Suppose $X$ is an elliptic curve over $\Bbb Q$ defined by a Weierstrass equation. We may consider this with integer coefficients and then reduce modulo $p$, and if $p$ is nice enough with respect to $X$ we get an elliptic curve which we'll call $X'$ over the field $\Bbb F_p$. Something I've heard/read a few times is that the $n$-torsion of $X(\Bbb Q)$ injects in to the $n$-torsion of $X'(\Bbb F_p)$ as long as $n$ is coprime to $p$.
I'm trying to understand why this is so. In my search, I've found Matt Emerton's answer to a very closely related question. It seems that there's a strategy which proceeds via some facts about finite flat group schemes, but he doesn't say much about it in the linked post (or maybe he does and I'm not getting it). Is there any way to show this claim about the injection of torsion just from knowing that the kernel is a finite flat group scheme? I have an ok background with schemes, but I have not yet spent much serious time with elliptic curves, so I would appreciate answers that focus more on the scheme-theoretic aspects.
Best Answer
Emerton’s answer is, I think, on a less elementary level than that: it’s addressing the case of what $p$-torsion becomes mod $p$. This case is rather simpler: the kernel of the $n$-torsion extends to a smooth finite scheme $G$ over $\mathbb{Z}_{(p)}$. By base changing to the ring of integers of a local field $K$ (with residue field $k$), we know that $G$ is finite smooth free of rank $m$ [in this case $m=n^2$] coprime to $p$ over $O_K$.
Let $C \subset G$ be a closed open subset and let $A=O(C)$ (free $O_K$-algebra of finite rank, with $A \otimes k$ reduced), we want to show that $A$, $A \otimes K$, and $A \otimes k$ have the same idempotents (which will imply the result – as I’ll explain afterwards).
The bijection from the idempotents of $A$ to the idempotents of $A \otimes k$ is well-known (it’s basically Hensel). The inclusion from the idempotents of $A$ to those of $A \otimes K$ is clearly well-defined and injective. But if $e \in A\otimes K$ is idempotent, then for some $a \in O_K$ (a power of the uniformizer, chosen to be minimal – suppose for the sake of contradiction that $a$ is noninvertible) $e’=ae \in A$ and $e’^2=ae’$. But this means that $e’^2$ vanishes in $A \otimes k$ so $e’$ was already divisible in $A$ by the uniformizer, so we could have chosen $a$ with a smaller valuation!
Why is this enough? Well, consider the residue disk $C$ of the unit point in the special fiber of $G$: $G$ is finite as a topological space, so it is the finite reunion of closures of points on the generic fiber, so $C$ is closed. But $G$ is finite as a topological space, so $C$ is constructible. Since $C$ is of course stable under generization, so $C$ is open.
So $C$ is a closed open subscheme of $G$ with its special fiber reduced to a point, so by the above its generic fiber must be connected. But $C$ is smooth over $O_K$ so its generic fiber is the spectrum of a finite reduced, connected $K$-algebra with a $K$-point (the unit), so it must be $K$, ie reduced to a point, so we are done.