Looking for all $p$ power cyclotomic units of $Q_p$ or the generator

Question

This is related to decomposition of $U_1=1+pZ_p$ in Iwasawa's Local class field theory book Chpt 2, Proposition 2.7

Let $k$ be a finite $Q_p$ extension, $d=[k:Q_p]$ and $q$ cardinality of residue field of $k$. Set $\pi$ uniformizer of valuation ring associated to $k$. Then $k^\times=<\pi>\oplus Z_{(q-1)}\oplus Z_{p^a}\oplus Z_p^d$ where $Z_{p^a}$ corresponds to all $p$-power roots of unity and $U_1=Z_p^d\oplus Z_{p^a}$. However $a$ is not concretely given.

$\textbf{Q:}$ I want to determine numerical values of $a$. The proof is not using constructive method by uses PID classification theorem and via $|U_1/U_n|<\infty$ for some large $n$ s.t. $U_n\cap W=1$ where $U_n$ is free of rank $d$ over $Z_p$(p-adic integer). The latter does not indicate exact size of $W$. How do I find this $a$ concretely in terms of $e,f,d$?

Answer 1

$$K^* = O_K^\times \times \pi^\Bbb{Z}$$ $$O_K^\times= \langle \zeta_{p^f-1}\rangle \times 1+\pi O_K$$ $$H = (1+\pi O_K)_{tors}$$ Since $H$ is torsion it contains some roots of unity, since they are equal to $1\bmod (\pi)$ it means they are $p^r$-th roots of unity, ie. $H = \langle \zeta_{p^a}\rangle$.

Because $\lim_{n\to \infty} (1+\pi x)^{p^n}=1$ then $(1+\pi x)^b$ is well-defined for $b\in \Bbb{Z}_p$ and $(1+\pi O_K)/H$ is a torsion free $\Bbb{Z}_p$-module, it is finitely generated because $1+p^2 O_K$ is of finite index in $1+\pi O_K$ and $\log$ is a $\Bbb{Z}_p$-module isomorphism $1+p^2 O_K\to p^2 O_K$,

that is to say $$(1+\pi O_K)/H= \prod_{j=1}^{[O_K:\Bbb{Z}_p]} (1+\pi x_j)^{\Bbb{Z}_p} H$$

$$1+\pi O_K = H \times \prod_{j=1}^{[O_K:\Bbb{Z}_p]} (1+\pi x_j)^{\Bbb{Z}_p} $$

$p^f$ is the size of $O_K/(\pi)$ and $\zeta_{p^a}$ is the largest $p^r$-root of unity in $K$.