# A short route to the main theorems of Complex Multiplication.

Class field theory classifies the abelian extension of a number (or more generally, global) field using data intrinsic to the number field – line bundles over it. This is a very powerful theory and can be used to answer a great many classical questions in algebraic number theory.

Nevertheless, the general theory offers little to no clues about generating the abelian extensions directly. In the simplest case of $\mathbb Q$, we can in fact explicitly parametrize the abelian extensions in terms of a simple analytic function, the exponential. In short, every abelian extension of $\mathbb Q$ is contained in a cyclotomic field $\mathbb Q(e^{2\pi i/n})$ for some $n$ determined by the ramification in the abelian extension. This is a classical result predating class field theory known as the Kronecker-Weber theorem.

The problem for more general number fields looks hopelessly hard except in a few special cases. One of these is the case of quadratic imaginary number fields (that is, fields of the form $\mathbb Q(\sqrt{-d})$ for $d > 0$ and square-free. In this case, the abelian extensions are generated by torsion points on a a certain Elliptic curve. This is in analogy to the case of $\mathbb Q$ where the abelian extensions were generated by torsion points on the one dimensional group scheme $\mathbb G_m = \mathrm{Spec}\ \mathbb Q[x,x^{-1}]$. This is known as complex multiplication.

Despite the parallels, the theory of complex multiplication is a lot harder than Kornecker-Weber. In fact, I believe there is no known proof of complex multiplication that does not already assume the main theorems of class field theory (in the case of quadratic imaginary number fields). For example, Silverman takes around 100 pages to develop the theory completely.

In spite of the difficult nature of the proofs, the results are very elegant and justify the effort that goes into the proofs. Hilbert has a famous quote towards this:

The theory of complex multiplication is not only the most beautiful part of mathematics but also of the whole of science.

I believe that there is a way to prove the main theorems of complex multiplication, undercutting much of the hard work done in Silverman’s book and I would like to explain that in this post.

The first few sections set up the general theory and follow the standard sources fairly closely. Skip directly to section 3 for the novel part.

Warning: I have not seen anyone else prove complex multiplication along these lines and it is possible that I am missing some subtlety. Proceed with caution!

Prerequisites: I will assume familiarity with the basic theory of Elliptic Curves (as in Silverman’s first book) and the main theorems of global Class Field Theory via ideals (but not their proofs!). I will gloss over many easy to prove results, especially in the early stages. A complete reference is, as mentioned above, Silverman’s second book on elliptic curves.

I would like to begin by recalling the main theorems of Class field theory but there is quite a bit of jargon used so perhaps it is better to simply link to an excellent summary by Poonen here . In particular, the statements of section 3 (global class field theory) will be most relevant to us.

Let me now jump right into defining complex multiplication. The link to abelian extensions will not be explicit initially but hopefully, the general idea becomes clear quickly.

An Elliptic curve defined over $\mathbb C$ is said to have complex multiplication if it’s endomorphism ring is strictly larger than $\mathbb Z$. Recall that, in characteristic $0$, the endomorphism ring of an elliptic curve is always an order in some quadratic imaginary number field.

We would like to study Elliptic curves whose endomorphism ring is exactly equal to $R$ for $R$ an order in a quadratic imaginary number field $K$. In fact, we will assume for simplicity that $R$ is a maximal order but the general case is not very different. It will help to first talk about some generalities on the endomorphism ring of Elliptic Curves in characteristic $0$.

### 1. Complex Models of Elliptic Curves and their Endomorphisms:

Recall that every elliptic curve over $\mathbb C$ has a model of the form $\mathbb C/\Lambda$ where $\Lambda$ is a lattice in the complex plane. The addition on the curve is given by addition in the complex plane modulo the lattice. Two such models are isomorphic are isomorphic precisely when the lattices differ by a complex scalar. That is:

$\mathbb C/\Lambda_1 \cong \mathbb C/\Lambda_2 \iff \Lambda_1 = z\Lambda_2$  for some $z \in \mathbb C^\times$.

More generally, maps between elliptic curves (that preserve the origin) are given by scalar multiplication by a non zero complex number. Using the explicit model (or through more abstract arguments), it is not very difficult to prove that the endomorphism ring of an elliptic curve (over $\mathbb C$) is either $\mathbb Z$ or an order in a quadratic imaginary number field.

As alluded to above, we will focus on the latter case and in particular, we will assume that the endomorphism ring is the maximal order $R$ in a quadratic imaginary number field $K$.

Suppose $E = \mathbb C/\Lambda$ is an elliptic curve with endomorphism ring $R$. This makes $\Lambda$ into a $R$ module and it can be shown fairly easily that $\Lambda$ is a projective, invertible $R$-module (=fractional ideal).

Conversely, any fractional ideal $\mathfrak a$ of $R$ is a lattice in $\mathbb C$ under one of the complex embeddings of $K$ and gives rise to the elliptic curve $\mathbb C/\mathfrak a$. Note that the isomorphism class of this elliptic curve depends only on the ideal class of the fractional idea in the class group of $R$. An elementary (but slightly complicated) argument in the theory of imaginary quadratic number fields shows that $\mathbb C/\mathfrak a$ has endomorphism ring $R$ precisely when $\mathfrak a$ is an invertible $R$-module.

In fact, even more is true. We can define a multiplication of ideals on lattices in the following way:

$\mathfrak a\Lambda = \{a\lambda : a \in \mathfrak a, \lambda \in \Lambda\}$

and it is easily checked that $\mathfrak a\Lambda$ is a lattice with the same endomorphism ring as $\Lambda$. This lets us define an action of the class group on the set of elliptic curves with complex multiplication by $R$:

$\mathfrak a*(\mathbb C/\Lambda) = \mathbb C/(\mathfrak a^{-1}\Lambda).$

Taking the inverse is a convention that will make our lives simpler later on. It is a simple verification now that this action is faithful and transitive. If we denote by $\mathscr E(R)$ the set of elliptic curves with complex multiplication by $R$, we will have shown:

Theorem 1: $\mathscr E(R)$ is a torsor under the class group of $R$. In particular, $\mathscr E(R)$ is a finite group and has $h = |Cl(R)|$ elements.

Remark 1: Note that $\mathfrak a*E$ comes with an isogeny $[\mathfrak a]: E \to \mathfrak a*E$ that sends $z \to z$ in the complex model. It is easily verified that this isogeny has degree $N(\mathfrak a)$. The isogeny is only defined up to isomorphism and it will be important later to remove this ambiguity.

This is our first hint of a connection between class field theory and Elliptic curves.

### 2. Galois action on CM Curves:

So far, we have only used the complex model of Elliptic Curves. However, out real interest in arithmetic and to get anywhere, we would like the Elliptic curves we are interested in to be defined over a number field. This is indeed true.

For $\sigma \in \mathrm{Aut}\ (\mathbb C/{\mathbb Q})$ and any Elliptic Curve $E$, note that $E^\sigma$ has the same endomorphism ring as $E$. In particular, this implies that $\mathrm{Aut}\ (\mathbb C/{\mathbb Q})$ acts on $\mathscr E(R)$.

We would like to translate this into a statement about algebraic numbers. The j-invariant of an Elliptic curve lets us do this. Recall that the isomorphism class of an Elliptic curve over $\mathbb C$ is exactly determined by it’s $j$-invariant. Also, it is clear from the definition of the j-invariant that for $\sigma \in \mathrm{Aut}\ (\mathbb C/{\mathbb Q})$ and any Elliptic curve over $E$:

$\sigma(j(E)) = j(E^\sigma).$

Since $\mathscr E(R)$ is a finite set, we have shown:

Theorem 2: For $E$ an Elliptic Curve with CM by $R$, $j(E)$ is an algebraic number of degree at most $h = |Cl(R)|$.

In particular, $E$ is defined over $\overline{\mathbb Q}$ and we have finally moved into arithmetic.

Now, if we are being optimistic, we might hope that $j(E)$ has degree exactly equal to $h$. We might even go so far as to hope that $K(j(E))$ is the Hilbert Class Field of $K$.

If this were true, this would have some remarkable consequences. For instance, it would imply that the action of $\mathrm{Gal}\ (\overline{\mathbb K}/\mathbb K)$ on $\mathscr E(R)$ is transitive and that the $j(E)$ are Galois conjugates for $E \in \mathscr E(R)$. In fact all this and more is true as we will prove shortly.

### 3. Galois Representations attached to CM Curves:

To prove the above claims, we will construct a homomorphism $G_K = \mathrm{Gal}\ (\overline K/K)$ using Theorem 1 and show that it is indeed the Artin map from Class Field Theory. So far, this is exactly how Silverman does it.

However, the difference in our approach will be that we will not prove that our representation is a group homomorphism. We will define it only as a set map. This will allow us to bypass the (hard) result of Silverman that $(\mathfrak *E)^\sigma = \mathfrak a^\sigma *E$.

Defining the map is easy: Fix an $E$ with CM by $R$. As we showed above, for any $\sigma \in G_K$, $E^\sigma$ also has CM by $R$. Therefore, by Theorem 1, there is a unique ideal class $F_E(\sigma) \in Cl(R)$ such that $E^\sigma = F_E(\sigma)* E$.

Thus, we have defined a set map $F_E: G_K \to Cl(R)$. Note that we know very little about this map. We expect that it is a group homomorphism independent of $E$ and equal to the Artin map, however we can prove none of it!

We will prove everything at once by showing that it is equal to the Artin map. The key idea is that, since $G_K$ is Hausdorff as a topological group, we can show that two continuous maps from $G_K$ are equal by showing that they are equal on a dense set.

If we give $Cl(R)$ the discrete topology, then $F_E$ is automatically continuous and we will only need to show that it is equal to the Artin map on a dense set. This is easier than it sounds!

Let $P$ be the set of primes of $K$ that split completely over $Q$ minus a finite set of primes to be specified later. This a dense set of primes in $K$ (exercise!). Let $G$ be the set of Frobenius elements in $G_K$ corresponding to $P$. Of course, the Frobenius elements are only defined upto conjugacy but our proof will implicitly show that $F_E$ is constant over a conjugacy class.

With this notation, we have:

Theorem 3[Main Theorem of CM]: $F_E$ is equal to the Artin map on the set $G$. That is, for $\mathfrak p \in P$, $F_E(\mathfrak p) = [\mathfrak p]$. Equivalently:

$E^{Frob_\mathfrak p} = \mathfrak p * E.$

Proof: The first problem that faces us is that we do not know much about the action of $Frob_\mathfrak p$ on $E$. Let us fix a prime $\pi$ in the algebraic closure over $\mathfrak p$ so that $Frob_\mathfrak p = F(\pi/\mathfrak p)$ is specified in it’s conjugacy class.

Then, we do know quite a lot  about the action of $F(\pi/\mathfrak p)$ on the reduction of $E$ modulo $\pi$. Therefore, let us define $\widetilde E$ to be the reduction of $E$ modulo $\pi$. Our first goal should be to prove that $\widetilde E^{Frob(\pi/\mathfrak p)}$ is isomorphic to $\mathfrak p * E$.

To do this, recall that there is an isogeny $[\mathfrak p]: E \to \mathfrak p * E$ defined using the complex model. Since both $E$ and $\mathfrak p*E$ are defined over a number field, descent (or the theory of Elliptic Curves) tells us that $[\mathfrak p]$ is defined over $\overline{\mathbb Q}$ and thus, we can reduce it modulo $\pi$ to get $\widetilde{[\mathfrak p]}: \widetilde E \to \widetilde{\mathfrak p*E}$.

We know that $[\mathfrak p]$ is of degree $p = N(\mathfrak p)$. Since $\mathfrak p$ splits completely over $\mathbb Q$, $p$ is prime. To show that that $\widetilde{\mathfrak p*E} \cong \widetilde E^{F(\pi/\mathfrak p)}$, it is sufficient to show that $\widetilde{[\mathfrak p]}$ is inseparable and is therefore the Frobenius isogeny.

To show inseparability, note that we can find an ideal $\mathfrak a$ coprime to $p$ such that $\mathfrak a\mathfrak p = (f)$ is principal and $f$ is rational and divisible by $p$. This implies that the composition $\widetilde{[\mathfrak a]\circ[\mathfrak p]}$ is inseparable. We would be done if we could show that $[\mathfrak a]$ is inseparable (after reduction) but we can use the same trick again:

We can find yet another ideal $\mathfrak b$ such that $\mathfrak a\mathfrak b = (g)$ is principal and $g$ is rational and coprime to $p$. Therefore, the reduction of $[\mathfrak{ab}]$ is separable and so is $[\mathfrak a]$ after reduction!

Thus, we have shown that $E^{F(\pi/\mathfrak p)}$ is isomorphic to $[\mathfrak p]*E$ after reduction but of course, they might not be isomorphic in characteristic $0$. The key idea now is to note that there are only finitely many possibilities for $\mathfrak p$ where this can happen. If $E,E'$ are isomorphic post reduction, then this implies that $j(E) \equiv j(E') \pmod{\mathfrak p}$ and since there are only finitely many curves in $\mathscr E(R)$, there are only finitely many possibilities for $\mathfrak p$.

We can just define $P$ to avoid these finitely many possibilities and will be done!

$\Box.$

Thus, we have shown that $F_E$ is equal to the Artin map from Class Field Theory. This immediately implies that $F_E$ is independent of $E$, is a surjective group homomorphism and has kernel exactly equal to the Hilbert Class Field of $K$. On the other hand, the kernel is also equal to $K(j(E))$ and therefore, we have shown that $K(j(E))$ is the Hilbert Class Field and has degree $h = Cl(R)$ over $K$.