Consider an elliptic curve in the short Weierstrass form $$ y^2 = x^3 + bx + c, $$ defined over rational numbers ($b,c$ are integers). My goal is to provide an example of congruence relations on $b$ and $c$ which will provide a trivial torsion subgroup $T(E(\mathbb{Q}))$.

We know that, for example, by Lutz-Nagell that by considering square divisors of the determinant $\Delta$ we can find possible points of finite order. However, this does not give any congruence relations on $b$ and $c$.

Another idea is to use the reduction modulo $p$, where $p$ is a prime which does not divide $2\Delta$. Then we know that $|T(E(\mathbb{Q}))|$ divides $|E'(F_p)|$, where $E'(F_p)$ is reduced modulo $p$ curve over a field of $p$ elements. This seems to be more helpful, but I still have no idea how to find such relations. Could you provide any hints, please?

  • 1
    $\begingroup$ What if you reduce modulo 2 different primes $p$? Try to get the two $|E'(\mathbb F_p)|$ to be relatively prime. Find an elliptic curve with trivial torsion, and then find a set of primes that guarantees it to have trivial torsion. $\endgroup$ – Holden Lee Mar 24 '14 at 21:45
  • $\begingroup$ @HoldenLee Yes, I'm actually trying to do this. However, I don't know what to say about |E'(F_p)|, since we only know "approximate" number of points over $F_p$. $\endgroup$ – Juloko Mar 24 '14 at 21:48
  • $\begingroup$ You can always count the number of points over $\mathbb F_p$: just substitute in all the values of $x$ modulo $p$ and see whether there are solutions for $y$. (It's good to learn how to use a computer to do this.) It may take some trial and error to find a curve with no torsion, but you can always look up an example in a table: homepages.warwick.ac.uk/staff/J.E.Cremona/ftp/data/INDEX.html. $\endgroup$ – Holden Lee Mar 24 '14 at 22:09
  • $\begingroup$ @HoldenLee But I want to obtain some relations "in general", without knowing exactly what are $b$ and $c$. I.e. I want obtain some congruent conditions on integers $b$ and $c$ for which $T(E(\mathbb{Q}))$ is wittingly trivial. $\endgroup$ – Juloko Mar 24 '14 at 22:24
  • $\begingroup$ @HoldenLee Such relations must not cover all the cases when $T(E(\mathbb{Q}))$ is trivial. I just want to find an example such conditions where we can say this for sure. $\endgroup$ – Juloko Mar 24 '14 at 22:43

Holden Lee had the right idea in the comments. Here is how it can be implemented. Let $E/\mathbb{Q}:y^2=x^3+x+1$. The discriminant is $-2^4\cdot 31$, so it only has bad reduction at $2$ and $31$. Moreover,

  • When $p=5$, the curve $E/\mathbb{F}_5$ has $9$ points.

  • When $p=7$, the curve $E/\mathbb{F}_7$ has $5$ points.

Now let $E_{A,B}/\mathbb{Q}: y^2=x^3+Ax+B$ be any curve with $A,B\in\mathbb{Z}$ that satisfies $$A\equiv B\equiv 1 \bmod 35.$$ In particular, $A\equiv B\equiv 1 \bmod 5$ and $\bmod 7$. Several remarks:

  1. $E_{A,B}$ is an elliptic curve. For this we need to check that $\Delta=-16(4A^3+27B^2)\neq 0$. It suffices to realize that $-16(4A^3+27B^2)\equiv -16(4+27)\equiv -16\cdot 31\not\equiv 0 \bmod 5$ (or $\bmod 7$). In particular, $\Delta\neq 0$. Thus, $E_{A,B}$ is smooth.

  2. It also follows from our previous calculation that $\Delta\not\equiv 0 \bmod 5$ or $\bmod 7$. Hence, $E_{A,B}$ has good reduction at $5$ and $7$.

  3. Since $E_{A,B}\equiv E \bmod 5$ and $\bmod 7$, it follows that $E_{A,B}(\mathbb{F}_5)$ and $E(\mathbb{F}_5)$ have the same cardinality (equal to $9$), and so do $E_{A,B}(\mathbb{F}_7)$ and $E(\mathbb{F}_7)$ (equal to $5$).

  4. Thus, the prime-to-$5$ subgroup of $E_{A,B}(\mathbb{Q})_\text{tors}$ embeds into $E(\mathbb{F}_7)$, which has order $5$. This implies that if there is rational torsion, it must have order $5$. But the prime-to-$7$ part of $E_{A,B}(\mathbb{Q})_\text{tors}$ embeds into $E(\mathbb{F}_5)$, which has order $9$, so there is no $5$ torsion either. Hence, $E_{A,B}(\mathbb{Q})_\text{tors}$ is trivial.


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