Corollary 4

Under the hypotheses of Corollary 3, if L is a separable extension of K, each of the $\hat{L}_i$ is a separable extension of $\hat{K}$. If further L is a Galois extension of K with Galois group $\mathscr{G}$ and $\mathscr{G}_i$ denotes the decomposition group of the ideal of $v'_i$ in B (Chapter V, § 2, no. 2, Definition 2), then $\hat{L}_i$ is a Galois extension of $\hat{K}$ whose Galois group is isomorphic to $\mathscr{G}_i$.

Clearly $\hat{L}_i = \hat{K}(L)$; hence, if L is separable over K, $\hat{L}_i$ is separable over $\hat{K}$ (Algebra, Chapter V, § 7, no. 6, Proposition 10). Suppose now that L is Galois. Every automorphism $\sigma \in \mathscr{G}_i$ is continuous on L with the topology defined by $v'_i$, the fact that no two of the ideals of the $v'_i$ are comparable with respect to inclusion (§ 7, no. 2, Corollary 1 to Theorem 1) necessarily implying that $v'_i = v'_i \circ \sigma$ by definition of $\mathscr{G}_i$; hence $\sigma$ may be extended by continuity to a $\hat{K}$-automorphism $\hat{\sigma}$ of $\hat{L}_i$. This proves that the number of $\hat{K}$-automorphisms of $\hat{L}_i$ is at least equal to $\mathrm{Card}(\mathscr{G}_i)$. But as the valuations $v'_i$ are pairwise conjugate under $\mathscr{G}$ (Chapter V, § 2, no. 3, Proposition 6), $s = (\mathscr{G}_i)$, whence

and on the other hand $n = sn$, by Corollary 3; this proves that $\hat{L}_i$ is a Galois extension of $\hat{K}$ and that the extensions by continuity of the automorphisms $\sigma \in \mathscr{G}_i$ are the only $\hat{K}$-automorphisms of $\hat{L}_i$.

Remark (2). Part of the above results extends to the case of valuations on a not necessarily commutative field K (cf. § 3, no. 1). Let L be an extension field of K and let $v'$ be a valuation on L, $v$ its restriction to K and A' and A the respective rings of the valuations $v'$ and $v$; then there is defined a ramification index $e(v'/v)$ as in no. 1; on the other hand, $\kappa(A)$ is identified with a subfield of $\kappa(A')$ and the (left) residue rank of $v'$ with respect to $v$ is defined to be the number $f(v'/v)$ equal to the dimension of the left vector $\kappa(A)$-space $\kappa(A')$, if this dimension is finite, and $+\infty$ in the opposite case. Then, if L is a left vector K-space of finite dimension $n$, Lemma 2 of no. 1 and its proof go over unchanged. Moreover, if K is complete with respect to $v$, the assertions of Corollary 2 to Theorem 2 of no. 5 (other than the existence of $u'$) are also valid ($n$ denoting the dimension of L as a left vector K-space) with the following proof:

In the first place the topology defined by $v'$ on L is Hausdorff and compatible with its left vector K-space structure and hence two extensions of $v$ to L give the same topology on L ($§ 5$, no. 2, Proposition 4), which proves that these extensions are the same up to equivalence ($§ 6$, no. 2). We show next that, if $m = m(A)$, $A'/mA'$ is a left vector $(A/m)$-space of dimension $e(v'/v) f(v'/v)$. Writing $e = e(v'/v)$, we may assume that $v(K^*) = Z$ and $v'(L^*) = e^{-1}Z$; let $u'$ be an element of L such that $v'(u') = e^{-1}$ and $u$ an element of K such that $v(u) = 1$; hence $u = z{u'}^e$, where $z \in L$ is such that $v'(z) = 0$. As m is generated by $u$ (as a left or right ideal of A), $mA' = {u'}^eA' = A'{u'}^e$ and it suffices to prove that, for $0 \leq k \leq e - 1$, $A'{u'}^k/A'{u'}^{k+1}$ is a left vector $(A/m)$-space of dimension $f(v'/v)$. But $t \mapsto t{u'}^k$ is an isomorphism of the left A-module A onto the left A-module $A'{u'}^k$ mapping $A'u'$ to $A'{u'}^{k+1}$ and which therefore gives by taking quotients an $(A/m)$-isomorphism of $A'/A'u'$ onto $A'{u'}^k/A'{u'}^{k+1}$, whence our assertion by definition of $f(v'/v)$, $u'$ generating the maximal ideal of $A'$. The proof is completed as when K and L are commutative (the fact that a finitely generated torsion-free A-module is free being proved as in $§ 3$, no. 6, Lemma 1).