I quote the following problem from the chapter Hilbert's Ramification Theory of Jurgen Neukirch
Let $L/K$ is a Galois extension with prime ideal $\mathfrak{P}$, unramified over $K$, then there is one and only one automorphism $\Phi_{\mathfrak{P}}\in\operatorname{Gal}L/K$, s.t. $\Phi_{\mathfrak{P}}(a)\equiv a^q \pmod{\mathfrak{P}}$, where $q$ is the size of the residue field $O_K/\mathfrak{P}\cap O_K$. It is called the Frobenius Automorphism.
Now, if we take $K=\mathbb{Q}$ and $p=\mathfrak{P}\cap \mathbb{Z}$, I know that the residue field $O_L/\mathfrak{P}$ is a finite extension over $\mathbb{F}_p$, so, $a\mapsto a^p$ is a frobenius automorphism in $O_L/\mathfrak{P}$ and it generates all the other frobenius automorphisms. But the definition of the automorphism from the exercise seems different. Am I missing anything? How can it be a frobenius automorphism? The frobenius automorphism in the sense of finite field is not matching with this.
Best Answer
For a number field $L$ and a prime ideal $\mathfrak{P}\subset O_L$ and $L/K$ Galois. There is some automorphism $\in Gal(L/K)$ whose reduction is the Frobenius. If the extension is unramified at $\mathfrak{P}$ then it is unique and we can call it the Frobenius.
It is not true that $q$ is the degree of the extension. It is the cardinality of $O_K/(\mathfrak{P}\cap O_K)$, which is also $p^f$ where $f=[O_K/(\mathfrak{P}\cap O_K):\Bbb{F}_p]$ and $p$ is the characteristic. The order of the Frobenius (here I assume the extension is unramified at $\mathfrak{P}$) is $[O_L/\mathfrak{P}:O_K/(\mathfrak{P}\cap O_K)]$.