Last time, we were thinking about an abelian extension {L/K}, and we constructed, for most of the primes {v} of {K} a well-defined associated Frobenius element {\sigma_v \in Gal(L/K)}. To do this, we used the canonical isomorphism between the decomposition group {G_w} of some prime {w} extending {v} (a subgroup of {Gal(L/K)}) and the Galois group {Gal(k(w)/k(v))} of the residue fields. Since this is a Galois group of finite fields, it is generated by a Frobenius element {x \mapsto x^{|k(v)|}}, whose image in {G_w} gave us the distinguished element we wanted. In this post we extend this map linearly and state the Artin reciprocity law. We then briefly sketch its relation to the law of quadratic reciprocity.

Now, if {S} is a finite set of primes of {K} which includes all the archimedean primes and all those which ramify in {L/K}, we have an associated Frobenius element for each {v} not in {S}. Since {Gal(L/K)} is abelian, a natural thing to do is to extend the set map {M_K - S \rightarrow Gal(L/K)} to a map of abelian groups. Indeed, take {I^S} to be the free abelian group on {M_K - S}, and we get a morphism

\displaystyle F_{L/K}: I^S \rightarrow Gal(L/K).

This might be mistaken for idle fooling around, but let us think about what {I^S} is. An element of {I^S} is a finite {{\mathbb Z}}-linear combination of primes {v \not\in S}. These are all finite primes, so they correspond to prime ideals, and since our ring is commutative, it’s not totally unnatural to interpret our linear combinations as specifying the factors in a product (so for example, take {v+2v'} to mean {\mathfrak{p}_v\mathfrak{p}_{v'}^2}). Aha! We recall that any fractional ideal can be factorised uniquely into primes in a Dedekind domain. So {I^S} is in fact none other than the group of fractional ideals which are coprime to {S}: one of the central objects of algebraic number theory. Great!

At this point, Tate proves that if we have a square of number fields {K \subset K', L \subset L'} with {L'/K'} and {L/K} abelian, and {S'} is the set of primes of {K'} extending those of {S}, then we get a commuting diagram with F_{L'/K'}, F_{L/K} on rows and the norm map and natural restriction map on columns. It is an important result because we want our maps to behave nicely with respect to changes of field and a similar diagram forms a central part of the main theorem of global class field theory. The proof is just to check on basis elements, and remarking that Frobenius elements are determined by their effect on residue fields. I don’t think it’s worth giving the details here.

What we do next is very interesting. Until now we have been thinking about our number fields as basically abstract things which have Galois groups and primes. Indeed, the easiest way to set up our {F_{L/K}} map was to define it on primes, and then extend linearly to get the group of fractional ideals (coprime to a finite set of primes). But of course, now we have that, recall that any nonzero element {a} of the field {K} defines a principal maximal ideal {(a)}. Therefore, for any element {a \in K^*} which is coprime to all the ramified primes, we can associate a Frobenius element by taking {F_{L/K}((a))}.

Startlingly, it turns out that under slightly stronger assumptions than being merely coprime to all the ramified primes, this element is always the identity element. This is a form of Artin reciprocity, our first really deep theorem. It turns out that there is some {\epsilon>0} such that the following happens. Suppose that {|a-1|_v < \epsilon} for all {v \in S}. Then {F_{L/K}((a)) = 1}.

Note that the condition {|a-1|_v < \epsilon} is saying that {a \equiv 1} modulo a sufficiently large power of each ramified prime, and seems at first sight to be reasonably prohibitive at the archimedean primes. However, there is a nice trick to get round this and reduce this apparently hefty analytic constraint to a nice algebraic one. Firstly, we note that if {n=[L:K]} then obviously {F_{L/K}((b)^n) = F_{L/K}((b))^n = 1}. Hence, if we can write {a=b^n a'} for {a'} close to {1} in all valuations, we will still get the same result.

Suppose {a} is in fact an {n}th power in each {K_v^*}, so {a=c_v^n} for some {c_v \in K_v}. Since {S} is finite, we can use the weak approximation theorem (a sort of generalised Chinese remainder theorem that also works at archimedean places) to find {b \in K} within any distance {\eta>0} of {c_v^{-1}} in the norm associated to {v}, for each {v \in S}. Then such a {b} has the property that

 

\displaystyle |ab^n-1|_v = |a|_v|b^{n} - c_v^{-n}|_v \leq \eta |a|_v n (|c_v^{-1}|_v + \eta)^{n-1},

and choosing {\eta} sufficiently small (such that the RHS above drops below {\epsilon}, we deduce that {F_{L/K}((a))=F_{L/K}((ab^n))=1}.

So in fact it will suffice for {a} to be a {n=[L:K]}th power in each {K_v} for {v} ramified or archimedean. But this condition can be simplified even further. Indeed, at the archimedean primes, where the completion is {{\mathbb R}} or {{\mathbb C}}, we need only be concerned when it is {{\mathbb R}} and {n} is even, in which case we get the condition that {a>0}. Also, at the non-archimedean primes, Hensel’s lemma tells us that being an {n}th power in {K_v} is equivalent to being an {n}th power in {k(v)}. Thus we obtain the following more concrete version of Artin reciprocity.

Let {a \in K^*} be an {n}th power modulo every ramified prime, and, if {n} is even, positive in every real embedding. Then {F_{L/K}((a))=1}.

I should re-emphasise that we haven’t proved this, and that the proof is in general hard. However, in some special cases, we can prove Artin reciprocity directly. For example, in the extension {{\mathbb Q}(\zeta)/{\mathbb Q}} where {\zeta} is a primitive {m}th root of unity, it is not difficult to show that {F(p)\zeta = \zeta^p} for all {p} not dividing {m}, whence {F(a)\zeta = \zeta^a} for all {a} coprime to {m}. So if we insist that {a \equiv 1 \text{ mod } m} (taking a fairly liberal value of {\epsilon}), then since an element of {Gal({\mathbb Q}(\zeta)/{\mathbb Q})} is determined by the image of {\zeta}, {F(a)=1}, as required. Tate mentions that similar direct proofs, in which we can explicitly keep tabs on the effects of each {F(v)}, are possible for abelian extensions of complex quadratic fields by making use of elliptic curves with complex multiplication, but that it seemed likely there was no analagous direct proof for general abelian extensions of number fields.

To get a flavour for the power of Artin reciprocity, we shall now use it to deduce Gauss’ celebrated law of quadratic reciprocity. Doing so will not be totally trivial because we have to figure out how to relate our abstract general result to a special case, but I hope the reader will see that in terms of mathematical content the law of quadratic reciprocity should be seen as in some sense a very special and easy consequence of Artin reciprocity.

Let {a \in {\mathbb Z}} not be a square. We shall consider the Artin map {F} associated to the quadratic extension {{\mathbb Q}(\sqrt{a})/{\mathbb Q}}, taking all the fractional ideals with unramified support to elements of {Gal({\mathbb Q}(\sqrt{a})/{\mathbb Q}) = \{-1, 1\}}. Define the Legendre symbol by the following equation ({p} is a rational prime)

\displaystyle F((p))(\sqrt{a}) = \left(\frac{a}{p} \right)\sqrt{a}.
In other words, the Legendre symbol keeps track of how an associated Frobenius element acts on {\sqrt{a}} by recording the root of unity picked up. In fact, in our case, this is enough to determine precisely what the Frobenius element involved is. This obviously has the standard multiplicative properties and is well-defined in {a} modulo {P}. In fact, it is not difficult to check that it satisfies the Euler criterion

\displaystyle \left(\frac{a}{p} \right) \equiv a^{(p-1)/2} \mod p.
Indeed, this follows very quickly from the characterisation of the Frobenius element {\sigma_w} associated to a prime {w} as that which satisfies {\sigma_w(x) \equiv x^{Nv} \mod \mathfrak{p}_w}. Therefore, it must be the classical Legendre symbol.

Now what does Artin reciprocity in {{\mathbb Q}(\sqrt{a})/{\mathbb Q}} tell us? Suppose {p} and {q} are rational primes that are congruent modulo {8a_0}, where {a_0} is the odd part of {a}. Then {pq^{-1}} is a 2-adic square and a square modulo every odd prime dividing {a} (and it’s positive in any embedding into {{\mathbb R}}). The reciprocity law tells us that therefore if {p \equiv q \mod 8a_0},

\displaystyle \left(\frac{a}{p} \right) = \left(\frac{a}{q} \right).

And from this formula, the classical result of quadratic reciprocity is easily deduced.

So in this post we obtained a map from the subgroup of fractional ideals which don’t touch the ramified primes into the Galois group of {L/K}, and stated an important property of this map, namely Artin reciprocity, which vastly generalises the classical reciprocity laws. This property can be restated in the following language. Let {K} be a number field, and {G} a topological abelian group. A homomorphism {\psi: I^S \rightarrow G} is called admissible if for any neighbourhood {U} of the identity, there exists an {\epsilon > 0} such that whenever {a \in K^*} is coprime to all the ramified primes,

\displaystyle |a-1|_v<\epsilon \text{ for all } v \in S \Rightarrow \psi((a)) \in U.

Then Artin reciprocity is simply the statement that (viewing {Gal(L/K)} with the discrete topology) {F_{L/K}} is admissible for any abelian extension {L/K} of number fields. Next time, we will define the Artin map in terms of ideles, which are an important modification of the ideal class group which put archimedean primes and discrete primes on a proper equal footing, and in whose topology the property of being admissible becomes partly a continuity statement. We will then discuss some basic ideas to lay the foundations on which we will finally be able to state the main theorems of global class field theory in their full generality.