# The Frobenius automorphism of a number field extension

I wish to prove the existence of the Frobenius automorphism of an extension of algebraic number fields.

Let $$E/K$$ be a normal extension of algebraic number fields and let $$\mathfrak{P} \subset \mathcal{O}_L$$ be a prime ideal that does not ramify over $$K$$. Then there is a unique $$K$$-automorphism $$\sigma_{\mathfrak{P}} \in \textrm{Gal}(E/K)$$ satisfying: $$$$\sigma_{\mathfrak{P}}(\alpha) \equiv \alpha^q\ (\textrm{mod}\ \mathfrak{P}) \qquad \forall \alpha \in \mathcal{O}_L$$$$ with $$q = [\mathcal{O}_E/\mathfrak{P} : \mathcal{O}_K/\mathfrak{p}]$$. The decomposition group $$G_{\mathfrak{P}} < G$$ is cyclic and generated by $$\sigma_{\mathfrak{P}}$$ - the Frobenius automorphism.

I am aware of the existence of the Frobenius automorphism in the Galois group of an extension of finite fields:

Let $$q=p^n$$, $$n,p \in \mathbb{N}$$, $$p$$ prime, and let $$\mathbb{F}_q$$ and $$\mathbb{F}_p$$ be the unique finite fields with cardinality $$q$$ and $$p$$ respectively. Then $$\mathbb{F}_q / \mathbb{F}_p$$ is a Galois extension, and there exists an automorphism $$\sigma \in \textrm{Gal}(\mathbb{F}_q / \mathbb{F}_p)$$ satisfying $$\sigma(\alpha)=\alpha^p$$ for all $$\alpha \in \mathbb{F}_q$$.

Showing that $$\sigma$$, as defined above, is an automorphism, is simple. But here the behaviour of $$\sigma$$ on the entirety of $$\mathbb{F}_q$$ is given.

The problem in the case of number fields is that if we want $$\sigma_{\mathfrak{P}} \in \textrm{Gal}(E/K)$$ to satisfy $$$$\sigma_{\mathfrak{P}}(\alpha) \equiv \alpha^q\ (\textrm{mod}\ \mathfrak{P}) \qquad \forall \alpha \in \mathcal{O}_L$$$$ how can we control its behaviour on the rest of $$L$$ and ensure that $$\sigma_{\mathfrak{P}}$$ indeed is an automorphism on $$L$$?

The above appears as exercise 2 of chap. I §9 of J. Neukirch's $$\textit{Algebraische Zahlentheorie}$$.

$$\textbf{Addendum:}$$ I spent months trying to figure this out. It turns out there is a typo. See here.

• You should probably try to give the English title of a book when it has been translated (and also the edition if you are referring to a specific exercise). Commented Dec 13, 2019 at 0:10

Let $$G_{\mathfrak{P}}$$ be the corresponding decomposition group. We have a quotient morphism $$q: G_{\mathfrak{P}} \rightarrow Gal((\mathcal{O}_E/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p}))$$ and we want to prove it is an isomorphism.

Now, $$G$$ acts transitively over the set of prime ideals of $$E$$ above $$\mathfrak{p}$$. Let $$r$$ denote the cardinality of this set, and $$G_{\mathfrak{P}}$$ is the stabilizer of $$\mathfrak{P}$$.

Standard group action theory yields $$|G_{\mathfrak{P}}|=|G|/r=[E:K]/r$$. By standard study of number field extensions, and because there is no ramification, $$[E:K]=r[(\mathscr{O}_E/\mathfrak{P}):(\mathcal{O}_K/\mathfrak{p})]$$, so $$q$$ is a group morphism between two groups of same cardinality.

Therefore, we want to show that $$q$$ is injective, and we are done.

So let $$\sigma \in G_{\mathfrak{P}}$$ be in the kernel of $$\sigma$$: then $$\sigma$$ moves elements of $$\mathcal{O}_E$$ by elements of $$\mathfrak{p}\mathcal{O}_E$$. Assume that $$\sigma \neq id$$.

Let $$n \geq 1$$ be maximal such that for each $$x \in \mathcal{O}_E$$, $$\sigma(x)-x \in \mathfrak{p}^n\mathcal{O}_E$$.

Let $$p$$ be a uniformizer in $$K$$ for $$\mathfrak{p}$$, it is a uniformizer for $$E$$ and $$\mathfrak{P}$$ as well. Define, for $$x \in \mathcal{O}_E$$, $$\tau(x)=\frac{\sigma(x)-x}{p^n} \in \mathcal{O}_E$$. It is easy to see that $$x \in \mathcal{O}_E^{\times} \longmapsto \tau(x)/x \in \mathcal{O}_E$$ induces, by quotient, a morphism $$(\mathcal{O}_E/\mathfrak{P})^{\times} \rightarrow \mathcal{O}_E/\mathfrak{P}$$.

Because the cardinals are coprime, this morphism must be trivial, that is, $$\sigma(x) \in x+p^n\mathfrak{P}$$ for each $$x \in \mathcal{O}_E^{\times}$$. By multiplying by $$p$$ enough times, it follows that $$(\sigma-id)(\mathcal{O}_E) \subset \mathfrak{p}^{n+1}$$, a contradiction.

Therefore $$q$$ is injective and we are done.

• I am afraid that I fail to see how $\textrm{Gal}((\mathcal{O}_E/\mathfrak{P})/(\mathcal{O}_K/\mathfrak{p})) \cong G_{\mathfrak{P}}$ implies the existence of a $\sigma_{\mathfrak{P}} \in G$ satisfying the above. Commented Jan 17, 2020 at 19:03
• Consider the inverse image by $q$ of the Frobenius automorphism in the residual fields. Commented Jan 17, 2020 at 23:59