Equivalent notions of congruence for elliptic curves over $\mathbb{Q}$

Question

Let $E_1$ and $E_2$ be elliptic curves over $\mathbb{Q}$ and $f_i$ the eigencuspform of weight $2$ attached to $E_i$. Express $f_1=\sum a_i q^i$ and $f_2=\sum b_i q^i$.

Suppose that the residual Galois representations $E_1[p]\simeq E_2[p]$. Note that $E_1$ and $E_2$ have bad reduction at the same primes.

It is easy to see that $a_l\equiv b_l\mod{p}$ at a prime $l$ of good reduction for both elliptic curves (since the residual representations are unramified at $l$). Does it follow that $a_l\equiv b_l\mod{p}$ at a prime $l$ of bad reduction for both elliptic curves. If so how do we show this? Perhaps this comes down to showing that $E_1$ has the same kind of bad reduction at $p$ as $E_2$. I know that if $E_1$ has additive reduction then so does $E_2$, but I'm not sure about distinguishing between split multiplicative and non-split multiplicative reduction.

Answer 1

I found myself wondering the same thing a couple of weeks ago. Even with the restriction that $p > 2$ and the residual representation is irreducible, it does not follow that $E_{1}$ and $E_{2}$ have the same primes of bad reduction, nor that $a_{\ell}(E_{1}) \equiv a_{\ell}(E_{2}) \pmod{p}$ for primes of bad reduction.

For example, take $p = 5$ and $E_{1} : y^{2} + xy + y = x^{3} + 4x - 6$ (aka $X_{0}(14)$). The mod $5$ Galois representation attached to $E_{1}$ is surjective. For any elliptic curve $E$, Rubin and Silverberg write down an isomorphism $X_{E}(5) \cong \mathbb{P}^{1}$, the modular curve parametrizing elliptic curves $F$ so that $F[5]$ and $E[5]$ are isomorphic via and isomorphism respecting the Weil pairing. For $E = E_{1}$, one can take $E_{2} : y^{2} + xy = x^{3} - x^{2} - 4492x + 126416$. This curve has $E_{1}[5] \cong E_{2}[5]$, but $E_{2}$ has bad reduction at $5$ and $E_{1}$ has good reduction at $5$. It is not surprising that one can fail to have $a_{\ell}(E_{1}) \equiv a_{\ell}(E_{2}) \pmod{p}$ when one of $E_{1}$ and $E_{2}$ has good reduction at $\ell$ and the other doesn't.

One can also have $a_{\ell}(E_{1}) \not\equiv a_{\ell}(E_{2}) \pmod{p}$ when both curves have bad reduction at $\ell$. Looking in the same family (of curves directly $5$-congruent to $E_{1}$) one can find curves $E_{2}$ and $E_{3}$ that have multiplicative reduction at $199$ where one curve has $a_{199}(E_{2}) = -1$ and the other has $a_{199}(E_{3}) = 1$. The examples I found are a bit horrendous: $$ E_{2} : y^2 + xy = x^3 - x^2 - 183192520591565859491828595856323163822228663229886627562x + 1062123179152064591727007023640066384957014639073044998393946451273297951607053468340$$ and $$ E_{3} : y^2 + xy + y = x^3 - 187739104428369548177010660620161617529156118721103898180850571x + 1108087782673654453656602976287807709340970590450649179565293267500311360376488672021879300982. $$

Answer 2

Note that a conjecture attributed to Frey and Mazur says that for each elliptic curve $E_1$, there exists a constant $C$ depending on $E_1$, such that for every prime $p>C$ and for every elliptic curve $E_2$ defined over $\mathbb{Q}$ with $E_1[p]\simeq E_2[p]$ as Galois modules, one has $E_1$ isogenous to $E_2$. In particular, $E_1$ and $E_2$ have the same $L$-functions and hence the same coefficients $a_\ell$ (including at the bad primes).

It is even conjectured that $C$ can be taken independently of $E_1$. Moreover there is no known pair of non-isogenous elliptic curves $(E_1,E_2)$ such that $E_1[p]\simeq E_2[p]$ for some prime $p>17$. If you want to learn more about this conjecture you can have a look at the recent preprint by Cremona and Freitas "Global methods for the symplectic type of congruences between elliptic curves": https://arxiv.org/abs/1910.12290