We’ll drop
\(\Gal(L/K)\) from the notation, because it’s the same group throughout the proof. Keep in mind that an unramified extension is always Galois and cyclic, so we can apply periodicity of Tate groups (
Theorem 3.4.1) and Herbrand quotients.
Consider the short exact sequence
\begin{equation*}
0 \to U_L \to A_L \to A_L/U_L \to 0\text{.}
\end{equation*}
In this sequence, \(A_L/U_L\) is isomorphic to \(Z = \im(v)\) with trivial group action, so \(H^0_T(Z) = Z/\Norm(Z)\) is cyclic of order \([L:K]\) generated by \(\pi_L\) and \(H^{-1}_T(Z) = \ker(\Norm)\) is trivial. The long exact sequence in Tate groups looks like
\begin{gather*}
1 = H^{-1}_T(A_L/U_L) \to H^{0}_T(U_L) \to H^{0}_T(A_L) \\
\to H^{0}_T(A_L/U_L) \to H^{1}_T(U_L) \to H^1_T(A_L) = 1
\end{gather*}
and by the long exact sequence, the two ends of the middle arrow \(H^0_T(A_L) \to H^0_T(A_L/U_L)\) have the same order. It is thus enough to show that \(H^1_T(U_L) = 1\text{,}\) as then this map will be an isomorphism.
Here is where we use that
\(L/K\) is unramified, not just cyclic: any uniformer of
\(K\) is also a uniformizer of
\(L\text{,}\) which allows us to split the surjection
\(A_L \to A_L/U_L\) of
\(\Gal(L/K)\)-modules (as in
Corollary 4.2.6). This in turn means that
\(H^1_T(U_L)\) is a direct summand of
\(H^1_T(A_L)\text{,}\) and the latter vanishes by the class field axiom.