Proposition 2

*Let $K$ be afield, $v$ a valuation on $K$, $\hat{K}$ the completion of $K$ with respect to $v$, $\hat{v}$ the continuous extension of $v$ to $\hat{K}$ and $L$ a finite extension of $K$ of degree $n$. (a) Let $v'$ be a valuation on $L$ extending $v$; let $\hat{L}_{v'}$ denote the completion of $L$ with respect to $v'$ and $8'$ the continuous extension of $v'$ to $\hat{L}_{v'}$; identifying $\hat{K}$ with the closure of $K$ in $\hat{L}_{v'}$,

(b) Every set of pairwise independent valuations on $L$ extending a non-improper valuation $v$ is finite. Let $v'_1, \ldots, v'_s$ denote pairwise independent valuations on $L$ extending v such that every valuation on L extending v is dependent on one of the v_i'; let L_i be the field L with the topology defined by v_i' and $\hat{L}_i$ its completion; we write $n_i = [\hat{L}_i : \hat{K}]$. Then the canonical mapping

(extending by continuity the diagonal mapping $L \to \prod_{i=1}^s L_i$) is surjective, its kernel is the Jacobson radical of $\hat{K} \otimes_K L$ and

Let us first prove (a). Suppose that v is not improper. As v and $\hat{v}$ (resp. $v'$ and $\hat{v}'$) have the same order group and the same residue field (§ 5, no. 3, Proposition 5 (b) and (f)), (4) holds. We deduce (6) from it by means of Lemma 2. Finally the vector sub-$\hat{K}$-space of $\hat{L}_{v'}$ generated by L is closed (§ 5, no. 2, Corollary to Proposition 4) and everywhere dense and hence equal to $\hat{L}_{v'}$; this shows (5).

We now pass to (b). We may still assume that v is not improper. Let $(v_1', \ldots, v_r')$ be any finite family of pairwise independent valuations on L extending v. The image of L in $\prod_{i=1}^r L_i$, under the diagonal mapping is everywhere dense (§ 7, no. 2, Theorem 1) and $\prod_{i=1}^r L_i$ is dense in $\prod_{i=1}^r \hat{L}_i$. Hence the canonical image of $\hat{K} \otimes_K L$ in $\prod_{i=1}^r \hat{L}_i$ is everywhere dense. On the other hand this image is a vector sub-$\hat{K}$-space of $\prod_{i=1}^r \hat{L}_i$; as $\prod_{i=1}^r \hat{L}_i$ is of finite dimension over $\hat{K}$ by (5), the image of $\hat{K} \otimes_K L$ is closed (§ 5, no. 2, Corollary to Proposition 4) and hence equal to $\prod_{i=1}^r \hat{L}_i$. As the dimension of $\hat{K} \otimes_K L$ over K is n, $\sum_{i=1}^r n_i \leq n$. This shows in particular that the integer r is bounded above by n and shows the first assertion of (b).

We now take $(v_1', \ldots, v_s')$ as in the statement. The fact that

is surjective and relation (7) have already been shown. It remains to verify that the kernel n of $\phi$ is the Jacobson radical $t$ of $\hat{K} \otimes_K L$. As $\prod_{i=1}^s \hat{L}_i$ is semi-simple, $t \subset n$. On the other hand, for every maximal ideal m of $\hat{K} \otimes_K L$, the quotient field $L(m) = (\hat{K} \otimes_K L)/m$ is a composite extension of $\hat{K}$ and L over K (Algebra,

Chapter VIII, § 8, Proposition 1). There exists a valuation w on L(m) extending $\vartheta$ (§ 3, no. 3, Proposition 5); the restriction $v'$ of w to L extends $v$. As $[L(m): \hat{K}]$ is finite, $L(m)$ is complete with respect to w ($§ 5$, no. 2, Proposition 4). Now the closure of L in $L(m)$ is a field containing $\mathbf{K}$ and L and hence is equal to $L(m)$. Consequently $L(m)$ is identified with the completion $\hat{L}_{v'}$ and m is the kernel of the canonical mapping of $\hat{K} \otimes_K L$ onto $\hat{L}_{v'}$. Now, by hypothesis, there exists an index $i$ such that $v'$ and $v'_i$ are dependent; whence $L_{v'} = L_i$ ($§ 7$, no. 2, Proposition 3). Thus $n \subset m$, which proves that $n \subset r$ and completes the proof.