# Seminar on More Algebraic Number Theory

This seminar was held in #mathematics on Freenode (with a relay set up to EFnet) on **24th of September 2005** at 22:00 EDT by **nerdy2**.

## Seminar Log

`
22:11:07 <@nerdy2> ok, so we want to talk number theory?
22:11:14 <@Chandra> yes sir :)
22:11:17 < driggers> yap
22:11:21 < Galois> I think class field theory was the plan
22:12:28 <@nerdy2> yes, although the original title said "introductory" or some such, so i don't have too grand plans
22:12:51 <@nerdy2> ramanujan specifically suggested artin reciprocity and quadratic reciprocity as an example of this
22:12:55 < yrlnry> "Theory of the positive integers les than 7."
22:13:43 <@nerdy2> class field theory is really just a statement that two (usually finite, or profinite) groups are isomorphic
22:14:37 <@nerdy2> unfortunately, to define one of the groups is a bit complicated
22:15:07 < Galois> you mean ray class groups?
22:15:08 <@nerdy2> hmm, relaying /topic changes, interesting :)
22:15:42 <@nerdy2> relaybot, those being the same as galois groups of the class fields, or what is the same the idele class group being the same as the galois group
22:15:49 <@nerdy2> sorry, galois:
22:16:01 < Galois> yeah that's the way, all in one shot with id\`eles
22:16:06 < Galois> but it's not "introductory"
22:16:19 <@nerdy2> yes, so i'll mention both :)
22:17:02 <@nerdy2> i gave a talk some time ago in efnet #mathematics, and talked about valuations, places, adeles, ideles and the like.... i'll review this stuff very quickly
22:18:35 <@nerdy2> take a number field K [a finite extension of Q], we are interested in absolute values on K
22:19:05 <@nerdy2> ostrowski's theorem says that on Q the only absolute values are the p-adic valuations |p^a m/n| = p^-a, and the usual absolute value [up to equivalence]
22:20:20 <@nerdy2> so we call equivalence classes of nontrivial absolute values "primes"
22:20:20 <@nerdy2> or "places"
22:20:20 <@nerdy2> ostrowski's theorem says we aren't too far off, for Q, there is one finite place for each integral prime, and one "infinite" place
22:21:43 <@nerdy2> more generally, the "infinite" places correspond to embeddings into C [modulo complex conjugation] so we know there are r+s of them, where r,s are the number of real embeddings, conjugate pairs of complex embeddings of K respectively
22:22:08 <@nerdy2> and the finite places of K again actually correspond to primes of K in the more ordinary sense [prime ideals of the ring of integers in K]
22:22:30 <@nerdy2> everyone with me so far? :)
22:23:41 <@nerdy2> ok, so we have absolute values, this means we can complete... given a place v, K_v denotes the completion at v (i.e. with respect to any of the absolute values in v)
22:24:31 <@nerdy2> for Q, this again gives us the familiar fields Q_p [for each integral prime p] and R
22:25:37 <@nerdy2> number fields are an example of global fields [the only other global fields are function fields in one variable over finite fields] and what we get when we complete a number field [more generally a global field] at a place is known as a local field
22:25:57 < DaMancha> !
22:26:00 <@nerdy2> class field theory has two forms : local and global, which are respectively statements about local and global fields
22:26:03 <@nerdy2> yes?
22:26:28 < DaMancha> when you complete at a "prime" p, is K_p wrt all the members of the equiv class?
22:26:56 <@nerdy2> well equivalent valuations give the same completion, so you can complete with respect to any absolute value in the class
22:27:35 <@nerdy2> equivalence is just the relation |.|_1 = |.|^c_2 for some real number c, so they trivially give the same completion
22:27:42 < Galois> more generally, equivalent valuations give the same topology (in fact this is the definition of equivalence for valuations), and same topology means same completion
22:28:10 < dublisk> !
22:29:09 <@nerdy2> yes?
22:29:35 < dublisk> could you clarify the definition of infinite vs prime for an arbitrary valuation on K? infinite means by definition it gives embedding into C?
22:30:29 <@nerdy2> so we have absolute values (which i haven't defined, but the axioms you can guess), and there are two options for nontrivial ones (the trivial one being |0| = 0, |x| = 1 for x neq 0)
22:31:10 <@nerdy2> one option (archimedean): { |n| : n in Z } is unbounded --- then the place/absolute value is called archimedean
22:31:37 <@nerdy2> second option (nonarchimedean): { |n| : n in Z } is bounded -- then we say the place/abs. value is nonarchimedean
22:32:11 <@nerdy2> now the nonarchimedean places correspond in a one-to-one way with prime ideals of K
22:32:35 <@nerdy2> hence the term prime in general, and in particular these are the "finite" primes
22:32:42 <@nerdy2> the nonarchimedean places are the "infinite" primes
22:32:52 < Galois> it might be easier just to say that infinite means { |n| : n in Z } is unbounded. That's nice and clean and makes sense (infinite == unbounded)
22:33:33 <@nerdy2> sorry, the archimedean places are the "infinite" primes
22:33:42 < dublisk> by primes of K I assume you always mean primes of O_K
22:33:57 <@nerdy2> dublisk, yes, prime ideals of O_K
22:34:04 <@nerdy2> for "finite" primes
22:34:43 <@nerdy2> dublisk, and yes infinite primes correspond to complex embeddings (any complex embedding gives you an absolute value.... )
22:36:56 <@nerdy2> so one thing this gets you is the product formula : prod |x|_v = 1 for all x in K [you have to take the product over /all/ places v of K, and take the right absolute value |.|_v in each class... there's a canonical choice in some sense]
22:37:28 <@nerdy2> [for Q, the "canonical" ones are the p-adic absolute value normalized as above | p^a m/n| = p^-a and of course the usual real absolute value]
22:37:37 <@nerdy2> [so exercise: see that the product formula holds for Q]
22:39:43 < landen> Kummer ban on some number of caps in a row?
22:40:39 <@nerdy2> [then from there another neat thing: say you want to know about O_K^*, the group of units... then |x|_v = 1 if x is a unit and v is a finite prime, so if you wanted to care about valuations you only have to worry about the infinite primes, so you could define a map O_K^* -> R^(r + s) by x -> (log |x|_v) where you range over infinite primes v -- of which there are r+s of them)
22:41:47 <@nerdy2> [now this is a group homomorphism whose image lies in the plane x_1 + ... + x_{r+s} = 0, so if you look at the image there you can prove that it's a complete lattice in this plane of dimension r+s-1, and the roots of unity go to 0, so you see that O_K^* = Z^(r+s-1) (+) finite group as an abstract abelian group]
22:42:05 <@nerdy2> so the infinite primes do play some interesting role here
22:42:21 <@nerdy2> landen, when do i ever use caps :)
22:42:33 <@Chandra> nerdy2 - don't worry, go on
22:42:35 <@ramanujan> people were spamming #math, don't worry about that.
22:43:05 <@nerdy2> ok, so the adeles are likewise you get from taking a product of K_v's essentially
22:43:28 <@nerdy2> you don't just want to take the product though
22:44:48 <@nerdy2> for any finite set containing the infinite primes, S, we define A_{K,S} = the subset of product K_v consisting of (x_v) such that x_v is in O_{K_v} for v not in S
22:45:30 <@nerdy2> now if you can parse that, it means just we are taking things in the product which are actually integral /almost everywhere/ (i.e. except for finitely many finite primes -- we can't really put any condition on the infinite primes)
22:45:46 <@nerdy2> and A_K = dir lim A_{K,S}
22:46:41 <@nerdy2> so it's just the union of the A_{K,S}'s but with a funky topology
22:47:14 <@nerdy2> this is a nice space [you can do analysis on it]
22:47:25 <@nerdy2> and there's a natural map from K to A_K
22:47:42 <@nerdy2> which embeds K as a discrete cocompact subgroup
22:48:05 <@nerdy2> so you can do fourier analysis on A_K/K, but kit hasn't spoken up so we'll not mention this again :)
22:48:09 < _llll_> !
22:48:12 <@nerdy2> yes?
22:48:19 < _llll_> what's cocompact?
22:48:56 <@nerdy2> it means the quotient is compact (so A_K/K is compact)
22:50:39 <@nerdy2> ok, now the ideles are just the invertible elements in the adeles, J_K = A_K^* (A_K being a ring)
22:51:12 <@nerdy2> but it also gets a funky topology... it's the one it gets as a subgroup of A_K x A_K via the embedding x -> (x, 1/x)
22:51:33 <@nerdy2> and K^* of course embeds in J_K, and the quotient is known as the idele class group C_K = J_K/K^*
22:51:44 <@nerdy2> now you may in fact wonder what's the point of all this :)
22:51:47 < supremum> !
22:52:02 <@nerdy2> and this is the group i mentioned above that's hard to define but is in isomorphism with another group
22:52:03 <@nerdy2> yes?
22:52:14 < supremum> what if x=0 in x -> (x, 1/x)
22:53:23 <@nerdy2> J_K is the group of invertible elements, so x isn't 0 in J_K :)
22:53:37 < supremum> ohh ;)
22:53:39 <@nerdy2> the map is J_K -> A_K x A_K, x -> (x, 1/x)
22:53:46 <@nerdy2> (where J_K is the invertible elements of A_K)
22:54:08 <@nerdy2> this is an injection, and J_K gets the topology from this map (from viewing it as a subset of A_K x A_K)
22:55:15 <@nerdy2> ok, so now that we've defined this we can state artin reciprocity...
22:55:44 <@nerdy2> [and this is the point after all... this tells us about abelian extensions of K which is what class field theory is about]
22:56:17 <@nerdy2> suppose you take a finite abelian extension L/K, then there is a map C_K / N C_L -> G(L/K) which is an isomorphism!
22:56:40 < supremum> !
22:56:46 <@ramanujan> !
22:56:48 <@nerdy2> yes?
22:56:57 <@ramanujan> is C_K related to the class group of K?
22:57:06 < supremum> why is it an isomorfism?
22:57:21 <@nerdy2> ramanujan, yes it is, but it contains much more info, we'll talk about this in a sec
22:57:56 <@nerdy2> ramanujan, essentially the ideal class group only contains info about /unramified/ abelian extensions and the idele class group (or equivalently the ray class groups) allow ramification
22:58:32 <@nerdy2> so it's a bigger group
22:59:11 <@nerdy2> the why it is an isomorphism is long
22:59:36 <@nerdy2> this theorem is the main theorem of (global) class field theory, and it takes a few pages to prove
22:59:36 < supremum> do you have some references to where is its provend?
22:59:47 < supremum> ahh ok
22:59:49 < dublisk> !
23:00:01 <@nerdy2> any book that covers global class field theory, for example cassels & frohlich, or neukirch, or ...
23:00:10 <@nerdy2> dublisk, yes?
23:00:10 < dublisk> N C_L is the normal subgroup generated by C_L ?
23:00:24 <@nerdy2> milne also has class field theory notes available online which is a good resource
23:00:35 <@nerdy2> dublisk, no, i hadn't yet described what N meant :)
23:00:57 <@nerdy2> N is the norm map, there's a map N : C_L -> C_K
23:02:11 <@nerdy2> (the existence of the norm map isn't hard... each place w of L lies over some place v of K, and so L_w/K_v is a finite abelian extension and there's a norm map L_w^* -> K_v^*, so now you patch these together and make sure nothing goes wrong)
23:02:25 < DaMancha> !
23:02:29 <@nerdy2> yes?
23:02:44 < DaMancha> is it difficult to show you hit all of Gal(L/K)?
23:04:44 <@nerdy2> it's not the hardest part, no
23:06:15 <@ramanujan> are the isomrophisms C_K/NC_L =~ Gal(L/K) comptabile with inclusions into larger fields? i.e: can we take inverse limits and somehow get an idea of the structure of G_K^{ab}?
23:06:23 <@nerdy2> one way to do it is to reduce to cyclic groups, calculate the order of C_K / N C_L [using herbrand quotients say] and then see that it's greater than the order of G(L/K)
23:06:30 <@nerdy2> ramanujan, yes
23:07:15 <@nerdy2> ramanujan, you can make the appropriate statements just about G(K^ab/K)
23:08:28 <@nerdy2> ok so far? :)
23:08:51 <@nerdy2> point is, there's this random group and this random map which happens to be an isomorphism
23:09:07 <@Chandra> atleast ramanujan is
23:09:10 <@nerdy2> however the group is easy to understand (it's really just built up from data about K) ...
23:09:47 < dublisk> !
23:10:02 <@nerdy2> so if you want to take ramanujan's point of view and look at the map to G(K^ab/K) and if you use the fact that the norm subgroups of C_K are just the subgroups of finite index, so we are just using data from K!
23:10:05 <@nerdy2> on the left anyways
23:10:11 <@nerdy2> the right still includes all info from abelian extensions
23:10:13 <@nerdy2> dublisk, yes?
23:10:13 < dublisk> the isomorphism is a homeomorphism as topological groups with the profinite topology on the Galois group?
23:11:02 <@nerdy2> dublisk, i stated it for finite abelian extensions, in this case: both are finite groups :)
23:11:13 < dublisk> oh, finite
23:13:04 <@nerdy2> so if you wanted ramanujan's statement you would just take inverse limits of everything and it'd still work
23:13:30 <@nerdy2> and if you then use the fact that norm subgroups of C_K are just open subgroups of finite index, the thing on the left side does encompass only data from K
23:13:39 <@nerdy2> whereas the thing on the right side contains all information about all abelian extensions
23:13:57 <@nerdy2> so this isomorphism is magical: we can deduce information about abelian extensions of K from data about K alone!
23:15:15 <@nerdy2> now that i've probably lost all of you, let's bring up local class field theory where the group which has just data from the field is easier to understand
23:16:02 <@nerdy2> suppose we have a local field K_v (so it doesn't get confusing, we'll pretend that it's the completion of the global field at some place v -- but this info is extraneous)
23:16:27 <@nerdy2> furthermore assume it's nonarchimedean [because extensions of R are not very interesting! :)]
23:16:36 <@nerdy2> (so we completed at a finite prime)
23:16:57 <@nerdy2> so we should be thinking of Q_p for some prime p or a finite extension thereof
23:17:26 <@nerdy2> now, the same thing holds, there is a class field theory for K_v, i.e. we can get all information about abelian extensions of K from data about K alone
23:17:46 <@nerdy2> sorry extension of K_v from data bout K_v alone
23:18:29 <@nerdy2> specifically if L_w is a finite extension of K_v, then there is a map K_v^* / N L_w^* -> G(L_w/K_v)
23:18:37 <@nerdy2> sorry, finite abelian extension
23:18:42 <@nerdy2> and then this map is an isomorphism
23:20:25 <@nerdy2> so now this is an easier case of the above, the left side is really something that only deals with info from K_v [it's the multiplicative group mod a subgroup -- further if you take the inverse limit to get the fuller picture, and use the fact that these subgroups are just the ones which are open and of finite index, you do indeed get something that only depends on K_v, and an isomorphism \hat{K^*_v} -> G(K_v^ab/K) where \h
23:20:26 <@nerdy2> at{K^*_v} is the completion of K^*_v with respect to subgroups of finite index]
23:21:23 <@nerdy2> so in this case the group on the left is pretty easy to understand, so we immediately get a knowledge about abelian extensions...
23:23:49 <@nerdy2> and in fact, these theories fit together pretty well... in the global case we have an idele class group (which is built up out of various K_v^*'s) and we have local CFT which has a bunch of maps from K_v^*'s, if we patch them together in the naive way we get the map from C_K which is the global CFT isomorphism
23:23:57 < ramanujan> ! there is also a local kronecker-weber type theorem, right? something that describes local abelian
extensions "explicitly"?
23:24:03 <@nerdy2> yes
23:24:28 <@nerdy2> lubin-tate theory is what i suspect you mean
23:24:55 < ramanujan> I was expecting something simpler, like all abelian extensions being contained in cyclotomic ones?
23:26:10 <@nerdy2> for local fields, if you have unramified extensions, CFT is pretty easy... the norm map is surjective on units, and so the quotient K_v^* / N L_w^* is a cyclic group, generated by any uniformizer pi of K_v, and so you just have to describe where this goes, and this goes to the frobenius element F the unique thing such that Fx = x^p mod pi
23:26:28 <@nerdy2> (where p is the characteristic of the quotient of the ring of integers)
23:27:09 < ramanujan> in the unramified case, what I said is surely true, right?
23:27:29 < ramanujan> oops, nm. unramified extensions are abelian anyways
23:28:24 <@nerdy2> unramified extensions can be put in cyclotomic fields, yes
23:28:54 < ramanujan> sorry I was being unclear above. What I meant was: unramified extensions are abelian and are cyclotomic
because of hensel, basically. go on
23:28:57 <@nerdy2> but for ramified ones you have to add torsion points of elliptic curves
23:29:39 <@nerdy2> [which is what lubin-tate theory is]
23:29:42 <@nerdy2> anyways...
23:30:07 <@nerdy2> so unramified local extensions are pretty easy [i've told you what the reciprocity map is, ramanujan told you they are subextensions of cyclotomic extensions]
23:31:06 <@nerdy2> likewise unramified global class field theory is not so bad, there the group on the left actually becomes the ideal class group of K, so you get an isomorphism Cl(O_K) -> G(K^nr/K) where K^nr is the maximal unramified abelian extension of K
23:31:54 <@nerdy2> so as an example, take K = Q(sqrt(-5)), we know Cl(O_K) = Z/2, and in fact the maximal unramified extension of K is K(i), an extension of degree 2
23:32:33 <@nerdy2> likewise in the unramified case we can describe the map pretty explicitly, primes go to frobenius elements
23:33:11 <@nerdy2> so far, so good? :)
23:33:41 < Chandra> actions won't get relayed
23:33:54 < Chandra> i accidentally removed the feature
23:33:54 <@nerdy2> topic changes, but no actions :)
23:33:57 < ramanujan> I nodded, go on.
23:34:43 <@nerdy2> note that for global fields we may also have ramification at the infinite primes, and by "unramified" above i include the infinite primes
23:34:58 < ramanujan> ! what's ramification for infinite primes?
23:35:30 < ramanujan> I guess the question is: Is C/R unramified or totally ramified?
23:37:33 <@nerdy2> let me check
23:39:07 <@nerdy2> to be consistent with what i said, i think C/R should be ramified?
23:41:04 <@nerdy2> no, unramified :)
23:41:11 <@nerdy2> i was doing something wrong the first time :)
23:41:56 <@nerdy2> anyways, ray class groups are a way to get easier to handle groups on the left
23:42:03 <@nerdy2> (generalizations of the ideal class group)
23:43:49 <@nerdy2> instead of taking ideals, you take ideals relatively prime to some finite set, and then quotient out by elements of K satisfying certain conditions with respect to that finite set
23:44:42 <@nerdy2> of course then you have to prove that for any finite abelian extension, there is a finite set that works
23:45:17 <@nerdy2> but it's a little easier to handle these groups
23:46:26 <@nerdy2> anyways, with all that abstract nonsense, let's now talk about quadratic reciprocity :)
23:48:17 <@nerdy2> take a prime p, and look at K = Q(zeta_p), this has a unique subfield of deg 2 over Q, F = Q(sqrt((-1)^((p-1)/2)p)), to make things easier, let's write p* = (-1)^((p-1)/2) p, then F = Q(sqrt(p*))
23:49:41 <@nerdy2> sorry, let's assume p is an odd prime
23:50:01 < ramanujan> primes are always odd, dude.
23:50:34 <@nerdy2> now, we have our reciprocity map [either in terms of ray class groups or the idele class group] for these extensions
23:50:55 <@nerdy2> now since i mentioned we build global reciprocity out of local, we can figure out something
23:51:54 <@nerdy2> for any q an odd prime different from p [hence not ramified!] q (viewed as uniformizer in Q_q) goes to F_q in G(K/Q), the frobenius element, the thing which makes F_q x = x^q mod q
23:52:11 <@nerdy2> we know what this is, in fact it's not hard to list automorphisms in G(K/Q)
23:52:28 <@nerdy2> F_q = (the map which sends zeta_p to zeta_p^q)
23:53:56 <@nerdy2> so for K/Q anyways we already know the galois group, G(K/Q) = (Z/p)^*, and this lets us go the other way and figure out exactly what the group on the left hand side is
23:54:33 <@nerdy2> and now we can use functoriality and all that good stuff to figure out when somethings a square, remember that's what quadratic reciprocity is about
23:54:53 <@nerdy2> note that F corresponds to the subgroup of (Z/p)^* of index 2, i.e. the group of squares [which is why i mentioned him]
23:55:26 <@nerdy2> so we have the map which sends q to essentially q in (Z/p)^* = G(K/Q), and we want to know when it's in this subgroup of index 2
23:57:10 <@nerdy2> so we need the map C_Q / N C_K -> C_F / N C_K which is compatible (via reciprocity) with the inclusion of galois groups
23:57:33 <@nerdy2> now there's an obvious map in this direction, and i claim it's indeed the one
23:57:43 <@nerdy2> so now we have to find out where q goes
23:57:48 <@nerdy2> in this map...
23:59:15 <@nerdy2> sorry the idele class group map is C_F / N C_K -> C_Q / N C_K (as the galois group inclusion is G(K/F) -> G(K/Q))
23:59:23 <@nerdy2> and yes, there is an obvious map here
23:59:42 <@nerdy2> afk for a sec :)
00:00:24 <@Chandra> *stretch
00:01:46 <@nerdy2> ok
00:03:20 <@nerdy2> so we want to look at the image of this map, and know whether or not q is in the image...
00:03:37 <@nerdy2> (q being in the image if and only if q is a square mod p, as we've already seen by reciprocity)
00:05:29 <@nerdy2> but this is a local question, the "obvious" map is the norm map... suppose p* is a square mod q, then there's only one prime of F lying over q [the polynomial over Q_q already splits], and the norm map is surjective, and q is in the image and therefore [by reciprocity] is a square mod p
00:07:37 <@nerdy2> conversely, suppose p* is not a square mod q, then there's one prime of F lying above q again, so F_? = Q_q(sqrt(p*)) lying over Q_q, and here the norm map doesn't have q in the image
00:07:57 <@nerdy2> so yay, p* is a square mod q iff q is a square mod p
00:08:01 <@nerdy2> this is quadratic reciprocity!
00:08:55 <@nerdy2> any questions?
00:09:02 <@nerdy2> the proof is short assuming global cft :)
00:10:11 <@nerdy2> that should be more than enough material for now, 2 hours by my count :)
00:10:38 <@nerdy2> cubic reciprocity is also pretty easy assuming cft, maybe someone should do that :)
00:11:01 < lsmsrbls> thank you, nerdy. : )
00:11:31 < shazam> yes, good job :)
00:11:43 * dublisk claps
00:12:06 < ramanujan> thanks
00:12:10 < ramanujan> that helped out
00:12:18 < ramanujan> s/out/a lot
00:12:19 < DaMancha> a very interesting look at some of the tools cft provides - thanks.
00:12:36 < ramanujan> okay, now questions :)
00:12:55 < ramanujan> there are (at least) two approaches to local CFT: formal groups and cohomology
00:13:02 < ramanujan> can you (briefly) compare them?
00:13:05 <@nerdy2> [if it's not obvious, to prove cubic reciprocity, take primes congruent to 1 mod 3, and look at the cubic subfield of Q(zeta_p).... exercise for the reader]
00:13:10 < ramanujan> or, rather, what are each of them good for?
00:13:24 <@nerdy2> heh an essay question :)
00:13:25 < Ferakim> ok :)
00:14:33 <@nerdy2> ok, so formal groups is lubin tate theory
00:14:34 < ramanujan> heh, ok. i'll bug you on the AG channel then :)
00:14:37 <@nerdy2> it's what you talked about before
00:14:45 <@Chandra> thanks nerdy2
00:14:52 <@nerdy2> the advantage is that it is very explicit
00:15:04 < dublisk> formal groups leads into topology :)
00:15:09 <@Chandra> and mind, If i furnish the logs a bit more, and publish it?
00:15:23 <@nerdy2> (unramified extensions are subextensions of cyclotomic ones, extensions are harder but you get them by adjoining certain special values)
00:15:25 <@nerdy2> Chandra, no
00:15:50 <@nerdy2> so the lubin-tate approach has the advantage of being (relatively) elementary and explicit
00:15:56 <@Chandra> this is so far the longest and most indepth talk we had so far on freenode
00:16:30 <@Chandra> hopefully, we will get more of these
00:16:30 <@nerdy2> the cohomology is nice because you prove the same theorem once and use it over and over
00:16:48 <@nerdy2> e.g. you can mention a theorem (about class formations) of which global and local reciprocity are just a special case
00:16:52 <@nerdy2> this theorem involves cohomology
00:17:06 <@nerdy2> formal group laws are nice, but they don't extend to give things globally
00:17:19 <@nerdy2> whereas you can just as well use the group cohomology locally or globally
00:17:31 <@nerdy2> [and essentially prove exactly the same thing]
00:18:17 <@nerdy2> and besides you only have to deal with cohomology which involves groups, which are easy :)
00:19:09 <@Chandra> nerdy2 - mind checking /msg once?
00:19:12 <@nerdy2> and the facts you have to prove [in the local case] are dead easy
00:19:34 <@nerdy2> one of them is H^1(G, K^*) = 0 for G = G(L/K), L/K a finite abelian extension
00:19:54 <@nerdy2> sorry, H^1(G, L^*)
00:20:09 <@nerdy2> the other thing you have to do is find H^2(G, L^*), i.e. Br(L/K)
00:20:26 <@nerdy2> once you do this, it becomes a problem in group theory
00:21:20 <@nerdy2> this is easier than learning about formal groups :)
00:22:08 <@nerdy2> any other q's?
00:22:29 < ramanujan> is the canonical place for learning about formal groups the paper "Formal Complex Multiplication in Local
Fields"?
00:23:14 <@nerdy2> which paper is that
00:23:28 < ramanujan> the one by Lubin-Tate
00:23:36 <@nerdy2> serre's paper on local cft in cassels & frohlich is good
00:23:51 <@nerdy2> i haven't read lubin-tate
00:24:51 < ramanujan> okay. thanks, i'm about to start serre's article
00:25:16 < ramanujan> one really quick last question: where do I learn about cup products in group cohomology? serre's treatment
in local fields is rather terse
00:25:47 < dublisk> it comes from the fact that it is a hopf algebra or something
00:26:34 < dublisk> or alternatively just use the topological cup product
00:26:34 <Chandra> ramanujan - new talk already scheduled
00:26:40 <Chandra> again nerdy2 :)
00:26:54 <@nerdy2> hopf algebra?
00:27:31 < dublisk> I think hopf algebra = coalgebra
00:27:49 <@nerdy2> i know what it is, i don't know what you're suggesting
00:27:49 < ramanujan> in nerdy2's language, a hopf algebra is just affine group scheme :)
00:27:59 <@nerdy2> ramanujan, well there's various homological algebra books (weibel?) but it's most important to understand it in a few cases [low degree]
00:28:18 <@nerdy2> ramanujan, and the most important for cft is cup products in tate cohomology, which is just hard to understand anyways
00:28:52 < ramanujan> not for cyclic groups :D
00:29:09 <@nerdy2> not for cyclic groups?
00:29:21 <@nerdy2> oh, hard to understand, yea
00:29:33 <@nerdy2> more generally though :)
00:29:39 < dublisk> well alternatively H^*
00:29:51 < dublisk> H^*(G,k) = H^*(K(G,1),k)
00:30:04 <@nerdy2> yes,i got the topological suggestion :)
00:30:18 < dublisk> "The group algebra is a hopf algebra"
00:30:53 < dublisk> "hopf algebras have a cup product structure on their Ext ring"
00:31:33 <@nerdy2> how does Ext_ZG (Z, M) get a ring structure from the hopf algebra structure on ZG ?
00:31:52 < dublisk> dunno
00:32:01 < dublisk> http://www.msri.org/communications/ln/msri/2002/introcommalg/benson/1/index.html about 10:30
00:32:27 <@nerdy2> anyways, the naive thing works with the unnormalized bar resolution... but this isn't helpful :)
00:33:13 < dublisk> http://www.msri.org/communications/ln/msri/2002/introcommalg/benson/1/banner/04.html
00:34:00 < dublisk> I guess functoriality of Ext turns A -> A \tens A into a product
`