Tom Lovering's Blog

Why Modularity implies Fermat’s Last Theorem

January 8, 2012 in Mathematics | Leave a comment

In this post we shall sketch the reduction of Fermat’s Last Theorem to Wiles’ Theorem that every semistable elliptic curve over ${{\mathbb Q}}$ is modular. I believe the ideas are due mainly to Frey, with a helpful big black box courtesy of Serre and Ribet. My main reference is the article of Stephens from the 1995 conference.

Suppose, for a fixed prime ${p \geq 5}$ , that there are integers ${a,b,c}$ with ${a^p+b^p+c^p = 0}$ , ${abc \not=0}$ and, assuming wlog that ${a \equiv 3 \mod 4}$ and ${b}$ is even, form the elliptic curve

$\displaystyle E_0: y^2 = x(x-a^p)(x+b^p).$

Then, by a direct computation, ${E_0}$ is semistable with minimal discriminant ${\Delta_{E_0} = 2^{-8} (abc)^{2p}}$ and conductor ${N_{E_0}=\prod_{l|abc} l}$ .

Our strategy is to study the representation ${\rho_E: G_{\mathbb Q} \rightarrow GL_2(\mathbb{F}_p)}$ coming from the action of Galois on the ${p}$ -torsion points of ${E}$ over ${\bar{{\mathbb Q}}}$ (note that this is the same ${p}$ as in the previous paragraph).

In general, such Galois representations can be restricted to ${G_{{\mathbb Q}_l}}$ , considered as the decomposition group of the prime ${l}$ in ${G_{\mathbb Q}}$ (that this is only defined up to conjugacy, determined by a choice of embedding ${\bar{{\mathbb Q}} \rightarrow \bar{{\mathbb Q}}_l}$ is not important for us – we’ve already picked a basis for the torsion points anyway). The typical `good’ local behaviour at these primes are, for ${l \not= p}$ , that ${\rho_E}$ be unramified at ${l}$ (in the sense that the absolute inertia group acts trivially), and for ${l=p}$ a more general rather more technical condition which we describe by saying that ${\rho_E}$ is flat at ${p}$ .

Let us (nonstandardly) call a prime ${l}$ bad (for a representation ${\rho_E}$ ) if ${l \not= p}$ and ${\rho_E}$ is ramified at ${l}$ , or ${l=p}$ and ${\rho_E}$ fails to be flat at ${p}$ . The important fact we need is that bad primes for such representations coming from elliptic curves are all flagged up by the minimal discriminant.

Proposition 1 For ${E}$ a semistable elliptic curve, ${l}$ is a bad prime for ${\rho_E}$ iff ${p}$ does not divide ${ord_l(\Delta_E)}$ .

In particular, this implies for our curve ${E_0}$ that the only bad prime for ${\rho_{E_0}}$ is ${2}$ . People suspected that this possibility of such an `arithmetically simple’ representation coming from an elliptic curve should be pretty unlikely.

Now, as well as these representations on ${\mathbb{F}_p}$ coming from torsion points on an elliptic curve, it is possible to construct similar representations coming from certain modular forms. Indeed, from a newform ${f}$ of weight 2, whose Fourier coefficients and character are rational, it is possible to form a representation ${\rho_f: G_{\mathbb Q} \rightarrow GL_2(\mathbb{F}_p)}$ . Serre made a series of conjectures about which representations could be obtained from newforms, one of which was the following result, a hard theorem proved by Ribet.

Theorem 2 (Serre’s Epsilon Conjecture) If ${f}$ is a weight two newform of conductor ${N}$ and ${\rho_f}$ is absolutely irreducible, then letting ${N'}$ be the factor of ${N}$ obtained by throwing away all the non-bad primes which divide ${N}$ at most once, it is possible to find a different weight two newform ${g}$ of conductor ${N'}$ such that ${\rho_f \cong \rho_g}$ .

Finally, we need some results about modular curves. In fact, we just need the following basic fact. If ${E}$ is a modular curve, then the representation ${\rho_E}$ is equal to ${\rho_f}$ for some newform ${f}$ of conductor ${N_E}$ .

Proposition 3 The curve ${E_0}$ above (constructed from a contradiction of Fermat’s Last Theorem) is not modular.

First, one shows that ${\rho_{E_0}}$ defined above is absolutely irreducible. Now, if ${E_0}$ it were modular, ${\rho_{E_0}}$ would be isomorphic to a representation ${\rho_f}$ coming from a newform of conductor ${N_{E_0}}$ . But looking at ${N_{E_0}}$ and the fact that ${2}$ is the only bad prime of ${\rho_f}$ , Ribet’s theorem implies that there is a newform ${g}$ of weight ${2}$ and conductor ${2}$ such that ${\rho_f \cong \rho_g}$ . But the space ${S_2(\Gamma_0(2))}$ has dimension ${0}$ , by the basic theory of modular forms, so no such newform exists.

Having proved proposition 3, recalling that ${E_0}$ was a semistable elliptic curve over ${{\mathbb Q}}$ , Wiles’ theorem that every semistable elliptic curve over ${{\mathbb Q}}$ is modular gives a contradiction and thus proves that there are no integers ${a,b,c}$ with ${a^p+b^p+c^p=0}$ , ${abc \not= 0}$ .

Hilbert 90 from 1-Dimensional Galois Descent

December 31, 2011 in Mathematics | 1 comment

In this short post I wish to record the proof of Hilbert’s Theorem 90 by interpreting it as the 1-dimensional case of Galois descent. The ideas all come from the `Arcata’ in SGA 4 1/2. Note that in some sense this completes the Kummer Theory argument in the previous post, except now I am assuming Galois Descent as a black box instead. Eventually, I intend to upload to this blog an article which will include a proof of Galois Descent (modelled on either the proof of Grothendieck’s fpqc descent or the more general monadic descent theory), so then finally everything will be in place . Alternatively, there are great articles about this elsewhere online. Maybe this is a good example of where you should generalise a problem so much it becomes easy.

Firstly, what is the statement of Galois Descent? Let ${L/k}$ be an algebraic Galois extension, with Galois group ${G}$ . Consider an ${L}$ -vector space ${V}$ . A semilinear ${G}$ -action on ${V}$ is a set of ${k}$ -endomorphisms ${\{\psi_\sigma\}}$ of ${V}$ indexed by elements of ${G}$ such that

$\displaystyle \psi_\sigma(\lambda v) = \sigma(\lambda) \psi_\sigma(v),$

and respecting the group operation on indices: ${\psi_{\tau \sigma} = \psi_\tau \circ \psi_\sigma}$ . The `semilinearity’ condition could be viewed as saying that ${\psi_\sigma}$ is an ${L}$ -linear map ${L\otimes_L V \rightarrow V}$ where the map used to tensor ${L}$ with itself is ${\sigma: L \rightarrow L}$ (rather than the identity map, which would be more conventional).

Given an ${L}$ -vector space ${V}$ equipped with a semilinear ${G}$ -action, we can consider the ${k}$ -vector space ${V^G = \{v \in V| \psi_\sigma(v) = v \ \forall \sigma \in G\}}$ of elements which are invariant under the semilinear action. Conversely, given a ${k}$ -vector space ${W}$ we can tensor up to get an ${L}$ vector space ${W \otimes_k L}$ and this has a natural semilinear ${G}$ -action given by letting ${G=Gal(L/k)}$ act in the natural manner on the second factor of the tensor product. If ${W}$ is a ${k}$ -subspace of an ${L}$ vector-space ${V}$ and this tensor product corresponds to a natural isomorphism ${W \otimes_k L \rightarrow v}$ , this gives us a semilinear ${G}$ -action on ${V}$ .

Theorem 1 (Galois Descent) Fix an ${L}$ -vector space ${V}$ of dimension ${n}$ . There is a natural bijection, given by the maps described above, between ${n}$ -dimensional ${k}$ -subspaces of ${V}$ and semilinear ${G}$ -actions of ${L}$ .

We will use this to prove Hilbert’s Theorem 90, viewing it as the 1-dimensional case of this theorem. In fact, given Hilbert 90 underpins Kummer Theory which underpins Class Field Theory, perhaps it is worth not-so-wildly speculating that more general Galois descent should be expected to underpin the study of nonabelian Galois representations, and therefore understanding these reductions is possibly important as a model for work on Langlands. Anyway, let us recall its statement.

Theorem 2 (Hilbert 90) Suppose ${\chi: Gal(L/k) \rightarrow L^*}$ is a cocycle: a function such that ${\chi(\tau\sigma) = \chi(\tau) \tau \chi(\sigma)}$ . Then it is a coboundary: ${\chi(\sigma) = \sigma(x)/x}$ for some ${x \in L*}$ . In terms of Galois cohomology, this can be simply stated as

$\displaystyle H^1(Gal(L/k), L^*) = 0.$

Proof: Consider ${V=L}$ as a 1-dimensional vector space over itself, and pick some distinguished nonzero vector ${v_0}$ . A cocycle ${\chi}$ can be interpreted as precisely a semilinear ${G}$ -action on ${V}$ . Indeed, given ${\chi}$ , we set ${\psi_\sigma(v_0) = \chi(\sigma) v_0}$ and extending semilinearly, obtain a semilinear ${G}$ -action. Indeed:

$\displaystyle \psi_{\tau \sigma}(\lambda v_0) = \tau\sigma(\lambda) \chi(\tau \sigma) v_0 = \tau\sigma(\lambda)\chi(\tau)\tau(\chi(\sigma)) v_0 = \psi_{\tau}(\sigma(\lambda) \chi(\sigma) v_0) = \psi_{\tau} \circ \psi_\sigma (\lambda v_0).$

Conversely, one may check that a semilinear ${G}$ -action gives rise, once we have fixed ${v_0}$ , to a coycle.

But Galois descent tells us that each such semilinear group action has a ${1}$ -dimensional ${k}$ -vector subspace, fixed by the action. Let ${v=\mu v_1}$ be a basis vector for this subspace (so ${\mu \not= 0}$ ).

Then for all ${\sigma \in G}$ ,

$\displaystyle \mu v_1 = \psi_\sigma(\mu v_1) = \sigma(\mu) \chi(\sigma) v_1,$

proving that ${\chi}$ is in fact a coboundary, corresponding to ${\mu \in L^*}$ . $\Box$

In fact, by doing a similar thing but working in ${n}$ dimensions and picking a larger basis at the start (and then working with matrices appropriately), it looks like we should be able to generalise Hilbert 90 to the following statement.

Proposition 3 (Generalised Hilbert 90?) Let ${L/k}$ be an algebraic Galois extension with Galois group ${G}$ , ${n}$ a positive integer. Then

$\displaystyle H^1(G, GL_n(L^*)) = 0.$

Presumably this is well-known, and maybe it gives rise to a natural generalisation of Kummer theory, but this post is now longer than I was planning anyway, so I think I’ll leave it there for now.

Basic Applications of Elementary Galois Cohomology

December 30, 2011 in Mathematics | Leave a comment

In this post we shall develop the very basics of Galois cohomology, see two basic but important applications: Kummer Theory and the Weak Mordell-Weil Theorem, and remark on the general shape of both arguments. Hopefully at a later point I will post on Neukirch’s `General’ Class Field Theory which is perhaps a more complex realisation of a similar philosophy. I guess the main theme is that even very basic Galois cohomology can be a powerful tool that allows you to `get your money’s worth’ out of Galois theory without having to think too hard.

Firstly, let us set up the general framework of Galois cohomology. In general we will have a profinite group ${G}$ and want to study the category of ${G}$ -modules, namely topological abelian groups ${A}$ with a continuous action by ${G}$ . There is an obvious functor on this category ${A \mapsto A^G = \{a \in A| \sigma(a) = a \ \ \forall \sigma \in G\}}$ , called `taking ${G}$ -invariants.’ In fact, for any subgroup (or even subset) ${H}$ of ${G}$ we have a similar functor ${A \mapsto A^H}$ . In the case where ${H}$ is a closed subgroup, these will be of interest to us. Thus in this post, in its most general setting Galois cohomology will be about studying ${G}$ -modules ${A}$ together with their submodules fixed by closed subgroups of ${H}$ .

Example (the most basic reason for the name): If we consider ${G=Gal(\bar{k}/k)}$ the absolute Galois group of a field ${k}$ , ${G}$ acts on ${\bar{k}}$ and by the fundamental theorem of Galois theory the closed subgroups of ${G}$ are in one to one correspondence with the subfields of ${\bar{k}}$ containing ${k}$ , so the Galois cohomology of ${\bar{k}}$ is just the study of the field extension ${\bar{k}/k}$ and its subextensions. In fact, in all of the examples that follow we will take ${G}$ to be an absolute Galois group, but it is useful to remember that the general theory that follows is entirely group-theoretical.

So let us go back to our functor ${A \mapsto A^G}$ . As a functor in a category of abelian groups it is reasonable to investigate its exactness properties. In fact, given a short exact sequence of ${G}$ -modules

$\displaystyle 0\rightarrow A' \rightarrow A \rightarrow A'' \rightarrow 0$

it is true that

$\displaystyle 0 \rightarrow A'^G \rightarrow A^G \rightarrow A''^G$

is exact, but the final map need no longer be surjective. However, by general homological algebra nonsense (including that the category of ${G}$ -modules has enough injectives) this is enough to get some right-derived functors going, and allow us to define Galois cohomology groups ${H^i(G, A)}$ with all the properties you would expect, but in particular such that ${H^0(G,A) = A^G}$ and such that the above short exact sequence gives rise to a long exact sequence of cohomology:

$\displaystyle 0 \rightarrow A'^G \rightarrow A^G \rightarrow A''^G \rightarrow H^1(G,A') \rightarrow \\ H^1(G,A) \rightarrow H^1(G,A'') \rightarrow H^2(G,A') \rightarrow ...$

In fact, for our examples we will only need to go up to the ${H^1}$ level, which already provides a powerful tool in virtue of the following very basic fact. If a ${G}$ -module ${A}$ is in fact invariant under ${G}$ , then ${H^1(G,A) \cong Hom(G,A)}$ . I.e. it is just group homomorphisms from ${G}$ to ${A}$ . In particular, if ${A'}$ is ${G}$ -invariant, the ${H^1(G,A')}$ term of the above sequence has a concrete interpretation, and will be of crucial importance.

We can already start thinking about our first application. Given an arithmetically interesting field ${k}$ , we want to study the relationship between the group ${k^* / k^{*n}}$ and its abelian extensions of exponent ${n}$ (i.e. every element of the Galois group has order dividing ${n}$ ). More specifically, we want to prove the following:

Theorem 1 (Kummer Theory) Let ${k}$ be a field containing all the ${n}$ th roots of unity. Then every degree ${n}$ cyclic extension of ${k}$ is of the form ${k(\sqrt[n]{a})}$ for some ${a \in k}$ . More generally, there is a bijective inclusion-preserving correspondence between abelian extensions of $k$ of exponent $n$ and subgroups of $k^*/k^{*n}$ given by ${L \mapsto \{l^n| l \in L^*\}}$ , and ${H \mapsto k(\sqrt[n]{H})}$ . There is a perfect pairing

$\displaystyle k^* / k^{*n} \stackrel{\cong}{\rightarrow} Hom(G_k, \mu_n(k)).$

How do we prove this theorem? Let us note several striking features of its statement. Firstly, for it to be nontrivial, the field involved has to actually have some abelian extensions of exponent ${n}$ . In particular, if ${k}$ is an algebraically closed field nothing interesting happens (and the theorem tells us that this is because every element of ${k}$ has an ${n}$ th root which is already in ${k}$ ). Secondly, it seems that to get things going, ${k}$ does need to make a small arithmetic sacrifice: the ${n}$ th roots of unity ${\mu_n(\bar{k})}$ must already be present in ${k}$ . If we want to use Galois cohomology, there really is only one short exact sequence we could think of writing down. Maybe try to guess what it is before reading it.

$\displaystyle 0 \rightarrow \mu_n(\bar{k}) \rightarrow \bar{k}^* \stackrel{(-)^n}{\rightarrow} \bar{k}^{*} \rightarrow 0.$

What happens when we take invariants of the absolute Galois group ${G_k = Gal(\bar{k}/k)}$ ? Galois cohomology tells us that we have a long exact sequence

$\displaystyle 0 \rightarrow \mu_n(k) \rightarrow k^* \stackrel{(-)^n}{\rightarrow} k^* \rightarrow H^1(G_k, \mu_n(\bar{k})) \\ \rightarrow H^1(G_k, \bar{k}^*) \stackrel{(-)^n}{\rightarrow} H^1(G_k, \bar{k}^*) \rightarrow ...$

So in particular, if we truncate the exact sequence at the ${H^1(G_k, \mu_n(k))}$ term, we get

$\displaystyle 0 \rightarrow k^*/k^{*n} \rightarrow H^1(G_k, \mu_n(\bar{k})) \rightarrow H^1(G_k, \bar{k}^*).$

This looks like we have got somewhere, provided we can compute or interpret the first cohomology groups. And here is where our assumption that ${k}$ contains all the ${n}$ th roots of unity comes into play. Given this, ${\mu_n(\bar{k})}$ is ${G_k}$ -invariant, so we have ${H^1(G_k, \mu_n(\bar{k}))}$ is in fact isomorphic as an abelian group to ${Hom(G_k, \mu_n(k))}$ .

Thus, certainly to prove the Kummer isomorphism it will suffice to prove that the next map in the exact sequence is the zero map, and in fact what we need is precisely Hilbert’s Theorem 90, which says that ${H^1(G_k, \bar{k}^*) = 0}$ .

Finally, the inclusion-preserving correspondence and the claim that cyclic extensions of exponent ${n}$ are generated by an ${n}$ th root follow after a little more work from the Kummer isomorphism. Indeed, a homomorphism ${G_k \rightarrow \mu_n(k)}$ gives a cyclic quotient of ${G_k}$ with exponent ${n}$ and unravelling the definitions we see that it is generated by the ${n}$ th root of the corresponding element in ${k^*/k^{*n}}$ . Furthermore, a collection of such homomorphisms gives rise to a map into a product of some ${\mu_n(k)}$ whose kernel corresponds to a field of exponent ${n}$ , etc.

So we are done, having quoted facts about Galois cohomology and Hilbert’s Theorem 90 (and suppressed several other less significant details). There are two questions we should ask ourselves. Firstly, what did the Galois cohomology do for us? It seems to me that it provided the key tool for passing from a slightly hard-to-handle object directly connected to the arithmetic of fields ( ${k^*/k^{*n}}$ ) to some more abstract and more group-theoretical objects (the ${H^1}$ groups) which turned out to be easier to compute. It did this by lifting all attention away from ${k^*}$ , instead replacing it with the pair ${(\bar{k}^*, G_k)}$ . The first object is in some sense arithmetically `trivial’, being algebraically closed, and the second is a group which we might hope is easier to manipulate than the complicated field ${k}$ and its algebraic extensions. Secondly, what didn’t it do for us? Most notably, it didn’t actually do the necessary computations involving ${\bar{k}}$ and ${G_k}$ . It just told us which ${H^1}$ s to compute and we had to go away and do that. It was clever of \text{us} (or Kummer) to realise we needed the hypothesis that ${k}$ contained the ${n}$ th roots of unity and this gave us a handle on one of the groups, and to control the other group we should really have proved Hilbert 90, a nontrivial theorem.

The general shape of this argument was rather pleasant however, and seems to be quite common in number theory. We have an operation on some geometrical object which is surjective over an algebraically closed field because inverting it involves solving algebraic equations (like taking ${n}$ th roots), but not necessarily surjective over arbitrary fields. Such a failure to be surjective reflects interesting arithmetic behaviour, and the cokernels involved can be important arithmetic invariants but difficult to get our hands on. Galois cohomology allows us to translate this difficult arithmetic-geometric behaviour of geometrical spaces over a field like ${{\mathbb Q}}$ into group-theoretical behaviour of the absolute Galois group together with the much simpler geometric behaviour of our space considered over an algebraically closed field.

Our second striking example is the Weak Mordell-Weil theorem on elliptic curves.

Theorem 2 (Weak Mordell-Weil) Let ${K}$ be a number field, ${E/K}$ an elliptic curve with full ${n}$ -torsion: ${E(K)[n] = E(\bar{K})[n]}$ . Then ${E(K)/nE(K)}$ is finite.

This might not superficially resemble Kummer Theory, but look at the data and the structure of the situation. We have an algebraic operation, the multiplication-by- ${n}$ map ${E \rightarrow E}$ , and in particular it is surjective over ${\bar{K}}$ but probably not surjective over ${K}$ . And, perhaps the biggest give-away of all, we have been told that the kernel of this map, the ${n}$ -torsion, is all ${G_K}$ invariant. This time, at least if you’re happy with elliptic curves and everything I’ve said so far, you should have no trouble working out which exact sequence to write down. Refer to the one above if it helps.

We consider

$\displaystyle 0 \rightarrow E(\bar{K})[n] \rightarrow E(\bar{K}) \stackrel{[n]}{\rightarrow} E(\bar{K}) \rightarrow 0.$

Taking ${G_K}$ -invariants, we get the usual long exact sequence:

$\displaystyle 0 \rightarrow E(K)[n] \rightarrow E(K) \stackrel{[n]}{\rightarrow} E(K) \rightarrow \\ H^1(G_K, E(K)[n]) \rightarrow H^1(G_K, E(\bar{K})) \stackrel{[n]}{\rightarrow} H^1(G_K, E(\bar{K})).$

and since ${E(K)[n] = E(\bar{K})[n]}$ , truncating at the usual place we can form the short exact sequence:

$\displaystyle 0 \rightarrow E(K)/nE(K) \rightarrow Hom(G_K, E(K)[n]) \rightarrow H^1(G_K, E(\bar{K}))[n] \rightarrow 0.$

Sadly, the group ${Hom(G_K, E(K)[n])}$ fails to be finite, even using the hypothesis that ${K}$ is a number field, so we do need slightly more arithmetic input, and it turns out that (certainly when ${\mu_n \subset K}$ ) we can use arithmetic arguments to prove that any homomorphism in the image of ${E(K)/nE(K)}$ factors through ${Gal(F/K)}$ where ${F}$ is the compositum of all degree ${n}$ cyclic extensions of ${K}$ which are unramified outside a finite set of primes ( ${2}$ and those primes dividing the discriminant of ${E}$ ). By Hermite’s theorem, ${F/K}$ is finite, so ${Hom(Gal(F/K), E(K)[n])}$ is finite, and hence we get to the theorem we wanted.

In fact, by doing slightly more work, one is able to use techniques known as ${n}$ -descent to get good bounds on the size of this image. I should mention that I only feel comfortable when ${\mu_n \subset K}$ , in which case Kummer theory works nicely to allow us to describe these homomorphisms explicitly via

$\displaystyle Hom(Gal(F/K), E(K)[n]) \cong Hom(Gal(F/K), {\mathbb Z}/n{\mathbb Z})^2 \subset (K^*/K^{*n})^2,$

but my suspicion is that things are in fact more general.

Again, as with the previous application, it is in some sense difficult to pin down exactly what Galois cohomology did for us. It translated our problem about the mysterious group ${E(K)/nE(K)}$ into one about an also slightly mysterious subgroup of the (at least superficially not too mysterious) ${Hom(G_K, E(K)[n])}$ , which then has to be attacked by further arithmetic arguments. However, it seems that in enacting this transformation, Galois cohomology redefines the problem in a way that does make it genuinely much easier to solve and I know of no proof that isn’t on one level or another via this very basic level of Galois cohomology.

The Etale Cohomology of Elliptic Curves

December 4, 2011 in Mathematics | Leave a comment

As a few people have requested them, here are the notes from my recent seminar on the Etale Cohomology of elliptic curves. They definitely contain a cheat or two, and aren’t particularly well formatted (and indeed aren’t really the same as what I actually said in the talk), but here they are anyway. I should credit Kolowski’s article “Trying to understand Deligne’s proof of the Weil Conjectures” from which almost all of the material was shamelessly pillaged.

EtaleEllCurve

Surjective Maps of Schemes (over non-algebraically closed fields)

November 27, 2011 in Mathematics | 3 comments

The notion of a surjective map of schemes is somewhat confusing. For example, consider an elliptic curve ${E}$ over ${{\mathbb Q}}$ of positive rank. The multiplication-by- ${m}$ map is obviously not surjective as a map ${E({\mathbb Q}) \rightarrow E({\mathbb Q})}$ , but in fact it is surjective as a map of schemes, because schemes carry baggage associated to the points we haven’t yet `added in’ by passing to an algebraic closure. In this post we investigate the condition of a map being surjective on schemes as related to its inducing a surjection on the geometric points. I looked briefly in Hartshorne and did not seem to find anything about this, and I still do not have the whole picture (I have only considered what happens on closed points), so if readers can suggest a proper reference I would be grateful. Also, this post is likely to contain errors which might need fixing, so I urge readers not to assume I know what I’m doing.

Proposition 1 Let ${\phi: Y \rightarrow X}$ be a map of schemes of finite type over ${k}$ , and ${k^a}$ an algebraic closure for ${k}$ . The induced map ${Hom_k(Spec \ k^a, Y) \rightarrow Hom_k(Spec \ k^a, X)}$ is surjective if and only if ${Y \rightarrow X}$ is surjective on closed points.

I think one direction is reasonably straightforward. Given a closed point ${x \in X}$ , take any geometric point ${f:Spec\ k^a \rightarrow X}$ with image ${x}$ , for example the point obtained from ${O_U \rightarrow O_{U,x} \rightarrow k(x) \rightarrow k^a}$ for some affine ${U}$ containing ${x}$ (since we are of finite type over ${k}$ and ${x}$ is maximal, ${k(x)}$ is a field of finite type over ${k}$ , so finite over ${k}$ by the Nullstellensatz). Then by assumption, there is a ${g:Spec\ k^a \rightarrow Y}$ with ${\phi g = f}$ , so in particular any closed point in the closure of ${g(0)}$ is a closed point of ${Y}$ lying above ${x}$ .

Let’s try to prove the other direction. So now we are given a geometric point ${f:Spec\ k^a \rightarrow X}$ and we want to lift it to such a point of ${Y}$ . Well, such an ${f}$ determines a unique pair ${(x,\phi)}$ with ${x\in X}$ and ${\psi: k(x) \rightarrow k^a}$ , and ${x}$ must be closed by a dimension argument, so let us choose a closed point ${y \in Y}$ with ${\phi(y)=x}$ . There is an induced map in the wrong direction ${k(x) \rightarrow k(y)}$ . However, since ${y}$ is closed, ${k(y)}$ is a finite extension of ${k}$ , so in particular we can generate a map ${\psi'k(y) \rightarrow k^a}$ . Furthermore, it is possible to do so in such a way as to ensure that ${k(x) \rightarrow k(y) \rightarrow k^a}$ is the map ${\psi}$ .

But then the pair ${(y,\psi')}$ determine a geometric point ${g:Spec\ k^a \rightarrow Y}$ in the obvious fashion, and it is clear from our construction that ${\phi g = f}$ , so we have proved that the map on geometric points is a surjection.

There are at least two things I still don’t know which would be good to know. Firstly, what conditions do I need to impose to guarantee that ${k^a}$ can be replaced with ${k^s}$ , a separable closure? There is some kind of condition needed to say that induced maps are separable (so I guess insisting that ${\phi}$ is unramified) but also that the extensions ${k(x)/k}$ are separable, which is probably some simple condition but I haven’t yet worked out what. Secondly, can I extend the result to surjectivity of the map ${Y \rightarrow X}$ on all points? Again, hopefully this is fairly easy messing around with taking closures, etc. but I still don’t seem to have managed it.

It would also be nice to know if the above argument is actually correct and/or necessary. It’s been assembled in a very ad-hoc way to convince me that things I was doing with étale covers of elliptic curves weren’t nonsense, so I can believe my sketchy knowledge of scheme theory could have allowed a flaw to pass. Note that the result claimed about multiplication by ${m}$ on an elliptic curve follows (on the closed points) immediately, because given any geometric point I can divide it by ${m}$ by solving an appropriate degree ${m^2}$ polynomial, hence I get a geometric point with co-ordinates in ${k^a}$ .

Global Class Field Theory 3

November 9, 2011 in Mathematics | Leave a comment

In the third post of this series, we shall very briefly introduce the formalism of ideles and restate the Artin reciprocity theorem in our new language, following the ideas of Chevalley. It is a little difficult to get used to the language of adeles and ideles, but they turn out to be a very natural setting in which to study a global field (like a number field) in relation to its completions with respect to absolute values. In fact, one of the reasons for their introduction was to facilitate a proof of global class field theory from local class field theory. However, the main results of global class field theory also become much cleaner to state, as we shall see next time when we finally get to them.

Firstly, we have to define these things! Fix a global field ${K}$ (a number field or a 1-dimensional ${\mathbb{F}_p}$ -function field). We want an adele to be an element of the product ${\prod_{v \in M_K} K_v}$ , where ${M_K}$ is the set of valuations on ${K}$ . In other words, we want to take one element of each completion of ${K}$ . However, there is far too much junk in this product! We are after all studying ${K}$ itself, and the diagonal map ${K \rightarrow \prod_{v \in M_K} K_v}$ is ridiculously far from being surjective. For example, a real life element of ${K}$ will have to be integral at all but finitely many places. In fact, once we’ve noted this, we are actually ready to define the adeles.

The ring of adeles ${\mathbb{A}_K}$ is the set of ${(x_v) \in \prod_{v} K_v}$ such that ${x_v \in \mathcal{O}_v}$ for all but finitely many ${v}$ (note that there are only finitely many archimedean places, so we can ignore them and the definition still makes sense).

This is a fairly hefty ring. It had better possess some kind of interesting topology if we’re going to be able to understand it. One would like to take the subspace topology coming from the product topology, but annoyingly a product of locally compact spaces need not be locally compact (think ${{\mathbb R}^{\mathbb N}}$ ), so if we want local compactness (which would be nice) we need a different topology. It turns out there is a fairly natural topology. We take as a base of open sets the products ${\prod_v U_v}$ such that ${U_v}$ is open for each ${v}$ and for all but finitely many ${v}$ , we have ${U_v=\mathcal{O}_v}$ . So this looks kinda similar to the product topology, but is more suited to situations where we have imposed a restriction like the one we did: it is called the restricted product topology.

With this shiny new topology, ${\mathbb{A}_K}$ becomes a locally compact topological ring (noting that ${\mathcal{O}_v}$ is in fact always compact and that if you’re “locally locally compact” then you’re locally compact), and that’s absolutely fantastic! However, life just keeps getting better for us. Remember how earlier we were talking about how we’re actually studying ${K}$ and so we want the image of the diagonal map ${K \rightarrow \mathbb{A}_K}$ to not be too insignificant. Well, we’re in unbelievably good luck! Firstly the image of ${K}$ is discrete (which seems like it should be true). However, there are so many elements of ${K}$ discretely scattered in ${\mathbb{A}_K}$ that in fact the quotient ${\mathbb{A}_K/K}$ is compact! The main ingredient of the proof is again the weak approximation theorem (which is very similar in flavour and sophistication, but also ubiquity, to the chinese remainder theorem).

There is some fairly cool measure theory going on here (in fact, if you develop this theory correctly, you can bypass the classical Minkowski proofs of finiteness of the class group and Dirichlet’s unit theorem). Locally compact groups have a unique (up to scaling) Haar measure, and being compact, ${\mathbb{A}_K/K}$ has finite quotient measure. Since the effect of multiplication by ${a \in K}$ on the measure of a subset of ${\mathbb{A}_K}$ can be shown to be precisely the product of the normalised valuations, this gives an instant and satisfying proof of the classical product formula for all norms on ${K}$ . We shall not explore these issues here, but some other post may, and they are definitely worth attention.

Anyway, back to the task in hand. Now we have the adeles. How about the ideles? Well, they are just the multiplicative analogue. Indeed, the idele group of ${K}$ is just the group of units ${J_K = \mathbb{A}_K^*}$ . However, things aren’t quite as nice as that. We have no guarantee that inversion is continuous in the subspace topology, so we have to take a slightly refined topology: the subspace topology on ${\mathbb{A}_K^* \hookrightarrow \mathbb{A} \times \mathbb{A}}$ (where the embedding is ${x \mapsto (x,x^{-1})}$ ). Soon we will pass to a subgroup of the ideles which makes this issue go away, so it probably isn’t worth thinking about for too long.

So what do we know? It’s clear that the principal ideles ${K^* \hookrightarrow J_k}$ are discrete in this topology (since it’s stronger than the previous one). It can also be seen that in fact the ideles in this topology are just the elements ${(x_v) \in \prod_v K_v^*}$ such that ${x_v \in \mathcal{O}^*_v}$ for all but finitely many ${v}$ , with a restricted product topology like that on the adeles, so they really are exactly the multiplicative analogue (even their topology is!).

But always we should get a nagging feeling that we should be studying real life elements of ${K}$ and maybe that ${J_K}$ is just a bit too big to be studying. What other properties do elements of ${K}$ have? Well, they satisfy a product formula

$\displaystyle \prod_v |x|_v = 1.$

We cannot guarantee this for every idele. We therefore consider the map

$\displaystyle c: (x_v) \mapsto \prod_v |x_v|_v$

which gives a continuous homomorphism ${J_k \rightarrow {\mathbb R}_{>0}^*}$ , and let ${J_K^1}$ be its kernel.

This set ${J_K^1}$ turns out to be often a more natural object of study. It contains all of the principal ideles, as we remarked above. It also is a closed subgroup of ${J_K}$ and has the surprising property that its topology is equal to its subspace topology inherited from ${\mathbb{A}_K}$ . However, the main reason it is a good thing to study comes from its receiving the same seal of approval as that of the ideles:

Theorem 1 The topological group ${J^1_K/K^*}$ is compact.

This is all great fun, at least for me, but lots of readers have either now probably stopped reading or started wondering how any of this is remotely related to class field theory. Well, there is a natural map ${\alpha}$ from ${J_K}$ to the ideal group ${I_K}$ of ${K}$ , obtained by just trying to construct the fractional ideal which feels most likely to `be generated by’ our idele. Indeed, we define

$\alpha: \displaystyle (x_v) \mapsto \prod_{v \text{ finite}} \mathfrak{p}_v^{v(x_v)}.$

This makes sense because all but finitely many of the ${x_v}$ are units in their valuation rings.

Notice how by fudging things into the archimedean valuations (which don’t get detected by this map) we get something better. In fact, restricting to ${J_K^1}$ we still get a surjection ${J_K^1 \rightarrow I_K \rightarrow 0}$ . This has all kinds of cool consequences. Firstly, including the finiteness of the class group ( ${I_K/K^*}$ is discrete, and the above surjection implies it is compact, whence finite).

So it’s worth bearing in mind we have this nice ${J_K^1}$ thing lying around. However, for class field theory we will decide to work with ${J_K}$ instead. I am not entirely sure why – I guess it is ultimately a bigger object, and given we are able to prove stuff about it, we might as well use it (since the things proved are presumably in some sense `more general’). Having not seen a proof myself it is difficult to say, but I shall try to post back here when I do find out (or you might want to tell me why we use ${J_K}$ and not ${J_K^1}$ if you know).

Now, let us take ${L/K}$ a finite abelian extension. The classical Artin map ${F_{L/K}: I^S \rightarrow G(L/K)}$ can be composed with our canonical map ${\alpha: J_K \rightarrow I_K}$ to obtain a map ${\psi_{L/K}: J_K \rightarrow G(L/K)}$ which from now on is what we mean when we say the phrase `Artin map’. The reciprocity law can be re-written in the following form:

Theorem 2 (Artin Reciprocity, idelic form) There exists a continuous map ${\psi_{L/K}: J_K \rightarrow Gal(L/K)}$ with the following properties. The kernel of ${\psi}$ must contain ${K^*}$ , and any idele ${x}$ which takes the value ${1}$ at all the ramified and archimedean primes must satisfy ${\psi(x)=F_{L/K}(\alpha(x))}$ .

In fact, the condition of being trivial on the principal ideles makes it natural to define the idele class group to be ${C_K = J_K/K^*}$ and regard the Artin map as a continuous homomorphism ${C_K \rightarrow Gal(L/K)}$ . It is this language which turns out to seem optimal for stating the main results of global class field theory in their greatest generality, and that is what we will do next time. Indeed, we shall see that the main result of global class field theory is the now simple statement that there is an astonishing correspondence between the finite index open subgroups of ${C_K}$ and the finite abelian extensions of ${K}$ .

Zeta functions of varieties over a finite field

November 1, 2011 in Mathematics | Leave a comment

In this short post we look at the general definition of a zeta function for a scheme of finite type over ${Spec {\mathbb Z}}$ and record its relationship with the zeta function of an algebraic variety ${X}$ over a finite field ${\mathbb{F}_q}$ , defined as a kind of generating function for the number of points of ${X}$ over all finite extensions of ${\mathbb{F}_q}$ .

Let ${X}$ be a scheme of finite type over ${Spec {\mathbb Z}}$ . In fact, for concreteness, we start by looking at a single affine chart: imagining ${X=Spec A}$ as a geometrical space whose ring ${A}$ of functions is a quotient of some ring ${{\mathbb Z}[t_1,...,t_n]}$ of polynomials (in finitely many variables).

Then for every maximal ideal ${m}$ of ${A}$ , the quotient ${A/m}$ is in fact a finite field, so we can define the norm ${N_m = |A/m|}$ to get a positive integer ${\geq 2}$ associated with ${m}$ . We can use this to define the zeta function associated with ${X}$ to be the following formal Dirichlet series, by analogy with the classical Euler product formula.

$\displaystyle \zeta_X(s) = \prod_{m \text{ a maximal ideal of }A} \frac{1}{1-N_m^{-s}}.$

Notice that taking ${X=Spec {\mathbb Z}}$ we recover the classical Riemann zeta function, and taking the ring of integers of a number field recovers the zeta function of a number field. What happens if we take ${X}$ to be a variety over a finite field ${\mathbb{F}_q}$ ?

Well, ${log \zeta_X(s) = - \sum_{x} log(1-N_x^{-s})}$ where the sum is over all closed points ${x \in X}$ .

What is a closed point of ${X}$ ? What is an ${\mathbb{F}_{q^r}}$ -rational point of ${X}$ ? This is one of the confusing subtleties of scheme theory: these concepts are related but not exactly the same. Since the more classical definition of a zeta function is in terms of rational points, we will have to make this leap.

In fact, an ${\mathbb{F}_{q^r}}$ -rational point (which we will also call a `geometric point’, perhaps slightly nonstandardly) is simply a map ${Spec(\mathbb{F}_{q^r}) \rightarrow X}$ . This makes sense for two reasons. Firstly, a point in ordinary geometry can be described as just a map from a one point space (or as a zero-simplex, if you like), so our definition seems plausible. Secondly, and more convincingly, in the case where ${X}$ is affine, for example the elliptic curve over ${{\mathbb Z}}$ , ${y^2=x^3-x}$ , such a map corresponds to a map of rings

$\displaystyle {\mathbb Z}[x,y]/(y^2-x^3+x) \rightarrow \mathbb{F}_{q^r}.$

In other words, it corresponds to a choice of values for ${x}$ and ${y}$ in ${\mathbb{F}_{q^r}}$ satisfying the governing equation – exactly what we should mean by specifying a ${\mathbb{F}_{q^r}}$ -rational point.

On the other hand, a closed point is a maximal ideal of the ring of functions of an affine neighbourhood. Whenever we have a map ${Spec\mathbb{F}_{q^r} \rightarrow X}$ , it corresponds to a map of rings whose image is a field. In other words, a map whose kernel is a maximal ideal. Therefore, every ${\mathbb{F}_{q^r}}$ -rational point is associated with a unique closed point ${x}$ . In fact, we get an induced map ${k(x) \rightarrow \mathbb{F}_{q^r}}$ , where ${k(x)}$ is the residue field at ${x}$ . Conversely, any such map can be extended and determines a geometric point.

So finally we have arrived at the relationship between closed points and geometric points. Namely, each closed point ${x}$ determines the a set of ${\mathbb{F}_{q^r}}$ -rational points ${X_x}$ which can be canonically identified with ${Hom_{\mathbb{F}_q}(k(x), \mathbb{F}_{q^r})}$ . But by basic Galois theory, this says that, if ${|k(x)|=q^d}$ , ${X_x}$ is nonempty iff ${d|r}$ , in which case ${|X_x|=d}$ .

So let us return to our zeta function. Writing ${N_x=q^d}$ , we can expand

$\displaystyle -log(1-N_x^{-s}) = \sum_{m \geq 1} \frac{q^{-sdm}}{m}.$

Now, set ${T=q^{-s}}$ and bearing in mind where we are trying to get (to an expression involving the numbers ${|X_x|}$ ), we can define ${N(m,x)}$ to be ${d}$ if ${d|m}$ and ${0}$ otherwise, and rewrite this sum as

$\displaystyle \sum_{m \geq 1} N(m,x)\frac{T^m}{m} = \sum_{m \geq 1} |Hom_{\mathbb{F}_q}(k(x), \mathbb{F}_{q^r})|\frac{T^m}{m}.$

Now, take the sum over all closed points ${x}$ , and conclude that

$\displaystyle log \zeta_X(s) = \sum_{m \geq 1} (\sum_x |Hom_{\mathbb{F}_q}(k(x), \mathbb{F}_{q^r})|)\frac{T^m}{m} = \sum_{m \geq 1} |X(\mathbb{F}_{q^m})|\frac{T^m}{m}.$

This gives the familiar expression for the zeta function as an object whose logarithmic derivative is a generating function for the number of points of ${X}$ over all the finite field extensions of ${\mathbb{F}_q}$ , as appears in the statement of the Weil Conjectures and is well studied. One can check further that given the analysis above, the `Riemann Hypothesis’ for varieties of dimension ${d}$ over a finite field asserts that all the zeroes are on the lines ${Re(s) = 1/2,3/2,....,(2d-1)/2}$ and the poles are on the lines ${Re(s) = 0,1,2,...,d}$ . Noting that ${{\mathbb Z}}$ is an object of dimension 1, the analogy with the classical Riemann hypothesis could not be clearer.

Global Class Field Theory 2

October 21, 2011 in Mathematics | Leave a comment

Last time, we were thinking about an abelian extension ${L/K}$ , and we constructed, for most of the primes ${v}$ of ${K}$ a well-defined associated Frobenius element ${\sigma_v \in Gal(L/K)}$ . To do this, we used the canonical isomorphism between the decomposition group ${G_w}$ of some prime ${w}$ extending ${v}$ (a subgroup of ${Gal(L/K)}$ ) and the Galois group ${Gal(k(w)/k(v))}$ of the residue fields. Since this is a Galois group of finite fields, it is generated by a Frobenius element ${x \mapsto x^{|k(v)|}}$ , whose image in ${G_w}$ gave us the distinguished element we wanted. In this post we extend this map linearly and state the Artin reciprocity law. We then briefly sketch its relation to the law of quadratic reciprocity.

Now, if ${S}$ is a finite set of primes of ${K}$ which includes all the archimedean primes and all those which ramify in ${L/K}$ , we have an associated Frobenius element for each ${v}$ not in ${S}$ . Since ${Gal(L/K)}$ is abelian, a natural thing to do is to extend the set map ${M_K - S \rightarrow Gal(L/K)}$ to a map of abelian groups. Indeed, take ${I^S}$ to be the free abelian group on ${M_K - S}$ , and we get a morphism

$\displaystyle F_{L/K}: I^S \rightarrow Gal(L/K).$

This might be mistaken for idle fooling around, but let us think about what ${I^S}$ is. An element of ${I^S}$ is a finite ${{\mathbb Z}}$ -linear combination of primes ${v \not\in S}$ . These are all finite primes, so they correspond to prime ideals, and since our ring is commutative, it’s not totally unnatural to interpret our linear combinations as specifying the factors in a product (so for example, take ${v+2v'}$ to mean ${\mathfrak{p}_v\mathfrak{p}_{v'}^2}$ ). Aha! We recall that any fractional ideal can be factorised uniquely into primes in a Dedekind domain. So ${I^S}$ is in fact none other than the group of fractional ideals which are coprime to ${S}$ : one of the central objects of algebraic number theory. Great!

At this point, Tate proves that if we have a square of number fields ${K \subset K', L \subset L'}$ with ${L'/K'}$ and ${L/K}$ abelian, and ${S'}$ is the set of primes of ${K'}$ extending those of ${S}$ , then we get a commuting diagram with $F_{L'/K'}, F_{L/K}$ on rows and the norm map and natural restriction map on columns. It is an important result because we want our maps to behave nicely with respect to changes of field and a similar diagram forms a central part of the main theorem of global class field theory. The proof is just to check on basis elements, and remarking that Frobenius elements are determined by their effect on residue fields. I don’t think it’s worth giving the details here.

What we do next is very interesting. Until now we have been thinking about our number fields as basically abstract things which have Galois groups and primes. Indeed, the easiest way to set up our ${F_{L/K}}$ map was to define it on primes, and then extend linearly to get the group of fractional ideals (coprime to a finite set of primes). But of course, now we have that, recall that any nonzero element ${a}$ of the field ${K}$ defines a principal maximal ideal ${(a)}$ . Therefore, for any element ${a \in K^*}$ which is coprime to all the ramified primes, we can associate a Frobenius element by taking ${F_{L/K}((a))}$ .

Startlingly, it turns out that under slightly stronger assumptions than being merely coprime to all the ramified primes, this element is always the identity element. This is a form of Artin reciprocity, our first really deep theorem. It turns out that there is some ${\epsilon>0}$ such that the following happens. Suppose that ${|a-1|_v < \epsilon}$ for all ${v \in S}$ . Then ${F_{L/K}((a)) = 1}$ .

Note that the condition ${|a-1|_v < \epsilon}$ is saying that ${a \equiv 1}$ modulo a sufficiently large power of each ramified prime, and seems at first sight to be reasonably prohibitive at the archimedean primes. However, there is a nice trick to get round this and reduce this apparently hefty analytic constraint to a nice algebraic one. Firstly, we note that if ${n=[L:K]}$ then obviously ${F_{L/K}((b)^n) = F_{L/K}((b))^n = 1}$ . Hence, if we can write ${a=b^n a'}$ for ${a'}$ close to ${1}$ in all valuations, we will still get the same result.

Suppose ${a}$ is in fact an ${n}$ th power in each ${K_v^*}$ , so ${a=c_v^n}$ for some ${c_v \in K_v}$ . Since ${S}$ is finite, we can use the weak approximation theorem (a sort of generalised Chinese remainder theorem that also works at archimedean places) to find ${b \in K}$ within any distance ${\eta>0}$ of ${c_v^{-1}}$ in the norm associated to ${v}$ , for each ${v \in S}$ . Then such a ${b}$ has the property that

$\displaystyle |ab^n-1|_v = |a|_v|b^{n} - c_v^{-n}|_v \leq \eta |a|_v n (|c_v^{-1}|_v + \eta)^{n-1},$

and choosing ${\eta}$ sufficiently small (such that the RHS above drops below ${\epsilon}$ , we deduce that ${F_{L/K}((a))=F_{L/K}((ab^n))=1}$ .

So in fact it will suffice for ${a}$ to be a ${n=[L:K]}$ th power in each ${K_v}$ for ${v}$ ramified or archimedean. But this condition can be simplified even further. Indeed, at the archimedean primes, where the completion is ${{\mathbb R}}$ or ${{\mathbb C}}$ , we need only be concerned when it is ${{\mathbb R}}$ and ${n}$ is even, in which case we get the condition that ${a>0}$ . Also, at the non-archimedean primes, Hensel’s lemma tells us that being an ${n}$ th power in ${K_v}$ is equivalent to being an ${n}$ th power in ${k(v)}$ . Thus we obtain the following more concrete version of Artin reciprocity.

Let ${a \in K^*}$ be an ${n}$ th power modulo every ramified prime, and, if ${n}$ is even, positive in every real embedding. Then ${F_{L/K}((a))=1}$ .

I should re-emphasise that we haven’t proved this, and that the proof is in general hard. However, in some special cases, we can prove Artin reciprocity directly. For example, in the extension ${{\mathbb Q}(\zeta)/{\mathbb Q}}$ where ${\zeta}$ is a primitive ${m}$ th root of unity, it is not difficult to show that ${F(p)\zeta = \zeta^p}$ for all ${p}$ not dividing ${m}$ , whence ${F(a)\zeta = \zeta^a}$ for all ${a}$ coprime to ${m}$ . So if we insist that ${a \equiv 1 \text{ mod } m}$ (taking a fairly liberal value of ${\epsilon}$ ), then since an element of ${Gal({\mathbb Q}(\zeta)/{\mathbb Q})}$ is determined by the image of ${\zeta}$ , ${F(a)=1}$ , as required. Tate mentions that similar direct proofs, in which we can explicitly keep tabs on the effects of each ${F(v)}$ , are possible for abelian extensions of complex quadratic fields by making use of elliptic curves with complex multiplication, but that it seemed likely there was no analagous direct proof for general abelian extensions of number fields.

To get a flavour for the power of Artin reciprocity, we shall now use it to deduce Gauss’ celebrated law of quadratic reciprocity. Doing so will not be totally trivial because we have to figure out how to relate our abstract general result to a special case, but I hope the reader will see that in terms of mathematical content the law of quadratic reciprocity should be seen as in some sense a very special and easy consequence of Artin reciprocity.

Let ${a \in {\mathbb Z}}$ not be a square. We shall consider the Artin map ${F}$ associated to the quadratic extension ${{\mathbb Q}(\sqrt{a})/{\mathbb Q}}$ , taking all the fractional ideals with unramified support to elements of ${Gal({\mathbb Q}(\sqrt{a})/{\mathbb Q}) = \{-1, 1\}}$ . Define the Legendre symbol by the following equation ( ${p}$ is a rational prime)

$\displaystyle F((p))(\sqrt{a}) = \left(\frac{a}{p} \right)\sqrt{a}.$
In other words, the Legendre symbol keeps track of how an associated Frobenius element acts on ${\sqrt{a}}$ by recording the root of unity picked up. In fact, in our case, this is enough to determine precisely what the Frobenius element involved is. This obviously has the standard multiplicative properties and is well-defined in ${a}$ modulo ${P}$ . In fact, it is not difficult to check that it satisfies the Euler criterion

$\displaystyle \left(\frac{a}{p} \right) \equiv a^{(p-1)/2} \mod p.$
Indeed, this follows very quickly from the characterisation of the Frobenius element ${\sigma_w}$ associated to a prime ${w}$ as that which satisfies ${\sigma_w(x) \equiv x^{Nv} \mod \mathfrak{p}_w}$ . Therefore, it must be the classical Legendre symbol.

Now what does Artin reciprocity in ${{\mathbb Q}(\sqrt{a})/{\mathbb Q}}$ tell us? Suppose ${p}$ and ${q}$ are rational primes that are congruent modulo ${8a_0}$ , where ${a_0}$ is the odd part of ${a}$ . Then ${pq^{-1}}$ is a 2-adic square and a square modulo every odd prime dividing ${a}$ (and it’s positive in any embedding into ${{\mathbb R}}$ ). The reciprocity law tells us that therefore if ${p \equiv q \mod 8a_0}$ ,

$\displaystyle \left(\frac{a}{p} \right) = \left(\frac{a}{q} \right).$

And from this formula, the classical result of quadratic reciprocity is easily deduced.

So in this post we obtained a map from the subgroup of fractional ideals which don’t touch the ramified primes into the Galois group of ${L/K}$ , and stated an important property of this map, namely Artin reciprocity, which vastly generalises the classical reciprocity laws. This property can be restated in the following language. Let ${K}$ be a number field, and ${G}$ a topological abelian group. A homomorphism ${\psi: I^S \rightarrow G}$ is called admissible if for any neighbourhood ${U}$ of the identity, there exists an ${\epsilon > 0}$ such that whenever ${a \in K^*}$ is coprime to all the ramified primes,

$\displaystyle |a-1|_v<\epsilon \text{ for all } v \in S \Rightarrow \psi((a)) \in U.$

Then Artin reciprocity is simply the statement that (viewing ${Gal(L/K)}$ with the discrete topology) ${F_{L/K}}$ is admissible for any abelian extension ${L/K}$ of number fields. Next time, we will define the Artin map in terms of ideles, which are an important modification of the ideal class group which put archimedean primes and discrete primes on a proper equal footing, and in whose topology the property of being admissible becomes partly a continuity statement. We will then discuss some basic ideas to lay the foundations on which we will finally be able to state the main theorems of global class field theory in their full generality.

Analytic proof that cyclotomic polynomials are irreducible

October 17, 2011 in Mathematics | Leave a comment

Apparently this is due to Kronecker – the idea is to prove that cyclotomic polynomials are irreducible indirectly using the following two analytic facts:

Fact 1 (Dirichlet’s theorem, special case): The Dirichlet density of primes which are congruent to $1$ modulo $m$ is precisely $\frac{1}{\phi(m)}$ .

Fact 2 (due to Kronecker, but a super special case of Cebotarev’s density theorem): The set of primes which split completely in an extension $L/\mathbb{Q}$ has Dirichlet density $\frac{1}{[L:\mathbb{Q}]}$ .

And now we just stick these facts together. Let $L=\mathbb{Q}(\zeta_m)$ and take $f$ to be the minimal polynomial of $\zeta_m$ . Then a rational prime $p$ splits iff $f$ splits modulo $p$ , which only happens (assuming $p$ does not divide $m$ ) if there is a primitive $m$ th root of unity in $\mathbb{F}_p$ . For this to be the case we must have $m|p-1$ . Thus by fact 1, at most density $\frac{1}{\phi(m)}$ primes split in this extension. Hence by fact 2, $deg f = [L:\mathbb{Q}] \geq \phi(m)$ , and since the RHS is the degree of the $m$ th cyclotomic polynomial, we have equality and in fact $f$ is the $m$ th cyclotomic polynomial, which must therefore be irreducible.

Global Class Field Theory 1

October 16, 2011 in Mathematics | 2 comments

In this series of posts I shall try to sketch (at least statements of) the basic results of global class field theory, following John Tate’s notes from the 1966 Brighton Conference. I begin by emphasising that I am writing them mainly in order to try to learn the material myself rather than really as an exposition to help others. As such, I expect to make countless errors, which expert readers are encouraged to point out.

It was conjectured by Hilbert that there exists, for a number field ${K}$ living inside a fixed algebraic closure, a unique totally unramified Galois extension ${L/K}$ with ${Gal(L/K)}$ isomorphic to the ideal class group of ${K}$ . Such an extension is obviously abelian, and in fact Hilbert conjectured further that it is the maximal totally unramified abelian extension of ${K}$ . Of course, Hilbert was right, and proving his conjecture is one of our main tasks.

In general, we will be studying a field ${K}$ which is usually a number field (or sometimes the function field of a curve over a finite field). Our overall goal seems to be to come up with a theory relating certain groups connected with the ideal class group of ${K}$ to the Galois groups of extensions of ${K}$ . More specifically, we will define the idele group ${J_K}$ of ${K}$ , which fits into an exact sequence

$\displaystyle 0 \rightarrow J_K^{S_\infty} \rightarrow J_K \rightarrow I_K \rightarrow 0,$

where ${I_K}$ is the group of fractional ideals and where ${J_K^{S_\infty}}$ is the subgroup of ideles all of whose `prime factors’ correspond to archimedean valuations. In some sense therefore, we want to imagine archimedean valuations as `primes at infinity’ and then ideles are a way to allow ideals to be divisible by these primes. With these defined, our main objects of study will be Artin maps

$\displaystyle \psi_{L/K}: J_K \rightarrow Gal(L/K),$

which give information about the splitting of primes in ${L/K}$ and provide the desired link between ideal class groups and the Galois groups.

In this first post we take the first tentative steps towards defining the Artin map, but will make no further mention of ideles yet. Let us set up some terminology. Fix ${K}$ a global field (a finite extension of ${{\mathbb Q}}$ or some ${\mathbb{F}_q(T)}$ ). A prime of ${K}$ is defined to be an equivalence class of nontrivial valuations on ${K}$ . Primes ${v}$ are either discrete (they correspond to a prime ideal ${\mathfrak{p}_v = \{x| v(x) \geq 1\}}$ , by satisfying an ultrametric law) or archimedean. We let ${M_K}$ denote the set of all primes of ${K}$ , ${S^\infty = S^\infty_k}$ the set of archimedean primes and ${M_K^*}$ the set of discrete primes.

Now take ${L/K}$ a finite abelian extension, and let ${M_K^u}$ be the subset of ${M_K^*}$ consisting of unramified discrete primes. Our goal in this post will be to find a natural map

$\displaystyle F_{L/K}: M_k^u\rightarrow Gal(L/K),$

which we will extend in future posts to the Artin map ${\psi_{L/K}}$ . To begin, let us start with some basic facts about how ${G=Gal(L/K)}$ acts on ${M_k}$ and the corresponding local fields.

We define the action of ${G}$ on ${M_L}$ by ${|\sigma x|_{\sigma w} := |x|_w}$ . With this definition we get a left action ${\sigma(\tau w) = (\sigma \tau) w}$ , and if ${v}$ is discrete it corresponds to the obvious action on prime ideals

$\displaystyle \mathfrak{p}_{\sigma w} = \{x| \sigma w(x) \geq 1\} = \{\sigma(x)| w(x) \geq 1 \} = \sigma \mathfrak{p}_w,$

and similarly ${\sigma}$ moves the local rings around in a natural fashion.

Now, there is a natural restriction map ${M_L \rightarrow M_K}$ , and since ${\sigma(x) = x}$ for all ${x \in K}$ , ${|x|_{\sigma w} = |x|_w}$ for all ${x \in K, \sigma \in G}$ . Therefore the image of a Galois orbit in ${M_L}$ is just a single element of ${M_K}$ . In fact we will show that ${G}$ acts transitively on the primes over a fixed ${v \in M_K}$ , so since any such ${v}$ can be extended to ${L}$ in at least one way, the restriction map induces a correspondence:

$\displaystyle \{ G\text{-orbits of } M_L \} \leftrightarrow M_k.$

How can we prove this transitivity? Well, let ${G_w = \{\sigma \in G| \sigma w = w\}}$ be the decomposition group of ${w}$ . The orbit of ${w}$ will simply be ${w=w_1, w_2,...,w_r}$ where ${r=(G:G_w)}$ , with one prime ${w_i}$ corresponding to each coset ${\sigma_i G_w}$ . For each ${w_i}$ , an element of ${G_w}$ gives an isometric automorphism, which extends to an automorphism of the completion ${L_{w_i} \rightarrow L_{w_i}}$ and thus we have a natural embedding

$\displaystyle G_w \hookrightarrow Aut(L_{w_i}/K_v).$

Hence we get, by basic field theory, the bounds

$\displaystyle |G| = r|G_w| \leq \sum_{i=1}^r |Aut(L_{w_i}/K_v)| \leq \sum_{i=1}^r [L_{w_i}: K_v]$

Suppose there are some other primes ${w_{r+1},...,w_s}$ also extending ${v}$ . Crucially, the natural map ${L \otimes_K K_v \rightarrow \prod_{i=1}^s L_{w_i}}$ must be surjective, as an application of the weak approximation theorem. This implies that

$\displaystyle \sum_{i=1}^s [L_{w_i}:K_v] \leq [L:K] = |G|.$

Putting our two estimates together, we get that equality holds at every stage, so in particular ${s=r}$ (the action is transitive).

Note we have also proved that ${L_w/K_v}$ is Galois and has Galois group naturally isomorphic to ${G_w}$ . In fact, this group is about to play a key role in another capacity. Under the additional assumptions that ${w}$ is discrete and unramified over ${v}$ , this Galois group is canonically isomorphic to the Galois group of residue fields ${Gal(k(w)/k(v))}$ (where ${k(v) = \mathcal{O}_v/\mathfrak{p}_v}$ ). So we have canonical isomorphisms

$\displaystyle G \supseteq G_w \cong Gal(L_w/K_v) \cong Gal(k(w)/k(v)).$

But the residue fields are just finite fields, so letting ${q=|k(v)|}$ , ${Gal(k(w)/k(v))}$ is generated by ${Frob_q: x \mapsto x^q}$ . Since the above isomorphisms are canonical, this corresponds to a canonical generating element ${\sigma_W}$ of ${G_w}$ , which is as such called the Frobenius automorphism associated with ${w}$ , characterised by the property ${\sigma_w(a) \equiv a^q (\text{mod }\mathfrak{p}_w)}$ for all ${a \in L}$ .

Wait! What just happened? We took a prime ${w}$ of ${M_L^*}$ (assumed not to be ramified) and its decomposition group contained a special element ${\sigma_w}$ . In other words, we defined a natural mapping ${M_L^* \rightarrow Gal(L/K)}$ . Recall that we were looking for a map ${M_K^u \rightarrow Gal(L/K)}$ , so it feels like we are almost done.

But just a minute ago we proved that the Galois orbits of ${M_L}$ are in bijective correspondence with ${M_K}$ , and this correspondence obviously respects the properties of being discrete and unramified, so the obvious thing to hope is that ${\sigma_w = \sigma_{\tau w}}$ for all ${\tau \in G}$ , since we can then just define ${\sigma_v}$ to be this element (the Frobenius automorphism associated with every prime above ${v}$ ).

Oh, but this is obvious, because ${Gal(L/K)}$ is abelian, and by the characterising property of ${\sigma_w}$ , ${\sigma_{\tau w}}$ must be conjugate to it. So we are done. Provided ${L/K}$ is abelian, we have a lovely well-defined map called `take the associated Frobenius automorphism of any prime above ${v}$ ‘

$\displaystyle F_{L/K}: M_K^u \rightarrow Gal(L/K).$

Next time, we will extend this to a group homomorphism and state the Artin reciprocity law, our first deep theorem which tells us a lot about how these maps work. However, before we finish it is worth remarking how the above map relates to the splitting of an unramified prime ${v}$ in ${L}$ .

Indeed, we saw in the above discussion about the action of ${G}$ on ${M_L}$ that the set of primes ${w}$ dividing ${v}$ is isomorphic to ${(G:G_w)}$ . Now, ${G_w}$ is actually generated by ${\sigma_w}$ , and hence by the symmetry of the action, we know how ${v}$ splits. If ${\sigma_w}$ has order ${f}$ , ${v}$ must split into ${|G|/f}$ distinct factors, each of degree ${f}$ . So knowing ${\sigma_v = F_{L/K}(v)}$ tells us how ${v}$ splits and gives us a generator for the decomposition group. This is an very nice situation to be in, and hints at some of the wide-reaching consequences of the theory.