Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $E/\mathbb{Q}$ be an elliptic curve and suppose that the trace of Frobenius values are such that $a_{p}(E) \equiv 0 \pmod{2}$ for all odd primes avoiding the conductor. Is it the case that $E$ contains a nontrivial rational two torsion point? Why?

share|cite|improve this question

2 Answers 2

up vote 12 down vote accepted

Yes. Let $P(x) \in {\mathbb Q}[X]$ be the cubic whose roots are the $x$-coordinates of the $2$-torsion points. The hypothesis says that $P$ has a root mod $p$ for all but finitely many primes $p$. If $P$ were irreducible then its Galois group would contain a $3$-cycle, and then there would be infinitely many $p$ such that $P$ remains irreducible mod $p$ (using Čebotarev). Hence $P$ is reducible. Since its degree is only $3$, it follows that $P$ has a rational root, whence $E$ has a rational $2$-torsion point, QED.

share|cite|improve this answer

Alternative method without using the cubic equation (of course not fundamentally different): the Galois representation $\rho: G_{\mathbb Q} \rightarrow GL_2(\mathbb Z/2 \mathbb Z)$ on the points of $2$-torsion of $E$ satisfies tr $\rho(Frob_p)=0$ for every odd prime $p$ by hypothesis, hence tr $\rho=0$ by Chebotarev. (edited:) It has obviously determinant $1$. Hence the semi-simplification of $\rho$ is $1 \oplus 1$ (a semi-simple representation of dim 2 is characterized by its trace and determinant, even in characteristic 2). Therefore $\rho$ fixes a line, and the non-zero point of this line is a rational 2-torsion point of $E$.

share|cite|improve this answer
The fact that $\omega_2 = 1$ is immediate because the determinant is identically $1$ on ${\rm GL}_2({\bf Z}/2{\bf Z})$... – Noam D. Elkies Aug 29 '14 at 2:17
Yes, of course. I edit. – Joël Aug 29 '14 at 4:30
If you try this for $2$ replaced by $\ell$, you get that the trace is $p+1 \mod \ell$ and the determinant is $p$, so the semi-simplification is $1+\chi$ where $\chi$ is the cyclotomic character. But it isn't clear which one is the subobject. Indeed, I think both are possible: I get that the number of $\mathbb{F}_p$ points on $y^2 = x^3-3$ is always divisible by $3$, but there is no $3$-torsion point over $\mathbb{Q}$. – David Speyer Aug 29 '14 at 15:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.