This is the lemma we shall use in the proof of the reciprocity law, to reduce the cyclic case to the cyclotomic case:

Lemma 4 (Artin) Let {L/k} be a cyclic extension of degree {n} and {\mathfrak{p}} a prime of {k} unramified in {L}. Then we can find a field {E}, a subextension of {L(\zeta_m)}, with {E \cap L = k} such that in the lattice of fields


we have:

1. {\mathfrak{p}} splits completely in {E/k}

2. {E(\zeta_m) = L(\zeta_m)}, so that {LE/E} is cyclotomic

3. {\mathfrak{p}} is unramified in {LE/k}

Moreover, we can choose {m} such that it is divisible only by arbitrarily large primes.

The proof of this will use the previous number-theory lemmas and the basic tools of Galois theory.

So, first of all, we know that {E} is a subextension of some {L(\zeta_m)}. We don’t know what {m} is, but pretend we do, and will start carrying out the proof. As we do so, we will learn more and more about what {m} has to be like, and eventually choose it.

2.1. The lattice condition

The first thing we will want is for {k(\zeta_m)} and {L} to satisfy {L \cap k(\zeta_m) =  k}, so there is a lattice

The reason is that we then have a nice description of {G(L(\zeta_n)/k)} (as a product of Galois groups {G(L/k) \times G(k(\zeta_m)/k)}), and we will define {E} as a fixed field of a certain subgroup.

I claim that this condition will happen once {m} is divisible only by sufficiently large primes. For in this case, {L \cap \mathbb{Q}(\zeta_m)} is unramified everywhere over {\mathbb{Q}} (because {k} has fixed, finite ramification and the cyclotomic part is ramified precisely at the primes dividing {m}). It is a theorem, depending on the Minkowski bound for the discriminant, that any number field different from {\mathbb{Q}} is ramified somewhere, so it follows that {L \cap \mathbb{Q}(\zeta_m) = \mathbb{Q}}. Now in the lattice of fields

we have that {[k(\zeta_m): L \cap k(\zeta_m)] = [L(\zeta_m): L] =  \phi(m)}, the last equality in view of the lattice of fields

However, we also know (by the same reasoning) that {[k(\zeta_m):k] =\phi(m)}. It follows that {k = L  \cap k(\zeta_m)}, as claimed, when {m} is not divisible by the primes of {\mathbb{Q}} ramified in {L}.

2.2. Unramification in {LE}

Since we are going to define {E} as a subfield of {L(\zeta_m)}, this one is easy: just choose {m} not divisible by the rational prime {\mathfrak{p}} prolongs.

2.3. Complete splitting

We are now going to choose a subgroup {H \subset G(L/k) \times  G(k(\zeta_m)/k)}, whose fixed field will be {E}.

Now since {\mathfrak{p}} splits completely in {E/k}, we know that we are going to require {H} to contain the decomposition group of some extension of it, i.e. the group generated by the Frobenius elements

\displaystyle  (\mathfrak{p}, L/k) \times  (\mathfrak{p}, k(\zeta_m)/k).

This, however, is still not enough. We need {L \cap E  =k} and {L(\zeta_m) = E(\zeta_m)}.

2.4. {L \cap E}

So {L \cap E} is the fixed field of the subgroup of {G = G(L/k) \times G(k(\zeta_m)/k)} generated by {H} and by {G(L(\zeta_m)/L) = 1 \times  G(k(\zeta_m)/k)}. If we want {L \cap E =  k}, we will need

\displaystyle  H( 1 \times G(k(\zeta_m)/k)) = G.

We can arrange this if we choose a generator {\sigma} of {G(L/k)} and some {\tau \in G(k(\zeta_m)/k)} and add {\sigma  \times \tau} to {H}, so that {H} is generated by two elements. So do this. We will choose {\tau} and {m} later.

2.5. {LE} is cyclotomic; choice of {m,\tau}

This is the most subtle part of the proof, and where the number theory developed earlier will come in handy. We will prove that {L(\zeta_m) = E(\zeta_m)}, so that {LE/E} is a subextension of a cyclotomic extension.

The field {E(\zeta_m) = Ek(\zeta_m)} is the fixed field of the intersection of {H} and {G(L(\zeta_m)/k(\zeta_m)) = G(L/k) \times 1}, and we therefore want this intersection to be the trivial subgroup. We have defined {H} to be generated by

\displaystyle  (\mathfrak{p}, L/k) \times  (\mathfrak{p}, k(\zeta_m)/k) , \ \sigma \times \tau

where {\tau \in G(k(\zeta_m)/k) =  (\mathbb{Z}/m\mathbb{Z})^*} is yet to be chosen. Suppose this subgroup intersected with {1 \times  G(k(\zeta_m/k)}; we’d then be able to write for some {i,j},

\displaystyle   (\mathfrak{p}, k(\zeta_m)/k)^i = \tau^j,

which is to say that the cyclic groups in {(\mathbb{Z}/m\mathbb{Z})^*} generated by {{\mathbb N} \mathfrak{p}} (since this is how {\mathfrak{p}} acts on {\zeta_m, \zeta_m  \rightarrow \zeta_m^{{\mathbb N} \mathfrak{p}}}) and {\tau} intersect.

OK. Now it’s starting to become clear how this intersects (groan) with our previous number-theoretical shenanigans.

So, let {a = {\mathbb N} \mathfrak{p}}. Choose {m} divisible only by super large primes and such that {a} has order in {(\mathbb{Z}/m\mathbb{Z})^*} dividing {n}, and such that there also exists {b \in  (\mathbb{Z}/m\mathbb{Z})^*} of order dividing {n} such that the cyclic groups generated by {a,b} respectively are disjoint. We can do this by yesterday’s lemma.

We have now chosen {m}. For {\tau,} take it to correspond to {b}. I claim now that {H \cap G(L/k) \times 1 =  \{1\}}, which will prove the last remaining claim in Artin’s lemma.

Indeed, if we had some element in the intersection, then first of all we’d be able to write

\displaystyle   (\mathfrak{p}, k(\zeta_m)/k)^i = \tau^j,

as mentioned earlier; this is because we’d have an expression {(\mathfrak{p}) \times (\mathfrak{p}))^i = (\sigma \times \tau)^j \times  (\rho \times 1)} for some {\rho} in the intersection of the two groups. This implies that {  (\mathfrak{p}, k(\zeta_m)/k)^i = \tau^j=1} by independence and {i,j} are consequently divisible by {n}. But then, because {G(L/k)} has order {n}, it follows that {(\mathfrak{p}) \times (\mathfrak{p}))^i = (\sigma \times  \tau)^j =1}, and {\rho =1} too. Thus the intersection of the two subgroups is the trivial subgroup as claimed. This proves Artin’s lemma.

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