Frobenius automorphism ($\mathrm{Frob}$)

Let $K$ be a non-Archimedean local field with residue field $k_K$ of cardinality $q$. Arithmetic Frobenius automorphism acts on the algebraic closure of the residue field by $x\mapsto x^q$. On $k_K$ itself this is the identity; on a degree-$f$ residue extension it has order $f$.

For the degree-$f$ unramified extension $L/K$, reduction identifies its Galois group with the residue-field Galois group, so there is a unique lifted automorphism with this residue action. It generates $\mathrm{Gal}(L/K)$. Geometric Frobenius is its inverse. For example, in an unramified quadratic extension of ${\mathbb{Q}}_p$, arithmetic Frobenius lifts $x\mapsto x^p$ on ${\mathbb{F}}_{p^2}$, rather than acting nontrivially on ${\mathbb{F}}_p$.

In a finite Galois extension $L/K$, choose a prime ideal $\mathfrak{P}$ above an unramified $\mathfrak{p}$. The Frobenius automorphism at $\mathfrak{P}$ is characterized on its residue field by $x\mapsto x^{N\mathfrak{p}}$. Changing $\mathfrak{P}$ conjugates this element, so its conjugacy class is well-defined. It is the identity class exactly at a completely split prime.

Let a monic integer polynomial have distinct roots in its splitting field. Outside the finitely many primes dividing its polynomial discriminant or ramifying in that field, its roots reduce to distinct residue roots. The Frobenius automorphism permutes them by $a\mapsto a^p$; a reduced root belongs to ${\mathbb{F}}_p$ exactly when it is fixed. Distinctness of reduction identifies this with a fixed root of the Galois permutation itself. Combining this observation with a derangement in a transitive group action and the Chebotarev density theorem proves that an irreducible polynomial of degree at least two has rootless reduction at a positive Dirichlet density of primes. The degree bound is essential: every monic linear integer polynomial has a root modulo every prime.