Suppose $E/ \mathbf{Q}$ is an elliptic curve with additive reduction at a prime $p$. Is there an easy way to tell if $E$ is a quadratic twist of an elliptic curve $E'/\mathbf{Q}$ with good reduction at $p$? I have asked one or two experts about this, without a satisfying answer...
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In other words, given an elliptic curve $E/\mathbb{Q}_p$ with additive reduction, you wish to know whether there is a quadratic extension $F/\mathbb{Q}_p$ such that $E/F$ has good reduction. Of course the $j$-invariant must be integral. Let $\Phi_p$ be the Serre-Tate group which is the Galois group of the minimal extension of $\mathbb{Q}_p^{unr}$ such that $E$ acquires good reduction. In Serre's paper "Propriétés galoisiennes des points d'ordre fini des courbes elliptiques", page 312, there is a table of what $\Phi_p$ is. You wish to know when $\Phi_p$ is cyclic of order 2., If $p\neq 2,3$ then there exists a quadratic extension $F$ that makes $E$ having good reduction if and only if the Kodaira type is $I_0^{*}$, i.e. if and only if the order of $v_p(\Delta)$ in $\mathbb{Z}/12\mathbb{Z}$ is 2. (as in Robin Chapman's answer). If the equation is minimal, this is equivalent to $v_p(\Delta) = 6$. If $p=2$ or 3, then it is still true that we must have $v_p(\Delta)\cdot 2\equiv 0\pmod{12}$ to guarantee that you can twist to good reduction. This excludes certain Kodaira types. To find the complete answer one would have to analyse this better. |
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Another approach is the following: let $f$ be the newform giving rise to $E$, form the twist $f_{\chi}$ where $\chi$ is the quadratic char. of conductor $p$, and now find the conductor of $f_{\chi}$ and see whether $p$ divides it or not. |
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If $p\ge5$ then $E$ has equation $y^2=x^3+Ax+B$ with $p\mid A$ and $p\mid B$. A quadratic twist alters the discriminant, essentially $4A^3+27B^2$, by a sixth power, so for it to have good reduction $v_p(4A^3+27B^2)=6k$ where $k\in\mathbb{Z}$. Then the quadratic twist $y^2=x^3+p^{-2k}Ax+p^{-3k}B$ will work as long as $v_p(A)\ge 2k$ and $v_p(B)\ge 3k$. Otherwise any quadratic twist making the discriminant a $p$-unit will have coefficients which are non $p$-integral so no quadratic twist will have good reduction. The cases $p=3$ or $p=2$ will be harder :-) ADDED Even in these awkward characteristics the same argument shows that $v_p(4A^3+27B^2)$ being a multiple of $6$ is a necessary condition. |
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