# Elliptic curves with trace of Frobenius values always congruent to 0 modulo 2

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?

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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.
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 $E$.
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
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