 Solution

Use the following simple-root Hensel lemma. Let $R$ be a complete discrete valuation ring, with maximal ideal $\mathfrak{m}$, and let $f\in R[X]$. If $f(a_0)\in \mathfrak{m}$ and $f^{\prime }(a_0)\in R^{\ast }$, there is a unique $a\in R$ with $a\equiv a_0(\mathrm{mod}\mathfrak{m})$ and $f(a)=0$.

For the proof, define Newton's method iterates $a_{r+1}=a_r-f(a_r)/f^{\prime }(a_r)$. Each denominator stays a unit, because all iterates are congruent to $a_0$. Taylor's polynomial identity givesIf an iterate is already a root, the process stops. Otherwise these valuations tend to infinity, so the iterates form a Cauchy sequence in the complete ring. Its limit $a$ has the required residue and satisfies $f(a)=0$. If $a,b$ were two such roots, thenThe parenthesis is a unit, forcing $a=b$.

Now let $K/{\mathbb{Q}}_p$ be finite, with residue field $k={\mathbb{F}}_q$. For every $m\ge 1$ choose a monic irreducible polynomial $\bar{f}\in k[X]$ of degree $m$ and lift it to a monic $f\in {\mathcal{O}}_K[X]$. The lift is irreducible over $K$, since a factorization would reduce to a factorization of $\bar{f}$, by the monic form of Gauss's lemma over the valuation ring. If $L=K(\alpha )$ for a root, its residue field contains $k(\bar{\alpha })$ of degree $m$. The inequality $e(L/K)f(L/K)\le [L:K]=m$ forces residue degree $m$ and ramification index one. Thus it is an unramified extension of degree $m$.

Conversely, in any unramified extension of degree $m$, choose a generator of its residue field and lift its separable minimal polynomial. Hensel's lemma supplies a root with that residue. Its degree is at least $m$ and hence it generates the extension. The same unique lifting of all residue conjugates shows that the extension is Galois andOne can see uniqueness inside a fixed algebraic closure explicitly: the roots of $X^{q^m-1}-1$ lift all nonzero residue elements, and a lifted primitive residue element generates the extension. Hencewith arithmetic Frobenius acting by $\zeta \mapsto {\zeta }^q$. The case $q^m-1=1$ simply gives $K_1=K$. These are all finite unramified extensions; $K_m\subseteq K_r$ exactly when $m\mid r$. Their union is the maximal unramified extension, whose Galois group is $\hat{\mathbb{Z}}$.

Finally prove norm-surjectivity on units using the Herbrand quotient. For a cyclic group $G$ of order $m$, writeThe quotient is multiplicative on short exact sequences, equals one for finite modules, equals $m$ on the trivial module $\mathbb{Z}$, and equals one on induced regular modules. For a finite module, the equality follows directly from $|M^G||(\sigma -1)M|=|M|=|\ker N||NM|$.

Let $U_L={\mathcal{O}}_L^{\ast }$. For sufficiently large $r$, the equivariant p-adic logarithm identifies $U_L^r=1+{\mathfrak{p}}_L^r$ with ${\mathfrak{p}}_L^r$ additively. In the unramified case this is an induced regular lattice. Indeed choose a normal basis of the finite residue extension and lift its generating element. Its $m$ Galois conjugates reduce to a basis and hence form an ${\mathcal{O}}_K$-basis of ${\mathcal{O}}_L$; their coordinate determinant is a unit. Since a uniformizer of $K$ is also one of $L$, multiplication by its $r$th power gives the same regular-module structure on ${\mathfrak{p}}_L^r$. For a regular module, invariants are norm images and a vector of coefficient sum zero is a cyclic difference, so both indicated Tate cohomology groups vanish. Consequently $h_G(U_L^r)=1$. The quotient $U_L/U_L^r$ is finite, so $h_G(U_L)=1$.

Apply multiplicativity toIt gives $h_G(L^{\ast })=m$. Hilbert 90 makes ${\hat{H}}^{-1}(G,L^{\ast })$ trivial, so $[K^{\ast }:N_{L/K}{L}^{\ast }]=m$. On the other handso the quotient maps onto $\mathbb{Z}/m\mathbb{Z}$. As its order is already $m$, its unit kernel is trivial. A norm which is a unit comes from a unit, by the same valuation formula. ThereforeThe standard normal-basis theorem for finite fields and the convergent logarithm/exponential on sufficiently deep principal units are the ingredients used in evaluating the Herbrand quotient; the norm conclusion itself has been deduced here.