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Suppose $E_1$ and $E_2$ are elliptic curves defined over $\mathbb{Q}$. Now we know that both curves are isomorphic over $\mathbb{C}$ iff they have the same $j$-invariant.

But $E_1$ and $E_2$ could also be isomorphic over a subfield of $\mathbb{C}$. As is the case for $E$ and its quadratic twist $E_d$. Now the question general is.

$E_1$ and $E_2$ defined over $\mathbb{Q}$ and isomorphic over $\mathbb{C}$. Let $K$ the smallest subfield of $\mathbb{C}$ such that $E_1$ and $E_2$ become isomorphic over $K$. What can be said about $K$. Is it always a finite extension of $\mathbb{Q}$. If so, what can be said about the extension $K|\mathbb{Q}$.

My second question is something goes something like in the opposite direction. I start again with quadratic twists. Let $E$ be an elliptic curve over $\mathbb{Q}$ and consider the quadratic extension $\mathbb{Q}|\mathbb{Q}(\sqrt{d})$. Describe the curves over $\mathbb{Q}$(or isomorphism classes over $\mathbb{Q}$) which become isomorphic to $E$ over $\mathbb{Q}(\sqrt{d})$. I think the answer is $E$ and $E_d$. Again I would like to know what happens if we take a larger extension.

Let $E$ be an elliptic curve over $\mathbb{Q}$ and $K|\mathbb{Q}$ a finite extension. Describe the isomorphism classes of elliptic curves over $\mathbb{Q}$ which become isomorphic to $E$ over K.

I have no idea what is the right context to answer such questions.

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up vote 4 down vote accepted

Question 1: Putting both curves in say, Legendre Normal Form (or else appealing the lefschetz principle) shows that if the two curves are isomorphic over $\mathbf{C}$ then they are isomorphic over $\overline{\mathbf{Q}}$. Now we could say that for instance $E_2$ is an element of $H^1(G_{\overline{Q}}, Isom(E_1))$ where we let $Isom(E_1)$ be the group of isomorphisms of $E_1$ as a curve over $\mathbf{Q}$ (as in Silverman, to distinguish from $Aut(E_1)$, the automorphisms of $E_1$ as an Elliptic Curve over $\mathbf{Q}$, that is, automorphisms fixing the identity point). However, $E_2$ is also a principle homogeneous space for a unique curve over $\mathbf{Q}$ with a rational point, which of course has to be $E_2$, so the cocycle $E_2$ represents could be taken to have values in $Aut(E_1)$. Now $Aut(E_1)$ is well known to be of order 6,4 or 2 depending on whether the $j$-invariant of $E_1$ is 0, 1728 or anything else, respectively. Moreover the order of the cocycle representing $E_2$ (which we now see must divide 2, 4 or 6) must be the order of the minimal field extension $K$ over which $E_1$ is isomorphic to $E_2$. So $K$ must be degree 2,3,4 or 6 unless I've made an error somewhere.

Question 2: If you restrict your focus to just elliptic curves, yes your idea is right. If it's a quadratic extension, you have exactly 1 non-isomorphic companion. If you have a higher degree number field, you have nothing but composites of the quadratic case unless your elliptic curve has j invariant 0 or 1728.

Notice I am very explicitly using your choice of the word elliptic curve for both of these answers.

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The answer is a bit more complicated if $j=0,1728$ because the corresponding elliptic curves have a bigger automorphism group, so I'll leave those out and let you (or others) deal with this case. If $j \ne 0,1728$, then the automorphism group of $E$ is of order $2$ and all other elliptic curves isomorphic to $E$ over an extension are quadratic twists. It seems that you know what happens in this case. This is well-discussed in Silverman's book.

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Seconding the pointer to Silverman's book. Once you learn the basic definitions of Galois cohomology, you can check as an exercise that the set of isomorphism classes in your second question, when K is a Galois extension of Q, is in natural bijection with

H^1(Gal(K/Q), Aut(E)(K)).

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Hopefully I'll have some time later to ellaborate, but for now - here is a great reference I wish somebody had shown me when I started out, for how to attack questions of this type:

look at chapter 17 (and 18).

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I will take a look. Thanks. – wood Feb 3 '10 at 20:41

A concrete explanation: for elliptic curves defined using short Weierstrass equations $E_i : y^2 = x^3 + a_ix + b_i$ over $K$ (not of characteristic $2$ or $3$), all isomorphisms over $L$ (an extension of $K$) are just given by $f(x,y) = (\lambda^2 x,\lambda^3 y)$ for some $\lambda \in L^\times$, so we then need $\lambda^4 a_1 = a_2$ and $\lambda^6 b_1 = b_2$ (from the expression for $j$). In this case, basic algebra shows that if $j \neq 0,1728$ then the isomorphism involves extracting a square root, so can take place in a degree 2 extension of $\mathbb{Q}$. If $j=1728$, then $b_i=0$ and we just need to extract a fourth root, so we have (at most) a degree $4$ extension over which $E_1$ and $E_2$ become isomorphic. Similarly, if $j=0$, we can need up to a degree $6$ extension.

You can arrange so that you need to extract the root of any given element in $K^\times$, this describes the behaviour over various extensions.

I believe that the situation in characteristic $2$ or $3$ gets rather trickier, and you might need an extension of degree up to $24$.

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As for your first question: if you think of your elliptic curves as plane cubics (Weierstrass'model) the isomorphism between them is a polynomial function. Polynomials include only finitely many coefficients and the isomorphism is defined over the field generated by them, which is finite over $\Bbb Q$.

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Finitely generated $\ne$ finite. There are pairs of algebraic varieties (not elliptic curves) with families of isomorphisms. Heck, if you ignore the fixed origin, even elliptic curves have them (think translations) – Felipe Voloch Feb 3 '10 at 16:23

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