Corollary

— Let E be an extension of K and let $(E_i)_{i \in I}$ be a family of subextensions of E such that $E = K \left( \bigcup_{i \in I} E_i \right)$. Suppose that each extension $E_i$ is abelian, then the same is true of E.

In the first place E is a Galois extension of K (V, p. 57, Prop. 1). If the field F is defined as in Prop. 1, then we have $E_i \subset F$ for all $i \in I$, whence $E = F$.

An extension E of K is said to be an abelian closure of K if it is an abelian extension of K and every abelian extension of K is isomorphic to a subextension of E. Prop. 1 implies the existence of an abelian closure of K: for let $K_s$ be a separable closure of K, with Galois group $\Gamma$ and let $(\overline{\Gamma}, \overline{\Gamma})$ be the closure of the derived group of $\Gamma$; denote by $K_{ab}$ the field of invariants of $(\overline{\Gamma}, \overline{\Gamma})$; since every separable algebraic extension of K is isomorphic to a subextension of $K_s$ (V, p. 45, Cor.). Prop. 1 shows that $K_{ab}$ is an abelian closure of K. The Galois group of $K_{ab}$ over K is canonically isomorphic to $\Gamma / (\overline{\Gamma}, \overline{\Gamma})$. Let us now show the uniqueness of abelian closures: let E, E' be two abelian closures of K; by definition there exist K-homomorphisms $u : E \to E'$ and $v : E' \to E$ and Prop. 1 (V, p. 52) implies that $v(u(E)) = E$ and $u(v(E')) = E'$, hence $u$ is a K-isomorphism of E onto E'. Any other K-isomorphism of E onto E' is of the form $u_1 = \sigma_0 \circ u$ with $\sigma_0 \in \mathrm{Gal}(E'/K)$; since $\mathrm{Gal}(E'/K)$ is commutative, the isomorphism $\sigma \mapsto u \circ \sigma \circ u^{-1}$ of $\mathrm{Gal}(E/K)$ onto $\mathrm{Gal}(E'/K)$ is independent of $u$; it is called the canonical isomorphism of $\mathrm{Gal}(E/K)$ onto $\mathrm{Gal}(E'/K)$.