Ramification groups of the splitting field of Xp minus p over the p-adic numbers (${\mathbb{Q}}_p({\zeta }_p,\sqrt[p]{p})/{\mathbb{Q}}_p$)

For odd $p$, the splitting field $K={\mathbb{Q}}_p({\zeta }_p,\sqrt[p]{p})$ is totally ramified of degree $p(p-1)$. Withthe lower ramification groups are

To verify the filtration, write ${\alpha }^p=p$ and $\lambda ={\zeta }_p-1$. The Eisenstein polynomials for $\alpha$ and $\lambda$ give coprime degrees $p$ and $p-1$, so their compositum has degree $p(p-1)$. Its element $\varpi =\lambda /\alpha$ has $p$-adic valuation $1/[p(p-1)]$, forcing total ramification. For an automorphism ${\sigma }_{a,b}$ with $\alpha \mapsto {\zeta }_p^a\alpha$ and ${\zeta }_p\mapsto {\zeta }_p^b$,Its residue is $b$, so the normalized displacement valuation is one if $b\ne 1$. If $b=1$ and $a\ne 0$, that valuation is $1+v_K({\zeta }_p^{-a}-1)=p+1$. The uniformizer criterion for lower ramification groups now gives every group displayed above.

If $a^3=3$ and ${\zeta }_3$ is a primitive cube root of unity, then ${\pi }^3=1+2{\zeta }_3$ and ${\pi }^6=-3$. The Eisenstein polynomial $X^6+3$ therefore presents the entire degree-six splitting field, and $\pi$ is a uniformizer. Its normalised valuation gives $v_L(3)=6$, $v_L(a)=2$ and $v_L({\zeta }_3-1)=3$. The order-three automorphisms displace $\pi$ with valuation four, whereas transpositions displace it with valuation one.