Some Fields of Definition Results

Suppose {E/\mathbb{C}} is an elliptic curve. We will need some standard facts. First, if {\sigma: \mathbb{C}\rightarrow \mathbb{C}} is a field automorphism, then {E^\sigma} will denote the elliptic curve formed by acting on the coefficients of a Weierstrass equation by {\sigma}. We have an isomorphism {End(E)\stackrel{\sim}{\rightarrow} End(E^\sigma)} by {f\mapsto f^\sigma}. If {E} has CM by the full ring of integers {R_K} in a quadratic imaginary {K}, then {j(E)\in \overline{\mathbb{Q}}}. This follows by exploiting the fact that {j(E^\sigma)=j(E)^\sigma} to argue that {j(E)} lies in a finite extension of {\mathbb{Q}}.

The natural map {Ell_{\overline{\mathbb{Q}}}(R_K)\rightarrow Ell(R_K)} from isomorphism classes of curves over {\overline{\mathbb{Q}}} with CM by {R_K} to isomorphism classes over {\mathbb{C}} is an isomorphism. This follows easily from the fact that {j(E)\in\overline{\mathbb{Q}}}. It turns out that if we use our normalized isomorphism {[\cdot]: R\rightarrow End(E)}, we get the nice relation {[\alpha]_E^\sigma=[\alpha^\sigma]_{E^\sigma}} for all {\alpha\in R} and {\sigma\in Aut(\mathbb{C})}.

If {E} is defined over {L} and has CM by {R} in {K}, then every endomorphism of {E} is defined over {LK}. It is useful to see why this is. Let {\sigma\in Aut(\mathbb{C})} fix {L}. Then by definition {E^\sigma=E}, so using the normalized {[\cdot]} relation we get that {[\alpha]^\sigma=[\alpha^\sigma]} for all {\alpha\in R}. Thus if {\sigma} also fixes {K}, then {[\alpha]} will be fixed by {\sigma} and hence by Galois descent every endomorphism descends to {LK}.

We can use this result to show that if {E} has CM by the full ring of integers {R_K}, then {[\mathbb{Q}(j(E)): \mathbb{Q}]\leq h_K} the class number of {K}. As an example application, without knowing the Weierstrass equation for our curve we can argue that the unique curve with CM by {\mathbb{Z}[i]} must be defined over {\mathbb{Q}}. This is because {h_K=1}, so the degree of the extension is {1} and hence {j(E)\in\mathbb{Q}}. Of course, we already knew that because we said last time it is given by {y^2=x^3+x}. Likewise, we could make the same argument for the curve with CM by {\mathbb{Z}[\rho]} where {\rho=e^{2\pi i/3}}.

Now we get to our first real relation between CM and class field theory. Suppose {E/\mathbb{C}} has CM by {R_K}. The field {L=K(j(E), E_{tors})} generated by the {j}-invariant and the torsion points is an abelian extension of {K(j(E))}. The proof uses our field of definition result above about the endomorphisms. If {H=K(j(E))}, then we can reduce to showing that each {L_m=H(E[m])} is abelian over {H}. Using the standard Galois representation, we see that {Gal(L_m/H)} injects into {GL_2(\mathbb{Z}/m)}.

Since we have CM floating around we know every endomorphism is already defined over {H}. Using the fact that the image of {Gal(L_m/H)} commutes with {R_K}, we get that {Gal(L_m/H)} injects into {Aut_{R_k/mR_k}(E[m])\simeq (R_K/mR_K)^*} an abelian group and hence is abelian. This result about being abelian I think is an exercise in Silverman’s first book.

Fix a quadratic imaginary field {K}. Let’s return to the idea that we have a simply transitive action of {Cl(R_K)} on {Ell(R_K)} by {\frak{a}\cdot E_{\Lambda}=E_{\frak{a}^{-1}\Lambda}}. We also have a nice action of {Gal(\overline{K}/K)} on {Ell(R_K)} given by {\sigma\cdot E=E^{\sigma}}. Fix some {E\in Ell(R_K)}. This gives us a map {F: Gal(\overline{K}/K)\rightarrow Cl(R_K)} by sending {\sigma} to the unique element {\frak{a}} with the property that {\frak{a}\cdot E=E^\sigma}. It turns out that {F} is a group homomorphism that is independent of the choice of {E}. This is actually not so easy to prove, and the underlying philosophy of why it is hard is that the action of {Cl(R_K)} is analytic in nature since it acts on the lattice {\Lambda\subset \mathbb{C}}, whereas the Galois action is purely algebraic. To prove the independence of {E} requires translating between the two. Although it is sort of a fun proof, we’ll skip it in order to get to the meat of CM theory more quickly.


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