Put $\alpha=\sqrt[p]{p}$. The polynomial $X^p-p$ is <Eisenstein polynomial>[Eisenstein], so $\mathbb Q_p(\alpha)/\mathbb Q_p$ is totally ramified of degree $p$. The cyclotomic extension $\mathbb Q_p(\zeta_p)/\mathbb Q_p$ is totally ramified of degree $p-1$. Their coprime degrees make their intersection trivial, so
$$
K=\mathbb Q_p(\zeta_p,\alpha)
$$
has degree $p(p-1)$ and is totally ramified. It is the splitting field of $X^p-p$, hence Galois.

Normalize $v_K$ by $v_K(K^\times)=\mathbb Z$. Then
$$
v_K(\zeta_p-1)=p,
\qquad v_K(\alpha)=p-1,
$$
so
$$
\varpi=\frac{\zeta_p-1}{\alpha}
$$
is a <uniformizer>. Write an automorphism as
$$
\sigma_{a,b}(\zeta_p)=\zeta_p^a,
\qquad
\sigma_{a,b}(\alpha)=\zeta_p^b\alpha,
$$
where $a\in\mathbb F_p^\times$ and $b\in\mathbb F_p$. Since
$$
\frac{\sigma_{a,b}(\varpi)}{\varpi}
=\zeta_p^{-b}\frac{\zeta_p^a-1}{\zeta_p-1},
$$
the <uniformizer criterion for lower ramification groups> gives valuation one for $\sigma(\varpi)-\varpi$ when $a\ne1$, and valuation $p+1$ when $a=1$, $b\ne0$. Therefore
$$
G_0=G,qquad
G_i=\{\sigma_{1,b}:b\in\mathbb F_p\}\cong\mathbb Z/p\mathbb Z\quad(1\leq i\leq p),
$$
and $G_i=1$ for $i\geq p+1$. These are the <ramification groups of the splitting field of Xp minus p over the p-adic numbers>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 3 / b / Solution

{title2=$\mathbb Q_p(\zeta_p,\sqrt[p]{p})/\mathbb Q_p$}

For odd $p$, the splitting field $K=\mathbb Q_p(\zeta_p,\sqrt[p]{p})$ is totally ramified of degree $p(p-1)$. With
$$
\varpi=\frac{\zeta_p-1}{\sqrt[p]{p}},
$$
the lower ramification groups are
$$
G_0=\operatorname{Gal}(K/\mathbb Q_p),\qquad
G_i\cong\mathbb Z/p\mathbb Z\ (1\leq i\leq p),\qquad
G_i=1\ (i\geq p+1).
$$


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 Incoming links: Ramification groups of the splitting field of Xp minus p over the p-adic numbers