Proposition 11

*Let $K$ be a field, $f$ a non-ultrametric absolute value on $K$, $\hat{K}$ the completion of $K$ with respect to $f$, $\hat{f}$ the continuous extension off to $\hat{K}$ and $L$ a finite extension of $K$ of degree $n$.

(a) Let $f'$ be an absolute value on $L$ extending $f$; let $\hat{L}_{f'}$ denote the completion of $L$ with respect to $f'$ and let $\hat{K}$ be identified with the closure of $K$ in $\hat{L}_{f'}$; then $[\hat{L}_{f'} : K] \leq n$.

(b) The absolute values on $L$ extending $f$ are finite in number. If they are denoted by $f'_1, \ldots, f'_s$ and the completion of $L$ with respect to $f'_i$ by $\hat{L}_i$, the canonical mapping

is an isomorphism and

(10)

The proof is the same as that for the analogous assertions in Proposition 2 (no. 2). The references

§ 7, no. 2, Theorem 1; § 5, no. 2, Corollary to Proposition 4 should be replaced by the following

§ 7, no. 3, Theorem 2; Topological Vector Spaces, Chapter I, § 2, no. 3, Corollary 1 to Theorem 2.

Observe that two extensions off to L which define the same topology are equal (General Topology, Chapter IX, § 3, no. 2, Proposition 5). Finally, as $f$ is not ultrametric, K is of characteristic 0 and hence the Jacobson radical of $\hat{K} \otimes_K L$ is zero.

Remarks (1) Proposition 11 (b) shows that every composite extension of $\hat{K}$ and L over K is isomorphic to one of the completions $\hat{L}_i$ and that these are composite extensions no two of which are isomorphic. (2) We know that the completions $\hat{K}$ and $\hat{L}_i$ are isomorphic to $\mathbf{R}$ or $\mathbf{C}$ ($§ 6$, no. 4, Theorem 2). If $\hat{K}$ is isomorphic to $\mathbf{C}$, so is $\hat{L}_i$ for all $i$ and (10) shows that the number of extensions $f_i'$ is equal to n. If $\hat{K}$ is isomorphic to $\mathbf{R}$ (for example if $K = \mathbf{Q}$), let $r_1$ (resp. $r_2$) denote the number of indices $i$ such that $\hat{L}_i$ is isomorphic to $\mathbf{R}$ (resp. $\mathbf{C}$); then (10) may be written: