Seminar on Algebraic Number Theory

This seminar was held in #mathematics on EFnet on 24th of March 2004 at 19:00 EDT by nerdy2.

It covered rings of algebraic integers, class groups, factorisation, ramification, valuations, ideles, adeles.

Seminar log

[19:08] <ramanujan> well, Our speaker for the day is nerdy2 and he'll be talking about number theory
[19:09] <peterw> what is number theory?
[19:09] <nerdy2> number theory is the theory of the integers
[19:10] <nerdy2> it's a hodge-podge of various things, for example, one could ask if the equation x^2 + 5 = y^3 has any integral solutions....
[19:12] <nerdy2> as often happens in math, to answer questions about Z, we actually have to go beyond Z itself to answer the question ...
[19:12] <nerdy2> fermat's last theorem says that x^n + y^n = z^n has no nontrivial integral solutions for n > 2
[19:13] <nerdy2> and a naive way to prove it, is to look in Z[w] where w^n = -1, so that you can factor the equation as prod (x - w^i y) = z
[19:13] <nerdy2> z^n rather
[19:13] <nerdy2> and then the (x - w^i y) should be relatively prime, and since their product is an n-th power, they must each be n-th powers
[19:13] <nerdy2> and this shortly gives you a contradiction
[19:14] <nerdy2> the problem with this is that relatively prime doesn't really make sense in Z[w] for w arbitrary, it is only a unique factorization domain for n small
[19:14] <nerdy2> however if you ignore this, the proof still goes through as long as the unique factorization doesn't fail too badly
[19:15] <nerdy2> likewise, fermat's theorem that primes of the form x^2 + y^2 are exactly those congruent to 1 mod 4 relates to the arithmetic in Z[i], and the answer to the question i posed about x^2 + 5 = y^3 relates to the arithmetic of Z[sqrt(-5)]
[19:16] <nerdy2> so hopefully, everyone is absolutely convinced that we need to be able to effectively study rings like Z[i], or Z[sqrt(-5)] or Z[w]
[19:16] <nerdy2> in order to answer interesting questions that only involve integers
[19:16] <nerdy2> so with that in mind, if we are given a number field K, which means a field, which is a finite extension of Q....
[19:17] <nerdy2> we consider the ring O_K of (algebraic) integers in K
[19:17] <nerdy2> these are the elements of K which are roots of monic polynomials with integral coefficients
[19:18] <nerdy2> now, as with the example of Z[w], or Z[sqrt(-5)], this will often not be a unique factorization domain
[19:19] <nerdy2> for example, in sqrt(-5), we have 6 = 2*3 = (1 + sqrt(-5))(1 - sqrt(-5))
[19:19] <nerdy2> and you can show that 2,3,1+sqrt(-5),1-sqrt(-5) are all irreducible, but they can't be prime
[19:20] <nerdy2> however, these rings O_K do satisfy a property close enough to unique factorizatoin
[19:20] <nerdy2> factorization
[19:20] <nerdy2> we just have to factor in terms of ideals
[19:20] <nerdy2> that is every (nonzero, proper) ideal in O_K can be written (uniquely) as a product of prime ideals
[19:21] <nerdy2> so for example, in Z[sqrt(-5)], we have (6) = (2,1+sqrt(-5))(2,1-sqrt(-5))(3,1+sqrt(-5))(3,1-sqrt(-5))
[19:22] <nerdy2> so that the 2,3,1+sqrt(-5),1-sqrt(-5) aren't prime precisely because they break up into 2 ideals each!
[19:22] <nerdy2> so this of course raises many questions itself
[19:23] <nerdy2> 1) how do integers factor in this larger ring, since we've now made sense of factorization
[19:23] <nerdy2> 2) we know it's not a UFD, but how far is it from one?
[19:23] <nerdy2> i guess i'll state the answer to 2) first, but first i have to make sense of it
[19:24] <nerdy2> so consider all fractional ideals, that is nonzero finitely generated O_K-submodules of K
[19:25] <nerdy2> this is a fancy way to say we are considering things that look like aI where I is a nonzero ideal of O_K (f.g. since O_K is noetherian), and a is a nonzero element of K
[19:26] <nerdy2> call the set of all of these guys Div(O_K)
[19:26] <nerdy2> now, these form a group!
[19:26] <nerdy2> specifically if you have two such things M,N, you can form the product M*N, which is the submodule of K generated by the products of elements of M and N
[19:27] <nerdy2> (the inverse condition on being a group is exactly why we had to introduce fractional ideals, we had to have denominators)
[19:27] <nerdy2> and moreover, we have a map from the nonzero elements of K, K^* -> Div(K)
[19:27] <nerdy2> which sends a to the principal ideal aO_K
[19:27] <nerdy2> this is a group homomorphism
[19:27] <nerdy2> and it has image exactly the principal ideals
[19:28] <nerdy2> so the question of 'how far is O_K from being a unique factorization domain' which is really the same as 'how far is it from being a PID' can be rephrased as, what is the size of Cl(O_K) = Div(O_K)/K^*
[19:29] <nerdy2> the surprising answer is that Cl(O_K) is finite
[19:30] <nerdy2> i think a lot of what i do today will be stating things, and not proving them, although i can prove them if you wish :)
[19:31] <nerdy2> this thing, Cl(O_K) is called the ideal class group
[19:32] <nerdy2> and for Z[sqrt(-5)], the class group is a group of order 2
[19:32] <nerdy2> then we have the second question i posed, how do integers factor in O_K
[19:33] <nerdy2> now, integers factor into primes, so this is really the question of how do primes factor in O_K
[19:34] <nerdy2> so we know that pO_K = P_1^e_1 .... P_g^e_g for some prime ideals P_1, ..., P_g (distinct) of O_K
[19:35] <nerdy2> now each of the P_i contain pO_K, so O_K/P_i is an extension of Z/p
[19:35] <nerdy2> and in O_K nonzero primes are maximal, so O_K/P_i is a field, which is a finite extension of the finite field Z/p
[19:35] <nerdy2> write f_i for it's degree
[19:36] <nerdy2> now, if K/Q is galois, the galois group permutes the primes P_i transitively... this means that all of the e_i are equal (say to e) and all of the f_i are equal (say to f)
[19:36] <nerdy2> in general we have sum e_i f_i = [K:Q] (so if K/Q is galois, efg = [K:Q])
[19:37] <nerdy2> so if any of the e_i are > 1 we say that p is ramified in K, and otherwise p is unramified
[19:37] <nerdy2> the somewhat magical thing is that most primes don't ramify
[19:37] <nerdy2> in fact, we can find out exactly which primes ramify...
[19:38] <nerdy2> indeed, the primes which ramify are exactly those which divide something called the discriminant
[19:39] <nerdy2> which one can calculate as follows: O_K is a free finitely generated abelian group of rank n = [K:Q], take a basis x_1, ..., x_n of it
[19:40] <nerdy2> let s_1, ..., s_n be the different embeddings of K into the algebraic closure Q-bar (if K/Q is galois, this means just the elements of the galois group) and form the matrix D = (s_i x_j)
[19:40] <nerdy2> then (det D)^2 is an element of Q and is the discriminant
[19:40] <nerdy2> and it turns out that a prime ramifies if and only if it divides this
[19:41] <nerdy2> for example, for Z[3^(1/4)], the discriminant is -6192 if i recall, only divisible by 2,3 so the primes which ramify in this extension are exactly 2 and 3
[19:41] <nerdy2> for Z[sqrt(d)] when d=2,3 mod 4 and is squarefree, the discriminant is 4d, so the primes which ramify are exactly those dividing d and 2
[19:41] <nerdy2> so in particular in Z[sqrt(-5)] the primes which ramify are 2 and 5
[19:43] <nerdy2> we can delve into the question of how primes factor in rings of integers more deeply when i cover class field theory in an upcoming lecture
[19:44] <nerdy2> class field theory answers the question of which primes split completely in an extension (i.e. the g is as large as possible)
[19:44] <nerdy2> in Z[i] it turns out that these are exactly the ones corresponding to the primes = 1 mod 4, and this is exactly the same question as which primes can be written as a sum of two squares
[19:44] <nerdy2> of course, the 'class field theory' needed here amounts to nothing more than quadratic reciprocity
[19:45] <nerdy2> so, if we know a little bit about how multiplication works, the next natural question is what does the group of units in O_K look like
[19:46] <nerdy2> so if we fix an embedding Q-bar subset C, we have s_1, ..., s_n, which embed K into C now
[19:47] <nerdy2> and for each s_i, we have two options, either the image lies in R or it doesn't (in which case, after complex conjugation we get another s_i)
[19:48] <nerdy2> so we call the number whose image is in R r_1, and we call the number whose image doesn't 2r_2 (it must be even since complex conjugation acts on them)
[19:48] <nerdy2> and it turns out that the group of units in the ring O_K actually depends heavily on these numbers r_1 and r_2
[19:49] <nerdy2> dirichlet's unit theorem says that it is a finitely generated abelian group with rank r_1 + r_2 - 1 !!
[19:49] <nerdy2> so that for Z[sqrt(3)], since there are 2 real embeddings and no nonreal ones, its group of units is f.g. with rank 1, so it looks like Z + torsion as an abelian group
[19:49] <nerdy2> in particular, there are infinitely many units
[19:50] <nerdy2> finding units in Z[sqrt(3)] is exactly solving the equation a^2 - 3b^2 = 1 (a pell's equation) in the integers, and this is classical
[19:51] <nerdy2> in particular, for this ring, there is one unit which generates the torsion free part of the unit group (it and its inverse generate)
[19:53] <nerdy2> so i think i'll take a short break and then introduce the ring of adeles :)
[19:55] <Kummer> well this is 30 minutes late but I might as well mention that Div(O_K) has a very geometric meaning in algebraic geometry
[19:56] <Kummer> Recall that Div(O_K) is the fractional ideals of K. In geometry, ideal == point so Div(E) for a curve E would be the free group generated by the points of the curve
[19:56] <Kummer> on an elliptic curve, Div(E) is combinations of points of E, and the ideal _classes_ Cl(E) is the group of points on E
[19:57] <Kummer> number theorists like to think of elliptic curves as a geometric version of ideal factorization
[19:57] <Kummer> algebraic geometers think the reverse: they see factorization in number fields as an algebraic version of a curve!
[19:58] <Kummer> "to ramify" means "to branch" in English -- if you didn't know that, now you do
[19:58] <Kummer> drini you get a group but in general it's not as nice as just the set of points on the curve
[19:58] <Kummer> for example with hyperelliptic curves, the group of ideals equals the set of _pairs_ of points on the curve
[19:58] <nerdy2> yea, you should think of the extension K/Q as really a covering of curves (or riemann surfaces) but it has branch points
[19:58] <nerdy2> and these finitely many points are the primes which ramify
[19:59] <Kummer> I'm just trying to emphasize that ramify has a geometric meaning here. The ideal factorizations really do branch into two.
[20:00] <nerdy2> i think under your characterization i think like an algebraic geometer :)
[20:02] <nerdy2> i'm tempted to bring up sn and cn and dn :)
[20:02] <nerdy2> but i think i'll let that lie for now
[20:02] <nerdy2> so anyways, we'll chat a little about adeles
[20:03] <nerdy2> we want to introduce topology into the picture, so we start with a number field, and consider a 'valuation' on K
[20:04] <nerdy2> this is an absolute value |.|_v which means it satisfies |xy| = |x||y|, |x| = 0 iff x = 0, and |x+y| <= max{|x|,|y|}
[20:05] <nerdy2> more precisely, we'll consider them equivalent if they introduce the same topology (this means that one will be a power of another)
[20:05] <nerdy2> and we'll throw out the trivial one (|x| = 0 for x=0, 1 otherwise)
[20:06] <nerdy2> so, an equivalence class of nontrivial valuations will be called a prime of K
[20:06] <nerdy2> and you may think this completely nonsensical, but it does make some sense...
[20:06] <nerdy2> what are the primes of Q?
[20:07] <nerdy2> so the statement of ostrowski's theorem is that the primes of Q are exactly the p-adic valuation |p^m a/b| = p^-m (for a,b coprime to p) for primes p of Z, and the ordinary absolute value
[20:07] <nerdy2> so we have the usual primes, and we have an extra one thrown in
[20:07] <nerdy2> an 'infinite' or archimedean prime
[20:08] <nerdy2> we'll call the primes finite or nonarchimedean if the values |n| for n in Z are bounded (and archimedean if this isn't so)
[20:10] <nerdy2> for any valuation v on K, since we have a topology, we can consider the completion K_v of K with respect to v
[20:11] <nerdy2> by ostrowski's theorem, it will be a finite extension of Q_p (if v is finite and restricts to the p-adic valuation on Q) or R (if v is infinite, and thus restricts to the archimedean valuation on Q)
[20:13] <nerdy2> now, a fancy restatement of chinese remainder theorem says that if we have valuations v_1, .., v_n, and x_i in K_{v_i} and e > 0, then we can find an x in K, such that |x - x_i| < e for all i
[20:13] <nerdy2> in fact much more than this is true as we'll see in a moment
[20:16] <[-K-]> the v_i's should be all distinct shouldn't they
[20:16] <nerdy2> yes
[20:17] <nerdy2> so i should rephrase my def'n of absolute value, i should require |x+y| <= 2 max{|x|,|y|} or |x+y| <= |x|+|y| or something like that
[20:17] <nerdy2> archimedean valuations will then satisfy the stronger |x+y|<= max {|x|,|y|}
[20:18] <nerdy2> anyways, with that modification, i can safely continue :)
[20:18] <nerdy2> now, we want to consider all of these guys together
[20:19] <nerdy2> so we might as well form the ring A_K = product K_v
[20:19] <nerdy2> unfortunately, the product topology is not what we want
[20:21] <nerdy2> so we form some subrings A_{K,S} whose union will be A_K
[20:21] <nerdy2> where S is a finite set of primes
[20:23] <nerdy2> containing the archimedean ones, A_{K,S} = prod(v in S) K_v x prod(v not in S) O_v
[20:23] <nerdy2> O_v is the ring of integers in the K_v, when v is finite
[20:24] <nerdy2> which means the elements x such that |x| <= 1
[20:24] <nerdy2> then give A_{K,S} the product topology, and topologize A_K as the union of the A_{K,S}
[20:25] <nerdy2> now, we have a natural inclusion map K -> A_K
[20:25] <nerdy2> which is injective of course
[20:25] <nerdy2> and we can say a few things about this: K is discrete in A_K, and A_K/K is compact
[20:25] <Evander> !
[20:25] <nerdy2> yes?
[20:26] <Evander> Sorry to lag behind, but you meant _non_archimedean valuations satisfy the stronger |x+y|<= max {|x|,|y|}, yes?
[20:26] <nerdy2> yes
[20:26] <Kummer> For those who may not find the map obvious, the map K -> A_K is defined by thinking of K \subset K_v and A_K = (as a set) Prod_{v is a valuation} K_v
[20:27] <nerdy2> (these facts allow us to do fourier analysis on these groups... but that's for another time)
[20:28] <Kummer> The statements "K is discrete in A_K, and A_K/K is compact" follow easily from the definition of the topology on A_K, but you need to get good with the definition before these are obvious.
[20:28] <nerdy2> yes
[20:29] <nerdy2> something else i will need if ever i lecture on class theory will be the ideles J_K, these are just the units in the ring A_K
[20:29] <nerdy2> however, not with the induced topology
[20:30] <nerdy2> units in anything get a natural topology (not necessarily the induced one), from the inclusion J_K -> A_K x A_K (x goes to (x,x^-1))
[20:31] <nerdy2> and K^* is included in J_K, and the quotient C_K = J_K/K^* is known as the idele class group
[20:32] <nerdy2> now the theorem that i alluded to earlier, the much stronger approximation theorem, partly answers the question 'why consider all the primes'....
[20:32] <nerdy2> if you take any prime v, and you omit it, and form the ring A' in the analogous way, just always omitting v, and embed K into A', then K is everywhere dense in A' !
[20:33] <[-K-]> how weird.
[20:33] <nerdy2> yea
[20:34] <nerdy2> i guess it should be stated that primes (in the valuation sense) act like primes (in the ring sense)
[20:35] <nerdy2> for a prime p (finite or infinite) of Q, you can consider which primes v lie above it in K, if the extension is galois, these are permuted transitively by the galois group...
[20:35] <nerdy2> you can define ramification for each of them as well...
[20:36] <nerdy2> anyways, i think this is a good short course, i think i could get to some interesting stuff now in a future lecture :)
[20:37] <Kummer> that's it?
[20:37] <nerdy2> so any questions ?
[20:37] <landen> clap clap
[20:37] <nerdy2> kummer, what else do you want?
[20:37] Action: [-K-] applauds
[20:37] Action: Evander applauds
[20:37] <Kummer> yes, why did you talk about adeles and ideles if you're not gonna do anything with them? :)
[20:37] Action: dioid applauds
[20:37] <[-K-]> !
[20:37] <nerdy2> kummer, cuz i want to get to class field theory eventually, but i didn't think i could get to it today :)
[20:37] <landen> Sir, is chapter 7 going to be on the quiz?
[20:37] <Evander> He talked about adeles?
[20:37] <Kummer> okay. I have 30 minutes, should I ramble about ideles?
[20:38] <nerdy2> yea sure
[20:38] <[-K-]> go for it
[20:38] <Kummer> A_K == adele group of K
[20:38] <landen> note 43 | 6192
[20:38] <Evander> Oh, ok
[20:38] <nerdy2> if you want tom ake my life easy, also tell them about tate cohomology
[20:38] <[-K-]> I was just curious what the connection was between C_K and Cl(O_K)
[20:38] <Kummer> remind me what C_K is
[20:38] <[-K-]> idele class group
[20:38] <nerdy2> Kummer, J_K/K^*
[20:38] <Kummer> ah! J_K
[20:38] <Kummer> /something
[20:38] <Kummer> okay great that's a good idea to talk about that
[20:38] <Kummer> ideles are like ideals on steroids.
[20:39] <[-K-]> hehe
[20:39] <Kummer> rememeber what an idele is, it's like a bunch of elements of K_v
[20:39] <Kummer> with finitely many exceptions, the v-th element of K_v is invertible mod v
[20:40] <Kummer> now think about, if you will, collecting the noninvertible coordinates
[20:40] <Kummer> let's say the v1-th coordinate is noninvertible, and lies inside the prime power v1^n1
[20:41] <Kummer> well hey that gives you an ideal v1^n1
[20:41] <Kummer> since there's only finitely many exceptions, the product of all these exceptions is a finite product v1^n1 * v2^n2 * .. * vk^nk
[20:41] <Kummer> so in this way an idele corresponds, very coarsely, to an ideal
[20:41] <[-K-]> just a question
[20:42] <Kummer> yes?
[20:42] <[-K-]> why does a unit of A_K correspond to a bunch of x_v's where finitely many are noninvertible
[20:42] <[-K-]> I would have thought all the x_vs have to be invertible
[20:42] <[-K-]> isn't the multiplication in A_K = Prod K_v pointwise?
[20:43] <nerdy2> so i defined A_K incorrectly :)
[20:44] <nerdy2> A_K = union A_{K,S}
[20:44] <nerdy2> use that one :)
[20:44] <[-K-]> ah ok
[20:44] <nerdy2> heh, how many other things did i get wrong? :)
[20:44] <landen> 43 | 6192
[20:45] <nerdy2> landen, ok, then it's -6912, sorry
[20:46] <Kummer> [-K-], the thing is, K_v is a field, so it has ridiculously many units (anything nonzero in fact)
[20:47] <[-K-]> yes
[20:47] <Kummer> the ideles, by definition, are the elements of Prod K_v satisfying the property that all but finitely many are invertible in O_v
[20:47] <[-K-]> ok
[20:47] <Kummer> these are invertible in A_K because A_K allows finitely many exceptions to the O_v requirement
[20:48] <[-K-]> this smells suspiciously of the weak / weak* topologies. But never mind.
[20:48] <Kummer> so to invert an idele, if the coordinate is O_v invertible then just take its O_v inverse, otherwise take its K_v inverse and chalk it up as an exception that you had to resort to K_v
[20:49] <Kummer> an idele i corresponds to an ideal I. the exceptions correspond exactly to the primes v that appear to nonzero power in the prime factorization of I.
[20:50] <[-K-]> so there's an map from A_K* to Div(O_K)?
[20:50] <Kummer> yeah, if you map infinite primes to 0
[20:50] <[-K-]> cool
[20:50] <Kummer> to get class groups, mod out by principal elements
[20:51] <[-K-]> so C_K -> Cl(O_k)
[20:51] <Kummer> ->> even
[20:51] <[-K-]> lovely
[20:51] <[-K-]> what's the kernel?
[20:51] <Kummer> it's actually just K*^ * (ideles generated by infinite primes)
[20:51] <Kummer> K^* even
[20:52] <[-K-]> right so the two things really are the same
[20:52] <Kummer> yes, but the infinite primes are useful
[20:53] <Kummer> I assume you know Dirichlet's theorem? Every congruence class mod n (except the stupid congruence classes) contains infinitely many primes.
[20:54] <Kummer> this is a statement about ideal classes in the n-th cyclotomic field. You can prove similar statements for idele classes. The infinite valuations give new possibilities.
[20:54] <[-K-]> you mean the Dirichlet Arithmetic Progressions Theorem
[20:54] <Kummer> yeah that one
[20:54] <[-K-]> yeah I came across it
[20:55] <Kummer> an infinite prime would classify by archimedean norm instead of by congruence mod n
[20:55] <Kummer> so you can prove that any angular sector of the complex plane has infinitely many Z[i]-primes
[20:56] <[-K-]> nice
[20:56] <[-K-]> I've forgotten. There are always only finitely many infinite primes in any given number field, aren't there
[20:57] <Kummer> yeah, cause K->C is a finite list of embeddings
[20:57] <Kummer> A reference for the Z[i] proof is p. 318 of Lang "Algebraic Number Theory"
[20:58] <Tohoku> Any references for algebraic number theory in general? (i.e: the stuff nerdy2 talked about)
[20:58] <[-K-]> I don't quite see how the sectors correspond to the _finitely many_ embeddings. But never mind :)
[20:58] <[-K-]> Tohoku: Stewart and Tall is good
[20:58] <Kummer> well you pick one single embedding, and take characters on that
[20:59] <Kummer> you may or may not how the Dirichlet's theorem on primes in an arithmetic progression is proved
[20:59] <[-K-]> that I don't know.
[20:59] <zeno> Dirichlet actually says a bit more, for example that \sum_{p = b (mod m)} 1/p = oo. Does that too have any sort of analogue in the ideles, if such is even meaningful?
[21:00] <Kummer> you take multiplicative maps from {set of primes} -> Z/nZ, find their L-functions, use mumbo jumbo about the residue thereof at s=1, and it happens
[21:00] <nerdy2> i never mentioned it, but has online notes on both alg. nt and class field theory
[21:01] <[-K-]> ah yes, it actually uses complex variable
[21:01] <Kummer> the angular stuff is more along the lines of, you take homomorphisms {set of ideles} -> S^1
[21:01] <Kummer> S^1 is parametrizing the angle of the prime of course
[21:01] <Kummer> now you might object that a homomorphism is never going to give you the characteristic function of an angular sector, but fourier series (!) takes care of that
[21:03] <Kummer> zeno I'm sure it does, but I don't know it
[21:03] <[-K-]> are these v's called "places", by any chance?
[21:03] <nerdy2> yea
[21:03] <[-K-]> excellent
[21:04] <Kummer> in fact, because I use it for my thesis, any explicit analytic estimate would imply new results in that line of work
[21:04] <[-K-]> I keep on attending talks on ANT here, and they all start by "let v be a place of K" and I go "huh" and go to sleep
[21:05] <[-K-]> for some reason everyone seems to be extremely interested in the Tate Shafarevich and Selmer groups, which seem to be defined using lots of products of homological stuff with vs all over the place. Ugh.
[21:05] <Kummer> haha that's Barry Mazur for you
[21:05] <[-K-]> it is? I thought it was John Coates and co. but there you go.
[21:06] <Kummer> William Stein's life mission is to prove finiteness of TS-group
[21:06] <Kummer> well, that and B-SD conjecture
[21:06] <[-K-]> Birch-Swinnerton-Dyer ^_^
[21:06] <[-K-]> (remove 2nd hyphen)
[21:06] <Kummer> anyway it's 6pm. I have to go to concert.
[21:07] <[-K-]> cya. Thx for all the info.
[21:07] <nerdy2> cya Kummer
[21:08] <[-K-]> nerdy: how come you've been learning all this ANT nonsense?
[21:08] <[-K-]> I thought you were just interested in agricultural mathematics :)
[21:09] <zeno> kernels, germs, sheaves...
[21:10] <zeno> foliations
[21:10] <Evander> Fields
[21:10] <[-K-]> sheaves, stacks, bundles and motives
[21:11] <nerdy2> heh
[21:11] <nerdy2> motives can be arithmetic beasts
[21:12] <nerdy2> and anything in alg geom is really a limit of arithmetic things :)
[21:12] <nerdy2> i dunno, i kinda like arithmetic stuff now
[21:12] <Tohoku> okay, I guess we can stop relaying.
[21:14] <nerdy2> much like you get cyclotomic fields by adjoining values of the function e^(2pi i x) ...
[21:14] <nerdy2> sure