Theorem 4

*a) The mapping $H \mapsto K(H^{1/n})$ is an inclusion preserving bijection of the set of subgroups of $K^*$ containing $K^{*n}$ onto the set of abelian subextensions of exponent dividing n of $\Omega$. The inverse mapping is $L \mapsto L'' \cap K^*$.

b) For every subgroup H of $K^*$ containing $K^{*n}$ the homomorphism

is bijective, and it is a homeomorphism when the group $\mathrm{Hom}(H/K^{*n}, \mu_n(K))$ is equipped with the topology of simple convergence.

c) Let H be a subgroup of $K^*$ containing $K^{*"n}$. For every $a \in H/K^{*n}$ let $\theta_a$ be an element of $\Omega$ such that $\theta_a^n$ is a representative of a in H. Then the $\theta_a, a \in H/K^{*n}$, form a basis of the vector K-space $K(H^{1/n})$. In particular,

A) For every abelian extension L of K of exponent dividing n let us put $H_L = L'' \cap K^*$. If $[L:K]$ is finite, the homomorphism $k'_L$ of $\mathrm{Gal}(L/K)$ into $\mathrm{Hom}(H_L, \mu_n(K))$ is bijective by Prop. 9 and V, p. 88, Prop. 8.

Since every abelian extension of K of exponent dividing $n$ is an increasing directed union of abelian subextensions of finite degree of exponent dividing $n$, we deduce by passage to inverse limits that $k'_L$ is a homeomorphism of topological groups for every abelian extension L of K dividing n.

B) Let L be an abelian extension of finite degree, of exponent dividing $n$, of K, and let $L' = K(H_L^{1/n})$; this is a subextension of L ; moreover $H_{L'}$ contains $H_L$ and so is equal to it. Since the homomorphisms $k'_L$ and $k'_{L'}$ are bijective by A) and $H_L = H_{L'}$, the groups Gal$(L/K)$ and Gal$(L'/K)$ have the same order and so are equal. This shows that $L' = L$, and hence that $L = K(H_L^{1/n})$. If L is an abelian extension of exponent dividing $n$ of K, we have $K(H_L^{1/n}) = L$, because $K(H_L^{1/n})$ is a subextension of L which contains every subextension of finite degree of L.

C) Let H be a subgroup of $K^*$ containing $K^{*n}$; put $L = K(H^{1/n})$, then this is an abelian extension of K of exponent dividing $n$ (V, p. 88, Lemma 5). We have $H \subset H_L$, whence we obtain an exact sequence of commutative groups annihilated by $n$

From this we obtain an exact sequence

where $\iota$ is the restriction homomorphism.

If we identify $\operatorname{Hom}(H_L/K^{*n}, \mu_n(K))$ with Gal$(L/K)$ by means of the isomorphism $k'_L$, the kernel of $\iota$ is identified with the set of $\sigma \in \operatorname{Gal}(L/K)$ such that $\sigma(\theta) = \theta$ for all $\theta \in H^{1/n}$. It follows that $\iota$ is injective, hence $\operatorname{Hom}(H_L/H, \mu_n(K))$ is reduced to (1) ; by Cor. 1 of V, p. 87 we thus have $H = H_L$. This completes the proof of a) and b).

D) Let us prove c). If $a, b \in H$, we have $\theta_a \theta_b / \theta_{ab} \in K$. It follows that the vector subspace of $K(H^{1/n})$ generated by the $\theta_a$ is stable under multiplication and so coincides with $K(H^{1/n})$. It only remains to show that the $\theta_a$ are linearly independent ; to do this we may evidently assume that $H/K^{*n}$ is finite ; then $[K(H^{1/n}) : K] = (\operatorname{Gal}(K(H^{1/n})/K) : \{1\}) = (H : K^{*n})$ by b) and Cor. 2 of V, p. 87 ; since the number of $\theta_a$ is equal to $(H : K^{*n})$ and they generate the vector K-space $K(H^{1/n})$, they are linearly independent.