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

$\displaystyle \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

$\displaystyle \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$.

William Stein 2004-05-06