Notation: Let $K/\mathbb{Q}$ be a quadratic number field; let $p\geq 3$ be a rational prime and let $\mathfrak{p}$ denote a prime lying above $p$; let $K_{\mathfrak{p}}$ denote the completion of $K$ with respect to the prime $\mathfrak{p}$; let $R$ denote the ring of integers of $K_{\mathfrak{p}}$.

Let $E/\mathbb{Q}$ be an elliptic curve define over $\mathbb{Q}$, however, we want to consider $E$ as an elliptic curve over $K$. Suppose that $v_{\mathfrak{p}}(j(E)) \geq 0$, which is equivalent to $j(E) \in R$. From Proposition VII.5.5 of Silverman's AEC, we know that $E$ has potential good reduction at $\mathfrak{p}$, meaning that there exists a finite extension $K'/K$ such that $E$ has good reduction modulo all primes $\mathfrak{p}_{K'}|\mathfrak{p}$ over $K'$.

Question: Can we pick $K'/K$ to be a normal extension such that $\gcd(p,[K':K]) = 1$?

Assumptions: I know that $E/\mathbb{Q}$ does not have any $\mathbb{Q}$-rational torsion of prime order, however, over the quadratic extension $K$, $E$ does gain a $K$-rational point of order 3,5 or 7. Also, one may assume that $K$ has class number 1, if that helps.

Example: In a paper of Frey, he constructs such extensions for elliptic curves defined over $\mathbb{Q}$ that have a $\mathbb{Q}$-rational point $P$ of order $p$ for $p\in \{5,7\}$, where $P$ is not contained in the kernel of reduction modulo $p$. In particular, he writes if $v_p(j(E)) \geq 0$, then $E$ has good reduction modulo all primes $\mathfrak{p}_N | p$ where $$N = \mathbb{Q}(\zeta_{12},\sqrt[12]{p}).$$ The case of $p = 3$ is dealt with separately.

Intial Idea: From the proof of Proposition VII.5.5, one can take $K'$ to be the extension of $K$ over which $E$ has a Weierstrass equation in Legendre form. From Proposition III.1.7 of Silverman's AEC, we have a procedure for writing $E$ in Legendre form using the defining cubic equation, however, it is not immediately clear to me how to use this construction to determine the degree of $K'$ over $K$.

Thank you in advance for you time!

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    $\begingroup$ Every abelian variety $A/K$ acquires stable reduction over the normal extension $K(A[12])/K$. This is an extension of bounded degree, when $\dim{A}$ is fixed. $\endgroup$ – Vesselin Dimitrov Nov 13 '15 at 19:53
  • $\begingroup$ @VesselinDimitrov Thank you for the comment. This does help me for the cases when $p = 5$ or $7$, however, I am still a bit confused about the case when $p=3$. Under my assumption, the elliptic curve $E/K$ has a $K$-rational points of order 3. Since $K(E[3]) \subseteq K(E[12])$, this would imply that $3$ divides the degree of the normal extension over which $E$ acquires stable reduction. $\endgroup$ – Jackson Morrow Nov 13 '15 at 21:10
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    $\begingroup$ I shouldn't think one can do this for p=3 in general. If the image of inertia at 3 in the ell-adic representation has order a multiple of 3 (and I don't see why this can't happen) then you'll need an extension of order a multiple of 3 to kill it. $\endgroup$ – eric Nov 13 '15 at 21:20
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    $\begingroup$ If you want to find an explicit example just browse through some table of elliptic curves trying the ones with integral j-invariant at 3 and conductor a multiple of 3, and then just compute how 3 ramifies in the field cut out by the 2-torsion. The moment you find an example where inertia has order a multiple of 3 you might be in trouble. $\endgroup$ – eric Nov 14 '15 at 0:15
  • $\begingroup$ @eric Thank you for you comments. After running through some examples, I do agree with you. In above mentioned work, Frey only considers the elliptic curve $y^2 = x^3 + 1$ for $p=3$, which is why I wrote that it is dealt with separately. However, it seems as though this is a special case. $\endgroup$ – Jackson Morrow Nov 14 '15 at 16:43

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