Seminar on Class Field Theory

This seminar was held in #mathematics on EFnet on 27th of April 2010 at 16:00 EDT by hochs.

Seminar log

<hochs> for what it's worth, i also typed up some chicken scratc log at http://dl.dropbox.com/u/3799589/LaTeX/math%20efnet.pdf
<hochs> ok it's 4:00 now, so i might as well start
<Fermat> (reminder, in case some one trolls - ignore it. I will try to be swift)
<hochs> ok so class field theory is about abelian extensions of local or global fields (there are two parts to this, local class field theory & global class field theory)
<hochs> and the attempt is to capture the extensions of these fields by the arithmetic of the fields alone
<hochs> so to speak
<hochs> for various reasons, number theorists like to study extensions of Q
<hochs> some motivation: if you're interested in primes of the form x^2 + y^2 = p, you'd write this as p = (x + iy)(x - iy) in the extension field Q(i) / Q
<hochs> and knowing the arithmetic properties of the corresponding "ring of integers" in Q(i) (in this case Z[i]) gives you the proof
<hochs> so ok, class field theory has two parts
<hochs> a quick review of what a local field K is: it's a field that is complete with respect to valuation v (which gives topology) that is also locally compact
<hochs> there are two classes of them, one is finite extension of Q_p (p-adics) or F_q(t) (rational function field in one variable over finite field)
<hochs> what local class field theory tells you is that there's a 1-1 correspondence between abelian extensions L of K and open subgroup of K of finite index
<hochs> the correspondence is given by L -> N_{L/K}L^* where N_{L/K} means the norm from L to K
<hochs> i should say *finite* abelian extension L of K for now
<hochs> all this is made very explicit by a mysterious map from K^* (multiplicative group of K) to Gal(L/K), the galois group of L over K
<hochs> which is called a local residue map, and has the following properties:
<hochs> I'll call this map phi_K : K^* -> Gal(L/K) for now
<hochs> 1) For any uniformizer pi of K and any finite abelian unramified extension L/K, phi_K(pi) acts on L as Frobenius of L/K
<hochs> a little explanation here: for L / K finite abelian unramified, Gal(L/K) is isomorphic to Gal(k_L / k), galois group of their residue fields
<hochs> (by Hensel basically)
<hochs> and in particular, in the local field case, residue fields are finite and so Gal(k_L / k) = Gal(L/K) is cyclic
<hochs> and the generator is what we call Frobenius = F_{L/K} notation
<lhrrwcc> uniformizer being the unique generator for the maximal ideal of the maximal ideal
<hochs> doesn't have to be unique
<hochs> you can multiply it by any unit
<hochs> and it'll be a uniformizer
<lhrrwcc> right
<hochs> and 2) for any abelian extension L / K, the map phi_K induces isomorphism K^* / NL^* ~= Gal(L/K)
<hochs> where N means the norm from L to K
<hochs> ok so that's basically the statements of local class field theory.
<hochs> the map K^* to Gal(L/K) is somewhat mysterious, and I plan to give a little recipe for constructing them
<hochs> in this seminar
<hochs> there are several approaches to proving local class field theory but I plan to discuss 2 approaches today
<hochs> one is the most common nowadays, using galois cohomology
<hochs> and this is probably the most stream-lined approach
<sek> (for the non cognoscenti, it may be an idea to spell out what the norm map is, and also to point out that an abelian extension is one such that the Galois group is an abelian group)
<hochs> the other is lubin-tate, which has advantage of being a bit more elementary but somewhat technical
<hochs> i see ok, so for L/K a galois extension with galois group G, norm N:L -> K is given by taking x in L and sending it to the product of all f(x) as f runs through elements of Gal(L/K)
<hochs> so part of what local class field theory tells you is that there's a one-to-one correspondence between abelian extension L (galois extension of K with abelian galois group) and certain subgroups of K^* (in multiplicative sense)
<hochs> and turns out that this is galois correspondence, i.e. it reverses inclusion, compositum of fields correspond to intersection of norm subgroups, and intersection of fields correspond to compositum of corresponding subgroups of K^*
<hochs> so that's the statement, now I'll discuss some approaches to proving these
<hochs> i'll discuss group cohomology method first, then discuss lubin-tate
<hochs> for those who already have a bit of background in elliptic curves and complex multiplication, lubin-tate is basically complex multiplication done-right on local fields
<hochs> you take certain special endormophism f, then show that there's a group law F_f for which f is an endomorphism, take the torsion points of f, get totally ramified extension K_{pi}, and manually define the action of K^* on each of G(K_nr /K) and G(K_{pi} / K)
<hochs> where K_nr denotes the maximal unramified extension of K in a fixed separable closure of K
<hochs> standard reference for this is the original annals paper by lubin-tate
<hochs> lubin-tate has the advantage of being very explicit about the *mysterious* map K^* -> Gal(K^{ab} / K) where K^{ab} is the maximal abelian extension
<hochs> and things like local-kronecker falls out quickly from this
<sek> !
<hochs> yes
<sek> what do you mean very explicit here? The one arising from Tate's theorem in galois coh is equally explicit if you write it all out no?
<hochs> it's explicit for unramified extension
<hochs> for totally ramified extension, it's not very clear
<hochs> for unramified extension, you can get the map explicitly by knowing the brauer group is H^2(K_nr / K)
<hochs> and H^2(L/K) will sit inside H^2(K_nr / K) in natural way
<hochs> (by inflation)
<hochs> and using description of reciprocity maps by characters, you can get explicitly that it sends uniformizer to the Frobenius
<hochs> but for other "bad" ramified extensions L/K, it doesn't seem clear, at least to me, how to get explicit map of Br K -> Q/Z by inv
<sek> thanks
<hochs> if you mean that you can follow through all the connecting morphisms and get the map K^* -> (which is inverse of cup pairing) Gal(L/K)
<hochs> in that sense yes it's explicit i suppose
<hochs> ok, going back to trying to get the map K^* -> Gal(L/K)
<sek> I did mean that yes, it's explicit but having done it in the past I wouldn't recommend it to anyone in the audience:)
<hochs> :)
<hochs> i've also struggled with all the connecting morphisms and explicitly following through all the transfer maps and so on
<hochs> and something which i also don't recommend
<hochs> ok, now back to K^* -> Gal(K^{ab} / K)
<hochs> sorry I keep typing K^* -> Gal
<hochs> we're actually going to show that for any abelian extension L / K, there's an isomorphism K^* / NL^* ~= Gal(L/K)
<hochs> and that these maps are all compatible for varying extensions L / K
<hochs> so you get a map K^* -> Gal(K^{ab} / K) by taking the limit
<hochs> (inverse limit)
<hochs> so how do we achieve this? quick review of group cohomology is in order
<hochs> recall that for a finite group G, and G-module A, H^q(G,A) is going to be derived functor of A -> A^G = Hom_{Z[G]}(Z,A)
<hochs> or Ext^q_{Z[G]} (Z, A)
<hochs> so H^0(G,A) = A^G
<hochs> and there's the usual long exact sequence for any short exact sequence 0 -> A -> B -> C -> 0
<hochs> note that if L/K is galois extension with galois group G, then G acts on L^* and with this setup, H^0(G, L^*) = K^*
<hochs> so we have a description of K^* in terms of 0-dimensional galois cohomology
<hochs> ok that's the cohomology side, and then there's homology side of all this
<hochs> btw by A^G i mean the set of elements in A fixed-pointwise by elements of G
<hochs> ok, homology: which is a derived functor of A -> A_G = A/(I_G A)
<hochs> where I_G is ideal of Z[G] generated by s-g
<hochs> or Tor^q_{Z[G]}(Z, A) since A_G = Z (tensor over Z[G])) A where G acts trivially on Z
<hochs> for the proof of local class field theory, you only need to understand these groups for low dimensions (q = 0, 1)
<hochs> the rest is automated by dimension shifting
<hochs> H_q(G,A) will be the q-th derived thing of this functor
<hochs> ex. H_1(G,Z) = G/G' where G' is the commutator subgroup of G
<hochs> this can be derived by using long exact sequence on 0 -> I_G -> Z[G] -> Z -> 0
<hochs> so what's the point of all this
<hochs> look, we described K^* as H^0(G,L^*) where G = G(L/K)
<sek> !
<hochs> on the other hand, we have H_1(G,Z) = G/G' = G for G abelian
<hochs> yes sek
<sek> sorry, just to clarify, when you say dimension shifting you actually mean a) look at the cyclic group case b) use degree shifting isomorphisms
<hochs> i mean b)
<hochs> degree shifting isomorphisms
<hochs> the acyclic objects in this category are induced modules, and they're easy to describe
<hochs> so degree shifting works pretty well and easily
<sek> such as tate's theorem?
<hochs> yep
<hochs> tate's theorem is proved manually for low dimension
<hochs> then the rest is killed by degree shift
<sek> ok, just to clarify that this is a particular scenario and that this magic doesn't work for any group cohomology of a module
<hochs> right
<sek> (sorry to keep interrupting:))
<hochs> np, interaction is good
<hochs> i prefer interaction so i know i'm not losing all my audience
<hochs> after I sketch the proofs and give statements of global class field theory, i'll apply them in very concrete setting
<hochs> for example i'll prove that prime p is of the form x^2 + 5y^2 iff p = 1,9 (mod 20)
<hochs> using class field theory
<hochs> and outline general method for determining primes of the form x^2 + ny^2
<hochs> as well as quadratic form theory (which is a very nice illustration of Hasse Principle)
<hochs> anyway going back to group cohomology, we have these definitions of H^q and H_q now
<hochs> cohomology and homology
<hochs> now there's a very nice device called Tate groups
<hochs> which connects these two
<hochs> the reason that we want to connect these two is that
<hochs> recall that we're trying to find a map K^* -> Gal(L/K)
<hochs> we determined K^* = H^0(G,L^*) and Gal(L/K) = G = H_1(G,Z)
<hochs> we need to somehow connect H^0 and H_1 together
<hochs> anyway going back, for quick review of tate groups
<hochs> for a finite group G, we have an element N = sum{s in G} which lies in Z[G]
<hochs> and for any A-module, this induces a homomorphism N:A -> A
<hochs> whose kernel contains (I_G)A and whose image is contained in A^G, inducing a map on N: H_0(G,A) -> H^0(G,A)
<hochs> and we define Ker N = hat{H}_0(G,A) and coker N = hat{H}^0(G,A) = (A^G) / (NA)
<hochs> now you can define an entire series of tate cohomology by hat{H}^q(G,A) = H^q(G,A) for q >= 1
<hochs> and hat{H}^(-1)(G,A) = hat{H}_0(G,A), and hat{H}^{-q} = H_{q-1}
<hochs> i.e. the tate H^q (with hat on) is same as usual H^q for q > 0
<hochs> and H^0 (with hat on) is H^0 mod out by the image of N
<hochs> and for q < 0, H^q will be homology groups
<hochs> and these guys fit nicely into long exact sequence for any exact sequence 0 -> A -> B -> C -> 0 of G-modules
<hochs> the connecting morphism from H_0(G,C) -> H^0(G,A) (with all hats on) is done by N, that is, you take [c] in H_0(G,C) and take a lifting B -> c in B, then multiply by N to get Nb, and the class of [Nb] is the image of connecting morphism in H^0(G,A)
<hochs> ok great, we connected cohomology and homology
<hochs> now there's something big missing
<hochs> for anyone who deals with cohomology
<hochs> we like cohomology because
<hochs> not because they're more complicated than homology, but because they come with cups
<hochs> and yes, you can define cup product on these tate groups (you can just define them on the complete resolution giving rise to these tate groups: such exists)
<hochs> satisfying all the usual properties
<hochs> now i'm going to state tate's theorem and be done with group cohomology and go back to local class field theory
<hochs> it says that under very special circumstance, cupping gives isomorphism
<hochs> well, i wonder if I should state it now. it's somewhat technical to write all out
<hochs> so i'll just say if you have an G-module A and a is in H^2(G,A) with H^1(G,A) = 0 and H^2(G,A) generated by a with |H^2(G,A)| = |G|, then H^n(G,Z) -> H^{n+2}(G,A) cupping by a is an isomorphism
<hochs> this is the punch-line, actually
<hochs> btw all my cohomology has hat on now
<hochs> this is the punch-line because we described H^0(G, L^*) (with hat on) = K^* / NL^*
<hochs> and H^{-2}(G,Z) = H_1(G,Z) = G/G' = G (for G = Gal(L/K) abelian extension)
<hochs> so as long as I can show that H^2(G, L^*) has the right order (in this case, |G| = n, say)
<hochs> then cupping by some generator of a of H^2(G,L^*) gives an isomorphism pairing K^* / NL^* with Gal(L/K)
<hochs> so one only needs to determine that H^2(G,L^*) is cyclic of the right order
<hochs> gee it's already been 55 mins, i'll be less technical from now on.. got so much more to cover
<hochs> anyway for grad students: determining the order of H^2(G,L^*) is done by reducing it to G cyclic, H^2(G,L^*) embeds in Br K, and one can determine Br K as H^2(K_nr / K)
<hochs> these tate cohomologies are easy to deal with by Herbrand Quotients for G cyclic
<hochs> ok so that's basically the sketch of proof
<hochs> for local class field theory
<hochs> global class field theory is obtained by pasting all these local residue maps (actually product of them all)
<hochs> so now we move onto K a global field
<hochs> which is either finite extension of Q or F_q(t) rational function field in one variable over finite field
<hochs> i'm going to be mainly interested in number field case (finite extension of Q)
<hochs> unlike in the local case where you have only one valuation / norm to deal with
<hochs> and hence only one topology,
<hochs> we have infinitely many norms / valuation on global field K
<hochs> for example, even in the field of Q case
<hochs> we have the p-adic valuation v_p
<hochs> which is nonarchimedean (meaning its completion is either R or C, meaning |n| is not bounded for integers n)
<Galois> !
<hochs> sorry
<hochs> i typed wrong thing
<hochs> nonarchimedean means |n| is bounded
<hochs> and completion is Q_p
<hochs> yes galois?
<Galois> What are the infinitely many valuations of F_q(t)?
<hochs> these are defined by irreducible polynomial p(t) in F_q[t]
<hochs> for any such irreducible p(t)
<Galois> OK but near the beginning of the seminar you wrote that F_q(t) is a local field
<hochs> oo that's a mistake someone should've corrected me earlier
<hochs> thank you galois
<Galois> So what's the local field?
<hochs> so for char K = p > 0, the local field will be k((T)) of *formal* power series
<lhrrwcc> I thought you meant to mean F_p((T))
<hochs> yes, formal power series
<Galois> Does that mean there's only one local field associated to F_q(t)?
<hochs> just as Z_p of p-adics is gotten from Z by "formal" power series, so to-speak
<hochs> no
<hochs> since it has many valuations, each completion gives different local fields
<lhrrwcc> [note: k((T)) is not local for char k = 0, ie: take k=Complex numbers, then k((T))/(T)=k which is infinite]
<Galois> OK, carry on
<hochs> actually i glossed over the function field case since for this seminar I mainly want to discuss number fields
<hochs> the fundamental inequality proofs will be harder for function field case, so one reason
<hochs> ok so K is a global field, (note: K finite extension of Q from now on - is in the back of my mind)
<hochs> these guys have infinitely many primes, but we can classify them as being either archimedean or nonarchimedean
<hochs> i'll denote by v a valuation of K and K_v by the completion of K with respect to the topology defined by v
<hochs> archimedean valuation v is the one where |n| is unbounded, i.e. K_v is either R or C
<hochs> and nonarchimedean valuation v is the one where |n| is bounded, i.e. K_v is Q_p for some prime p
<hochs> now I keep going back and forth between using multiplicative | | notation and additive v notation
<hochs> but all my | | will be normalized, in the sense of Haar measure
<hochs> so for a valuation v, |uniformizer of v|_v = 1/|(residue field of v)|
<hochs> now we like to set up some correspondence between finite abelian extension L / K and certain classes of objects
<hochs> in the local case, that "certain classes of objects" were open subgroups of finite index in K^* (this is the description of norm subgroups of K^*, turns out)
<hochs> called norm groups
<hochs> in the global case, we have infinitely many valuations and we somehow need an object that brings all these valuations together, and that's idele
<hochs> recall that for a global field K, adele of K is restricted topological product of all the K_v with respect to O_v (the ring of integers in K_V)
<hochs> well it's slightly funky topology, but basically it's defined so that locally it's just product of these things
<hochs> of K_v
<hochs> idele is the multiplicative side of this
<hochs> i'll denote A_K for adele of K
<hochs> and define J_K = (A_K)^*, the multiplicative group of it. the topology won't be induced from A_K, however
<hochs> the topology is forced so that the inverse map, x -> x^{-1} will be continuous
<hochs> which is induced by x -> (x,x^{-1}), if you will
<hochs> of product topology on A_K x A_K
<hochs> ok back to global CFT. every abelian extension L/K will actually correspond to open subgroup N of finite index in C_K = J_K / K^*, idele mod the principal ideles
<hochs> finite abelian extension L/K I mean.
<hochs> and like in the local setting, we'll have a map C_K -> Gal(L/K) that captures this correspondence
<hochs> note that the map no longer takes the form K^* -> Gal(L/K) like in the local setting,
<hochs> particular because K, global field now, now has more than 1 valuation
<hochs> this map has certain properties, and I'll explain that now
<hochs> first, the map is induced from J_K -> Gal(L/K), and has kernel containing K^* (that's why i could write C_K -> Gal(L/K))
<hochs> and it's somehow equivalent to the older formulation using ray groups,
<sek> !
<hochs> in the sense that for valuation v which is unramified at L, this map will send v to the corresponding frobenius of Gal(L_w / K_v) for any valuation w prolonging v
<hochs> yes sek
<sek> could you elaborate on why the global case having more valuations makes us consider the ideal class group instead of K^*. Morally that is, apart from the obvious kernel fact you just mentioned
<hochs> i see, ok
<hochs> historically actually this global class field theory business was conceived by artin to generalize reciprocity laws
<sek> in some sense, why is J_K the right thing? Other than the usual local-to-global philosophy, is there anything else?
<hochs> oh
<hochs> using the J_K thing is chevalley's idea
<hochs> originally it was by ray group, or subgroup of I_K the free abelian group on unramified valuations
<hochs> and a map ray group -> Gal(L/K)
<hochs> but these groups depend on L
<hochs> and need to be changed as L varies
<hochs> the "bad" primes - including archimedean and ramified ones are thrown out
<sek> right, so generalising reciprocity laws meant patching together valuation business via the various symbols
<hochs> yea
<hochs> so if anything, K^* should go to identity in Gal(L/K) (this is the reciprocity law)
<hochs> since in our formulation, J_K -> Gal(L/K) is going to be product of all the local residue maps
<hochs> in terms of hilbert symbol, product_{v ALL valuations} (a,b)_v = 1
<hochs> for any b in K^*
<sek> and the Hilbert symbols corresponds to quadratic extensions
<hochs> yes, quadratic is part of it
<hochs> the formulation using J_K makes stating the reciprocity law that product of (a,b)_v = 1 very natural i think
<lhrrwcc> yeah
<hochs> ok so we have this map J_K -> Gal(L/K) and its relation with local maps are very natural also
<hochs> for any valuation v of K, there's an injection K_v^* -> J_K
<hochs> and we can compose this to get a map K_v^* -> Gal(L/K)
<hochs> fact: the image of this map from K_v^* will land in Gal(L_w / K_v) for any valuation w prolonging v
<hochs> since Gal(L/K) is *abelian*, Gal(L_w / K_v) seen as a subgroup of Gal(L/K) doesn't depend on w
<hochs> in fact, Gal(L_w / K_v) as a subgroup of G(L/K) will just be the ones in Gal(L/K) that are continuous with respect to the topology defined by w
<hochs> i.e., those s in Gal(L/K) such that w(s(x)) = w(x) for any x in L
<hochs> and for varying w, but above a fixed v, these Gal(L_w / K_v) are conjugations of each other
<hochs> but in abelian setting, there's only one such
<hochs> xHx^(-1) = H in abelian group
<hochs> ok, so that's the relation between J_K -> Gal(L/K) and K_v^* -> Gal(L_w / K_v) for any w prolonging v
<hochs> so in fact, for any idele (x_v)_v in J_K, the map J_K -> Gal(L/K) sends (x_v)_v to product of all phi_v(x_v)
<hochs> where phi_v is the local residue map K_v^* -> Gal(L_w / K_v)
<hochs> and this globalized map J_K -> Gal(L/K) is called Artin map, I denote by phi:J_K -> Gal(L/K)
<hochs> the kernel of this map will be precisely K^* N_{L/K} J_L
<hochs> or in using the idele classes C_K, if C_K -> Gal(L/K) is artin map then its kernel is N_{L/K} C_L
<hochs> and these maps are all compatible for varying L
<hochs> so you again get a globalized artin map C_K -> Gal(K^{ab} / K)
<hochs> where K^{ab} is the maximal abelian extension of K
<hochs> we have then 1-1 correspondence between finite abelian extension L / K with certain subrgoups of C_K via L -> N_{L/K} C_L,
<hochs> and these "certain subgroups" turn out to be exactly the open subgroups of finite index in C_K
<hochs> this is so-called Existence Theorem
<hochs> these open subgroups N of finite index in C_K are called norm groups, and the corresponding L giving rise to N are called class fields belonging to N
<hochs> ok that's the statement of global class field theory
<hochs> with this, global kronecker-weber that max abelian extension of Q is union of all cyclotomic fields follows easily
<hochs> how to see this? Well we know what J_Q looks like
<hochs> the idele of Q, rationals
<hochs> it's going to be J_Q = Q^* x (positive reals with multiplication) x product of U_p's for varying rational prime p
<hochs> where U_p is the group of units of Z_p
<hochs> i'll leave this up as an exercise
<hochs> ok so global class field theory gives correspondence between open subgroups of J_Q of finite index and abelian extensions of Q
<hochs> if Q^{mc} is the maximal cyclotomic extension of Q, then one can check directly (again exercise) that the map J_Q -> Gal(Q^{mc} / Q) has kernel exactly (positive reals with multiplication) part of J_Q
<hochs> in particular, all the U_p's do NOT go to identity of Gal(Q^{mc} / Q)
<hochs> unless it's identity itself
<hochs> but (positive reals with multiplication) is *infinitely divisible*
<hochs> and such automorphisms don't exist in any galois group
<hochs> of abelian extensions
<hochs> so they're in the kernel of every J_Q -> Gal(L/Q) for any abelian extension L/Q
<hochs> so Q^{mc} corresponds to the minimal such norm group of C_Q
<hochs> but so does Q^{ab}
<hochs> so Q^{mc} = Q^{ab} is the maximal abelian extension of Q
<hochs> sorry, the kernel of J_Q -> Gal(Q^{mc} / Q) is exactly Q^* x (positive reals with multiplication)
<hochs> can't forget about those principal Q^* guys
<hochs> ok, let me do that example I promised to do earlier, about expressing p = x^2 + 5y^2
<hochs> now that we have the main statements of global class field theory
<hochs> as a motivation tool, try finding all primes of the form p = x^2 + y^2
<hochs> what's the most natural proof?
<hochs> use PID-ness of Z[i]
<hochs> Z[i] is euclidean, so is in particular a PID
<hochs> we also know that prime p = 1 (mod 4) ramify / split into product of distinct primes in Z[i]
<hochs> and p = 3 (mod 4) remain prime
<lhrrwcc> !
<hochs> yes
<lhrrwcc> p = 1 (mod 4) splits
<hochs> it doesn't remain prime
<hochs> because Z[i]/p ~= Z[x]/(p, x^2 + 1) ~= Z/p[x] / (x^2 + 1)
<lhrrwcc> indeed, but why you use the word ramify?
<hochs> and for p = 1 (mod 4), x^2 + 1 splits
<hochs> ahh
<lhrrwcc> 2 ramifies
<hochs> unramifies actually sorry
<lhrrwcc> thanks.
<hochs> only 2 ramifies
<hochs> the discriminant of Q(i) over Q is 4
<hochs> so everything except 2 is unramified
<hochs> thank you
<hochs> so an odd prime p has the form x^2 + y^2 = p iff p = (x + yi)(x - yi) iff p splits in Z[i] iff x^2 + 1 = 0 (mod p) has a solution
<hochs> iff p = 1 (mod 4)
<hochs> now try doing this for p = x^2 + 5y^2
<hochs> we run into a problem because Z[sqrt(-3)] (this is the ring of integers of Q(sqrt(-5))
<hochs> is not PID
<hochs> it has class number 2, in fact
<hochs> i'm not going to prove this now, but it can be done by some estimation of measures of norm groups
<hochs> you can find the reference in neukirch or jmilne for example <wli> Oh boy, class number I have to look up.
<hochs> ah class number of R is I / R^* where I is the free abelian group on all its nonarchimedean primes and R^* the principal ones <barf3> !hochs what is meant by the measure of a group in this context?
<Galois> you meant Z[sqrt(-5)]
<hochs> yes Z[sqrt(-5)], thank you
<hochs> barf3, there's a natural injection of number field K into products of C or R
<hochs> by embeddings of K into the complex
<lhrrwcc> [hint for the problem: 6 = 2*3 = (1+sqrt(-5))(1-sqrt(-5))]
<hochs> and the latter product has natural haar measure
<hochs> you can induce the measure from there <barf3> thank you hochs
<hochs> yes, 6 factors into like that
<hochs> so UFD doesn't hold
<hochs> ok so we can't say that p = x^2 + 5y^2 iff p splits in Z[sqrt(-5)]
<hochs> but we *can* say that p = x^2 + 5y^2 iff p splits into *principal* primes in Z[sqrt(-5)]
<hochs> as you know, class number sort of "measures" how far it is away from being a PID, and we can realize this more concretely by constructing what's called a hilbert-class-field L of Z[sqrt(-5)] such that the principal primes of Z[sqrt(-5)] are exactly those that split in L
<hochs> we'll construct this hilbert class field of Z[sqrt(-5)]
<hochs> then prove, by using class field theory, the statement I just made about principal primes of Z[sqrt(-5)]
<hochs> but before that a quick restroom break, i'll brb
<hochs> back
<hochs> ok we need to construct this magical field L
<hochs> in fact it's not even clear that such magical L exists
<hochs> let K = Q(sqrt(-5)) for convenience now <Jerry_2> hochs, barf3 had a question above
<hochs> oh sorry <barf3> oh sorry <barf3> i was realying that question for someone in freenode <barf3> but they withdrew the question
<hochs> collective mathematics employed here? <barf3> i'm not sure what they meant
<hochs> what moments <barf3> but anyway they changed their mind... <barf3> sorry
<hochs> ok np
<hochs> if L/K is any abelian extension, v a prime of K, and if i_v : K_v^* -> J_K is the canonical injection, then v splits completely in L if and only if i_v(K_v^*) is contained in K^* N_{L/K} J_L, and for non-archimedean v, v is unramified in L iff i_v(U_v) is contained in K^* N_{L/K} J_L
<hochs> the first is true because v splitting completely means that K_v = L_w for any w prolonging v, and this in turn means (by local class field theory) that i_v(K^v^*) is contained in the kernel of the residue map
<hochs> the latter is true because the group of units of local fields get sent surjectively (by local residue map) to the inertia subgroup of the galois group
<hochs> well maybe i shouldn't formulate hilbert class fields this way, ok scratch that
<hochs> you have J_K, the idele of K
<hochs> i'm going to construct a particular open subgroup of it of finite index
<hochs> which will correspond to a certain finite abelian extension L of K
<hochs> then i'm going to prove that the Cl K (the class group of Z[sqrT(-5)], slight abuse of terminology here) is isomorhpic to Gal(L/K) canonically
<hochs> which in particular says that the principal primes are precisely the ones that go to 1 in Gal(L/K) by artin map, which are the ones that split completely in L
<hochs> so what is this special open subgroup of J_K? I'll denote it as N, and N is going to be the product of K_v^* for all *archimedean primes* v, product the U_v for all *nonarchimedean* primes of v
<hochs> i'm using the term "prime, place, valuation" interchangeably here
<hochs> well i haven't used the term 'place' yet but they all mean the same thing in this context
<hochs> where U_v is the group of units for v
<hochs> ok, throw in the principal ones too
<hochs> i mean N = (K^*) product_{v archimedean} K_v^* product_{v nonarchimedean} U_v
<hochs> you can check that this is open subgroup of finite index in J_K
<hochs> and so corresponds to a certain abelian extension L / K
<hochs> now suppose Cl K = I_K / P_K is the class group of K, where I_K is the free abelian group of non-arch primes of K
<hochs> and P_K is the principal ones
<hochs> there's a canonical map J_K -> I_K sending (x_v)_v to sum of v(x_v) v for all non-arch v
<hochs> the kernel of which is precisely N
<hochs> i mean the kernel of J_K -> I_K/P_K
<hochs> is precisely N
<hochs> these principal ones are getting to me now
<hochs> so J_K / N is isomorphic to I_K / P_K = Cl K
<hochs> but on the other hand, J_K / N is isomorphic to Gal(L/K) by class field theory
<hochs> now one can check that every archimedean v of K splits completely in L
<hochs> whereas all the non-archimedeans v of K are unramified in L
<hochs> this actually follows from the beginning remark I made which I asked to scratch
<hochs> but it was just in a refrigerator until now
<hochs> since Cl K (our K = Q(sqrt(-5)), remember) has order 2
<hochs> Gal(L/K) will be of order 2
<hochs> i.e. L/K will be extension of order 2
<hochs> ok so we know this magical L exists and is of order 2 over K
<hochs> how do we find it?
<hochs> it's characterized by the property that very arch of K gets split completely in L and every non-arch v of K gets unramified in L
<hochs> well 5 is the only prime that gets ramified from Q to Q(sqrt(-5))
<hochs> and for example, 2 is the only prime that gets ramified from Q to Q(i)
<hochs> so if i take the compositum Q(sqrt(-5), sqrt(-1))
<hochs> then every nonarch prime in Q(sqrt(-5)) will get unramified at Q(sqrt(-5), sqrt(-1))
<hochs> and since these are all imaginary quadratic fields anyway, the arch primes give C as the completion and they're cleary all split completely
<hochs> so our L = Q(sqrt(-5), sqrt(-1))
<hochs> that is, a prime p of Q(sqrt(-5)) is principal iff p is split completely in Q(sqrt(-5),sqrt(-1))
<hochs> now the question about p = x^2 + 5y^2 is as easy as it were for p = x^2 + y^2
<lhrrwcc> !
<hochs> yes
<lhrrwcc> Can I take any other extension in which 2 ramifies?
<hochs> yes i think so
<hochs> can you check that quickly and let us all know
<lhrrwcc> I mean, you stated the correspondence
<lhrrwcc> between the composition field and the group (the galois correspondence)
<lhrrwcc> So I would imagine that any extension that makes all prime completely would work
<lhrrwcc> you may go on, please.
<hochs> yes i think that's right
<hochs> ok so
<hochs> p = x^2 + 5y^2 iff p splits into principal primes of Z[sqrt(-5)] iff p splits completely over Q(sqrt(-5), sqrt(-1)) iff Q_p(sqrt(-5), sqrt(-1)) = Q_p iff sqrt(-5) and sqrt(-1) exist in Q_p iff (by Hensel) sqrt(-5) and sqrt(-1) exist in Z/p iff -1,-5 are squares in Z/p
<hochs> and the latest condition is handled by quadratic reciprocity
<hochs> iff p = 1, or 9 (mod 20)
<hochs> in general, one can try to solve p = x^2 + ny^2 by determining hilbert class fields of Q(sqrt(-n))
<hochs> but that gets pretty out of control as class group grows 18:24:19] < lhrrwcc> !
<hochs> yes 18:24:28] < lhrrwcc> we always end with a quadratic reciprocity problem?
<hochs> nope
<hochs> because for example
<hochs> if class number is 3
<hochs> then you'll get cubic extensions
<lhrrwcc> ah, yeah
<hochs> and now you're doing cubic reciprocity
<lhrrwcc> thanks.
<hochs> np, also all the integers < 0 for which class number 1 is known
<hochs> thanks to Stark
<hochs> currently at UCSD
<hochs> hope i'm getting the literature right
<lhrrwcc> Yes, you are right.
<hochs> they're 3, 4, 7, 8, 11, 19, 43, 67, and 163.
<hochs> he used to be at MIT awhile back
<hochs> but he was having dual positions in both MIT and UCSD
<hochs> and time came when he had to decide between the two
<hochs> he chose UCSD (probably the weather)
<hochs> ok, so what to do in general
<lhrrwcc> Oh, i knew the result but for quadratic number fields
<hochs> ah :)
<lhrrwcc> he is also the one who proved it
<hochs> in general, if K is an imaginary quadratic field, its hilbert class field is going to be generated by j-invariants of the elliptic curves which have the ring of integers of K as its ring of endomorphisms
<hochs> and it gets pretty technical for primes of the form x^2 + ny^2 for large n
<hochs> maybe someone else who is interested can give a talk on complex multiplication
<hochs> yes, there's also this series of Stark Conjectures
<hochs> which seems to do with special values of L-functions
<zeno> As in Harold "Gauss-co-confirmer" Stark?
<hochs> yep
<hochs> that's the same Stark
<hochs> ok i'm just going to gloss over some things from now.. there's this quadratic form theory i mentioned
<hochs> you have a non-degenerate quadratic form f(x_1, ..., x_n) in n variables over a global field K, say
<hochs> when does it represent particular number c != 0 ?
<hochs> well this is equivalent to when does f(x_1, ..., x_n) - cX^2 represent 0
<hochs> because a quadratic form (non-degenerate) representing 0 actually represents everything
<hochs> hint: f(tX + Y) = t^2 f(X) + tA(X,Y) + f(Y)
<hochs> for non-degenerate A
<hochs> here's the main theorem over global field
<hochs> Let K be a global field, and f a non-degeneerate quadratic form in n variables over K which represents 0 in K_v for each prime v of K
<hochs> then guess what? f represents 0 in K
<hochs> this is an illustration of Hasse Principle, which says that if something is true locally then it's true globally
<hochs> ofc Hasse Principle fails for many things
<hochs> but for quadratic form theory it holds spectacularly
<hochs> i'm in need of some food now, so perhaps i should end soon
<hochs> the proof of this is by induction on n
<hochs> well i won't prove it now i think
<hochs> but as an application/exercise one can try to prove that every rational number is a sum of 4 rational squares
<hochs> and also a rational number c > 0 is a sum of 3 rational squares iff c = 4^n * d for some rational d with d != 7 (mod 8)
<hochs> hint: rewrite x^2 + y^2 + z^2 = c as a statement about x^2 + y^2 + z^2 - cT^2 representing 0, work locally
<hochs> the only problematic case is at Q_2, where you'll need d != 7 (mod 8)
<hochs> squares of Q_2^* are precisely those =1 (mod 8) also
<hochs> once you know this, it's an exercise in elementary algebra to show that the same is true if we place "rational numbers" with "integers" through out
<hochs> recipe of the proof for integers assuming it's true for rationals:
<hochs> so u know such c has form x^2 + y^2 + z^2 = c
<hochs> you need to find something with integral coordinates now
<hochs> x,y,z currently are rational
<hochs> c is integer of the form 4^n * d with d integer and d != 7 (mod 8)
<hochs> if (x,y,z) is integral then done
<hochs> otherwise take an integral point z in 3-space so that |z - x| is as small as possible
<hochs> draw a line joining x to z, show this is not tangent
<hochs> and so it intersects the sphere x^2 + y^2 + z^2 = c at some other rational point
<hochs> (looking like group law now)
<hochs> now this new approximation x' can be written with the common denominator strictly less than that of d
<hochs> so continuing this
<hochs> you get integer solution to x^2 + y^2 + z^2 = c
<hochs> in fact, if you're good with calculation then you can show that x' in the first step is already integral
<hochs> err rather, prone to doing calculations
<lhrrwcc> !
<hochs> yes
<lhrrwcc> That's similar strategy to find finite order points in elliptic curves, right?
<hochs> yep
<lhrrwcc> err, torsion points
<hochs> yes
<lhrrwcc> thanks.
<hochs> hmm ok so the proof of hasse principle for quadratic forms over global fields rely on induction
<hochs> but inductive step is kind of easy
<hochs> the hardest part is in low variable count again
<hochs> and there's a very slick proof that for quadratic extensions, something is a norm iff it is locally norm
<hochs> using cohomology of idele classes
<hochs> so there's an exact sequence 0 -> L^* -> J_L -> C_L -> 0 for any abelian finite extension L/K
<hochs> and the long exact sequence tells us that H^1(G, C_L) -> H^2(G, L^*) -> H^2(G, J_L) -> ...
<hochs> where G = Gal(L/K)
<hochs> there's some computation to show that H^1(G, C_L) = 0
<hochs> (this is part of what's called "second" fundamental inequality in the literature)
<hochs> computatoin involves some kummer theory stuff <relay bot> <fwe@freenode> i have p(h>h') if h and h are real gaussian would the same formula apply if they are rayleigh?
<hochs> so H^2(G, L^*) -> H^2(G, J_L) is an injection
<hochs> and short computation (using that cohomology commute with limits and that groups of units of unramified extension have trivial cohomology - i didn't prove this, but follows from it being a profinite group each of whose factor is isomorphic to residue field with addition)
<hochs> tells you that H^2(G,J_L) is sum of H^2(G(L_w / K_v), (L_w)^*) = H^2(L_w / K_v) as v runs through all primes of K
<hochs> in the cyclic case, we know H^2 = H^0 (with hat on)
<hochs> and H^0(G, L^*) = K^* / NL^*
<hochs> so what the injectivity is telling you (consequence of H^1(G,C_L) = 0) is that for cyclic extension L/K at least, an element in K is a norm iff it is norm in every L_w / K_v !
<hochs> ok that's about it, that's class field theory in nuts and bolt
<hochs> there's the analytic side to it also, which has to do with special values of L-functions
<hochs> and a pretty amazing theorem which is called Tchebotarev Density Theorem
<hochs> generalizing prime number theorem
<hochs> tells you about distribution of splitting of primes in extension fields
<hochs> Ok, that's Class Field Theory. Abelian Extension of Global / Local Fields.
<hochs> ok, so there's this thing you can prove about norm groups, which is this: if E / K is a finite extension and if L/K is the largest abelian extension contained in E, then N_{E/K} E^* = N_{L/K} L^*
<hochs> so we have no hope of determining non-abelian extensions using norm groups
<hochs> we need something else
<hochs> this, in the local case of non-abelian extensions, take the form of weil deligne representations
<hochs> well ofc there's also lubin-tate side of it
<hochs> anyway roughly, the correspondence in non-abelian case is {irreducible representations of GL_n(K)) <-> {n-dim representations of G_K}
<hochs> where G_K = Gal(K^{ab} / K) is the absolute galois group
<hochs> and K is our local field
<hochs> and in the global case, it's going to be realized by etale cohomology groups of certain *unitary* shimura varieties
<hochs> (which you can think of as generalization of global class field theory using complex multiplication)
<hochs> this is nonabelian lubin-tate
<sek> (and for the record, class field theory does describe 1-dimensional representations, so this is a generalisation)
<hochs> reference: Teruyoshi wrote some nice ones
<hochs> yep
<hochs> i think jerry can give a talk next time on this
<hochs> his advisor being dick gross
<hochs> ok i'm pooped for now, is there anything i missed?
<sek> it's Falting's theorem that tells you that in the local case all your representations arise from geometry right?
<hochs> yes
<hochs> that was a start for all these algebraic geometry madness i think
<sek> (and I guess his theorem includes the crystalline picture too)
<lhrrwcc> yeah
<lhrrwcc> there is a connection with determining the rational points on elliptic curves about reduction being good
<sek> do you mean over a local field lhrrwcc?
<sek> (if so then yes, Hensel's lemma)
<hochs> kek
<lhrrwcc> over number fields
<lhrrwcc> hochs, are you done sir?
<hochs> i think so
<lhrrwcc> Applauds!
<sek> *clap**clap*
<hochs> unless there's something i'm missing
<hochs> thank you
<hochs> that was 2 hour thing, certainly is slower to do this by IRC
<hochs> or not
<sek> hochs: you're missing the entire Langlands picture. Can we have another 24 hours of that please?
<hochs> i think i'm going to get something to eat finally
<hochs> and leave rest of the talk to jerry
<lhrrwcc> hochs, you typed this live?
<hochs> yes
<hochs> i guess it shows <wli> I'm missing Langlands, Jugendtraum, and the whole nine yards.
<hochs> since i wasn't fixing notations and things
<hochs> just talked whatever came to my mind
<hochs> but i'm no chuck norris? why is this joke so popular now
<hochs> next time: etale cohomology & weil conjecture?
<hochs> zariski topology sucks
<sek> I beg to differ
<sek> we can have a live debate sometime:)
<hochs> ahhh i need to meet my advisor tomorrow
<lhrrwcc> hochs, if zariski topology sucks, imagine m-Spec A
<hochs> need to get some work done now
<hochs> oh yes m-Spec A, major problem
<hochs> functoriality total failure
<lhrrwcc> But we can live with it :)
<sek> well thanks for the talk. I presume Fermat will be putting up the notes online
<lhrrwcc> Well, I am going to leave, very nice talk hochs. Congratulations.
<hochs> that's something that berkovich salvaged for rigid analytic spaces when Tate only worked with m-Spec of tate algebras
<hochs> for his uniformization of elliptic curves
<hochs> ok
<hochs> thank you all