This is obvious.

Let us take natural numbers $r$ and $s$ respectively not greater than $e(v'/v)$ and $f(v'/v)$. It suffices to show that $rs \leq n$. In view of the definition of $r$, there exist elements $x_i$ of $L$ ($1 \leq i \leq r$) such that $v'(x_i) \not\equiv v'(x_j)$ (mod. $\Gamma_v$) for $i \neq j$. In view of the definition of $s$, there exist elements $y_k$ ($1 \leq k \leq s$) of the ring $A'$ of $v'$ whose canonical images $\bar{y}_k$ in $\kappa(A')$ are linearly independent over $\kappa(A)$; obviously $v'(y_k) = 0$ for all $k$. We shall show that the $rs$ elements $x_i y_k$ are linearly independent over $K$, which will certainly establish the inequality $rs \leq n$.

Suppose then that there exists a non-trivial linear relation of the form

Let us choose the indices $j, m$ so that

for every ordered pair $(i, k)$; then $a_{jm} \neq 0$. If $i \neq j$, then $v'(a_{ik} x_i y_k) = v'(a_{jm} x_j y_m)$ is impossible for this would imply

contrary to the choice of the $x_i$. Multiplying (3) by $(a_{jm} x_j)^{-1}$, we obtain a relation of the form

and $v'(b_k) \geq 0, v'(z) > 0$. Whence, in $\kappa(A')$, a relation of the form $\sum_k \bar{b}_k \bar{y}_k = 0$. As $b_m = 1$, this contradicts the hypothesis made on $y_k$.

Let $(L_\alpha)$ be the family of finite sub-extensions of $L$; let us write $\Gamma_\alpha = v'(L_\alpha^*)$. The group $\Gamma_{v'}$ is the union of the right directed family consisting of the $\Gamma_\alpha$; as the groups $\Gamma_\alpha/\Gamma_v$ are finite (Lemma 2), $\Gamma_{v'}/\Gamma_v$ is a torsion group. The argument is similar to prove that $\kappa(A')$ is an algebraic extension of $\kappa(A)$.

This follows from Proposition 1 and the following lemma:

Clearly $H' \in \mathscr{G}'$ implies $H' \cap G \in \mathscr{G}$. Now let $H \in \mathscr{G}$; let $H'$ denote the set of $x' \in G'$ such that there exists $h \in H$ satisfying $-h \leq x' \leq h$; it is immediately verified that $H'$ is an isolated subgroup of $G'$; then $H' \cap G = H$ since $H$ is isolated; hence the mapping $H' \mapsto H' \cap G$ is surjective. Suppose finally that $G'/G$ is a torsion group; let $H'_1$ and $H'_2$ be two isolated subgroups of $G'$ such that $H'_1 \cap G = H'_2 \cap G$; then, for example, $H'_1 \supset H'_2$ (cf. § 4, no. 4); then $H'_1/H'_2$ is a totally ordered group and is isomorphic to a quotient group of $H'_1/(H'_1 \cap G)$ which itself is identified with a subgroup of $G'/G$; hence $H'_1/H'_2$ is a torsion group and therefore reduces to 0.

(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.

This follows since $\hat{K} \otimes_K L = L$.

is an isomorphism.

The Jacobson radical of $\hat{K} \otimes_K L$ is then zero (Algebra, Chapter VIII, § 7, no. 3, Theorem 1).

(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 $i = j$.

In the first case, let x be an element > 0 in G. The set of $y \in G$ such that $0 < y < x$ is infinite and hence there exist two elements of this set which are distinct and congruent modulo H; their difference is an element z of H such that $0 < z < x$. Hence every major subset which contains all the strictly positive elements of H contains x and hence all the elements > 0 of G.

In the second case, let $x$ be the least element >0 of G and let $n$ be the least integer >0 such that $nx \in H$. Clearly $n = (G : (G \cap H))$. On the other hand, writing $M(y)$ for the set of $z \in G$ such that $y \leq z \ (y \in G)$, it is immediately seen that the major sets of Definition 4 are just $M(x), M(2x), \ldots, M(nx)$.

In particular, $\varepsilon(G, H) \leq (G : H)$.

The ideals of $A'$ containing $mA'$ and distinct from $A'$ correspond to the major subsets of $\Gamma_w$ consisting of elements >0 and containing the elements >0 of $\Gamma_v$ (§ 3, no. 5, Corollary to Proposition 7). They are therefore equal in number to $\varepsilon(w/v)$ and, as they form a totally ordered set under inclusion, this number is equal to the length of the quotient ring $A'/mA'$. Now a module of length 1 over $A'$ is a 1-dimensional vector space over $A'/m'$ and hence a module of length $f(w/v)$ over $A$; hence, as $A'/mA'$ is of length $\varepsilon(w/v)$ over $A'$, it is of length $\varepsilon(w/v) f(w/v)$ over $A$, that is over $A/m$.

Let $A_i$ be the ring of $v'_i$; then $A_i = B_{m_i}$ where $m_i$ runs through the family of maximal ideals of $B$ (no. 3, Remark). Let $q_i$ be the saturation of $mB$ with respect to $m_i$ (Chapter II, § 2, no. 4). By Chapter V, Corollary 3 to Proposition 1, no. 1, § 2, the canonical homomorphism $B/mB \to \prod_{i=1}^s B/q_i$ is an isomorphism and $m_i$ is the only maximal ideal of B containing $q_i$. Hence $B/q_i$ is canonically isomorphic to $(B/q_i)_{m_i}$ (Chapter II, § 3, no. 3, Proposition 8), that is to

The equivalence of (a) and (b) follows from Lemma 1, § 3, no. 6. Clearly (b) implies (c) (Algebra, Chapter 11, § 1, no. 5, formula (19)). The equivalence of (c) and (d) follows from the Corollary to Proposition 5. It remains to show that (c) implies (b).

In general, if M is an A-module, we shall denote by V(M) the vector space $M/mM$ over $A/m$. Hypothesis (c) means that $\dim(V(B)) = n$. Let $x_1, \ldots, x_n$ be elements of B whose canonical images in V(B) form a basis of V(B) and let L $\subset B$ be the sub-A-module which they generate. As L is torsion-free and finitely generated, it is free (§ 3, no. 6, Lemma 1). We shall see that $B = L$. Let $y \in B$; we write $M = L + Ay$; this is also a free A-module. The canonical injections $L \to M \to B$ give canonical homomorphisms $V(L) \to V(M) \to V(B)$. As the ranks of L and M are $\leq n$, so are the dimensions of V(L) and V(M). Now, by hypothesis, $V(L) \to V(B)$ is surjective and $V(B)$ is n-dimensional hence $V(L)$ and $V(M)$ are n-dimensional and $V(L) \to V(M)$ is surjective. As M is finitely generated, $L \to M$ is surjective (Chapter II, § 3, no. 2, Corollary 1 to Proposition 4), whence $L = M, y \in L$ and $B = L$. Hence B is free.

Remark (1) If v is discrete, $\varepsilon(v'_i/v) = \varepsilon(v'_i/v)$ (no. 4) and condition (d) reduces to $\sum_{i=1}^s \varepsilon(v'_i/v) f(v'_i/v) = n.$

The integral closure B of A is then a free A-module of rank $n$, since A is a principal ideal domain (Chapter V, § 1, no. 6, Corollary 2 to Proposition 18).

adic topology (where $m = m(A)$); the ring A is complete, for it is closed in K. We conclude that, since $A'/mA'$ is a finite-dimensional vector ($A/m$)-space (no. 4, Proposition 4), A' is a finitely generated A-module (Chapter III, § 2, no. 9, Corollary 3 to Proposition 12). It is therefore free and $e(v'/v) f(v'/v) = n$ by

(no. 2, Proposition 2). *is bijective. For all $x \in L$, the characteristic polynomial $P_{C_{L/K}}(x; X)$ is equal to the product of the characteristic polynomials $P_{C_{\hat{L}_i/\bar{K}}}(x; X)$ ($1 \leq i \leq s$); in particular,

(The last relation in (9) is meaningful, for we may obviously assume that the $v_i'$, which are of height 1 by Corollary 2 to Proposition 1 of no. 1, take, as does v, their values in a subgroup of $\mathbf{R}$.)

As no two of the $v'_i$ are equivalent and they are of height 1, they are independent and Proposition 2 of no. 2 shows therefore that $e(v'_i/v) f(v'_i/v) \leq n_i$ for all $i$ and $\sum_{i=1}^s n_i \leq n$. The first assertion therefore follows from these inequalities and the relation $\sum_{i=1}^s e(v'_i/v) f(v'_i/v) = n$. Under the isomorphism $\phi$ the endomorphism $z \mapsto z(1 \otimes x)$ of $\hat{K} \otimes_K L$ (for $x \in L$) is transformed into the endomorphism of $\prod_{i=1}^s \hat{L}_i$ leaving invariant each of the factors and reducing on each factor to multiplication by $x$ ($L$ being canonically imbedded in its completion $\hat{L}_i$); whence the assertion relating to the characteristic polynomial of $x$ and the first two formulae of (9). Finally, let E be a finite quasi-Galois extension of $\hat{K}$, containing $\hat{L}_i$; as $\hat{K}$ is complete and $\hat{\nu}$ of height 1, there exists only one valuation (upto equivalence) $w$ on E extending $\hat{\nu}$ (no. 2, Corollary 1 to Proposition 2); then, for every $\hat{K}$-automorphism $\sigma$ of E, $w(\sigma(x)) = v'_i(x)$. Therefore