So we know many useful theorems that help characterize torsion points on elliptic curves over $\mathbb{Q}$ such as the Nagell–Lutz theorem which provides a useful way to find torsion points on $E/\mathbb{Q}$ and Mazur's theorem which characterizes the torsion subgroup of $E(\mathbb{Q})$.

However, does there exist any elliptic curve over $\mathbb{Q}$ with no torsion points other than the point at infinity? Are there any examples of such curves?


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    $\begingroup$ Curves without any non-trivial torsion have density one in the set of all elliptic curves over $\mathbb{Q}$. You can find many examples of such curves in Cremona's database. $\endgroup$
    – Zev
    Apr 25, 2012 at 6:33

2 Answers 2


Probably the set of $E/\mathbb{Q}$ with $E(\mathbb{Q})=\{O\}$ has density 1/2 (in an suitable way of ordering curves). A recent (and very deep) result of Manjul Bhargava and Arul Shankar proves that this set has positive density, so there are provably a lot of curves with the property that you request. There's a nice overview of the Bhargava-Shankar result, with some of the history, by Bjorn Poonen. It's freely available at


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    $\begingroup$ But somehow Bhargava-Shankar isn't really relevant to this question, right? He is only asking about curves with no non-trivial torsion, which is much easier. $\endgroup$
    – JSE
    Apr 28, 2012 at 3:27
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    $\begingroup$ Ah, I always read these things too fast. I thought he was asking for a curve that had no rational points at all, other than the point at infinity. So after ruling out nontrivial torsion, you still need to get rank 0. This alternative reading is an interesting question. Simply proving (say) that there are infinitely many curves with $E(\mathbb{Q})_{tors}=O$ is an elementary exercise via congruence conditions, as Landsburg noted. Or one can be fancier, e.g., if $E$ has no rational 2-torsion, then all but finitely many quadratic twists have no rational torsion. $\endgroup$ Apr 28, 2012 at 12:51
  • $\begingroup$ Dear Joe: Out of curiosity, can you prove your last statement without appealing to Mazur? I like the result, but at the same time it seems too simple a statement for me to require something so deep as Mazur's theorem... $\endgroup$
    – R.P.
    Nov 23, 2012 at 0:38
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    $\begingroup$ @ René: No, one doesn't need Mazur. One can prove it via a height argument, and probably also via an algebraic argument. The former is in a paper of mine, the latter in a paper of Setzer. For details, see my answer to the MO question mathoverflow.net/questions/76413. $\endgroup$ Nov 24, 2012 at 18:26

The curve $y^2+y=x^3-x$ has order 5 mod 2 and order 7 mod 3. Therefore the torsion injects into both $Z/5Z$ and $Z/7Z$ and hence is trivial. This is the first example in Husemoller's book on Elliptic Curves.

As Zev notes in his comment, this is the tip of a very large iceberg.


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