Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 136 3 Solution Created 2026-10-03 Updated 2026-10-05
Let be a finite Galois extension of non-Archimedean local fields, let , and normalize the discrete valuation by . The lower ramification numbering isHere is the inertia group and is the wild inertia group. For real , put ; thus for . Define the Herbrand function and its inverse byextending on . The upper ramification numbering is for .
Let over , with its uniformizer. The Artin–Schreier polynomial has derivative in characteristic , so all its roots are distinct. If is one root and another, then , so . Conversely every , , is a root. Thus contains all roots and is a splitting field of a separable polynomial; it is Galois. Every automorphism has the form , and its Galois group embeds in the additive group of .
There is no root in . If , then cannot have negative valuation. If , the ultrametric inequality gives , divisible by , whereas is not. The Galois group is consequently nontrivial. Its order divides the prime , so
Let be the ramification index. The equation forces , and henceAs , . Since , we get , residue-field degree one, and . Thus this is a totally ramified extension, andThis is a uniformizer. The uniformizer criterion for lower ramification groups states that for a totally ramified Galois extension, exactly when . Here every nonidentity automorphism satisfiesTherefore the ramification break of an Artin–Schreier pole occurs at , and the lower groups areThis includes , where the break is one. There is no off-by-one shift: the condition is .
For the upper groups, compute the Herbrand function explicitly:Hence the upper break is also and, with the stated real-index convention,As a consistency check, satisfies the Eisenstein polynomial . Its derivative in characteristic is , giving different exponent , equal to from the different exponent from ramification groups.