Let $p \geq 2$ be prime and $K=\mathbb Q(\zeta_p),~\zeta^{p}=1$ with ring of integers $\mathcal{O}_K$. we denote by $\mathfrak{p} \mid p$ the prime ideal of $K$ dividing $p$. Let $K_{\mathfrak{p}}$ be the $\mathfrak{p}$-completion of $K$. We can say that $K_{\mathfrak{p}}$ is a cyclotomic extension of the $p$-adic field $\mathbb Q_p$. Denote $\mathbb Z_p$ be the ring of $p$-adic integers in $\mathbb Q_p$, let $\mathcal{O}_{K_{\mathfrak{p}}}$ be ring of integers of $K_{\mathfrak{p}}$ and $\mathfrak{q}$ be the corresponding prime ideal in $K_{\mathfrak{p}}$.
What is the quotient group $\mathfrak{q}^2/\mathfrak{p}^2 \mathbb Z_p$ ?
What is the torsion subgroup of the quotient $\mathfrak{q}^2/\mathfrak{p}^2\mathbb Z_p$ ?
What is the order of the torsion subgroup of $\mathfrak{q}^2/\mathfrak{p}^2 \mathbb Z_p$ ?
My part:
Note that $K$ is ramified extension of $\mathbb Q$ with ramification index $p-1$ and hence $(p)=(\mathfrak{p}^{p-1}).$ Note that we can take $\mathfrak{p}=1-\zeta_p$.
Note also $\mathcal{O}_K=\mathbb Z[\zeta_p]$ and $\mathcal{O}_{K_{\mathfrak{p}}}=\mathbb Z_p[\zeta_p]$. And $\mathfrak{p}=\mathfrak{p} \mathbb Z[\zeta_p]$
The prime ideal $\mathfrak{q}$ can be written as $ (1-\zeta_p) \mathbb Z_p[\zeta_p]$, where $\pi=1-\zeta_p$ is uniformizer. Thus $$\mathfrak{q}^2/\mathfrak{p}^2 \mathbb Z_p \simeq \frac{(1-\zeta_p)^2 \mathbb Z_p[\zeta_p]}{(1-\zeta_p)^2 \mathbb Z[\zeta_p] \mathbb Z_p} \simeq \frac{\mathbb Z_p[\zeta_p]}{\mathbb Z[\zeta_p] \mathbb Z_p} \simeq \mathbb Z_p[\zeta_p]/\mathbb Z_p[\zeta_p]=\text{trivial}, $$ since $\mathbb Z \hookrightarrow \mathbb Z_p$. Is it trivial group really ?
I think I am making error at some step.
I appreciate your comments.