TITLE: remark in milne's class field theory notes
QUESTION [7 upvotes]: 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).

REPLY [14 votes]: 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.

REPLY [9 votes]: 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.