Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 123 2 b Solution Created 2026-10-03 Updated 2026-10-06
Let and let be a primitive cube root of unity. The splitting field is . The polynomial is an Eisenstein polynomial, so . Meanwhile shows that is quadratic: is not a square because its valuation is odd. A quadratic field cannot lie in the degree-three field. Consequently , and the action on the three roots identifies its Galois group with the symmetric group .
A useful uniformizer isSince , we obtainThus is a root of the Eisenstein polynomial . Its degree is six, so , the extension is totally ramified, and is a uniformizer. With ,This is the Eisenstein sextic presentation of the splitting field of X3 minus 3 over Q3.
Let , , and let , . Then has order three, has order two, and . Their actions on the uniformizer areThe two nonidentity elements of therefore satisfyEach transposition sends to for some , so its displacement has valuation one: is a unit, reducing to modulo the maximal ideal. The uniformizer criterion for lower ramification groups now givesFor real indices, this is on , on , and for . In particular the lower breaks are and .
On the index is two; beyond three it is six. Thus the Herbrand function isThe positive lower break becomes the upper break . ThereforeThe fractional upper break is allowed because the full extension is nonabelian. As a check, the different exponent from ramification groups is . The derivative of the minimal polynomial gives the same result: . This is the case of ramification groups of the splitting field of Xp minus p over the p-adic numbers.