Complex Multiplication 2

I’ll start with thanking Google/Chrome. I sent in a request to fix the CSS rendering issue that happened with the blog and the next day it was fixed (maybe a coincidence?).

We start with our standing assumptions. Let ${E}$ be an elliptic curve with CM by ${R_K}$ for some quadratic imaginary ${K}$ (with normalized iso ${[\cdot]: R_K\rightarrow End(R)}$). We will need the following fact. Suppose ${E}$ is defined over a number field ${L}$. Let ${\frak{p}}$ be a prime of ${L}$ and ${\overline{E}}$ the reduction of ${E}$ mod ${\frak{p}}$. An element ${\gamma}$ is in the image of the natural map ${End(E)\rightarrow End(\overline{E})}$ if and only if ${\gamma}$ commutes with every element in the image of the map.

The proof is a straightforward follow your nose type argument by cases. Caution: You must consider the case that ${End(\overline{E})}$ is an order in a quaternion algebra because ${\overline{E}}$ is defined over a field of positive characteristic. Recall that last time we said that ${H:=K(j(E))}$ is the Hilbert class field of ${K}$. Thus without loss of generality we may assume from here on that ${E}$ is defined over ${H}$ (because ${j(E)\in H}$).

Here is something that seemed bizarre to me the first time I saw it. Let ${\frak{p}}$ be a prime of degree ${1}$ of ${K}$ and ${\frak{b}}$ a prime of ${H}$ lying over ${\frak{p}}$. Suppose ${\frak{p}}$ has the property that the natural map ${E\rightarrow \frak{p}\cdot E}$ is an isogeny of degree ${p}$ (${\frak{p}}$ lies over ${p}$) and the reduction ${\overline{E}\rightarrow \overline{\frak{p}\cdot E}}$ is purely inseparable. This happens for all but finitely many ${\frak{p}}$. Then there is an isogeny ${\lambda: E\rightarrow E^{Frob_\frak{p}}}$ lifting the ${p}$-th power Frobenius. This should feel funny because we’re lifting the Frobenius map to characteristic ${0}$, i.e. the following diagram commutes:

$\displaystyle \begin{matrix} E & \rightarrow & E^{Frob_\frak{p}} \\ \downarrow & & \downarrow \\ \overline{E} & \rightarrow & \overline{E}^{(p)} \end{matrix}$

The proof is to use that we already know there is some ${\lambda: E\rightarrow \frak{p}\cdot E}$ whose reduction is purely inseparable of degree ${p}$ and hence factors as ${\overline{E}\rightarrow \overline{E}^{(p)}\rightarrow \overline{E^{Frob_\frak{p}}}=\overline{E}^{(p)}}$ where the second map of degree ${1}$ and hence an automorphism. We’re done if we can check that this automorphism lifts, but we can do this by checking that it commutes with everything in the image of ${End(E^{Frob_\frak{p}})\rightarrow End(\overline{E}^{(p)})}$ by the above lemma. There’s quite a bit of work checking things element-wise to finish this off.

A special case is the following. For all but finitely many primes ${\frak{p}}$ such that ${\psi_{H/K}(\frak{p})=1}$ (i.e. ${\frak{p}=\pi_\frak{p}R_K}$ is principal) the ${\pi_\frak{p}}$ is unique such that the endomorphism ${[\pi_\frak{p}]}$ descends to the ${p}$-the power Frobenius.

We want to now think about generating abelian extensions of ${K}$ using the torsion points of ${E}$. Fix a finite map ${h:E\rightarrow \mathbb{P}^1}$ defined over ${H}$ (called a Weber function for ${E/H}$). These are easy to come by. For example, if ${j(E)\neq 0, 1728}$, then if our model is ${y^2=x^3+Ax+B}$, then ${h(x,y)=x}$ works. Here is another beautiful relation between CM theory and class field theory. Recall ${E[\frak{a}]}$ are the ${\frak{a}}$-torsion points with respect to the normalized map ${[\cdot]}$, i.e. the set of ${P\in E}$ such that ${[\gamma]P=0}$ for all ${\gamma\in\frak{a}}$.

For any integral ideal ${\frak{a}}$ of ${R_K}$ the field ${K(j(E), h(E[\frak{a}]))}$ is the ray class field of ${K}$ of conductor ${\frak{a}}$. An immediate corollary is that ${K^{ab}=K(j(E), h(E_{tors}))}$. Hence we get the amazing fact that the maximal abelian extension of ${K}$ can be made by taking ${E=\mathbb{C}/R_K}$ and adjoining to ${K}$ the numbers ${j(E)}$ and the ${x}$-coordinates of the torsion points of the Weierstrass model (as long as ${j\neq 0, 1728}$). The proof of this is pretty long and involves all the stuff from earlier in the post, so we’ll skip it.

Last time we said that if we don’t just take ${x}$-coordinates and instead adjoin all torsion coordinates we get ${K(j(E), E_{tors})}$ which is an abelian extension of ${H}$. It is interesting that doing this does not in general produce an abelian extension of ${K}$. Another interesting corollary to this is that we get a situation in which we can adjoin all torsion points. Given any ${K}$ with class number ${1}$ we get a chain ${K^{ab}=K(j(E), h(E_{tors}))\subset K(E_{tors})\subset H^{ab}=K^{ab}}$ and hence ${K^{ab}=K(E_{tors})}$.