Put $\pi_n=\zeta_{p^n}-1$. The shifted cyclotomic polynomial
$$
\Phi_{p^n}(1+X)
=\frac{(1+X)^{p^n}-1}{(1+X)^{p^{n-1}}-1}
$$
is Eisenstein at $p$. Therefore it is irreducible, $\pi_n$ is a uniformizer, and
$$
[\mathbb Q_p(\zeta_{p^n}):\mathbb Q_p]
=\varphi(p^n)=p^{n-1}(p-1),
$$
so the extension is totally ramified. It is the splitting field of $\Phi_{p^n}$, and every automorphism is uniquely
$$
\zeta_{p^n}\longmapsto\zeta_{p^n}^{,a},
\qquad a\in(\mathbb Z/p^n\mathbb Z)^\times.
$$
This proves the <cyclotomic extension of a p-adic field> isomorphism
$$
\operatorname{Gal}(\mathbb Q_p(\zeta_{p^n})/\mathbb Q_p)
\cong(\mathbb Z/p^n\mathbb Z)^\times.
$$
Restriction in the cyclotomic tower corresponds to reduction of $a$, so taking the <inverse limit> gives
$$
\operatorname{Gal}(\mathbb Q_p(\zeta_{p^\infty})/\mathbb Q_p)
\cong\varprojlim_n(\mathbb Z/p^n\mathbb Z)^\times
=\mathbb Z_p^\times.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 136 / 4 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-136/4/b/solution