Split multiplicative reduction question

Question

Let $E/\mathbb{Q}$ be an elliptic curve and $E_{d}$ be the quadratic twist of $E$ by a squarefree integer $d$. Let $\ell$ be a prime of multiplicative reduction for $E$. If $(d, \ell) = 1$, then $\ell$ is a prime of multiplicative reduction for $E_{d}$. I read that $\ell$ is of split multiplicative reduction for $E$ if and only if $\ell$ is of split multiplicative reduction for $E_{d}$. Are there any sources for this?

Answer 1

The statement is false.

An elliptic curve $E$ with multiplicative reduction at $\ell$ has split reduction if and only if $-c_6$ is a square in $\mathbb{F}_\ell$. Now suppose that $\gcd(d,\ell)=1$, so that $E'=E_d$, the quadratic twist by $d$, also has multiplicative reduction. The corresponding coefficient for $E'=E_d$ is given by $c_6'= d^3 c_6$. Hence, $E'$ has split reduction if and only if $-c_6' = -d^3c_6$ is a square in $\mathbb{F}_\ell$. In particular, if $d$ is not a square mod $\ell$, then $d^3$ is not a square either, and $-c_6'$ is a square if and only if $-c_6$ is not a square.

For instance, let $E:y^2=x^3-x^2+35$. Then $E$ has split multiplicative reduction at $p=5$. Now:

• Let $d=11$. Since $11\equiv 1 \bmod 5$ is a square, the quadratic twist by $11$ also has split multiplicative reduction at $5$.

• Let $d=13$. Since $13\equiv 3 \bmod 5$ is not a square, the quadratic twist by $13$ has non-split multiplicative reduction at $5$.