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Let $E$ be an elliptic curve over $\mathbb{Q}$: $y^2+y=x^3-x$

I'm trying to find its torsion subgroup $E_{tors}$. Actually, I know that it is trivial. How to prove this? By using change of variable I've got:
$y^2=x^3-x+\frac{1}{4}$ or $y^2=x^3-16x+16$.

Also I calculated discriminant $\Delta=37$

Kartan12
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    Have you heard of the Nagell-Lutz theorem? – Mathmo123 Jun 13 '22 at 13:53
  • Nagell-Lutz theorem wasn't disscussed on leсtures. I went to Wiki and calculated that the only possible torsion points are (0, +-4), (1, +-1), (+-4, +-4), (4, +-4). Am I right? – Kartan12 Jun 13 '22 at 16:52
  • But I think that there exists proof without Nagell-Lutz theorem – Kartan12 Jun 13 '22 at 16:53
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    You could try the strategy here: the $m$-torsion injects into the points mod $p$ for all $m$ relatively prime to $p$. I found that $E(\mathbb{F}_3) \cong \mathbb{Z}/7\mathbb{Z}$ and $E(\mathbb{F}_5) \cong \mathbb{Z}/8\mathbb{Z}$, which is enough to conclude that the torsion is trivial. – Viktor Vaughn Jun 13 '22 at 17:50

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We can apply the strategy used in this post, which uses the following proposition (Proposition VII.3.1(b) of Silverman's The Arithmetic of Elliptic Curves.) Here $K$ is a local field with residue field $k$.

Proposition. Let $E/K$ be an elliptic curve and let $m \geq 1$ be an integer that is relatively prime to $\operatorname{char}(k)$. Assume further that the reduced curve $\newcommand{\Et}{\widetilde{E}} \newcommand{\Q}{\mathbb{Q}} \Et/k$ is nonsingular. Then the reduction map $$ E(K)[m] \to \Et(k) $$ is injective.

As mentioned in the linked post, then we have injections $$ E(\Q)[m]\hookrightarrow E(\mathbb{Q}_p)[m]\hookrightarrow \Et(\mathbb{F}_p) $$ for all primes $p \nmid \Delta$ and all $m$ relatively prime to $p$. We apply this for multiple of choices of $p$ in order to show that $E(\Q)_{\text{tors}}$ must be trivial. You've already computed that $\Delta = 37$, so we are free to choose any prime $p \neq 37$.

Taking $p = 2$, we find that $\newcommand{\F}{\mathbb{F}} \Et(\F_2)$ consists of the $5$ points $$ (0 : 1 : 0), (0 : 0 : 1), (0 : 1 : 1), (1 : 0 : 1), (1 : 1 : 1) \, . $$ This rules out all possible orders for torsion points except for $1, 2,$ and $5$. Now taking $p=3$, we find that $\newcommand{\F}{\mathbb{F}} \Et(\F_3)$ consists of the $7$ points $$ (0 : 1 : 0), (0 : 0 : 1), (0 : 2 : 1), (1 : 0 : 1), (1 : 2 : 1), (2 : 0 : 1), (2 : 2 : 1) \,, $$ which rules out all possible orders for torsion points except for $1, 3,$ and $7$. Thus we see that the only possible order is $1$, meaning that $E(\Q)_{\text{tors}} = \{(0:1:0)\}$.

Viktor Vaughn
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  • Thanks a lot! Why in the case of p=2 we don't rule out order 2, and p=3 the same with order 3? – Kartan12 Jun 13 '22 at 21:07
  • The proposition only tells us that the map is injective when $m$ is relatively prime to $p$. So if we just consider $p=2$, we don't know if $E(\mathbb{Q}_2)[2] \to \widetilde{E}(\mathbb{F}_2)$ is injective, so it's possible that there is a nontrivial $2$-torsion point that maps to the identity under this map. – Viktor Vaughn Jun 13 '22 at 21:13