# Injection of $E(\mathbb{Q})_{\text{tors}}$ into $\tilde{E}(\mathbb{F}_p)$?

I'm looking at Example VII.3.3.3 (p.193, 2nd ed.) of Silverman's The Arithmetic of Elliptic Curves. We have the elliptic curve $E:y^2=x^3+x$, with discriminant $\Delta=-64$, so there is good reduction for all primes $p\geq 3$. It is noted that $(0,0)$ is a point of order two in $E(\mathbb{Q})$, and that $$\tilde{E}(\mathbb{F}_3)=\{\mathcal{O},(0,0),(2,1),(2,2)\}\cong\mathbb{Z}/4\mathbb{Z}$$ $$\tilde{E}(\mathbb{F}_5)=\{\mathcal{O},(0,0),(2,0),(3,0)\}\cong(\mathbb{Z}/2\mathbb{Z})^2$$ Then it is said that

Since $E(\mathbb{Q})_{\text{tors}}$ injects into both of these groups, we see that $(0,0)$ is the only nonzero torsion point in $E(\mathbb{Q})$.

Now, my understanding of Proposition VII.3.1b (p.192, 2nd ed.) is that for any discretely valued local field $K$ with residue field $k$, the reduction map from $E(K)[m]$ to $\tilde{E}(k)$ is injective for all $\gcd(m,\text{char}(k))=1$, where $E(K)[m]$ denotes the $m$-torsion subgroup of $E(K)$. So, we are looking at the compositions $$E(\mathbb{Q})[m]\hookrightarrow E(\mathbb{Q}_3)[m]\hookrightarrow \tilde{E}(\mathbb{F}_3)\quad\text{ for all }3\nmid m$$ $$E(\mathbb{Q})[n]\hookrightarrow E(\mathbb{Q}_5)[n]\hookrightarrow \tilde{E}(\mathbb{F}_5)\quad\text{ for all }5\nmid n$$ It seems the best we can say (I think) is that the $3$-torsion-free part of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_3)$, and the $5$-torsion-free part of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_5)$. This still clearly implies that $E(\mathbb{Q})_{\text{tors}}$ must be of order 2 in this particular case, because $\tilde{E}(\mathbb{F}_5)$ is $3$-torsion-free and $\tilde{E}(\mathbb{F}_3)$ is $5$-torsion-free; but it seems to me that the reasoning in the block-quoted statement (at least without further explanation) is technically wrong.

If $p$ is a prime of good reduction for $E$, then is it true that all of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_p)$, or just that $E(\mathbb{Q})_{\text{tors}}[m]$ injects into $\tilde{E}(\mathbb{F}_p)$ for any $m$ relatively prime to $p$ (and hence the $p$-torsion-free part of $E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_p)$)?

• I don't see why all of the torsion shouldn't inject for sufficiently large $p$ by Nagell-Lutz. – Qiaochu Yuan May 8 '11 at 21:33
• I don't think it's a priori: if I read this correctly, the reduction modulo $3$ tells you there can be no 5-torsion, since $\gcd(5,\mathrm{char}(F_3))=1$; the reduction modulo $5$ tells you there can be no 3-torsion for the same reason. Put together, thess tells you that each of the reductions is actually an embedding of the entire torsion group, as there is no 5-torsion and there is no 3-torsion. – Arturo Magidin May 8 '11 at 21:40
• @Arturo: Thanks, this is precisely what I was wondering about. It wasn't clear to me if there were some minor extra steps here that were being elided, or if "$E(\mathbb{Q})_{\text{tors}}$ injects into $\tilde{E}(\mathbb{F}_p)$ for $p$ of good reduction" was some result I couldn't find. You are welcome to add this as an answer. – Zev Chonoles May 8 '11 at 21:48
• There could be something else (I don't have the book in front of me; it's in my office, I'm at home), I don't know. But that's how I would convince myself that it works. Generally, if you know things mod $p$ work except perhaps for the $p$-parts, and you know things mod $q$ work except perhaps for the $q$ parts, and mod $p$ shows no $q$ parts and mod $q$ shows no $p$-parts, then you know it works, period. – Arturo Magidin May 8 '11 at 22:01
• @Qiaochu: Ah, I think I've got your argument: Nagell-Lutz implies $E(\mathbb{Q})_{\text{tors}}$ is finite, hence for sufficiently large $p$ there is no $p$-torsion, hence the $p$-torsion-free part of $E(\mathbb{Q})_{\text{tors}}$ is all of $E(\mathbb{Q})_{\text{tors}}$ for sufficiently large $p$, etc. But I meant my question to be about arbitrary elliptic curves (which I now realize wasn't terribly clear); and I am under the (perhaps mistaken) impression that the statement of Nagell-Lutz is false when $E$ is presented as $$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$$ with $a_1\neq 0$ or $a_3\neq0$. – Zev Chonoles May 8 '11 at 22:07

I don't know if there is something else, but just from looking at what you quote, I would argue that the reduction modulo $3$ tells you that there can be no $5$-torsion (in fact, that there is only $2$-torsion and perhaps some $3$-torsion). Then the reduction modulo $5$ tells that there is some $2$-torsion, and perhaps some $5$ torsion. Putting the two together, once concludes that there is neither any $3$-torsion, nor any $5$-torsion, so that the reduction maps actually provide embeddings of the torsion group, and the argument proceeds from there.
Basically, if reductions modulo $p$ and modulo $q$ reveal only $\ell$-torsion, with $p$, $q$, and $\ell$ pairwise distinct primes, then you know that there can be only $\ell$-torsion and that the two reduction maps are in fact embeddings.
Let $K$ be a discretely valued field whose residue field is perfect of characteristic $p$, and let $E$ be an elliptic curve over $K$ with good reduction. If one supposes further that the absolute ramification index $e$ of $K$ is $< p-1$, then it follows that the reduction map is injective on the $K$-rational $p$-power torsion points of $E$. As you already noted, the reduction map is automatically injective on the prime-to-$p$ torsion, and so we conclude that (when $e < p-1$) the reduction map is injective on all the $K$-rational torsion.
There are various ways to see the claimed injectivity. One is via a consideration of the formal group, and another is via the theory of finite flat group schemes. The rough idea is that any non-trivial $p$-torsion point which reduces to the identity (and hence lies in the points of the formal group) is the solution to an Eisenstein polynomial of degree divisible by $p-1$. If $e < p-1$ then such a polynomial can have no roots in $K$, and so the non-trivial $K$-rational $p$-torsion points cannot lie in the formal group.
If one considers the particular case when $K = \mathbb Q$ or $\mathbb Q_p$ equipped with the $p$-adic valuation, where $p$ is an odd prime, then $e = 1 < p-1$, and this result applies. This is what Silverman is using.
I think this is discussed somewhere in Silverman, although I forget whether it is treated in the general form described above, or just in the particular case of odd primes $p$ in $\mathbb Q$.