Proposition 12

Let K be a field, f an absolute value on K, L a quasi-Galois extension of K and $f'$ and $f''$ two extensions off to L. Then there exists a K-automorphism s of L such that $f'' = f' \circ s$.

If $f$ is ultrametric, Corollary 1 to Proposition 7 (no. 6) shows that there exists a K-automorphism s of L such that $f''$ and $f' \circ s$ are equivalent absolute values; then there exists a real number $a > 0$ such that $f''(x) = (f'(s(x)))^a$ for all $x \in L$. If $f$ is not improper, take $x \in K^*$ such that $f(x) \neq 1$, which shows that $a = 1$. If $f$ is improper, so are $f'$ and $f''$ (Corollary 2 to Proposition 1, no. 1) and s may be taken to be the identity automorphism.

If $f$ is not ultrametric, there exist Q-isomorphisms $u', u''$ of L onto subfields of $\mathbf{C}$ and real exponents $a' > 0, a'' > 0$ such that $f'(x) = |u'(x)|^{a'}$ and

for all $x \in L$ ($§ 6$, no. 4, Theorem 2). Taking $x = 2$, it is seen that $a' = a''$. The restrictions of $u'$ and $u''$ to K extend by continuity to isomorphisms $u_1$ and $u_2$ of $\hat{K}$ onto $\mathbf{R}$ (resp. $\mathbf{C}$). Then $u_2 \circ \bar{u}_1^{-1}$ is an automorphism of the valued field $\mathbf{R}$ (resp. $\mathbf{C}$) and is therefore the identity (resp. the identity or the automorphism $c : \zeta \to \bar{\zeta}$). Replacing if need be $u'$ by $c \circ u'$, it is seen that the restrictions of $u'$ and $u''$ to K may be assumed to coincide. Identifying K with a subfield of $\mathbf{C}$ by means of this common restriction, $u'$ and $u''$ are K-isomorphisms of L onto subfields of $\mathbf{C}$. As L is a quasi-Galois extension of K, there exists a K-automorphism s of L such that $u'' = u' \circ s$; since $a' = a''$, we deduce immediately that $f'' = f' \circ s$.

Remark (3). If $\hat{K}$ is isomorphic to $\mathbf{R}$, Proposition 12 shows that all the completions $\hat{L}_t$ of $L$ (in the notation of Proposition 11) are isomorphic to one another. Thus, with the notation of Remark 2 above, either $r_1 = n$ and $r_2 = 0$, or $r_1 = 0$ and $2r_2 = n$.