Frobenius Elements

Suppose that $K/\mathbf{Q}$ is a finite Galois extension with group $G$ and $p$ is a prime such that $e=1$ (i.e., an unramified prime). Then $I=I_\mathfrak{p}=1$ for any $\mathfrak{p}\mid p$, so the map $\varphi$ of Theorem 14.1.5 is a canonical isomorphism $D_\mathfrak{p}\cong \Gal (\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)$. By Section 14.1.1, the group $\Gal (\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)$ is cyclic with canonical generator $\Frob _p$. The corresponding to $\mathfrak{p}$ is $\Frob _\mathfrak{p}\in D_\mathfrak{p}$. It is the unique element of $G$ such that for all $a\in\O _K$ we have

$$\Frob _\mathfrak{p}(a)\equiv a^p\pmod{\mathfrak{p}}.$$

(To see this argue as in the proof of Proposition 14.1.8.) Just as the primes $\mathfrak{p}$ and decomposition groups $D$ are all conjugate, the Frobenius elements over a given prime are conjugate.

Proposition 14.2.1   For each $\sigma\in G$, we have

$$\Frob _{\sigma\mathfrak{p}} = \sigma\Frob _\mathfrak{p}\sigma^{-1}.$$

In particular, the Frobenius elements lying over a given prime are all conjugate.

Proof. Fix $\sigma\in G$. For any $a\in\O _K$ we have $\Frob _\mathfrak{p}(\sigma^{-1}(a)) - \sigma^{-1}(a) \in \mathfrak{p}$. Multiply by $\sigma$ we see that $\sigma\Frob _\mathfrak{p}(\sigma^{-1}(a)) - a \in \sigma\mathfrak{p}$, which proves the proposition. $\qedsymbol$

Thus the conjugacy class of $\Frob _\mathfrak{p}$ in $G$ is a well defined function of $p$. For example, if $G$ is abelian, then $\Frob _\mathfrak{p}$ does not depend on the choice of $\mathfrak{p}$ lying over $p$ and we obtain a well defined symbol $\left(\frac{K/\mathbf{Q}}{p}\right) =\Frob _\mathfrak{p}\in G$ called the . It extends to a map from the free abelian group on unramified primes to the group $G$ (the fractional ideals of $\mathbf{Z}$). Class field theory (for  $\mathbf{Q}$) sets up a natural bijection between abelian Galois extensions of $\mathbf{Q}$ and certain maps from certain subgroups of the group of fractional ideals for  $\mathbf{Z}$. We have just described one direction of this bijection, which associates to an abelian extension the Artin symbol (which induces a homomorphism). The Kronecker-Weber theorem asserts that the abelian extensions of $\mathbf{Q}$ are exactly the subfields of the fields $\mathbf{Q}(\zeta_n)$, as $n$ varies over all positive integers. By Galois theory there is a correspondence between the subfields of $\mathbf{Q}(\zeta_n)$ (which has Galois group $(\mathbf{Z}/n\mathbf{Z})^*$) and the subgroups of $(\mathbf{Z}/n\mathbf{Z})^*$. Giving an abelian extension of $\mathbf{Q}$ is exactly the same as giving an integer $n$ and a subgroup of $(\mathbf{Z}/n\mathbf{Z})^*$. Even more importantly, the reciprocity map $p\mapsto \left(\frac{Q(\zeta_n)/\mathbf{Q}}{p}\right)$ is simply $p\mapsto p\in (\mathbf{Z}/n\mathbf{Z})^*$. This is a nice generalization of quadratic reciprocity: for $\mathbf{Q}(\zeta_n)$, the $efg$ for a prime $p$ depends in a simple way on nothing but $p\mod n$.