# BSD for a Large Class of Elliptic Curves

I’m giving up on the p-divisible group posts for awhile. I would have to be too technical and tedious to write anything interesting about enlarging the base. It is pretty fascinating stuff, but not blog material at the moment.

I’ve been playing around with counting fibration structures on K3 surfaces, and I just noticed something I probably should have been aware of for a long time. This is totally well-known, but I’ll give a slightly anachronistic presentation so that we can use results from 2013 to prove the Birch and Swinnerton-Dyer conjecture!! … Well, only in a case that has been known since 1973 when it was published by Artin and Swinnerton-Dyer.

Let’s recall the Tate conjecture for surfaces. Let ${k}$ be a finite field and ${X/k}$ a smooth, projective surface. We’ve written this down many times now, but the long exact sequence associate to the Kummer sequence

$\displaystyle 0\rightarrow \mu_{\ell}\rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m\rightarrow 0$

(for ${\ell\neq \text{char}(k)}$) gives us a cycle class map

$\displaystyle c_1: Pic(X_{\overline{k}})\otimes \mathbb{Q}_{\ell}\rightarrow H^2_{et}(X_{\overline{k}}, \mathbb{Q}_\ell(1))$

In fact, we could take Galois invariants to get our standard

$\displaystyle 0\rightarrow Pic(X)\otimes \mathbb{Q}_{\ell}\rightarrow H^2_{et}(X_{\overline{k}}, \mathbb{Q}_\ell(1))^G\rightarrow Br(X)[\ell^\infty]\rightarrow 0$

The Tate conjecture is in some sense the positive characteristic version of the Hodge conjecture. It conjectures that the first map is surjective. In other words, whenever an ${\ell}$-adic class “looks like” it could come from an honest geometric thing, then it does. But if the Tate conjecture is true, then this implies the ${\ell}$-primary part of ${Br(X)}$ is finite. We could spend some time worrying about independence of ${\ell}$, but it works, and hence the Tate conjecture is actually equivalent to finiteness of ${Br(X)}$.

Suppose now that ${X}$ is an elliptic K3 surface. This just means that there is a flat map ${X\rightarrow \mathbb{P}^1}$ where the fibers are elliptic curves (there are some degenerate fibers, but after some heavy machinery we could always put this into some nice form, we’re sketching an argument here so we won’t worry about the technical details of what we want “fibration” to mean). The generic fiber ${X_\eta}$ is a genus ${1}$ curve that does not necessarily have a rational point and hence is not necessarily an elliptic curve.

But we can just use a relative version of the Jacobian construction to produce a new fibration ${J\rightarrow \mathbb{P}^1}$ where ${J}$ is a K3 surface fiberwise isomorphic to ${X}$, but now ${J_\eta=Jac(X_\eta)}$ and hence is an elliptic curve. Suppose we want to classify elliptic fibrations that have ${J}$ as the relative Jacobian. We have two natural ideas to do this.

The first is that etale locally such a fibration is trivial, so you could consider all glueing data to piece such a thing together. The obstruction will be some Cech class that actually lives in ${H^2(X, \mathbb{G}_m)=Br(X)}$. In fancy language, you make these things as ${\mathbb{G}_m}$-gerbes which are just twisted relative moduli of sheaves. The class in ${Br(X)}$ is giving you the obstruction the existence of a universal sheaf.

A more number theoretic way to think about this is that rather than think about surfaces over ${k}$, we work with the generic fiber ${X_\eta/k(t)}$. It is well-known that the Weil-Chatelet group: ${H^1(Gal(k(t)^{sep}/k(t), J_\eta)}$ gives you the possible genus ${1}$ curves that could occur as generic fibers of such fibrations. This group is way too big though, because we only want ones that are locally trivial everywhere (otherwise it won’t be a fibration).

So it shouldn’t be surprising that the classification of such things is given by the Tate-Shafarevich group:

Ш $\displaystyle (J_\eta /k(t))=ker ( H^1(G, J_\eta)\rightarrow \prod H^1(G_v, (J_\eta)_v))$

Very roughly, I’ve now given a heuristic argument (namely that they both classify the same set of things) that ${Br(X)\simeq}$ Ш ${(J_\eta)}$, and it turns out that Grothendieck proved the natural map that comes form the Leray spectral sequence ${Br(X)\rightarrow}$ Ш${(J_\eta)}$ is an isomorphism (this rigorous argument might actually have been easier than the heuristic one because we’ve computed everything involved in previous posts, but it doesn’t give you any idea why one might think they are the same).

Theorem: If ${E/\mathbb{F}_q(t)}$ is an elliptic curve of height ${2}$ (occuring as the generic fiber of an elliptic K3 surface), then ${E}$ satisfies the Birch and Swinnerton-Dyer conjecture.

Idea: Using the machinery alluded to before, we spread out ${E}$ to an elliptic K3 surface ${X\rightarrow \mathbb{P}^1}$ over a finite field. As of this year, it seems the Tate conjecture is true for K3 surfaces (the proofs are all there, I’m not sure if they have been double checked and published yet). Thus ${Br(X)}$ is finite. Thus Ш${ (E)}$ is finite. But now it is well-known that if Ш${ (E)}$ being finite is equivalent to the Birch and Swinnerton-Dyer conjecture.

# Newton Polygons of p-Divisible Groups

I really wanted to move on from this topic, because the theory gets much more interesting when we move to ${p}$-divisible groups over some larger rings than just algebraically closed fields. Unfortunately, while looking over how Demazure builds the theory in Lectures on ${p}$-divisible Groups, I realized that it would be a crime to bring you this far and not concretely show you the power of thinking in terms of Newton polygons.

As usual, let’s fix an algebraically closed field of positive characteristic to work over. I was vague last time about the anti-equivalence of categories between ${p}$-divisible groups and ${F}$-crystals mostly because I was just going off of memory. When I looked it up, I found out I was slightly wrong. Let’s compute some examples of some slopes.

Recall that ${D(\mu_{p^\infty})\simeq W(k)}$ and ${F=p\sigma}$. In particular, ${F(1)=p\cdot 1}$, so in our ${F}$-crystal theory we get that the normalized ${p}$-adic valuation of the eigenvalue ${p}$ of ${F}$ is ${1}$. Recall that we called this the slope (it will become clear why in a moment).

Our other main example was ${D(\mathbb{Q}_p/\mathbb{Z}_p)\simeq W(k)}$ with ${F=\sigma}$. In this case we have ${1}$ is “the” eigenvalue which has ${p}$-adic valuation ${0}$. These slopes totally determine the ${F}$-crystal up to isomorphism, and the category of ${F}$-crystals (with slopes in the range ${0}$ to ${1}$) is anti-equivalent to the category of ${p}$-divisible groups.

The Dieudonné-Manin decomposition says that we can always decompose ${H=D(G)\otimes_W K}$ as a direct sum of vector spaces indexed by these slopes. For example, if I had a height three ${p}$-divisible group, ${H}$ would be three dimensional. If it decomposed as ${H_0\oplus H_1}$ where ${H_0}$ was ${2}$-dimensional (there is a repeated ${F}$-eigenvalue of slope ${0}$), then ${H_1}$ would be ${1}$-dimensional, and I could just read off that my ${p}$-divisible group must be isogenous to ${G\simeq \mu_{p^\infty}\oplus (\mathbb{Q}_p/\mathbb{Z}_p)^2}$.

In general, since we have a decomposition ${H=H_0\oplus H' \oplus H_1}$ where ${H'}$ is the part with slopes strictly in ${(0,1)}$ we get a decomposition ${G\simeq (\mu_{p^\infty})^{r_1}\oplus G' \oplus (\mathbb{Q}_p/\mathbb{Z}_p)^{r_0}}$ where ${r_j}$ is the dimension of ${H_j}$ and ${G'}$ does not have any factors of those forms.

This is where the Newton polygon comes in. We can visually arrange this information as follows. Put the slopes of ${F}$ in increasing order ${\lambda_1, \ldots, \lambda_r}$. Make a polygon in the first quadrant by plotting the points ${P_0=(0,0)}$, ${P_1=(\dim H_{\lambda_1}, \lambda_1 \dim H_{\lambda_1})}$, … , ${\displaystyle P_j=\left(\sum_{l=1}^j\dim H_{\lambda_l}, \sum_{l=1}^j \lambda_l\dim H_{\lambda_l}\right)}$.

This might look confusing, but all it says is to get from ${P_{j}}$ to ${P_{j+1}}$ make a line segment of slope ${\lambda_j}$ and make the segment go to the right for ${\dim H_{\lambda_j}}$. This way you visually encode the slope with the actual slope of the segment, and the longer the segment is the bigger the multiplicity of that eigenvalue.

But this way of encoding the information gives us something even better, because it turns out that all these ${P_i}$ must have integer coordinates (a highly non-obvious fact proved in the book by Demazure listed above). This greatly restricts our possibilities for Dieudonné ${F}$-crystals. Consider the height ${2}$ case. We have ${H}$ is two dimensional, so we have ${2}$ slopes (possibly the same). The maximal ${y}$ coordinate you could ever reach is if both slopes were maximal which is ${1}$. In that case you just get the line segment from ${(0,0)}$ to ${(2,2)}$. The lowest you could get is if the slopes were both ${0}$ in which case you get a line segment ${(0,0)}$ to ${(2,0)}$.

Every other possibility must be a polygon between these two with integer breaking points and increasing order of slopes. Draw it (or if you want to cheat look below). You will see that there are obviously only two other possibilities. The one that goes ${(0,0)}$ to ${(1,0)}$ to ${(2,1)}$ which is a slope ${0}$ and slope ${1}$ and corresponds to ${\mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$ and the one that goes ${(0,0)}$ to ${(2,1)}$. This corresponds to a slope ${1/2}$ with multiplicity ${2}$. This corresponds to the ${E[p^\infty]}$ for supersingular elliptic curves. That recovers our list from last time.

We now just have a bit of a game to determine all height ${3}$ ${p}$-divisible groups up to isogeny (and it turns out in this small height case that determines them up to isomorphism). You can just draw all the possibilities for Newton polygons as in the height ${2}$ case to see that the only ${6}$ possibilities are ${(\mu_{p^\infty})^3}$, ${(\mu_{p^\infty})^2\oplus \mathbb{Q}_p/\mathbb{Z}_p}$, ${\mu_{p^\infty}\oplus (\mathbb{Q}_p/\mathbb{Z}_p)^2}$, ${(\mathbb{Q}_p/\mathbb{Z}_p)^3}$, and then two others: ${G_{1/3}}$ which corresponds to the thing with a triple eigenvalue of slope ${1/3}$ and ${G_{2/3}}$ which corresponds to the thing with a triple eigenvalue of slope ${2/3}$.

To finish this post (and hopefully topic!) let’s bring this back to elliptic curves one more time. It turns out that ${D(E[p^\infty])\simeq H^1_{crys}(E/W)}$. Without reminding you of the technical mumbo-jumbo of crystalline cohomology, let’s think why this might be reasonable. We know ${E[p^\infty]}$ is always height ${2}$, so ${D(E[p^\infty])}$ is rank ${2}$. But if we consider that crystalline cohomology should be some sort of ${p}$-adic cohomology theory that “remembers topological information” (whatever that means), then we would guess that some topological ${H^1}$ of a “torus” should be rank ${2}$ as well.

Moreover, the crystalline cohomology comes with a natural Frobenius action. But if we believe there is some sort of Weil conjecture magic that also applies to crystalline cohomology (I mean, it is a Weil cohomology theory), then we would have to believe that the product of the eigenvalues of this Frobenius equals ${p}$. Recall in the “classical case” that the characteristic polynomial has the form ${x^2-a_px+p}$. So there are actually only two possibilities in this case, both slope ${1/2}$ or one of slope ${1}$ and the other of slope ${0}$. As we’ve noted, these are the two that occur.

In fact, this is a more general phenomenon. When thinking about ${p}$-divisible groups arising from algebraic varieties, because of these Weil conjecture type considerations, the Newton polygons must actually fit into much narrower regions and sometimes this totally forces the whole thing. For example, the enlarged formal Brauer group of an ordinary K3 surface has height ${22}$, but the whole Newton polygon is fully determined by having to fit into a certain region and knowing its connected component.

# More Classification of p-Divisible Groups

Today we’ll look a little more closely at ${A[p^\infty]}$ for abelian varieties and finish up a different sort of classification that I’ve found more useful than the one presented earlier as triples ${(M,F,V)}$. For safety we’ll assume ${k}$ is algebraically closed of characteristic ${p>0}$ for the remainder of this post.

First, let’s note that we can explicitly describe all ${p}$-divisible groups over ${k}$ up to isomorphism (of any dimension!) up to height ${2}$ now. This is basically because height puts a pretty tight constraint on dimension: ${ht(G)=\dim(G)+\dim(G^D)}$. If we want to make this convention, we’ll say ${ht(G)=0}$ if and only if ${G=0}$, but I’m not sure it is useful anywhere.

For ${ht(G)=1}$ we have two cases: If ${\dim(G)=0}$, then it’s dual must be the unique connected ${p}$-divisible group of height ${1}$, namely ${\mu_{p^\infty}}$ and hence ${G=\mathbb{Q}_p/\mathbb{Z}_p}$. The other case we just said was ${\mu_{p^\infty}}$.

For ${ht(G)=2}$ we finally get something a little more interesting, but not too much more. From the height ${1}$ case we know that we can make three such examples: ${(\mu_{p^\infty})^{\oplus 2}}$, ${\mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$, and ${(\mathbb{Q}_p/\mathbb{Z}_p)^{\oplus 2}}$. These are dimensions ${2}$, ${1}$, and ${0}$ respectively. The first and last are dual to each other and the middle one is self-dual. Last time we said there was at least one more: ${E[p^\infty]}$ for a supersingular elliptic curve. This was self-dual as well and the unique one-dimensional connected height ${2}$ ${p}$-divisible group. Now just playing around with the connected-étale decomposition, duals, and numerical constraints we get that this is the full list!

If we could get a bit better feel for the weird supersingular ${E[p^\infty]}$ case, then we would have a really good understanding of all ${p}$-divisible groups up through height ${2}$ (at least over algebraically closed fields).

There is an invariant called the ${a}$-number for abelian varieties defined by ${a(A)=\dim Hom(\alpha_p, A[p])}$. This essentially counts the number of copies of ${\alpha_p}$ sitting inside the truncated ${p}$-divisible group. Let’s consider the elliptic curve case again. If ${E/k}$ is ordinary, then we know ${E[p]}$ explicitly and hence can argue that ${a(E)=0}$. For the supersingular case we have that ${E[p]}$ is actually a non-split semi-direct product of ${\alpha_p}$ by itself and we get that ${a(E)=1}$. This shows that the ${a}$-number is an invariant that is equivalent to knowing ordinary/supersingular.

This is a phenomenon that generalizes. For an abelian variety ${A/k}$ we get that ${A}$ is ordinary if and only if ${a(A)=0}$ in which case the ${p}$-divisible group is a bunch of copies of ${E[p^\infty]}$ for an ordinary elliptic curve, i.e. ${A[p^\infty]\simeq E[p^\infty]^g}$. On the other hand, ${A}$ is supersingular if and only if ${A[p^\infty]\simeq E[p^\infty]^g}$ for ${E/k}$ supersingular (these two facts are pretty easy if you use the ${p}$-rank as the definition of ordinary and supersingular because it tells you the étale part and you mess around with duals and numerics again).

Now that we’ve beaten that dead horse beyond recognition, I’ll point out one more type of classification which is the one that comes up most often for me. In general, there is not redundant information in the triple ${(M, F, V)}$, but for special classes of ${p}$-divisible groups (for example the ones I always work with explained here) all you need to remember is the ${(M, F)}$ to recover ${G}$ up to isomorphism.

A pair ${(M,F)}$ of a free, finite rank ${W}$-module equipped with a ${\phi}$-linear endomorphism ${F}$ is sometimes called a Cartier module or ${F}$-crystal. Every Dieudonné module of a ${p}$-divisible group is an example of one of these. We could also consider ${H=M\otimes_W K}$ where ${K=Frac(W)}$ to get a finite dimensional vector space in characteristic ${0}$ with a ${\phi}$-linear endomorphism preserving the ${W}$-lattice ${M\subset H}$.

Passing to this vector space we would expect to lose some information and this is usually called the associated ${F}$-isocrystal. But doing this gives us a beautiful classification theorem which was originally proved by Diedonné and Manin. We have that ${H}$ is naturally an ${A}$-module where ${A=K[T]}$ is the noncommutative polynomial ring ${T\cdot a=\phi(a)\cdot T}$. The classification is to break up ${H\simeq \oplus H_\alpha}$ into a slope decomposition.

These ${\alpha}$ are just rational numbers corresponding to the slopes of the ${F}$ operator. The eigenvalues ${\lambda_1, \ldots, \lambda_n}$ of ${F}$ are not necessarily well-defined, but if we pick the normalized valuation ${ord(p)=1}$, then the valuations of the eigenvalues are well-defined. Knowing the slopes and multiplicities completely determines ${H}$ up to isomorphism, so we can completely capture the information of ${H}$ in a simple Newton polygon. Note that when ${H}$ is the ${F}$-isocrystal of some Dieudonné module, then the relation ${FV=VF=p}$ forces all slopes to be between 0 and 1.

Unfortunately, knowing ${H}$ up to isomorphism only determines ${M}$ up to equivalence. This equivalence is easily seen to be the same as an injective map ${M\rightarrow M'}$ whose cokernel is a torsion ${W}$-module (that way it becomes an isomorphism when tensoring with ${K}$). But then by the anti-equivalence of categories two ${p}$-divisible groups (in the special subcategory that allows us to drop the ${V}$) ${G}$ and ${G'}$ have equivalent Dieudonné modules if and only if there is a surjective map ${G' \rightarrow G}$ whose kernel is finite, i.e. ${G}$ and ${G'}$ are isogenous as ${p}$-divisible groups.

Despite the annoying subtlety in fully determining ${G}$ up to isomorphism, this is still really good. It says that just knowing the valuation of some eigenvalues of an operator on a finite dimensional characteristic ${0}$ vector space allows us to recover ${G}$ up to isogeny.

# A Quick User’s Guide to Dieudonné Modules of p-Divisible Groups

Last time we saw that if we consider a ${p}$-divisible group ${G}$ over a perfect field of characteristic ${p>0}$, that there wasn’t a whole lot of information that went into determining it up to isomorphism. Today we’ll make this precise. It turns out that up to isomorphism we can translate ${G}$ into a small amount of (semi-)linear algebra.

I’ve actually discussed this before here. But let’s not get bogged down in the details of the construction. The important thing is to see how to use this information to milk out some interesting theorems fairly effortlessly. Let’s recall a few things. The category of ${p}$-divisible groups is (anti-)equivalent to the category of Dieudonné modules. We’ll denote this functor ${G\mapsto D(G)}$.

Let ${W:=W(k)}$ be the ring of Witt vectors of ${k}$ and ${\sigma}$ be the natural Frobenius map on ${W}$. There are only a few important things that come out of the construction from which you can derive tons of facts. First, the data of a Dieudonné module is a free ${W}$-module, ${M}$, of finite rank with a Frobenius ${F: M\rightarrow M}$ which is ${\sigma}$-linear and a Verschiebung ${V: M\rightarrow M}$ which is ${\sigma^{-1}}$-linear satisfying ${FV=VF=p}$.

Fact 1: The rank of ${D(G)}$ is the height of ${G}$.

Fact 2: The dimension of ${G}$ is the dimension of ${D(G)/FD(G)}$ as a ${k}$-vector space (dually, the dimension of ${D(G)/VD(G)}$ is the dimension of ${G^D}$).

Fact 3: ${G}$ is connected if and only if ${F}$ is topologically nilpotent (i.e. ${F^nD(G)\subset pD(G)}$ for ${n>>0}$). Dually, ${G^D}$ is connected if and only if ${V}$ is topologically nilpotent.

Fact 4: ${G}$ is étale if and only if ${F}$ is bijective. Dually, ${G^D}$ is étale if and only if ${V}$ is bijective.

These facts alone allow us to really get our hands dirty with what these things look like and how to get facts back about ${G}$ using linear algebra. Let’s compute the Dieudonné modules of the two “standard” ${p}$-divisible groups: ${\mu_{p^\infty}}$ and ${\mathbb{Q}_p/\mathbb{Z}_p}$ over ${k=\mathbb{F}_p}$ (recall in this situation that ${W(k)=\mathbb{Z}_p}$).

Before starting, we know that the standard Frobenius ${F(a_0, a_1, \ldots, )=(a_0^p, a_1^p, \ldots)}$ and Verschiebung ${V(a_0, a_1, \ldots, )=(0, a_0, a_1, \ldots )}$ satisfy the relations to make a Dieudonné module (the relations are a little tricky to check because constant multiples ${c\cdot (a_0, a_1, \ldots )}$ for ${c\in W}$ involve Witt multiplication and should be done using universal properties).

In this case ${F}$ is bijective so the corresponding ${G}$ must be étale. Also, ${VW\subset pW}$ so ${V}$ is topologically nilpotent which means ${G^D}$ is connected. Thus we have a height one, étale ${p}$-divisible group with one-dimensional, connected dual which means that ${G=\mathbb{Q}_p/\mathbb{Z}_p}$.

Now we’ll do ${\mu_{p^\infty}}$. Fact 1 tells us that ${D(\mu_{p^\infty})\simeq \mathbb{Z}_p}$ because it has height ${1}$. We also know that ${F: \mathbb{Z}_p\rightarrow \mathbb{Z}_p}$ must have the property that ${\mathbb{Z}_p/F(\mathbb{Z}_p)=\mathbb{F}_p}$ since ${\mu_{p^\infty}}$ has dimension ${1}$. Thus ${F=p\sigma}$ and hence ${V=\sigma^{-1}}$.

The proof of the anti-equivalence proceeds by working at finite stages and taking limits. So it turns out that the theory encompasses a lot more at the finite stages because ${\alpha_{p^n}}$ are perfectly legitimate finite, ${p}$-power rank group schemes (note the system does not form a ${p}$-divisible group because multiplication by ${p}$ is the zero morphism). Of course taking the limit ${\alpha_{p^\infty}}$ is also a formal ${p}$-torsion group scheme. If we wanted to we could build the theory of Dieudonné modules to encompass these types of things, but in the limit process we would have finite ${W}$-module which are not necessarily free and we would get an extra “Fact 5” that ${D(G)}$ is free if and only if ${G}$ is ${p}$-divisible.

Let’s do two more things which are difficult to see without this machinery. For these two things we’ll assume ${k}$ is algebraically closed. There is a unique connected, ${1}$-dimensional ${p}$-divisible of height ${h}$ over ${k}$. I imagine without Dieudonné theory this would be quite difficult, but it just falls right out by playing with these facts.

Since ${D(G)/FD(G)\simeq k}$ we can choose a basis, ${D(G)=We_1\oplus \cdots \oplus We_h}$, so that ${F(e_j)=e_{j+1}}$ and ${F(e_h)=pe_1}$. Up to change of coordinates, this is the only way that eventually ${F^nD(G)\subset pD(G)}$ (in fact ${F^hD(G)\subset pD(G)}$ is the smallest ${n}$). This also determines ${V}$ (note these two things need to be justified, I’m just asserting it here). But all the phrase “up to change of coordinates” means is that any other such ${(D(G'),F',V')}$ will be isomorphic to this one and hence by the equivalence of categories ${G\simeq G'}$.

Suppose that ${E/k}$ is an elliptic curve. Now we can determine ${E[p^\infty]}$ up to isomorphism as a ${p}$-divisible group, a task that seemed out of reach last time. We know that ${E[p^\infty]}$ always has height ${2}$ and dimension ${1}$. In previous posts, we saw that for an ordinary ${E}$ we have ${E[p^\infty]^{et}\simeq \mathbb{Q}_p/\mathbb{Z}_p}$ (we calculated the reduced part by using flat cohomology, but I’ll point out why this step isn’t necessary in a second).

Thus for an ordinary ${E/k}$ we get that ${E[p^\infty]\simeq E[p^\infty]^0\oplus \mathbb{Q}_p/\mathbb{Z}_p}$ by the connected-étale decomposition. But height and dimension considerations tell us that ${E[p^\infty]^0}$ must be the unique height ${1}$, connected, ${1}$-dimensional ${p}$-divisible group, i.e. ${\mu_{p^\infty}}$. But of course we’ve been saying this all along: ${E[p^\infty]\simeq \mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$.

If ${E/k}$ is supersingular, then we’ve also calculated previously that ${E[p^\infty]^{et}=0}$. Thus by the connected-étale decomposition we get that ${E[p^\infty]\simeq E[p^\infty]^0}$ and hence must be the unique, connected, ${1}$-dimensional ${p}$-divisible group of height ${2}$. For reference, since ${ht(G)=\dim(G)+\dim(G^D)}$ we see that ${G^D}$ is also of dimension ${1}$ and height ${2}$. If it had an étale part, then it would have to be ${\mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$ again, so ${G^D}$ must be connected as well and hence is the unique such group, i.e. ${G\simeq G^D}$. It is connected with connected dual. This gives us our first non-obvious ${p}$-divisible group since it is not just some split extension of ${\mu_{p^\infty}}$‘s and ${\mathbb{Q}_p/\mathbb{Z}_p}$‘s.

If we hadn’t done these previous calculations, then we could still have gotten these results by a slightly more general argument. Given an abelian variety ${A/k}$ we have that ${A[p^\infty]}$ is a ${p}$-divisible group of height ${2g}$ where ${g=\dim A}$. Using Dieudonné theory we can abstractly argue that ${A[p^\infty]^{et}}$ must have height less than or equal to ${g}$. So in the case of an elliptic curve it is ${1}$ or ${0}$ corresponding to the ordinary or supersingular case respectively, and the proof would be completed because ${\mathbb{Q}_p/\mathbb{Z}_p}$ is the unique étale, height ${1}$, ${p}$-divisible group.

# Frobenius Semi-linear Algebra: 1

Today I want to explain some “well-known” facts in semilinear algebra. Here’s the setup. For safety we’ll assume ${k}$ is algebraically closed of characteristic ${p>0}$ (but merely being perfect should suffice for the main point later). Let ${V}$ be a finite dimensional vector space over ${k}$. Consider some ${p}$-semilinear operator on ${V}$ say ${\phi: V\rightarrow V}$. The fact that we are working with ${p}$ instead of ${p^{-1}}$ is mostly to not scare people. I think ${p^{-1}}$ actually appears more often in the literature and the theory is equivalent by “dualizing.”

All this means is that it is a linear operator satisfying the usual properties ${\phi(v+w)=\phi(v)+\phi(w)}$, etc, except for the scalar rule in which we scale by a factor of ${p}$, so ${\phi(av)=a^p\phi(v)}$. This situation comes up surprisingly often in positive characteristic geometry, because often you want to analyze some long exact sequence in cohomology associated to a short exact sequence which involves the Frobenius map or the Cartier operator. The former will induce a ${p}$-linear map of vector spaces and the latter induces a ${p^{-1}}$-linear map.

The facts we’re going to look at I’ve found in three or so papers just saying “from a well-known fact about ${p^{-1}}$-linear operators…” I wish there was a book out there that developed this theory like a standard linear algebra text so that people could actually give references. The proof today is a modification of that given in Dieudonne’s Lie Groups and Lie Hyperalgebras over a Field of Characteristic ${p>0}$ II (section 10).

Let’s start with an example. In the one-dimensional case we have the following ${\phi: k\rightarrow k}$. If the map is non-trivial, then it is bijective. More importantly we can just write down every one of these because if ${\phi(1)=a}$, then

$\displaystyle \begin{array}{rcl} \phi(x) & = & \phi(x\cdot 1) \\ & = & x^p\phi(1) \\ & = & ax^p \end{array}$

In fact, we can always find some non-zero fixed element, because this amounts to solving ${ax^p-x=x(ax^{p-1}-1)=0}$, i.e. finding a solution to ${ax^{p-1}-1=0}$ which we can do by being algebraically closed. This element ${b}$ obviously serves as a basis for ${k}$, but to set up an analogy we also see that ${Span_{\mathbb{F}_p}(b)}$ are all of the fixed points of ${\phi}$. In general ${V}$ will breakup into parts. The part that ${\phi}$ acts bijectively on will always have a basis of fixed elements whose ${\mathbb{F}_p}$-span consists of exactly the fixed points of ${\phi}$. Of course, this could never happen in linear algebra because finding a fixed basis implies the operator is the identity.

Let’s start by proving this statement. Suppose ${\phi: V\rightarrow V}$ is a ${p}$-semilinear automorphism. We want to find a basis of fixed elements. We essentially mimic what we did before in a more complicated way. We induct on the dimension of ${V}$. If we can find a single ${v_1}$ fixed by ${\phi}$, then we would be done for the following reason. We kill off the span of ${v_1}$, then by the inductive hypothesis we can find ${v_2, \ldots, v_n}$ a fixed basis for the quotient. Together these make a fixed basis for all of ${V}$.

Now we need to find a single fixed ${v_1}$ by brute force. Consider any non-zero ${w\in V}$. We start taking iterates of ${w}$ under ${\phi}$. Eventually they will become linearly dependent, so we consider ${w, \phi(w), \ldots, \phi^k(w)}$ for the minimal ${k}$ such that this is a linearly dependent set. This means we can find some coefficients that are not all ${0}$ for which ${\sum a_j \phi^j(w)=0}$.

Let’s just see what must be true of some fictional ${v_1}$ in the span of these elements such that ${\phi(v_1)=v_1}$. Well, ${v_1=\sum b_j \phi^j(w)}$ must satisfy ${v_1=\phi(v_1)=\sum b_j^p \phi^{j+1}(w)}$.

To make this easier to parse, let’s specialize to the case that ${k=3}$. This means that ${a_0 w+a_1\phi(w)+a_2\phi^2(w)=0}$ and by assumption the coefficient on this top power can’t be zero, so we rewrite the top power ${\phi^2(w)=-(a_0/a_2)w - (a_1/a_2)\phi(w)}$.

The other equation is

$\displaystyle \begin{array}{rcl} b_0w+b_1\phi(w) & = & b_0^p\phi(w)+b_1^p\phi^2(w)\\ & = & -(a_0/a_2)b_1^pw +(b_0^p-(a_1/a_2)b_1^p)\phi(w) \end{array}$

Comparing coefficients ${b_0=-(a_0/a_2)b_1^p}$ and then forward substituting ${b_1=-(a_0/a_2)^pb_1^{p^2}-(a_1/a_2)b_1^p}$. Ah, but we know the ${a_j}$ and this only involves the unknown ${b_1}$. So since ${k}$ is algebraically closed we can solve to find such a ${b_1}$. Then since we wrote all our other coefficients in terms of ${b_1}$ we actually can produce a fixed ${v_1}$ by brute force determining the coefficients of the vector in terms of our linear dependence coefficients.

There was nothing special about ${k=3}$ here. In general, this trick will work because it only involves the fact that applying ${\phi}$ cycled the vectors forward by one which allows us to keep forward substituting all the equations from the comparison of coefficients to get everything in terms of the highest one including the highest one which transformed the problem into solving a single polynomial equation over our algebraically closed field.

This completes the proof that if ${\phi}$ is bijective, then there is a basis of fixed vectors. The fact that ${V^\phi=Span_{\mathbb{F}_p}(v_1, \ldots, v_n)}$ is pretty easy after that. Of course, the ${\mathbb{F}_p}$-span is contained in the fixed points because by definition the prime subfield of ${k}$ is exactly the fixed elements of ${x\mapsto x^p}$. On the other hand, if ${c=\sum a_jv_j}$ is fixed, then ${c=\phi(c)=\sum a_j^p \phi(v_j)=\sum a_j^p v_j}$ shows that all the coefficients must be fixed by Frobenius and hence in ${\mathbb{F}_p}$.

Here’s how this is useful. Recall the post on the fppf site. We said that if we wanted to understand the ${p}$-torsion of certain cohomology with coefficients in ${\mathbb{G}_m}$ (Picard group, Brauer group, etc), then we should look at the flat cohomology with coefficients in ${\mu_p}$. If we specialize to the case of curves we get an isomorphism ${H^1_{fl}(X, \mu_p)\simeq Pic(X)[p]}$.

Recall the exact sequence at the end of that post. It told us that via the ${d\log}$ map ${H^1_{fl}(X, \mu_p)=ker(C-I)=H^0(X, \Omega^1)^C}$. Now we have a ridiculously complicated way to prove the following well-known fact. If ${E}$ is an ordinary elliptic curve over an algebraically closed field of characteristic ${p>0}$, then ${E[p]\simeq \mathbb{Z}/p}$. In fact, we can prove something slightly more general.

By definition, a curve is of genus ${g}$ if ${H^0(X, \Omega^1)}$ is ${g}$-dimensional. We’ll say ${X}$ is ordinary if the Cartier operator ${C}$ is a ${p^{-1}}$-linear automorphism (I’m already sweeping something under the rug, because to even think of the Cartier operator acting on this cohomology group we need a hypothesis like ordinary to naturally identify some cohomology groups).

By the results in this post we know that the structure of ${H^0(X, \Omega^1)^C}$ as an abelian group is ${\mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$ where there are ${g}$ copies. Thus in more generality this tells us that ${Jac(X)[p]\simeq Pic(X)[p]\simeq H^0(X, \Omega^1)^C\simeq \mathbb{Z}/p\oplus \cdots \oplus \mathbb{Z}/p}$. In particular, since for an elliptic curve (genus 1) we have ${Jac(E)=E}$, this statement is exactly ${E[p]\simeq \mathbb{Z}/p}$.

This point is a little silly, because Silverman seems to just use this as the definition of an ordinary elliptic curve. Hartshorne uses the Hasse invariant in which case it is quite easy to derive that the Cartier operator is an automorphism (proof: it is Serre dual to the Frobenius which by the Hasse invariant definition is an automorphism). Using this definition, I’m actually not sure I’ve ever seen a derivation that ${E[p]\simeq \mathbb{Z}/p}$. I’d be interested if there is a lower level way of seeing it than going through this flat cohomology argument (Silverman cites a paper of Duering, but it’s in German).

# Serre-Tate Theory 2

I guess this will be the last post on this topic. I’ll explain a tiny bit about what goes into the proof of this theorem and then why anyone would care that such canonical lifts exist. On the first point, there are tons of details that go into the proof. For example, Nick Katz’s article, Serre-Tate Local Moduli, is 65 pages. It is quite good if you want to learn more about this. Also, Messing’s book The Crystals Associated to Barsotti-Tate Groups is essentially building the machinery for this proof which is then knocked off in an appendix. So this isn’t quick or easy by any means.

On the other hand, I think the idea of the proof is fairly straightforward. Let’s briefly recall last time. The situation is that we have an ordinary elliptic curve ${E_0/k}$ over an algebraically closed field of characteristic ${p>2}$. We want to understand ${Def_{E_0}}$, but in particular whether or not there is some distinguished lift to characteristic ${0}$ (this will be an element of ${Def_{E_0}(W(k))}$.

To make the problem more manageable we consider the ${p}$-divisible group ${E_0[p^\infty]}$ attached to ${E_0}$. In the ordinary case this is the enlarged formal Picard group. It is of height ${2}$ whose connected component is ${\widehat{Pic}_{E_0}\simeq\mu_{p^\infty}}$. There is a natural map ${Def_{E_0}\rightarrow Def_{E_0[p^\infty]}}$ just by mapping ${E/R \mapsto E[p^\infty]}$. Last time we said the main theorem was that this map is an isomorphism. To tie this back to the flat topology stuff, ${E_0[p^\infty]}$ is the group representing the functor ${A\mapsto H^1_{fl}(E_0\otimes A, \mu_{p^\infty})}$.

The first step in proving the main theorem is to note two things. In the (split) connected-etale sequence

$\displaystyle 0\rightarrow \mu_{p^\infty}\rightarrow E_0[p^\infty]\rightarrow \mathbb{Q}_p/\mathbb{Z}_p\rightarrow 0$

we have that ${\mu_{p^\infty}}$ is height one and hence rigid. We have that ${\mathbb{Q}_p/\mathbb{Z}_p}$ is etale and hence rigid. Thus given any deformation ${G/R}$ of ${E_0[p^\infty]}$ we can take the connected-etale sequence of this and see that ${G^0}$ is the unique deformation of ${\mu_{p^\infty}}$ over ${R}$ and ${G^{et}=\mathbb{Q}_p/\mathbb{Z}_p}$. Thus the deformation functor can be redescribed in terms of extension classes of two rigid groups ${R\mapsto Ext_R^1(\mathbb{Q}_p/\mathbb{Z}_p, \mu_{p^\infty})}$.

Now we see what the canonical lift is. Supposing our isomorphism of deformation functors, it is the lift that corresponds to the split and hence trivial extension class. So how do we actually check that this is an isomorphism? Like I said, it is kind of long and tedious. Roughly speaking you note that both deformation functors are prorepresentable by formally smooth objects of the same dimension. So we need to check that the differential is an isomorphism on tangent spaces.

Here’s where some cleverness happens. You rewrite the differential as a composition of a whole bunch of maps that you know are isomorphisms. In particular, it is the following string of maps: The Kodaira-Spencer map ${T\stackrel{\sim}{\rightarrow} H^1(E_0, \mathcal{T})}$ followed by Serre duality (recall the canonical is trivial on an elliptic curve) ${H^1(E_0, \mathcal{T})\stackrel{\sim}{\rightarrow} Hom_k(H^1(E_0, \Omega^1), H^1(E_0, \mathcal{O}_{E_0}))}$. The hardest one was briefly mentioned a few posts ago and is the dlog map which gives an isomorphism ${H^2_{fl}(E_0, \mu_{p^\infty})\stackrel{\sim}{\rightarrow} H^1(E_0, \Omega^1)}$.

Now noting that ${H^2_{fl}(E_0, \mu_{p^\infty})=\mathbb{Q}_p/\mathbb{Z}_p}$ and that ${T_0\mu_{p^\infty}\simeq H^1(E_0, \mathcal{O}_{E_0})}$ gives us enough compositions and isomorphisms that we get from the tangent space of the versal deformation of ${E_0}$ to the tangent space of the versal deformation of ${E_0[p^\infty]}$. As you might guess, it is a pain to actually check that this is the differential of the natural map (and in fact involves further decomposing those maps into yet other ones). It turns out to be the case and hence ${Def_{E_0}\rightarrow Def_{E_0[p^\infty]}}$ is an isomorphism and the canonical lift corresponds to the trivial extension.

But why should we care? It turns out the geometry of the canonical lift is very special. This may not be that impressive for elliptic curves, but this theory all goes through for any ordinary abelian variety or K3 surface where it is much more interesting. It turns out that you can choose a nice set of coordinates (“canonical coordinates”) on the base of the versal deformation and a basis of the de Rham cohomology of the family that is adapted to the Hodge filtration such that in these coordinates the Gauss-Manin connection has an explicit and nice form.

Also, the canonical lift admits a lift of the Frobenius which is also nice and compatible with how it acts on the above chosen basis on the de Rham cohomology. These coordinates are what give the base of the versal deformation the structure of a formal torus (product of ${\widehat{\mathbb{G}_m}}$‘s). One can then exploit all this nice structure to prove large open problems like the Tate conjecture in the special cases of the class of varieties that have these canonical lifts.

# Serre-Tate Theory 1

Today we’ll try to answer the question: What is Serre-Tate theory? It’s been a few years, but if you’re not comfortable with formal groups and ${p}$-divisible groups, I did a series of something like 10 posts on this topic back here: formal groups, p-divisible groups, and deforming p-divisible groups.

The idea is the following. Suppose you have an elliptic curve ${E/k}$ where ${k}$ is a perfect field of characteristic ${p>2}$. In most first courses on elliptic curves you learn how to attach a formal group to ${E}$ (chapter IV of Silverman). It is suggestively notated ${\widehat{E}}$, because if you unwind what is going on you are just completing the elliptic curve (as a group scheme) at the identity.

Since an elliptic curve is isomorphic to it’s Jacobian ${Pic_E^0}$ there is a conflation that happens. In general, if you have a variety ${X/k}$ you can make the same formal group by completing this group scheme and it is called the formal Picard group of ${X}$. Although, in general you’ll want to do this with the Brauer group or higher analogues to guarantee existence and smoothness. Then you prove a remarkable fact that the elliptic curve is ordinary if and only if the formal group has height ${1}$. In particular, since the ${p}$-divisible group is connected and ${1}$-dimensional it must be isomorphic to ${\mu_{p^\infty}}$.

It might seem silly to think in these terms, but there is another “enlarged” ${p}$-divisible group attached to ${E}$ which always has height ${2}$. This is the ${p}$-divisible group you get by taking the inductive limit of the finite group schemes that are the kernel of multiplication by ${p^n}$. It is important to note that these are non-trivial group schemes even if they are “geometrically trivial” (and is the reason I didn’t just call it the “${p^n}$-torsion”). We’ll denote this in the usual way by ${E[p^\infty]}$.

I don’t really know anyone that studies elliptic curves that phrases it this way, but since this theory must be generalized in a certain way to work for other varieties like K3 surfaces I’ll point out why this should be thought of as an enlarged ${p}$-divisible group. It is another standard fact that ${E}$ is ordinary if and only if ${E[p^\infty]\simeq \mu_{p^\infty}\oplus \mathbb{Q}_p/\mathbb{Z}_p}$. In fact, you can just read off the connected-etale decomposition:

$\displaystyle 0\rightarrow \mu_{p^\infty}\rightarrow E[p^\infty] \rightarrow \mathbb{Q}_p/\mathbb{Z}_p\rightarrow 0$

We already noted that ${\widehat{E}\simeq \mu_{p^\infty}}$, so the ${p}$-divisible group ${E[p^\infty]}$ is a ${1}$-dimensional, height ${2}$ formal group whose connected component is the first one we talked about, i.e. ${E[p^\infty]}$ is an enlargement of ${\widehat{E}}$. For a general variety, this enlarged formal group can be defined, but it is a highly technical construction and would take a lot of work to check that it even exists and satisfies this property. Anyway, this enlarged group is the one we need to work with otherwise our deformation space will be too small to make the theory work.

Here’s what Serre-Tate theory is all about. If you take a deformation of your elliptic curve ${E}$ say to ${E'}$, then it turns out that ${E'[p^\infty]}$ is a deformation of the ${p}$-divisible group ${E[p^\infty]}$. Thus we have a natural map ${\gamma: Def_E \rightarrow Def_{E[p^\infty]}}$. The point of the theory is that it turns out that this map is an isomorphism (I’m still assuming ${E}$ is ordinary here). This is great news, because the deformation theory of ${p}$-divisible groups is well-understood. We know that the versal deformation of ${E[p^\infty]}$ is just ${Spf(W[[t]])}$. The deformation problem is unobstructed and everything lives in a ${1}$-dimensional family.

Of course, let’s not be silly. I’m pointing all this out because of the way in which it generalizes. We already knew this was true for elliptic curves because for any smooth, projective curve the deformations are unobstructed since the obstruction lives in ${H^2}$. Moreover, the dimension of the space of deformations is given by the dimension of ${H^1(E, \mathcal{T})}$. But for an elliptic curve ${\mathcal{T}\simeq \mathcal{O}_X}$, so by Serre duality this is one-dimensional.

On the other hand, we do get some actual information from the Serre-Tate theory isomorphism because ${Def_{E[p^\infty]}}$ carries a natural group structure. Thus an ordinary elliptic curve has a “canonical lift” to characteristic ${0}$ which comes from the deformation corresponding to the identity.

# Moduli of Vector Bundles on Elliptic Curves

We’ve been talking about moduli problems, and one notoriously hard type of moduli problem is to “classify” vector bundles on some variety. Even when you restrict yourself to some special case like a specific surface and try to classify only vector bundles of certain rank or Chern class you run into trouble. To this day, these types of problems are a very active area of research.

It is fairly well-known (due to Grothendieck, but a problem in Hartshorne as well) that any finite rank vector bundle over ${\mathbb{P}^1}$ is just a finite direct sum ${\oplus_i\mathcal{O}(n_i)}$. The next interesting case would be to move up to genus ${1}$ curves. It turns out that Atiyah in the 1957 paper, Vector Bundles over an Elliptic Curve, worked out a classification. I just learned about this a month or two ago, and it is pretty cool so I’d like to briefly describe the idea.

Fix an algebraically closed field ${k}$ of characteristic ${0}$, and let ${E/k}$ be an elliptic curve. Define ${V(r,d)}$ to be the set of indecomposable vector bundles (up to isomorphism) on ${E}$ of rank ${r}$ and degree ${d}$, where degree just means the degree of the determinant. It is well known that ${V(1,0)}$ can be identified with ${E}$, because ${V(1,0)=Pic^0(E)}$ is the set of degree ${0}$ divisors. In particular, this says that the moduli space of degree ${0}$ divisors (line bundles) on ${E}$ is fine and representable by ${E}$.

Let’s continue with this idea of classifying degree ${0}$ vector bundles. It turns out there is a unique vector bundle ${T_r\in V(r,0)}$ with the property that ${\Gamma (E, T_r)\neq 0}$, i.e. there are non-trivial global sections. Now, recall how ${V(1,0)}$ works. Let ${0\in E}$ be the origin. Then our isomorphism ${E\rightarrow V(1,0)}$ is given by ${P\mapsto \mathcal{O}(-P)\otimes \mathcal{O}([0])}$. Our ${T_r}$ is going to play the role of ${\mathcal{O}([0])}$ here. In complete analogy we get a bijection ${V(1,0)\rightarrow V(r,0)}$ by ${L\mapsto L\otimes T_r}$.

This finishes off the case of degree ${0}$ vector bundles, because we get that ${E\simeq V(1,0)\simeq V(r,0)}$ (roughly speaking, of course there’s a lot more structure we need to know about to say something about the moduli spaces).

In more general situations we can use the same sort of trick. Note that we always have a bijection ${V(r,d)\rightarrow V(r, d+nr)}$ given by ${V\mapsto V\otimes \mathcal{O}(n[0])}$. Thus one reduction Atiyah makes right away is that (by the Euclidean algorithm) we only need to consider ${V(r,d)}$ for ${0\leq d < r}$.

Now suppose the rank and degree are non-zero, and ${n=\text{gcd}(r,d)}$. We can establish a bijection ${E\rightarrow V(r,d)}$, so that the composed map ${E\rightarrow V(r,d)\stackrel{det}{\rightarrow} V(1, d)\rightarrow E}$ is the map multiplication by ${n}$. The idea again is that one can find a certain vector bundle ${T_{r,d}\in V(r,d)}$ that is unique up to isomorphism from which all the others can be produced. Most people call this the Atiyah bundle. As a corollary, we see that if ${(r,d)=1}$, then the moduli space of indecomposable rank ${r}$ and degree ${d}$ vector bundles on ${E}$ is fine and again representable by ${E}$.

In modern language, we could work out that stable or semi-stable sheaves are the appropriate things to classify if we want a hope for the moduli space to be fine. It turns out that this condition of certain numerical invariants being coprime often implies that the sheaves are stable. For example, on an elliptic curve, a K3 surface, or even when dealing with relative moduli of sheaves on a K3 fibration. We may get to this another day.

# 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.

# 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.