# L-series of CM Elliptic Curves

This will be the last post in the CM elliptic curve series. Last time we covered the main theorem of complex multiplication. Today we’ll very, very briefly sketch one amazing use of the main theorem. We’ll first talk about how to associate a Grössencharacter to a CM elliptic curve and then use this to better describe the ${L}$-series of an elliptic curve.

Here’s why this will be amazing. Awhile ago we talked about ${L}$-series of varieties and various modularity conjectures. One of the huge, major theorems of modern number theory (which wasn’t proved in full until 2003, and built on all of Wiles and Taylor’s results) is the so-called Modularity Theorem. It says that an elliptic curve ${E/\mathbb{Q}}$ is modular and hence its ${L}$-series has an analytic continuation to all of ${\mathbb{C}}$.

This is still open (as far as I know) for elliptic curves over an arbitrary number field, but today we’ll see that we can use the theory we’ve built to show that any CM elliptic curve over any number field has an ${L}$-series that analytically continues to the whole plane.

Fix ${E/L}$ an elliptic curve over a number field with CM by ${R_K}$, the ring of integers in a quadratic imaginary field ${K}$. We use the main theorem of CM to do the following. Fix an idele ${x\in \mathcal{J}_L}$ and let ${s=N_{L/K}(x)\in\mathcal{J}_L}$. There is a unque ${\alpha\in K^*}$ such that ${\alpha R_K=(s)}$ and for any fractional ${\frak{a}}$ in ${K}$ and any analytic iso ${f: \mathbb{C}/\frak{a}\stackrel{\sim}{\rightarrow} E(\mathbb{C})}$ we get a commutative diagram:

$\displaystyle \begin{matrix} K/\frak{a} & \rightarrow & K/\frak{a} \\ \downarrow & & \downarrow \\ E(L^{ab}) & \rightarrow & E(L^{ab}) \end{matrix}$

What this gives us is a map ${\alpha_{E/L}: \mathcal{J}_L\rightarrow K^*\subset\mathbb{C}^*}$. Recall that a Grössencharacter of a number field ${L}$ is such a map that is trivial on ${L^*}$. We can alter this to the map ${\Psi_{E/L}:\mathcal{J}_L\rightarrow \mathbb{C}^*}$ by ${\Psi_{E/L}(x)=\alpha_{E/L}(x)(N_{L/K}(x^{-1}))_\infty}$. It turns out this is our desired Grössencharacter.

Recall how we formed the L-series of a variety over ${\mathbb{Q}}$. There is nothing special about ${\mathbb{Q}}$ going on in that constuction and so the same thing can be done for any elliptic curve ${E/L}$ where ${L}$ is a number field. Basically you piece it together as a product over primes of some expression involving the characteristic polynomial of the Frobenius elements acting on the cohomology of the reductions of ${E}$ mod these primes.

Given a Grössencharacter ${\Psi: \mathcal{J}_L\rightarrow \mathbb{C}^*}$ we can define the Hecke ${L}$-series to be ${\displaystyle L(s,\Psi)=\prod_{\frak{p}}(1-\Psi(\frak{p})q_\frak{p}^{-s})^{-1}}$, where ${q_\frak{p}}$ is the size of the residue field at ${\frak{p}}$. Hecke proved that this ${L}$-series has an analytic continuation to the complex plane.

Duering proved that if ${E/L}$ is an elliptic curve with CM by ${R_K}$, then ${L(E/L, s)=L(s, \Psi_{E/L})L(s, \overline{\Psi_{E/L}})}$. In fact, even better is that if ${K}$ is not contained in ${L}$, then the ${L}$-series of the elliptic curve over ${L}$ is precisely the Hecke ${L}$-series of the Grössencharacter attached to the base-changed elliptic curve ${E/KL}$. In either case, we see that it is much easier to prove that the ${L}$-series of an elliptic curve with CM by the ring of integers in a quadratic imaginary field has an analytic continuation.

# The Main Theorem of Complex Multiplication

Today we’ll state the Main Theorem of Complex Multiplication. We have our standing assumptions. Let ${K}$ be a quadratic imaginary field and ${R_K}$ the ring of integers. For a prime ${\frak{p}}$ of ${K}$ define ${K_\frak{p}}$ to be the completion at ${\frak{p}}$ and ${R_\frak{p}}$ the ring of integers. For a fractional ideal ${\frak{a}}$ of ${K}$ define ${\frak{a}_\frak{p}}$ to be the fractional ideal ${\frak{a}R_\frak{p}}$.

We will need a few facts about modules over ${R_K}$. We know ${R_K}$ is a Dedekind domain, so these will easily follow from the sequence of posts about the structure of modules over Dedekind domains. We will denote the ${\frak{p}}$-primary part of an ${R_K}$-module, ${M}$, to be ${M[\frak{p}^\infty]}$. This is just the set of elements of the module annihilated by some power of ${\frak{p}}$.

Back in our posts on this topic we only talked about finitely generated modules, but some things are true in more generality. For example, if ${M}$ is a torsion ${R_K}$-module, then the natural map ${\displaystyle \bigoplus_{\frak{p}}M[\frak{p}^\infty]\rightarrow M}$ is an isomorphism. This is essentially the part of the theorem that says if we attempt to decompose a module into a projective and torsion part, the torsion part has a nice primary decomposition. We also will need that for any fractional ${\frak{a}}$ the natural map ${(K/\frak{a})[\frak{p}^\infty]\rightarrow R_\frak{p}/\frak{a}_p}$ is an isomorphism. This comes from the inclusion ${K\hookrightarrow K_\frak{p}}$.

Putting the above to facts together gives us that ${\displaystyle K/\frak{a}\simeq \bigoplus_{\frak{p}}K_\frak{p}/\frak{a}_\frak{p}}$. Now let ${x\in \mathcal{J}_K}$ be an idele. The fractional ideal generated by this is ${\displaystyle (x)=\prod_\frak{p} \frak{p}^{ord_p(x_p)}}$. Given any fractional ideal ${\frak{a}}$ of ${K}$ we define ${x\frak{a}:=(x)\frak{a}}$. Since ${\displaystyle K/x\frak{a}\simeq \bigoplus_\frak{p} K_p/x_p\frak{a}_p}$ we can define multiplication by ${x}$ on all of ${K/\frak{a}\rightarrow K/x\frak{a}}$ piece by piece on the ${\frak{p}}$-primary parts ${(t_\frak{p})\mapsto (x_\frak{p}t_\frak{p})}$.

Let ${E/\mathbb{C}}$ be an elliptic curve with CM by ${R_K}$. Fix some ${\sigma\in Aut(\mathbb{C})}$. We were calling our surjection from idele class field theory ${\lambda: \mathcal{J}_K\rightarrow Gal(K^{ab}/K)}$. Choose some ${s\in \mathcal{J}_K}$ with the property that ${\lambda(s)=\sigma|_{K^{ab}}}$. Let ${\frak{a}}$ be a fractional ideal and fix a complex analytic isomorphism ${f:\mathbb{C}/\frak{a}\stackrel{\sim}{\rightarrow} E(\mathbb{C})}$.

The Main Theorem of Complex Multiplication tells us that with this setup there is a unique isomorphism ${f':\mathbb{C}/s^{-1}\frak{a}\stackrel{\sim}{\rightarrow} E^\sigma(\mathbb{C})}$ which makes the following diagram commute:

$\displaystyle \begin{matrix} \mathbb{C}/\frak{a} & \stackrel{s^{-1}}{\rightarrow} & \mathbb{C}/s^{-1}\frak{a} \\ \downarrow & & \downarrow \\ E(\mathbb{C}) & \stackrel{\sigma}{\rightarrow} & E^\sigma(\mathbb{C}) \end{matrix}$

This gives us the following simple way to translate between the analytic action of multiplying by ${s^{-1}}$ and the algebraic action of ${\sigma}$ by ${f(t)^{\lambda(s)}=f'(s^{-1}t)}$. The proof involves several reductions to a manageable case, and then a lot of work. As usual, the statements are all we are really interested in right now. The next two posts will be some applications of this, so we might prove a few more things there.

# 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 ${K^{ab}=K(E_{tors})}$.

# Complex Multiplication 1

Today we’ll really get into the class field theory of CM elliptic curves. Let ${K/\mathbb{Q}}$ be a quadratic imaginary field. Let ${E/\mathbb{C}}$ be an elliptic curve with the property that ${End(E)\simeq R_K}$, e.g. take ${E=\mathbb{C}/R_K}$. At the end of the last post we constructed a group homomorphism ${F:Gal(\overline{K}/K)\rightarrow Cl(R_K)}$ which was defined by the property that ${E^\sigma=F(\sigma)\cdot E}$.

Now we know that ${Cl(R_K)}$ is an abelian group, so in fact ${F}$ factors through (a now relabelled) ${F: Gal(K^{ab}/K)\rightarrow Cl(R_K)}$ where ${K^{ab}}$ is the maximal abelian extension of ${K}$. For the rest of the post we’ll sketch why a few things are true:

1) It turns out that ${H=K(j(E))}$ is the Hilbert class field of ${K}$.

2) Last time we stated that ${[\mathbb{Q}(j(E)): \mathbb{Q}]=[K(j(E)): K]=h_K}$, but given the previous statement, the proof is now easy from class field theory.

3) If ${E_1, \ldots, E_h}$ are a complete set of representatives of ${Ell(R_K)}$, then ${j(E_1), \ldots, j(E_h)}$ form a complete set of Galois conjugates for ${j(E)}$.

4) For any prime ${\frak{p}}$ of ${K}$, we have ${j(E)^{Frob_\frak{p}}=j(F(Frob_\frak{p})\cdot E)}$. Extending to all fractional ideals we get ${j(E)^{\psi_{H/K}(\frak{a})}=j(\frak{a}\cdot E)}$.

Let ${L=\overline{K}^{\ker F}}$. One can check directly that ${L=K(j(E))}$. This shows that ${L/K}$ is abelian. Let ${G=Gal(L/K)}$. By class field theory there is some associated conductor ${\frak{m}}$. Considering the composition with the Artin map ${I_\frak{m}\stackrel{\Psi_{L/K}}{\rightarrow} G\stackrel{F}{\rightarrow} CL(R_K)}$ one can see this is just the natural projection map. Now using the injectivity of ${F}$ one can extrapolate that ${\frak{m}=(1)}$ showing ${L/K}$ is unramified and hence contained in the Hilbert class field.

But class field theory tells us that ${I_\frak{m}}$ surjects onto ${Cl(R_K)}$ and hence ${F}$ is an isomorphism which shows ${H=L}$ and hence ${K(j(E))}$ is the Hilbert class field of ${K}$. We’ve already discussed the second item. The third just follows from identifying ${Ell(R_K)}$ with ${\{j(E_1), \ldots, j(E_h)\}}$ and using the fact that ${Ell(R_K)}$ is a ${Cl(R_K)}$-torsor. The fourth just comes from the fact that ${I_\frak{m}}$ consists of all fractional ideals because ${\frak{m}=(1)}$.

I know, this was pretty skimpy on details, but the point of this series should be to see some of the ideas and results from basic CM theory for elliptic curves and not full blown explanations. The results in this post are really cool in my mind because it takes these very classical purely field theory questions and converts them to geometric questions about elliptic curves and vice-versa. Given a quadratic imaginary ${K}$, it is easy to cook up an elliptic curve ${E}$ with CM by ${R_K}$. If you want to know the maximal unramified abelian extension of ${K}$, we now know we need only figure out ${j(E)}$ because ${H=K(j(E))}$.

# 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.

# Elliptic Curves over C

Today we’ll start exploring the fascinating topic of complex multiplication. I’ll assume familiarity with basic elliptic curve theory. For today we’ll review some of the basic theory when our elliptic curve ${E}$ is defined over ${\mathbb{C}}$. In this case, recall that an elliptic curve has complex multiplication by ${R}$ if ${End(E)\simeq R}$ is an order in a quadratic imaginary field ${End(E)\otimes \mathbb{Q}}$. A quick reminder about the terminology. Recall that ${E\simeq \mathbb{C}/\Lambda}$ for some lattice ${\Lambda}$. Thus if we take a complex number ${\alpha}$ with the property that ${\alpha\Lambda\subset \Lambda}$, then “multiplication by ${\alpha}$” descends to a well-defined map ${E\rightarrow E}$ and hence we get the term “complex multiplication” by ${R}$.

Define a normalized elliptic curve to be a pair ${(E, [\cdot ])}$ where ${[\cdot]: R\stackrel{\sim}{\rightarrow} End(E)}$ is the isomorphism preserving the invariant differential, i.e. ${[\alpha]^*\omega=\alpha \omega}$. Now let ${K=R\otimes \mathbb{Q}}$ and let ${R_K}$ be the ring of integers of ${K}$. One important thing to be able to do is classify the elliptic curves with complex multiplication by ${R_K}$. Given any non-zero ideal ${\frak{a}\subset R_K}$ we have by our theory of projective modules over Dedekind rings that ${\frak{a}}$ is a lattice in ${\mathbb{C}}$ and hence ${E_\frak{a}=\mathbb{C}/\frak{a}}$ is an elliptic curve.

Moreover, using the fact that ${\frak{a}}$ is an ideal, we get that ${E_\frak{a}}$ has complex multiplication by ${R_K}$. Clearly altering ${\frak{a}}$ to ${c\frak{a}}$ by a principal ideal will give a homothetic lattice and hence an isomorphic curve which leads to the natural guess that elements of the class group ${Cl(R_K)}$ might give the different isomorphism classes of curves with CM by ${R_K}$. Let’s denote the set of isomorphism classes of elliptic curves with CM by ${R_K}$ as ${Ell(R_K)}$. It is easily verified that the natural action ${Cl(R_K)\times Ell(R_K)\rightarrow Ell(R_K)}$ by ${\frak{a}\cdot E_\Lambda = E_{\frak{a}^{-1}\Lambda}}$ is well-defined and simply transitive.

Thus the number of curves over ${\mathbb{C}}$ with CM by ${R_K}$ is the class number of ${R_K}$. As an example we could ask how many elliptic curves have complex multiplication by the Gaussian integers ${\mathbb{Z}[i]= \mathcal{O}_{\mathbb{Q}(i)}}$. Since the class number is ${1}$, there is a unique elliptic curve over ${\mathbb{C}}$ with complex multiplication by ${\mathbb{Z}[i]}$. This curve is well-known to have Weierstrass equation ${y^2=x^3-x}$, but from our above theory is also constructed as ${\mathbb{C}/(1\cdot\mathbb{Z}+i\cdot\mathbb{Z})}$.

Recall that ${E[n]}$ is the group of ${n}$-torsion points. Given ${\frak{a}}$ an ideal of ${R_K}$ we could also form ${E[\frak{a}]=\{P : [\alpha]P=0 \ \forall \alpha\in \frak{a}\}}$ with respect to the normalized ${[\cdot ]}$. Now since for any lattice ${\Lambda}$ we have ${\Lambda\subset \frak{a}^{-1}\Lambda}$ we get an isogenty ${E_\Lambda\rightarrow \frak{a}\cdot E_{\Lambda}}$. One can check that the kernel of this isogeny is exactly ${E[\frak{a}]}$ and that ${E[\frak{a}]}$ is a free ${R_K/\frak{a}}$-module of rank ${1}$ (less easy).

This allows us to compute that the degree of the isogeny ${E\rightarrow \frak{a}\cdot E}$ is ${N_{K/\mathbb{Q}}(\frak{a})}$ and for any ${\alpha\in \frak{a}}$ the map multiplication by ${\alpha}$ is an isogeny of degree ${|N_{K/\mathbb{Q}}(\alpha)|}$. This allows us to recover the classical fact that multiplication by ${n}$ is a degree ${n^2}$ isogeny (even an exercise in Hartshorne IV). One interesting thing about CM is that there are endomorphisms that have degree other than a square. Going back to our example we see that multiplication by ${1+i}$ on ${y^2=x^3-x}$ has degree ${2}$ since ${N(1+i)=1^2+1^2=2}$.

This was a short review and introduction. Next time we’ll move onto some more subtle issues that arise when we don’t work over ${\mathbb{C}}$.

# An Application to Elliptic Curves

Let’s do an application of our theorems about finitely generated projective modules over Dedekind domains. This is another one of those things that seems to be quite well known to experts, but it is not written anywhere that I know of. Suppose ${E}$ and ${F}$ are elliptic curves defined over a number field ${K}$ (this works in more generality, but this assumption will allow us to not break into lots of weird cases), and assume that the ${\ell}$-adic Tate modules are isomorphic for all ${\ell}$.

Recall briefly that the ${\ell}$-adic Tate module is just the limit over all the ${\ell^n}$ torsion points, i.e. ${\displaystyle T_\ell(E)=\lim_{\longleftarrow} E(\overline{K})[\ell^n]}$ as a ${G=Gal(\overline{K}/K)}$-module. We discussed this before in this post. An isogeny ${\phi: E\rightarrow F}$ defined over ${K}$ induces an action via pushforward ${T_\ell (E)\rightarrow T_\ell (F)}$ which is Galois equivariant. In fact, if ${\ell \nmid \text{deg}(\phi)}$, then it induces an isomorphism of Tate modules.

First define ${Hom(E,F)}$ to be the set of isogenies over ${K}$ (this could get me in trouble and has been the main delay in this post). If ${E}$ and ${F}$ are isogenous, then the natural action of ${End(E)}$ by composing turns ${Hom(E,F)}$ into a rank ${1}$ projective module over ${End(E)}$.

The question we want to ask ourselves is how much information do we get from the Tate module. It seems that surely this would not be enough information to recover the curve up to isomorphism, but recall that most elliptic curves do not have complex multiplication. Let’s start with that case. Suppose ${E}$ is non-CM so that ${End(E)\simeq \mathbb{Z}}$. The only endomorphisms are the isogenies given by multiplication by an integer.

The Tate conjecture formally says that there is an isomorphism ${Hom(E,F)\otimes_\mathbb{Z} \mathbb{Z}_\ell \stackrel{\sim}{\rightarrow}Hom_G(T_\ell(E), T_\ell(F))}$ (proved by Faltings in this case). This tells us that if you have some isomorphism ${T_\ell(E)\simeq T_\ell(F)}$, then there is an isogeny that induces it (maybe there is a less powerful tool to see this in this case). But we’ve now assumed that ${End(E)}$ is ${\mathbb{Z}}$, so ${Hom(E,F)}$ is not just a locally free module of rank ${1}$, but just plain free of rank ${1}$. All other isogenies are just composing this one with multiplication by an integer.

Let ${\phi}$ be the generator of ${Hom(E,F)}$. If ${deg(\phi)}$ is ${n}$, then all isogenies are divisible by ${n}$. Since we assume the Tate modules are isomorphic for all ${\ell}$, just pick some ${\ell}$ that divides ${n}$. Since Tate says there is an isogeny inducing the isomorphism we get a contradiction unless ${n=1}$. Thus ${deg(\phi)=1}$ and hence the generator is actually an isomorphism. This proves a fact I’ve seen stated, but haven’t seen written anywhere. If ${E}$ and ${F}$ are non-CM elliptic curves with isomorphic Tate modules for all ${\ell}$, then they must be isomorphic.

This should seem a little strange, because it basically says we can recover the curve up to isomorphism merely from knowing ${H_1}$. It turns out that weirder things can happen for CM curves, but we can use our structure theory from the last post to figure out what is going on. Suppose now that ${End(E)}$ is the full ring of integers in a quadratic imaginary field (the only other possibility is that it is merely an order in such a field).

It turns out that if ${E}$ and ${F}$ have isomorphic Tate modules for all ${\ell}$, then we can’t just conclude they are isomorphic. Here is a good way to think about this. We have that ${End(E)}$ is a Dedekind domain, and ${Hom(E,F)}$ is a rank ${1}$ projective module over it, so it is either generated by ${1}$ element and hence free in which case the same type of argument will show ${E}$ and ${F}$ must be isomorphic. The reason we get no information in the case where it is generated by two things is that these degrees can be coprime. In fact, they must be or else the same argument gives an isomorphism again.

This recently came up in something I was working on, and I couldn’t believe that I couldn’t find this fact stated anywhere (but several number theorists confirmed that this was something they knew). It might be because introductory books don’t want to assume the Tate conjecture, and anything that does assume the Tate conjecture assumes you can figure this out for yourself.