Galois-equivariance of $\overline{V_{\ell} f}$ follows from the one of $V_{\ell} f$ and $V_{\ell} r, V_{\ell} r'$. Namely, we have $$ V_{\ell} r ( \sigma \cdot t ) = \pi(\sigma) \cdot V_{\ell} r(t) $$ whenever $\sigma \in D_p, t \in V_{\ell} E$. This is because on the level of $P =(x,y) \in E[m] \subset E(\overline{\mathbb{Q}})$, we have $$r(\sigma \cdot (x,y)) = \pi(\sigma) \cdot r(x,y)$$ since $\pi(\sigma) : \bar{x} \mapsto \overline{\sigma(x)}$ is well-defined ($D_p$ is the stabilizer of a prime above $p$ in $\mathcal{O}_{\overline{ \mathbb{Q} }}$).
Note that if $f$ is defined between the minimal integral Weierstrass models of $E,E'$ using a ratio of polynomials having coefficients in $\Bbb Z \setminus p \Bbb Z$, then there is a well-defined reduction $\tilde f : \tilde E \to \tilde E'$ which is a morphism of elliptic curves, and cannot be zero (otherwise $f$ would map all points of $E$ to points having some $p$ in the denominator in the $x,y$-coordinates), hence is an isogeny — so no need of Tate's theorem here in that case.