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.