Theorem 1

Let K be a field, v a valuation on K and L a finite extension of K of degree n. Then:

(a) Every complete system $(v'_i)_{i \in I}$ of extensions of v to L is finite.

(b) $\sum_{i \in I} e(v'_i|v) f(v'_i|v) \leq n$ and a fortiori $\operatorname{Card}(I) \leq n$

(c) No two of the rings of the $v'_i$ are comparable with respect to inclusion.

Since the theorem is trivial if $v$ is improper, we shall assume that $v$ is not improper. Let $(v'_1, \ldots, v'_s)$ be any finite family of valuations on L extending $v$, no

THE RELATION $\sum_i e_i f_i \leq n$

two of which are equivalent. We shall first prove that $\sum_{i=1}^s e(v'_i/v)f(v'_i/v) \leq n$. This will prove (a) and (b).

We argue by induction on s and suppose therefore that the inequality has been established for the case of 0, 1, ..., s - 1 valuations. We distinguish two cases.

(1) Suppose that there exist at least two independent valuations $v'_i$. Then there exists (§ 7, no. 2, Remark 1) a partition $[1, s] = I_1 \cup \ldots \cup I_t$ of $[1, s]$ such that: (i) for $v'_i$ and $v'_j$ to be dependent, it is necessary and sufficient that $i$ and $j$ belong to the same $I_k$; (ii) Card$(I_k) < s$ for all $k$.

We choose in each $I_k$ an index $i(k)$. Let $\hat{L}_{i(k)}$ denote the completion of $L$ with respect to $v'_{i(k)}$ and $n(k) = [\hat{L}_{i(k)} : \hat{K}]$. For all $i \in I_k$, $v'_i$ defines on $L$ the same topology as $v'_{i(k)}$ (§ 7, no. 2, Proposition 3) and hence may be extended to a valuation $\bar{v}'_i$ on $\hat{L}_{i(k)}$ whose restriction to $\hat{K}$ is 0. Since no two of the $v'_i$ for $i \in I$, are equivalent, the same is true of the $\bar{v}'_i$. The induction hypothesis applied to the ordered pair $(\hat{K}, \hat{L}_{i(k)})$ shows, by virtue of Proposition 2 (a), formula (4), that $\sum_{i \in I_k} e(v'_i/v)f(v'_i/v) \leq n(k)$. As $\sum_{k=1}^t n(k) \leq n$ (Proposition 2 (b), formula (7)), certainly $\sum_{i=1}^s e(v'_i/v)f(v'_i/v) \leq n$.

(2) We now pass to the case where any two of the $v'_i$ are dependent. Let $A'_i$ be the ring of $v'_i$ ($1 \leq i \leq s$); writing $A$ for the ring of $v$, $A'_i \cap K = A$ for all $i$. Let $B'$ be the subring of $L$ generated by $A'_1, \ldots, A'_s$; we write $B = B' \cap K$; then $B \supseteq A$. Then $B$ is the ring of a valuation $w$ on $K$ and $B'$ the ring of a valuation $w'$ which is not improper and extends $w$ (§ 7, no. 2, Proposition 4); the field $\kappa(B')$ is an extension of $\kappa(B)$ of degree $f(w'/w)$. Consider the canonical images $\bar{A}'_i, \bar{A}$ of $A'_i$ and $A$ in $\kappa(B')$; then $\bar{A}$ is the ring of a valuation $\bar{v}$ on $\kappa(B)$ and the $\bar{A}'_i$ are rings of valuations $\bar{v}'_i$ on $\kappa(B')$ extending $\bar{v}$. As the $A'_i$ generate $B'$, the $\bar{A}'_i$ generate $\kappa(B')$ and hence the $\bar{v}'_i$ are not all dependent (§ 7, no. 2, Proposition 4). From the first part of the proof,

and hence

The proof of (a) and (b) will therefore be completed if we prove that

For this, we note that $v$ and $\bar{v}$ (resp. $v'_i$ and $\bar{v}'_i$) have the same residue field (§ 4, no. 1, Corollary to Proposition 2); this proves the first equation. For the second there is, by virtue of the Remark in §4, no. 3, the following commutative diagram, where the rows are exact sequences and the vertical arrows represent canonical injections:

We deduce that there is an exact sequence

by Chapter I, § 1, no. 4, Proposition 2, which proves the second formula (8).

To complete the proof of Theorem 1, it remains to prove (c). If the ring of $v_i'$ contains that of $v_j'$, $\Gamma_{v_i'}$ is identified with a quotient group $\Gamma_{v_j'}/H$, $H$ being an isolated subgroup (§ 4, no. 3). As the composite canonical mapping

is injective, $H \cap \Gamma_v = \{0\}$, whence $H = \{0\}$ (Lemma 3, no. 1). Then $v_i'$ and $v_j'$ are equivalent, whence $i = j$.