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 $\tilde{E}(\mathbb{F}_p)$)?

  • $\begingroup$ I don't see why all of the torsion shouldn't inject for sufficiently large $p$ by Nagell-Lutz. $\endgroup$ – Qiaochu Yuan May 8 '11 at 21:33
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin May 8 '11 at 21:40
  • $\begingroup$ @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. $\endgroup$ – Zev Chonoles May 8 '11 at 21:48
  • $\begingroup$ 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. $\endgroup$ – Arturo Magidin May 8 '11 at 22:01
  • $\begingroup$ @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 $a_3\neq0$. $\endgroup$ – 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 $\mathbb Q$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.