primes of multiplicative reduction for elliptic curves with rational $\ell$-torsion

Question

Very recently, I made an observation from scanning lists of elliptic curves on the LMFDB that leads me to the following (unproven) statement:

Fix $\ell \in \{5, 7\}$. Let $E$ be an elliptic curve over $\mathbb{Q}$ with a rational $\ell$-torsion generator. Then if $p \neq \ell$ is a prime at which $E$ has multiplicative reduction, and if $p$ appears in the prime factorization of the denominator of $j(E)$ with exponent not divisible by $\ell$, we have $p \equiv \pm 1$ (mod $\ell$).

I don't know how to get a spreadsheet of $j$-invariants of elliptic curves satisfying the hypothesis given above so I couldn't check this efficiently, but this was the case for each of the dozens of elliptic curves I checked. I noticed other strong trends as well, such as the fact that primes $p \neq \ell$ appearing in the factorization of the \textit{numerator} of $j(E)$ with exponent not a multiple of $\ell$ tend to satisfy $p \equiv \pm 1$ (mod $\ell$), but this is not true all of the time -- the exceptions are always smaller primes and always appear with exponent $3$ in the data I checked. Also, such a statement fails badly when I loosen the assumption about a rational $\ell$-torsion point to requiring only that the mod-$\ell$ Galois image be a Borel subgroup.

Is there any obvious proof or reason to expect that the main assertion above is true? Right now I don't see any way to prove it using rational points on modular curves, but I'm really not an expert on modular curves. No strategy using Neron models is jumping out at me either.

Answer 1

Let $E/\mathbb{Q}$ be an elliptic curve with a rational $\ell$-torsion point $P$ for $\ell\in\{5,7\}$.

Let $p$ be a prime different from $\ell$. Then the formal group $\hat{E}(p\mathbb{Z}_p)$ has no $\ell$-torsion, which shows that the reduction of $P$ is not the point at infinity.

Suppose $E$ has bad reduction at $p$. I claim that it follows that the reduction is multiplicative. This is because the group of connected components for additive reduction has order coprime to $\ell$ and hence $P$ would have to be a non-singular point. But then the reduction would have order $p$.

Now if $E$ has multiplicative reduction, there are two cases. Either the point $P$ reduces to the singular point or to a non-singular non-zero point.

In the frist case the group of components must now be a cyclic group of order divisible by $\ell$. Hence the reduction is split multiplicative and the Tamagawa number and the $p$-adic valuation of $j(E)$ are both divisible by $\ell$.

In the second case (that is what you observe), the point $P$ is a non-zero point in the group of non-singular points in $E(\mathbb{F}_p)$. This group is either cyclic of order $p-1$ or cyclic of order $p+1$ depending on whether the reduction is split or non-split. In the first case $\ell\mid p-1$ and in the second $\ell\mid p+1$.

Answer 2

Never mind -- I think I can answer this. The answer occurred to me as soon as I posted the question; I guess sometimes the process of writing it out for other people helps trigger the realization that there is a pretty straightforward answer.

Suppose that $E$ has multiplicative reduction at $p$ and $p$ appears in the factorization of the denominator of $j(E)$ with exponent $e$ with $(e, \ell) = 1$. If $E$ has non-split multiplicative reduction at $p$, we extend the base field $\mathbb{Q}$ to an appropriate quadratic extension, so we may assume that $E$ is defined over a number field $K$ and has split multiplicative reduction at a prime $\mathfrak{p}$ whose residue field $k$ has order either $p$ or $p^2$. Now the special fiber of the Neron model of $E$ at the prime $\mathfrak{p}$ has cyclic component group of order $e$ with the identity component being a copy of $\mathbb{G}_m / k$, so that the order of the special fiber as an algebraic group is $e(p-1)$ or $e(p^2-1)$. Since $\ell \neq p$, the hypothesized rational point of order $\ell$ reduces to a $k$-point of order $\ell$ in the special fiber of the Neron model, so we get $\ell | e(p-1)$ or $\ell | e(p^2-1)$. Since by hypothesis $\ell \not| e$, this implies that $p \equiv \pm 1$ (mod $\ell$).

This doesn't seem to explain the trend towards primes in the numerator of $j(E)$ being $\equiv \pm 1$ (mod $\ell$), but it answers my main question.