# Torsion points with exactly one singular prime (on elliptic curves)

Let $E$ be an elliptic curve over $\mathbb{Q}$. Let us fix a rational prime $\ell$ (odd, if you want) and an $\ell$-torsion point $P\in E[\ell]$. For a prime $\mathfrak{p}$ of $K=\mathbb{Q}(E[\ell])$, let us say that $P$ has nonsingular reduction at $\mathfrak{p}$, if $\tilde{P}\neq\tilde{O}$ in $\tilde{E}_\mathfrak{p}$ (the reduction of $E$ at $\mathfrak{p}$, so I implicitly assumed $E$ has good reduction at $p=\mathfrak{p}\cap\mathbb{Z}$), otherwise we say that $P$ has singular reduction at $\mathfrak{p}$. Let $S_P$ be the set of primes of singular reduction for $P$. With this setting, my question is

Can we have $S_P=\{\text{primes over }\ell\}$?

• $E[\ell]$ is etale over $\mathbb{Z}$ away from $\ell$ (and primes of bad reduction), so certainly $S_P\cap \{ \text{good primes}\} \subseteq \{\text{primes over }\ell\}$. And you can certainly have equality e.g. if $E$ is supersingular at $\ell$. – Piotr Achinger Mar 6 '17 at 12:23

The status of your "implicit assumption" isn't quite clear -- do you want all primes where $E$ has bad reduction to be in $S_P$, or only a subset of them?
If you take E to be Cremona's 27a1, $y^2 + y = x^3 - 7$, then $E$ has a rational 3-torsion point $(3, -5)$, and for all rational primes $p \ne 3$ of E, the mod $p$ reduction of $P$ is automatically non-trivial. At $p = 3$, the reduction is bad additive and $P$ maps to the singular point of the reduced curve. Does this fit your requirements?
(What's making this work is that the kernel of reduction modulo $p$ cannot contain non-trivial torsion points of prime-to-$p$ order.)
• Thank you very much. So, the confusion arose when I translate the real question I have in mind into using geometry. What I have really wondered is the following: if $P$ is an $\ell$-torsion point. whether its $x$-coordinate, say $x(P)$, has prime factor only $\ell$ or not. But, as is in your example, it is possible. – User0829 Mar 6 '17 at 12:36