# 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.