2
$\begingroup$

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?

$\endgroup$
  • 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
0
$\begingroup$

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.

$\endgroup$

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.