Suppose $E/ \mathbfQ$ is an elliptic curve with additive reduction at a prime $p$. Is there an easy way to tell if $E$ is really a quadratic twist of an elliptic curve $E'/\mathbfQ$ with good reduction at $p$? I've asked a single or two authorities about this,
Microsoft Office Enterprise 2007, with out a fulfilling remedy...
David Hansen
5,
Cheap Windows 7,504●12●30
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,
Microsoft Office 2010 Professional Plus, essentially $4A^3+27B^2$,
Office 2010 Home And Student, by a sixth power,
Microsoft Office 2007 Enterprise,
so for it to have good reduction $v_p(4A^3+27B^2)=6k$
where $k\in\mathbbZ$.
Then the quadratic twist $y^2=x^3+p^-2kAx+p^-3kB$
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 great 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.