Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 28 / 4 / b

Put $\alpha =\sqrt[p]{p}$ and $\lambda ={\zeta }_p-1$. The polynomial $T^p-p$ is Eisenstein, so $[{\mathbb{Q}}_p(\alpha ):{\mathbb{Q}}_p]=p$, while the preceding part gives degree $p-1$ for $E$. Both numbers divide $[K:{\mathbb{Q}}_p]$ by the tower law, and the compositum degree is at most their product. Coprimality forces $[K:{\mathbb{Q}}_p]=p(p-1)$. Since $K$ contains all the roots ${\zeta }_p^a\alpha$ of $T^p-p$, it is the splitting field of a separable polynomial and is Galois.

Extend $v_p$ with $v_p(p)=1$. Then $v_p(\alpha )=1/p$, $v_p(\lambda )=1/(p-1)$, andThe value group therefore forces the ramification index to be at least $p(p-1)$; it cannot exceed the extension degree. Thus $K/{\mathbb{Q}}_p$ is totally ramified and $\varpi$ is a uniformizer for the normalized valuation $v_K=p(p-1)v_p$.

Write its automorphisms as ${\sigma }_{a,b}(\alpha )={\zeta }_p^a\alpha$ and ${\sigma }_{a,b}({\zeta }_p)={\zeta }_p^b$, with $a\in {\mathbb{F}}_p$, $b\in {\mathbb{F}}_p^{\times }$. The degree count makes all $p(p-1)$ pairs occur, giving $G\cong {\mathbb{F}}_p⋊{\mathbb{F}}_p^{\times }$. Their action on the uniformizer isIf $b\ne 1$, the residue of this ratio is $b$, so subtracting one gives a unit and $v_K({\sigma }_{a,b}(\varpi )-\varpi )=1$. If $b=1$ and $a\ne 0$, the ratio is ${\zeta }_p^{-a}$, giving displacement valuation $1+v_K({\zeta }_p^{-a}-1)=p+1$. The identity has infinite displacement valuation.

By the uniformizer criterion for lower ramification groups, $\sigma \in G_i$ precisely when that displacement valuation is at least $i+1$. Hence the ramification groups of the splitting field of Xp minus p over the p-adic numbers have sizesFor $1\le i\le p$ the group itself is $\{{\sigma }_{a,1}:a\in {\mathbb{F}}_p\}\cong C_p$. The lower wild ramification break is $p$.