Coefficients of an elliptic curve for which the torsion group is trivial

Question

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?

Answer 1

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.