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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asked Apr 7, 2010 at 14:42
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3$\begingroup$ What kind of an answer are you looking for? If you know the equation and are near a computer then you can write down all the quadratic twists of the curve at p and then run Tate's algorithm on them (or simply compute their conductor at p) to see which if any have good reduction. That for me is completely "satisfying" because it works in practice, and if it's not satisfying for you then perhaps you could clarify what would satisfy you :-) $\endgroup$– Kevin BuzzardCommented Apr 7, 2010 at 17:43
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1$\begingroup$ Certainly you should check to see if the $j$ invariant is $p$-integral. That's a necessary condition. $\endgroup$– Victor MillerCommented Apr 8, 2010 at 0:01
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3 Answers
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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.
answered Apr 7, 2010 at 15:09
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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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$\begingroup$ If p is 2 one might have to try more than one quadratic twist, right? $\endgroup$– JSECommented Apr 7, 2010 at 18:19
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$\begingroup$ c.f. my "twists" in my comment ;-) $\endgroup$ Commented Apr 7, 2010 at 18:21
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$\begingroup$ Dear Jordan, I think I was implicitly taking $p$ to be odd! When $p = 2$, as you know, the conductor of the quadratic character can be 4 or 8, and there are 3 twists to make altogether. $\endgroup$– EmertonCommented Apr 7, 2010 at 18:42
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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.
answered Apr 8, 2010 at 12:22