Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 24 / 3

The geometric sum givesSince ${\Phi }_p(1)=p$, its factorization at the nonidentity $p$th roots of unity yieldsEach factor is a unit, and Wilson theorem gives $w\equiv (p-1)!\equiv -1(\mathrm{mod}{\pi }_K)$. Set $u=-w^{-1}$. Then $u$ is a principal unit andThe minus sign comes from the product of the nonzero elements of the residue field, not from an arbitrary choice of uniformizer.

For a principal unit $u$, apply the Hensel lemma to $F(T)=T^{p-1}-u$. The residue class one is a root, and $F^{\prime }(1)=p-1$ is a unit. Thus there is a unique $v\in 1+{\pi }_K{\mathcal{O}}_K$ withIn particular it is a unit of ${\mathcal{O}}_K$. Choose this $v$ for the $u$ just obtained and put $\alpha ={\pi }_K/v$. Then ${\alpha }^{p-1}=-p$, so ${\mathbb{Q}}_p(\alpha )\subseteq K$. The polynomial $T^{p-1}+p$ is Eisenstein, giving degree $p-1$ for the left-hand field. The p-adic cyclotomic extension $K$ also has degree $p-1$, so

Choose $\beta$ with ${\beta }^{p(p-1)}=-p$ and put $N=p(p-1)$. The Eisenstein polynomial $T^N+p$ shows that $L={\mathbb{Q}}_p(\beta )$ has degree $N$, is totally ramified, and has uniformizer $\beta$. Its subfield ${\mathbb{Q}}_p({\beta }^p)$ is the field $K$ from the previous part. Because $p$ is odd,Thus $L$ contains $a$ and ${\zeta }_p$, and contains all roots $a{\zeta }_p^j$ of $T^p-p$. Conversely $[{\mathbb{Q}}_p(a):{\mathbb{Q}}_p]=p$, while $[K:{\mathbb{Q}}_p]=p-1$. Their coprime degrees force their compositum to have degree $N$. It is contained in $L$ and hence equals it. Therefore $L$ is exactly the splitting field, not merely an extension containing it.

The Galois group has a normal subgroup $H=\mathrm{Gal}(L/K)$ of order $p$, acting by $\beta \mapsto {\zeta }_p^j\beta$. The $(p-1)$st roots of unity already lie in ${\mathbb{Q}}_p$ by the Hensel lemma. The maps $\beta \mapsto \omega \beta$, with ${\omega }^{p-1}=1$, supply a complement of order $p-1$. Its conjugation acts faithfully on $H$, so $G\cong C_p⋊{\mathbb{F}}_p^{\times }$.

Normalize $v_L(\beta )=1$. Total ramification and the uniformizer criterion for lower ramification groups reduce the calculation to $i_G(\sigma )=v_L(\sigma \beta -\beta )$. For nonidentity $\sigma \in H$,since $e(L/K)=p$ and $v_K({\zeta }_p^j-1)=1$. For $\sigma \notin H$, write $\sigma \beta =\omega {\zeta }_p^j\beta$ with $\omega \ne 1$. Its multiplier has residue $\overline{\omega }\ne 1$, so $i_G(\sigma )=1$. The lower ramification numbering is thereforeAll later groups are trivial. The wild lower break is $p$, not one. In the upper ramification numbering, the Herbrand function sends this break to $p/(p-1)$: $G^0=G$, $G^u=H$ for $0<u\le p/(p-1)$, and $G^u=1$ above it.

As an independent consistency check, the different exponent from ramification groups is $(N-1)+p(p-1)=2N-1$. The derivative of the Eisenstein polynomial gives the same answer, $v_L(N{\beta }^{N-1})=N+(N-1)$, since $v_L(p)=N$.