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In the introduction of his class field theory notes Milne mentions that some famous mathematicians failed to ask if the Artin isomorphism is canonical (between $Gal(L/K)$ and $C_m/H$ where $H$ is generated by the split primes in $L$). Does this mean:

1)in category theory terms: there is a natural transformation between the functors from abelian extensions over K to abelian groups given by $Gal(?/K)$ and $C_m/H?$ (where H? is generated by the primes split over $?/K$).

2)or some kind of vaguer statement about whether we need to make choices along the definition of the map.

or maybe 2) is precisely encoded in the definition of 1).

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Your question will be much easier to read if you use LaTeX: mathoverflow.net/faq#latex –  Ben Webster Jan 28 '10 at 8:19
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latex added ... –  Rado Jan 28 '10 at 8:33

3 Answers 3

up vote 9 down vote accepted

The point is that it is one thing to show that two mathematical objects are isomorphic; it is another (stronger) thing to give a particular isomorphism between them. A rather concrete instance of this is in combinatorics, where if $(A_n)$ and $(B_n)$ are two families of finite sets, one could show that $\# A_n = \# B_n$ by finding formulas for both sides and showing they are equal, but it is preferred to find an actual family of bijections $f_n: A_n \rightarrow B_n$.

This is not just a matter of fastidiousness or a general belief that constructive proofs are better. When considering functorialities between various isomorphic objects, the choice of isomorphism matters. For instance, often one wants to put various isomorphic objects into a diagram and know that the diagram commutes: this of course depends on the choice of isomorphism.

In the case of class field theory, these functorialities take the form of maps between the abelianized Galois groups / norm cokernel groups / idele class groups of different fields. The isomorphisms of class field theory can be shown to be the unique ones which satisfy various functoriality properties (and some "normalizations" involving Frobenius elements), and this uniqueness is often just as useful in the applications of CFT as the existence statements.

All of this, by the way, is explained quite explicitly in Milne's (excellent) notes: you just have to read a bit further. See for instance Theorem 1.1 on page 20: "There exists a unique homomorphism...with the following properties [involving Frobenius automorphisms and functoriality]..."

As a final remark: it is important to note that the word "canonical" in mathematics does not have a canonical meaning. To say that two objects are canonically isomorphic requires further explanation (as e.g. in the Theorem I mentioned above). Even the "unique isomorphisms" that one gets from universal mapping properties are not unique full-stop [generally!]; they are the unique isomorphisms satisfying some particular property.

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The unique isomorphisms from the initial and terminal object are actually unique, and all of the universality properties can be expressed by choosing appropriate initial and terminal objects in the various comma categories (not restricted to slice categories). –  Harry Gindi Jan 28 '10 at 10:15
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Sure, but the "particular property" in Pete's answer goes into the definition of the comma category. –  José Figueroa-O'Farrill Jan 28 '10 at 11:37
    
Thanks, for some reason I thought that there always was an explicit isomorphism (something along the frobenius map), but it was not clear (or proven) whether it is canonical. Although, it still seems the "functoriality" you mention above is (at least partially) captured by the nat.transformation definition if the functors are defined properly. In any case, I will try to digest CFT better first and then go back to the foundational questions. –  Rado Jan 29 '10 at 1:26

1. Here is what Tate says in his account of the General Reciprocity Law in the AMS volume on Hilbert's problems :

With this work of Takagi the theory of abelian extensions --- "class field theory" --- seemed in some sense complete, yet there was still no general reciprocity law. It remained for Artin to crown the edifice with such a theorem. He conjectured in 1923 and proved in 1927 that there is a natural isomorphism $$ > C_K/N_{L|K}C_L\buildrel\sim\over\to\operatorname{Gal}(L|K) > $$ which is characterised by the fact that...

And a little later :

How did Artin guess his reciprocity law ? He was not looking for it, not trying to solve a Hilbert problem. Neither was he, as would seem so natural to us today, seeking a canonical isomorphism, to make Takagi's theory more functorial. He was led to the law by trying to show...

Read him.

2. Here is a toy example --- not unrelated to class field theory --- of how a bijection can be more natural than others. Let $p$ be a prime number and let $K$ be finite extension of $\mathbb{Q}_p$ containing a primitive $p$-th root of $1$. There are only finitely many degree-$p$ cyclic extensions $L|K$, and there are only finitely many vectorial lines in the $\mathbb{F}_p$-space $K^\times/K^{\times p}$. In fact the two sets have the same number of elements, but the only natural bijection is $$ L\mapsto\operatorname{Ker}(K^\times/K^{\times p}\to L^\times/L^{\times p}), $$ of which the reciprocal bijections can be written $D\mapsto K(\root p\of D)$.

It follows that the number of degree-$p$ cyclic extensions $L|K$ is the same as the number of hyperplanes in $K^\times/K^{\times p}$. But is there a natural bijection between these two sets ? You will agree that $L\mapsto N_{L|K}(L^\times)/K^{\times p}$ is as natural a bijection as there can be.

One last point : Given a hyperplane $H\subset K^\times/K^{\times p}$, how do you recover the degree-$p$ cyclic extension $L|K$ such that $H=N_{L|K}(L^\times)/K^{\times p}$ ? Answer : use the natural reciprocity isomorphism $K^\times/K^{\times p}\to\operatorname{Gal}(M|K)$, where $M|K$ is the maximal elementary abelian $p$-extension, to identify $H$ with a subgroup of $\operatorname{Gal}(M|K)$, and take $L=M^H$.

Addendum (2011/11/21) In Recountings (edited by Joel Segel, A K Peters Ltd, Natick, Mass.), Arthur Mattuck recounts a conversation with Emil Artin about his reciprocity law:

I will tell you a story about the Reciprocity Law. After my thesis, I had the idea to define $L$-series for non-abelian extensions. But for them to agree with the $L$-series for abelian extensions, a certain isomorphism had to be true. I could show it implied all the standard reciprocity laws. So I called it the General Reciprocity Law and tried to prove it but couldn't, even after many tries. Then I showed it to the other number theorists, but they all laughed at it, and I remember Hasse in particular telling me it couldn't possibly be true.

Still, I kept at it, but nothing I tried worked. Not a week went by --- for three years ! --- that I did not try to prove the Reciprocity Law. It was discouraging, and meanwhile I turned to other things. Then one afternoon I had nothing special to do, so I said, `Well, I try to prove the Reciprocity Law again.' So I went out and sat down in the garden. You see, from the very beginning I had the idea to use the cyclotomic fields, but they never worked, and now I suddenly saw that all this time I had been using them in the wrong way --- and in half an hour I had it.

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thanks for the interesting toy example. When you say vectorial lines in $K^\times / K^{\times p}$ is there a way of visualizing this or the algebra is all we have left at this point. I am guessing there is non, but then again p-adic numbers can be visualized quite nicely as branched trees. –  Rado Jan 29 '10 at 6:12
    
By "vectorial line" I mean a line passing though the origin in the $\mathbb{F}_p$-space $K^\times/K^{\times p}$, as opposed to an "affine line" in this vector space of dimension $2+[K:\mathbb{Q}_p]$ over $\mathbb{F}_p$. It is perhaps a pedantic way of saying "order-$p$ subgroup". As for the dimension, see Part V of arXiv:0711.3878. –  Chandan Singh Dalawat Jan 29 '10 at 9:51

My understanding of Milne's comment is as follows (note: my history here is second-hand, so it may contain mistakes): when class field theory was first proved, it was not by actually producing an isomorphism between class group and Galois group, but rather by checking that they had the same number of elements of any given order. Apparently, for many years it never occurred to anyone to find a particular isomorphism, but in fact there is one which most observers agree is "canonical". In retrospect, it seems outrageous (and thus worthy of comment) that no one seemed to be bothered by the lack of an actual constructed isomorphism at the time.

In fact, there are examples out there is mathematics of groups which are provably isomorphic, but don't have any preferred isomorphism (a finite abelian group and its Pontryagin dual, for example).

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To add to Ben's statement: However, in the case of the pontryagin double-dual (of a finite group), a canonical isomorphism does exist to the double-dual. This is because there is a canonical injection into the double-dual, and by proving that the group is isomorphic to its dual, you also prove an isomorphism between the dual and the double dual. However, in the case of finite groups, every injective homomorphism between isomorphic groups is in fact an isomorphism. This is exactly the same proof as for finite dimensional vector spaces. –  Harry Gindi Jan 28 '10 at 8:43
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A canonical isomorphism with the double dual holds for all locally compact abelian groups. –  Pete L. Clark Jan 28 '10 at 8:53
    
I acknowledge my error and I should have added that the groups need to be abelian. Also, the standard proof for finite abelian groups uses the structure theorem at least once to reduce to the case of cyclic groups, so the proof of finite case doesn't generalize well to the general case. –  Harry Gindi Jan 28 '10 at 11:05
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No, that's not true either: one does not need the structure theorem for finite abelian groups in order to prove this. See e.g. math.uga.edu/~pete/4400algebra2point5.pdf, where the canonical isomorphism of an abelian group with its second dual is proved in Section 4 and the structure theorem, which lies deeper, is not touched until Section 5. (I mention this for your edification, not to beat you down.) –  Pete L. Clark Jan 28 '10 at 16:56

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