Lemma 5

— $K(A^{1/n})$ is an abelian extension of K of exponent dividing n.

Since the polynomials $X^n - a, a \in A$ are separable over K, $L = K(A^{1/n})$ is a separable, hence a Galois extension of K. Let $\sigma \in \mathrm{Gal}(L/K)$ and let $0 \in \Omega$ be such that $\theta^n \in A$. We have $\sigma(\theta)^n = \theta^n$; hence there exists $\zeta \in \mu_n(\Omega) = \mu_n(K)$ such that $\sigma(\theta) = \zeta \theta$; this implies that $\sigma^n(\theta) = \zeta^n \theta = 0$, whence $\sigma^n = 1$. If $\sigma'$ is another element of $\mathrm{Gal}(L/K)$ there exists $\zeta' \in \mu_n(K)$ such that $\sigma'(\theta) = \zeta' \theta$, whence $\sigma'\sigma(\theta) = \zeta \zeta' \theta = \sigma\sigma'(\theta)$ and so $\sigma'\sigma = \sigma\sigma'$.