For a finite point , use the valuation with local parameter ; at infinity use . Its extensions to are permuted transitively by . The completed residue field is , because is algebraically closed, so the local decomposition group equals its inertia group.
Write . Characteristic zero makes all local ramification tame. One can choose a uniformizer in the completion for which
Indeed, initially is times a unit; that unit has an th root in , by its algebraically closed residue field and the invertibility of . Absorb it into . The local automorphisms are then , , and form a cyclic group of order . Define .
To specify an element rather than a subgroup, fix a compatible choice of primitive roots of unity . Define by its uniformizer character
and let be its conjugacy class in . Over , take , corresponding to positive local continuation around . At an unramified point, and .
Changing the uniformizer does not affect this character. If with a unit, inertia fixes its residue, so
The character is injective: a nonidentity finite-order automorphism with derivative one would have for , and its th iterate would have coefficient at that first nonzero displacement. Characteristic zero contradicts finite order. Thus the chosen primitive character identifies a unique generator.
For another place , conjugation gives , preserves its order, and carries to , since fixes the constants and hence . This proves that and the branch-cycle conjugacy class are independent of the place and uniformizer. Without the stated roots-of-unity convention, only the cyclic inertia-subgroup conjugacy class is intrinsic: replacing by another primitive root replaces the generator by a coprime power, which need not lie in the same conjugacy class.