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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?

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  • $\begingroup$ Two elliptic curves in Weierstrass form with $j(E) = j(E')$ are isomorphic over $\overline{\mathbb{Q}}$, and $E_a : y^2 = x^3 - \frac{27 a}{a-1728} x-\frac{27 a}{a-1728} $ has $j(E_a) = a$. If $E_a$ has CM by a subring of $\mathbb{Q}(\sqrt{-d})$ then you'll take $\mathbb{Q}(\sqrt{-d},a)$ as the field of definition of $E_a$ because $\text{End}(E_a)$ is defined over it. $\endgroup$ – reuns Jan 15 '18 at 16:11
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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 $K$.

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    $\begingroup$ Slight correction: $K(j(E))$ is the maximal abelian everywhere unramified extension of $K$ if the endomorphism ring ${\mathop{\rm End}}_{\bf C}(E)$ is the full ring of integers of $K$, but not in general (when the extension $K(j(E))/K$ is still abelian but can be ramified). $\endgroup$ – Noam D. Elkies Jan 14 '18 at 22:57
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    $\begingroup$ @NoamD.Elkies Thanks. I corrected the misstatement. $\endgroup$ – Joe Silverman Jan 14 '18 at 23:15

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