A question on Ray class groups and the relative degree of a prime I have been stuck on a detail in the proof of Theorem 4 in R. Gold, The non-triviality of certain $\mathbb{Z}_l$-extensions (see also the the lemma and the following statement on page 3).  Here is the basic setup (I am changing the prime $l$ to the prime $p$):
Let $k$ be an imaginary quadratic field and $p \in \mathbb{Z^+}$ an odd prime such that $p\mathcal{O}_k = \mathfrak{p}\overline{\mathfrak{p}}$.  Denote $I_{\overline{\mathfrak{p}}}$ to be the set of fractional ideals of $k$ which are co-prime to $\overline{\mathfrak{p}}$ and $S_{\overline{\mathfrak{p}}^2} \subseteq I_{\overline{\mathfrak{p}}}$ the principle ideals in $k$ generated by elements $\alpha \equiv 1 \pmod{\overline{\mathfrak{p}}^2}$.  Let $\zeta_p^2$ be a primitive $p^2$-th root of unity and $k_1$ the unique subfield of $k(\zeta_p^2)$ such that $[k_1 : k] = p$ ($k_1$ is the first layer of the cyclotomic $\mathbb{Z}_p$-extension of $k$).  Let $L$ be the maximal unramified extension of $k_1$ which is Abelian over $k$, and $T(\mathfrak{p})$ the inertia group of $\mathfrak{p}$ for $L/k$. Denote $E$ to be the fixed field of $T(\mathfrak{p})$.  From the lemma on page 3 we have $Gal(L/k) \cong T(\mathfrak{p}) \times T(\overline{\mathfrak{p}})$, so $\overline{\mathfrak{p}}$ is the unique ramified prime in $E/k$* (Please correct me if I'm wrong. I think there may be a typo in the statement following the proof of the lemma in Gold.)
Now, in the proof of Theorem 4 Gold establishes that $E$ is in fact the fixed field of the $p$-primary subgroup of the Ray class group of conductor $\overline{\mathfrak{p}}^2$ and also establishes that $(\mathcal{O}_k/\overline{\mathfrak{p}}^2)^{\times}_p \cong (I_{\overline{\mathfrak{p}}}/S_{\overline{\mathfrak{p}}^2})_p$, where $A_p$ denotes the $p$-primary subgroup of the group $A$.
Here is where I'm stuck: Why is it that the relative degree of $\mathfrak{p}$ in $E/k$ (I take this to mean $[\mathcal{O}_E/\mathfrak{P} : \mathcal{O}_k/\mathfrak{p}]$ with $\mathfrak{P}\cap k = \mathfrak{p}$) equal to the order of $\mathfrak{p}$ in $(I_{\overline{\mathfrak{p}}}/S_{\overline{\mathfrak{p}}^2})_p$?
 A: I will try to restate The Phoenix's comment and fill in some more details.  Let $\mathfrak{P}$ be a prime in $\mathcal{O}_E$ above $\mathfrak{p}$.  Denote $Z = Z(\mathfrak{P}/\mathfrak{p})$ and $T = T(\mathfrak{P}/\mathfrak{p})$ to be the decomposition and inertia subgroups of $Gal(E/k)$ respectively.  By assumption $\mathfrak{p}$ is unramified in $E$ so that $T(\mathfrak{P}/\mathfrak{p})$ is trivial.  From the exact sequece
\begin{align}
1 \to T \to Z\to Gal((\mathcal{O}_E/\mathfrak{P})/ (\mathcal{O}_k/\mathfrak{p}))\to 1
\end{align}
we have that $Z \cong Gal((\mathcal{O}_E/\mathfrak{P})/ (\mathcal{O}_k/\mathfrak{p}))$.  Now, $Z$ is cyclic and generated by the Frobenius automorphism $\sigma_{\mathfrak{p}}$ which is determined by $\sigma_{\mathfrak{p}}(x) \equiv x^{N(\mathfrak{p})} \pmod{\mathfrak{P}}$, where $N$ is the norm from $k$ to $\mathbb{Q}$, which is equal to $p^f =p^{ [\mathcal{O}_k/\mathfrak{p}:\mathbb{Z}/p\mathbb{Z}]}$.  From Gold we have the isomorphism $Gal(E/k) \cong (I_{\overline{\mathfrak{p}}}/S_{\overline{\mathfrak{p}}^2})_p$ which is induced by the Artin map sending $\mathfrak{p}$ to $\sigma_{\mathfrak{p}}$.  But then the order of $\mathfrak{p}$ in $(I_{\overline{\mathfrak{p}}}/S_{\overline{\mathfrak{p}}^2})_p$ is the order of $\sigma_{\mathfrak{p}}$ in $Gal(E/k)$, which is in turn the order of the image of $\sigma_p$ in $Gal((\mathcal{O}_E/\mathfrak{P})/ (\mathcal{O}_k/\mathfrak{p}))$ which is $[\mathcal{O}_E/\mathfrak{P}:\mathcal{O}_k/\mathfrak{p}]$.
