It is now time to begin the final descent towards the Artin reciprocity law, which states that for an abelian extension ${L/k}$, there is an isomorphism $\displaystyle J_k/k^* NJ_L \simeq G(L/k).$

We will actually prove the Artin reciprocity law in the idealic form, because we have only defined the Artin map on idelas. In particular, we will show that if ${\mathfrak{c}}$ is a suitable cycle in ${k}$, then the Artin map induces an isomorphism $\displaystyle I(\mathfrak{c}) / P_{\mathfrak{c}} N(\mathfrak{c}) \rightarrow G(L/k).$

The proof is a bit strange; as some have said, the theorems of class field theory are true because they could not be otherwise. In fact, the approach I will take (which follows Lang’s Algebraic Number Theory, in turn following Emil Artin himself).

So, first of all, we know that there is a map ${I(\mathfrak{c}) \rightarrow G(L/k)}$ via the Artin symbol, and we know that it vanishes on ${N(\mathfrak{c})}$. It is also necessarily surjective (a consequence of the first inequality). We don’t know that it factors through ${P_{\mathfrak{c}}}$, however.

Once we prove that ${P_{\mathfrak{c}}}$ (for a suitable ${\mathfrak{c}}$) is in the kernel, then we see that the Artin map actually factors through this norm class group. By the second inequality, the norm class group has order at most that of ${G(L/k)}$, which implies that the map must be an isomorphism, since it is surjective.

In particular, we will prove that there is a conductor for the Artin symbol. If ${x}$ is sufficieintly close to 1 at a large set of primes, then the ideal ${(x)}$ has trivial Artin symbol. This is what we need to prove.

Our strategy will be as follows. We will first analyze the situation for cyclotomic fields, which is much simpler. Then we will use some number theory to reduce the general abelian case to the cyclotomic case (in a kind of similar manner as we reduced the second inequality to the Kummer case). Putting all this together will lead to the reciprocity law.

Cyclotomic extensions

Today, we will begin by analyzing what goes on in the cyclotomic case. By the “functoriality” of the Artin map, we can start by looking at ${\mathbb{Q}(\zeta_n)/\mathbb{Q}}$ and analyzing the Artin map (for which, as we shall see, there is an explicit expression)

It is known that the only primes ramified in thi sextension are the primes dividing ${n}$. I shall not give the proof here; the interested reader can refer for instance to Washington’s Introduction to Cyclotomic Fields. The proof is not very difficult. In addition, I shall use the fact that the Galois group is the multiplicative group ${(\mathbb{Z}/n\mathbb{Z})^*}$, where the residue class of ${r}$ induces the automorphism ${\zeta_n \rightarrow \zeta_n^r}$.

So let ${p \not\mid n}$. I claim that the Frobenius element ${\sigma_p=\sigma_{(p)}}$ is just the automorphism ${\zeta_n \rightarrow \zeta_n^p}$. Indeed, it is the “raise to the ${p}$-th power” automorphism on the residue field, so ${\overline{ \sigma_p (\zeta_n)} = \overline{ \zeta_n^p}}$. In addition, ${\sigma_p(\zeta_n)}$ is an ${n}$-th root of unity. But I claim that the primitive ${n}$-th roots of unity are distinct modulo any prime of the ring of integers in ${\mathbb{Q}(\zeta_n)}$ prolonging ${(p)}$; indeed, $\displaystyle \prod_k (1 - \zeta_n^k) = n \not\equiv 0 \mod p,$

so no ${n}$-th root of unity (other than 1 itself) is congruent to 1 modulo a prime extending ${(p)}$. It follows that if ${m}$ is a rational integer prime to ${n}$, then ${((m), \mathbb{Q}(\zeta_n)) }$ corresponds to ${\zeta_n \rightarrow \zeta_n^m}$.

In particular, if ${m>0}$ is a rational number with no prime dividing ${n}$ in the numerator or denominator, and ${m \equiv 1 \mod n}$, then ${((m), \mathbb{Q}(\zeta_n)) = 1}$. So there exists a conductor in the case of ${\mathbb{Q}(\zeta_n)/\mathbb{Q}}$, and the reciprocity law is true in this case.

More generally,

Theorem 1 The reciprocity law is true for an extension ${L/k}$ if ${L}$ is contained in a cyclotomic extension of ${k}$.

By the consistency property, if there exists a conductor for the Artin symbol for a superextension of ${L}$, then that conductor works for ${L}$ too. So it is enough to show that there is a conductor for the Artin symbol in the case of ${L = k(\zeta_n)}$.

Now if ${x \in k^*}$ is close to 1 at all primes of ${k}$ dividing ${n}$ and totally positive at the real places, then ${N^k_{\mathbb{Q}}(x)}$ is positive and close to 1 at the primes dividing ${n}$. So $\displaystyle ( (x), k(\zeta_n)/k)) = ( (N^k_{\mathbb{Q}}(x), \mathbb{Q}(\zeta_n)/\mathbb{Q}) = 1.$

We will next discuss a lemma of Artin which allows the general case to be reduced to the cyclotomic case.