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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.

Apparently this is due to Kronecker – the idea is to prove that cyclotomic polynomials are irreducible indirectly using the following two analytic facts:

Fact 1 (Dirichlet’s theorem, special case): The Dirichlet density of primes which are congruent to $1$ modulo $m$ is precisely $\frac{1}{\phi(m)}$.

Fact 2 (due to Kronecker, but a super special case of Cebotarev’s density theorem): The set of primes which split completely in an extension $L/\mathbb{Q}$ has Dirichlet density $\frac{1}{[L:\mathbb{Q}]}$.

And now we just stick these facts together. Let $L=\mathbb{Q}(\zeta_m)$ and take $f$ to be the minimal polynomial of $\zeta_m$. Then a rational prime $p$ splits iff $f$ splits modulo $p$, which only happens (assuming $p$ does not divide $m$) if there is a primitive $m$th root of unity in $\mathbb{F}_p$. For this to be the case we must have $m|p-1$. Thus by fact 1, at most density $\frac{1}{\phi(m)}$ primes split in this extension. Hence by fact 2, $deg f = [L:\mathbb{Q}] \geq \phi(m)$, and since the RHS is the degree of the $m$th cyclotomic polynomial, we have equality and in fact $f$ is the $m$th cyclotomic polynomial, which must therefore be irreducible.

In this series of posts I shall try to sketch (at least statements of) the basic results of global class field theory, following John Tate’s notes from the 1966 Brighton Conference. I begin by emphasising that I am writing them mainly in order to try to learn the material myself rather than really as an exposition to help others. As such, I expect to make countless errors, which expert readers are encouraged to point out.

It was conjectured by Hilbert that there exists, for a number field ${K}$ living inside a fixed algebraic closure, a unique totally unramified Galois extension ${L/K}$ with ${Gal(L/K)}$ isomorphic to the ideal class group of ${K}$. Such an extension is obviously abelian, and in fact Hilbert conjectured further that it is the maximal totally unramified abelian extension of ${K}$. Of course, Hilbert was right, and proving his conjecture is one of our main tasks.

In general, we will be studying a field ${K}$ which is usually a number field (or sometimes the function field of a curve over a finite field). Our overall goal seems to be to come up with a theory relating certain groups connected with the ideal class group of ${K}$ to the Galois groups of extensions of ${K}$. More specifically, we will define the idele group ${J_K}$ of ${K}$, which fits into an exact sequence

$\displaystyle 0 \rightarrow J_K^{S_\infty} \rightarrow J_K \rightarrow I_K \rightarrow 0,$

where ${I_K}$ is the group of fractional ideals and where ${J_K^{S_\infty}}$ is the subgroup of ideles all of whose prime factors’ correspond to archimedean valuations. In some sense therefore, we want to imagine archimedean valuations as primes at infinity’ and then ideles are a way to allow ideals to be divisible by these primes. With these defined, our main objects of study will be Artin maps

$\displaystyle \psi_{L/K}: J_K \rightarrow Gal(L/K),$

which give information about the splitting of primes in ${L/K}$ and provide the desired link between ideal class groups and the Galois groups.

In this first post we take the first tentative steps towards defining the Artin map, but will make no further mention of ideles yet. Let us set up some terminology. Fix ${K}$ a global field (a finite extension of ${{\mathbb Q}}$ or some ${\mathbb{F}_q(T)}$). A prime of ${K}$ is defined to be an equivalence class of nontrivial valuations on ${K}$. Primes ${v}$ are either discrete (they correspond to a prime ideal ${\mathfrak{p}_v = \{x| v(x) \geq 1\}}$, by satisfying an ultrametric law) or archimedean. We let ${M_K}$ denote the set of all primes of ${K}$, ${S^\infty = S^\infty_k}$ the set of archimedean primes and ${M_K^*}$ the set of discrete primes.

Now take ${L/K}$ a finite abelian extension, and let ${M_K^u}$ be the subset of ${M_K^*}$ consisting of unramified discrete primes. Our goal in this post will be to find a natural map

$\displaystyle F_{L/K}: M_k^u\rightarrow Gal(L/K),$

which we will extend in future posts to the Artin map ${\psi_{L/K}}$. To begin, let us start with some basic facts about how ${G=Gal(L/K)}$ acts on ${M_k}$ and the corresponding local fields.

We define the action of ${G}$ on ${M_L}$ by ${|\sigma x|_{\sigma w} := |x|_w}$. With this definition we get a left action ${\sigma(\tau w) = (\sigma \tau) w}$, and if ${v}$ is discrete it corresponds to the obvious action on prime ideals

$\displaystyle \mathfrak{p}_{\sigma w} = \{x| \sigma w(x) \geq 1\} = \{\sigma(x)| w(x) \geq 1 \} = \sigma \mathfrak{p}_w,$

and similarly ${\sigma}$ moves the local rings around in a natural fashion.

Now, there is a natural restriction map ${M_L \rightarrow M_K}$, and since ${\sigma(x) = x}$ for all ${x \in K}$, ${|x|_{\sigma w} = |x|_w}$ for all ${x \in K, \sigma \in G}$. Therefore the image of a Galois orbit in ${M_L}$ is just a single element of ${M_K}$. In fact we will show that ${G}$ acts transitively on the primes over a fixed ${v \in M_K}$, so since any such ${v}$ can be extended to ${L}$ in at least one way, the restriction map induces a correspondence:

$\displaystyle \{ G\text{-orbits of } M_L \} \leftrightarrow M_k.$

How can we prove this transitivity? Well, let ${G_w = \{\sigma \in G| \sigma w = w\}}$ be the decomposition group of ${w}$. The orbit of ${w}$ will simply be ${w=w_1, w_2,...,w_r}$ where ${r=(G:G_w)}$, with one prime ${w_i}$ corresponding to each coset ${\sigma_i G_w}$. For each ${w_i}$, an element of ${G_w}$ gives an isometric automorphism, which extends to an automorphism of the completion ${L_{w_i} \rightarrow L_{w_i}}$ and thus we have a natural embedding

$\displaystyle G_w \hookrightarrow Aut(L_{w_i}/K_v).$

Hence we get, by basic field theory, the bounds

$\displaystyle |G| = r|G_w| \leq \sum_{i=1}^r |Aut(L_{w_i}/K_v)| \leq \sum_{i=1}^r [L_{w_i}: K_v]$

Suppose there are some other primes ${w_{r+1},...,w_s}$ also extending ${v}$. Crucially, the natural map ${L \otimes_K K_v \rightarrow \prod_{i=1}^s L_{w_i}}$ must be surjective, as an application of the weak approximation theorem. This implies that

$\displaystyle \sum_{i=1}^s [L_{w_i}:K_v] \leq [L:K] = |G|.$

Putting our two estimates together, we get that equality holds at every stage, so in particular ${s=r}$ (the action is transitive).

Note we have also proved that ${L_w/K_v}$ is Galois and has Galois group naturally isomorphic to ${G_w}$. In fact, this group is about to play a key role in another capacity. Under the additional assumptions that ${w}$ is discrete and unramified over ${v}$, this Galois group is canonically isomorphic to the Galois group of residue fields ${Gal(k(w)/k(v))}$ (where ${k(v) = \mathcal{O}_v/\mathfrak{p}_v}$). So we have canonical isomorphisms

$\displaystyle G \supseteq G_w \cong Gal(L_w/K_v) \cong Gal(k(w)/k(v)).$

But the residue fields are just finite fields, so letting ${q=|k(v)|}$, ${Gal(k(w)/k(v))}$ is generated by ${Frob_q: x \mapsto x^q}$. Since the above isomorphisms are canonical, this corresponds to a canonical generating element ${\sigma_W}$ of ${G_w}$, which is as such called the Frobenius automorphism associated with ${w}$, characterised by the property ${\sigma_w(a) \equiv a^q (\text{mod }\mathfrak{p}_w)}$ for all ${a \in L}$.

Wait! What just happened? We took a prime ${w}$ of ${M_L^*}$ (assumed not to be ramified) and its decomposition group contained a special element ${\sigma_w}$. In other words, we defined a natural mapping ${M_L^* \rightarrow Gal(L/K)}$. Recall that we were looking for a map ${M_K^u \rightarrow Gal(L/K)}$, so it feels like we are almost done.

But just a minute ago we proved that the Galois orbits of ${M_L}$ are in bijective correspondence with ${M_K}$, and this correspondence obviously respects the properties of being discrete and unramified, so the obvious thing to hope is that ${\sigma_w = \sigma_{\tau w}}$ for all ${\tau \in G}$, since we can then just define ${\sigma_v}$ to be this element (the Frobenius automorphism associated with every prime above ${v}$).

Oh, but this is obvious, because ${Gal(L/K)}$ is abelian, and by the characterising property of ${\sigma_w}$, ${\sigma_{\tau w}}$ must be conjugate to it. So we are done. Provided ${L/K}$ is abelian, we have a lovely well-defined map called `take the associated Frobenius automorphism of any prime above ${v}$

$\displaystyle F_{L/K}: M_K^u \rightarrow Gal(L/K).$

Next time, we will extend this to a group homomorphism and state the Artin reciprocity law, our first deep theorem which tells us a lot about how these maps work. However, before we finish it is worth remarking how the above map relates to the splitting of an unramified prime ${v}$ in ${L}$.

Indeed, we saw in the above discussion about the action of ${G}$ on ${M_L}$ that the set of primes ${w}$ dividing ${v}$ is isomorphic to ${(G:G_w)}$. Now, ${G_w}$ is actually generated by ${\sigma_w}$, and hence by the symmetry of the action, we know how ${v}$ splits. If ${\sigma_w}$ has order ${f}$, ${v}$ must split into ${|G|/f}$ distinct factors, each of degree ${f}$. So knowing ${\sigma_v = F_{L/K}(v)}$ tells us how ${v}$ splits and gives us a generator for the decomposition group. This is an very nice situation to be in, and hints at some of the wide-reaching consequences of the theory.