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Let $F$ be a number field and $\mathcal{O}$ an order in an imaginary quadratic field $K$. Assume $K\subseteq F$. In Lang's Elliptic Functions, it is shown that over that there is a bijection between, the isomorphism classes of elliptic curves over $\bar{F}$ and the class group of $\mathcal{O}$. This is done by defining an action the class group on lattices of $\mathbb{C}$. Moreover, any two elliptic curves $E,E'$ over $\bar{F}$ that have CM by $\mathcal{O}$ are isogenous over $\bar{F}$

However, I'm told this can be done F-rationally and in fact this is done by Deuring. But I don't read German. This is stated in Products of CM Elliptic Curves (equation 55, pg.20), but the proof is in service of much more complicated results. $$\{E'/F\; |\; E' \mbox{ is $F$-isogenous to $E$ and } \mathrm{End}(E)\cong \mathrm{End}(E') \} \cong \mathrm{Cl}(\mathcal{O})$$ I'm wondering if there is a simpler proof in the vein of defining an action of the class group on lattices of $F$ and if anyone can give me a reference.

More precisely, for an elliptic curve $E/F$ with CM by $\mathcal{O}$, let $E=E_1,\dots, E_h$ denote representatives for the $\bar{F}$ isomorphism classes of elliptic curves with CM by $\mathcal{O}$. For $j=2\dots h$, I want to find an elliptic curve $E'/F$ such that $E'$ is in the $\bar{F}$-isomorphism class of $E_j$ and $E'$ is isogenous to E over $F$.

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    $\begingroup$ You want $O$ to be the entire CM order (not just a subring of it) and the CM-type to be the given $K\hookrightarrow F$. For an invertible $O$-module $L$, the functor $S \rightsquigarrow L\otimes_O E(S)$ on $F$-schemes is represented by an elliptic curve denoted $L \otimes_O E$. Considerations with Tate modules imply the natural map $L\otimes_O{\rm{Hom}}_{F,O}(E,E')\to{\rm{Hom}}_{F,O}(E,L \otimes_O E')$ is an isomorphism for such $E$, $E'$, $L$. Via the operations $L\otimes_O(\cdot)$, ${\rm{Pic}}(O)$ acts simply transitively on the set of $O$-linear isomorphism classes of such $E$ over $F$. $\endgroup$ – nfdc23 Aug 9 '17 at 4:12
  • $\begingroup$ Start with the complex torus $\mathbb{C}/ \hat{O}$ with CM by $\mathcal{O}$ ($\hat{O}$ is a lattice in $K$ and $\mathcal{O} = \{ a \in K, a \hat{O} \subset \hat{O}\})$. If $I$ is a (fractional) ideal of $K$ then $I\hat{O}$ is a lattice in $K$ and $\mathbb{C}/I\hat{O}$ has CM by $\mathcal{O}$. There is a complex analytic map (an isogeny) $\mathbb{C}/\hat{O}\to \mathbb{C}/I\hat{O}$ iff $I$ is a (fractional) principal ideal. Then use $\wp(z)$ to translate this on the elliptic curve side. $\endgroup$ – reuns Aug 9 '17 at 10:42
  • $\begingroup$ This all makes sense to me but how does one shows that $\mathbb{C}/\hat{\mathcal{O}}$ and $\mathbb{C}/I\hat{\mathcal{O}}$ through $\wp(z)$ give elliptic curves that are defined over (or rather have models over $F$) and that the isogeny is $F$-rational? $\endgroup$ – Rdrr Aug 9 '17 at 17:42
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    $\begingroup$ @reuns: Strictly speaking, $I$ should be an invertible $\mathcal{O}$-module inside $K$ rather than a "(fractional) ideal of $K$" (which is usually taken to mean an invertible $O_K$-module inside $K$) to ensure that multiplying the homology lattice by $I$ preserves $\mathcal{O}$ as the entire endomorphism ring. Likewise, when you speak of $I$ being principal you mean as an $\mathcal{O}$-module and moreover that such principality encodes that the two analytic elliptic curves are $K$-linearly isomorphic (not just isogenous). Tracking $F$-rationality through elliptic functions is a bit involved. $\endgroup$ – nfdc23 Aug 10 '17 at 2:33

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