# 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

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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
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