TITLE: Fields of Definition of Elliptic Curves
QUESTION [6 upvotes]: I am currently studying the theory of complex multiplication and I find myself confused by the language in a lot of the literature.
In Silverman's Advanced Topics in the Arithmetic of Elliptic Curves, Silverman uses the term model without ever defining it (as far as I can see). What does he mean? F-isomorphism class?
In particular the proof of theorem II.2.3, he states
"We take a model for $E$ defined over $H=K(j(E))$". Why can one swap out $E$ with a model defined over $K(j(E))$.
Another example of this, is in Diamond, Darmon and Taylor's paper on Fermat's last theorem. In remark 1.3, it states that any elliptic curver $E/\mathbb{C}$ with CM is defined over an abelian extension of $K= \mathrm{End}_{\mathbb{C}}(E)\otimes E$. 
I assume they are citing the fact that $K(j(E))$ is an abelian extension of $K$, but why is $E$ defined over $K(j(E))$? Is this even the supposed abelian extension $E$ is defined over? If not, which one is it?

REPLY [11 votes]: What's usually meant when phrased this way is that within the $\overline K$-isomorphism class of $E$, there is an elliptic curve defined over $K(j(E))$. An indeed, for any elliptic curve $E$ defined over $\overline{\mathbb Q}$, there is an elliptic curve $E'$ defined over $\mathbb Q(j(E))$ that is $\overline{\mathbb Q}$-isomorphic to $E$. So that's the sort of model that one often takes. All this has nothing to do with CM, and is very elementary. 
If $E$ has CM with $K=\text{End}_{\mathbb C}(E)\otimes\mathbb Q$, then part of the basic theory of CM is that $K(j(E))$ is an abelian extension of $K$, and indeed if $\text{End}_{\mathbb C}(E)$ is the full ring of integers of $K$, then $K(j(E))$ is the maximal abelian everywhere unramified extension of $K$ (the Hilbert class field of $K$). This is far less elementary, but sufficiently well-known that DDT probably didn't feel it necessary to give a reference. The fact that we can find a model for $E$ over $K(j(E))$ follows from the previous paragraph. We can actually find such a model over $\mathbb Q(j(E))$, but probably they want the endomorphisms to also be defined over the field, and thus need to take the compositum with $K$.