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…
[Math] Additive reduction of elliptic curves
arithmetic-geometryelliptic-curvesnt.number-theory
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Part 1 is not true for rank 1 curves. If $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ then every prime trivially splits in $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))$. I will now give an explicit example of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$.
Let $E'$ be the curve with Cremona label 189b2, it is given by $y^2 + y = x^3 - 54x - 88$. One has that $E'(\mathbb{Q})\cong\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$. The free part is generated by $P':=(-6 : 4 : 1)$ and the torsion part is generated by $T':=(12 : -32 : 1)$. Now let $\phi:E' \to E$ be the isogeny whose kernel is generated by $T'$. Then $E$ is the elliptic curve with cremona label 189b3. Now $E(\mathbb{Q}) \cong \mathbb Z$ and with an explicit calculation one can show that $E(\mathbb{Q})$ is generated by $P:=\phi(P')$. So $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ as requested.
A computer search of an isogeny between rank 1 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$ of all elliptic curves up to conductor 1000 gave 225 counter examples. The example above is the one with smallest conductor. Code for performing this search can be found at https://sage.mderickx.nl/home/pub/9
Update on part 1:
I extended the search to rank > 1 curves and also found multiple examples of an isogeny between rank 2 curves $\phi:E' \to E$ such that $\mathbb{Q}(\phi^{-1}(E(\mathbb{Q})))=\mathbb{Q}$. An example is where $E'$ is the elliptic curve defined by $y^2 + xy + y = x^3 + x^2 - 71x - 196$ and the kernel of $\phi$ is generated by $(9 : -5 : 1)$. The Cremona label of $E'$ is '3315b2'. I did not find any examples of rank 3 after searching trough all elliptic curves of conductor < 100000.
ps. Note that my counter examples to part one are not counter examples to Lang-Trotter. The reason is that the Lang-Trotter conjecture is a conjecture about the density of the of the primes such that the reduction is surjective. In my examples both the conjectured density and the actual density are both 0.
Part 3: Let $E$ be a non CM rank 1 elliptic curve that is the only one in it's isogeny class. Then the only isogenies to $E$ are the multiplication by $n$ maps. For concreteness I let $E$ be the curve among all curves with these properties of smallest conductor. This curve $E$ is given by $y^2 + y = x^3 - x$ and $E(\mathbb Q)= \mathbb Z$ is generated by $P=(0 : -1 : 1)$. Now let $p=23$ then $\\#E(\mathbb F_p)=22$ but the order of $P$ after reduction is $11$ so that the index is $2$. This means that the obstruction cannot come from an isogeny because this would mean that it comes from some multiplication by $n$ map and hence that the index should not be squarefree.
In the article of Lang and Trotter where they state their conjecture they give a criterion in terms of $\mathbb Q(l^{-1}E(\mathbb Q))$ that is equivalent to $l$ being a divisor of the index. If you read that obstruction carefully you will realize that its really easy to cook up counter examples to part $3$, in particular using their criterion one can show that the set of $p$ such that reduction mod $p$ is a counter example has positive density for the above $E$.
$y^2=x^3-p$ has reduction type $II$, $y^2=x^3-1/p$ has reduction type $II^*$.
$y^2=x^3-p^2$ has reduction type $IV$, $y^2=x^3-1/p^2$ has reduction type $IV^*$.
Moreover, these examples are universal, in that everything of those fiber types looks like those equations up to linear change of variables and higher-order terms.
I believe this is the source of the names.
Similarly, $y^2=x^3-px$ is $III$, and $y^2=x^3-x/p$ is $III^*$.
This is only for $p\neq 2,3$.
Best Answer
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.