Question about quadratic twists of elliptic curves Let $E$ be an elliptic curve and $d$ be a squarefree integer. If $E'$ and $E$ are isomorphic over $\mathbb{Q}(\sqrt{d})$, must $E'$ be a quadratic twist of $E$?
 A: Yes, this is true, for elliptic curves over any field $K$ of characteristic different from $2$, with $d \in K^{\times}$.  [The argument I give also stays away from characteristic $3$.  In fact you can work in characteristic $2$ as well if you make some small modifications.]
Recall first that two elliptic curves over $K$ are isomorphic over the algebraic closure $\overline{K}$ iff they have the same $j$-invariant.  See e.g. Theorem 5.10 in these notes.  The proof shows that a change of variables from one Weierstrass elliptic curve
$E: y^2 = x^3 + Ax+B$
to a second Weierstrass elliptic curve
$E': y^2 = x^3 + A'x + B'$
with the same $j$-invariant is given by
$(x,y) = (u^2X,u^3Y)$
with $u = (A/A')^{\frac{1}{4}} = (B/B')^{\frac{1}{6}}$,
provided $A'B' \neq 0$.  In this case we find $u^4, u^6 \in K$ hence $u^2 \in K$, say $u^2 = d$.  Thus $E$ and $E'$ become isomorphic over the quadratic extension $K(\sqrt{d})$ and if you look for just a little bit at the equations you'll see that $E'$ is the quadratic twist of $E$ by $d$.
The cases $B' = 0$ -- equivalently $j = 1728$ and $A' = 0$ -- equivalently $j = 0$ -- really are slightly different.  In these cases we really may need to pass to a degree $4$ or degree $6$ extension in order for $E$ and $E'$ to become isomorphic and we speak of quartic and sextic twists in this case.  But once you analyze carefully what happens -- see the exercises following Theorem 5.10 -- you can see that the isomorphism between $E$ and $E'$ can be performed over a quadratic extension if and only if $u^2$ (resp. $u^3$) lies in $K$, so setting $d = u^{\frac{1}{4}}$ (resp. $d = u^{\frac{1}{6}}$) still gives you a quadratic twist.
