Clearly $B_{p_i} \subset A_i$. To prove the converse inclusion we need the following lemma:

Assuming this lemma for a moment, we show how it implies that $A_1 \subset B_{p_1}$. Let $x$ be a non-zero element of $A_i$. We apply the lemma to $x$ and valuations $v_i$ associated with the $\mathbf{A}_i$. Then $v_i(z) \geq 0$ and $v_i(zx) \geq 0$ for all $i$, hence $z \in B$ and $zx \in B$. As $v_1(x) \geq 0$, $v_1(z) = 0$ and hence $z \notin p_1$. Hence $x = zx/z \in B_{p_1}$. The field of fractions of $B$ then contains $A_1$ and hence is $K$.

We now pass to the proof of the lemma. Let $I$ be the set of indices $i$ such that $v_i(x) \geq 0$. For all $i \in I$, let $\bar{x}_i$ denote the canonical image of $x$ in $\kappa(\mathbf{A}_i)$. For all $i \in I$ we construct a polynomialf, as follows: if there exists a polynomial $g(X)$ of the form (1) such that $g(\bar{x}_i) = 0$ in $\kappa(\mathbf{A}_i)$, we takef, to be such a polynomial; otherwise we take $f_i = 1$. Then we write $f(X) = 1 + X^2 \prod_{i \in I} f_i(X)$. It is obviously a polynomial of the form (1). If $i \in I$, then $f(x) \in A_i$ and alsof $(\bar{x}_i) \neq 0$ by construction; hence $f(x) \notin m(\mathbf{A}_i)$, $v_i(f(x)) = 0$ and $v(z) = 0$. If $i \notin I$, then $v_i(x) < 0$, whence $v_i(f(x)) = kv_i(x)$ (§ 3, no. 1, Proposition 1) and

(since $k \geq 2$). Whence the lemma.

If $p_i \subset p_j$ for $i \neq j$, $\mathbf{A}_i = B_{p_i} \supset B_{p_j} = \mathbf{A}_j$. It is then sufficient to apply Chapter II, § 3, no. 5, Corollary to Proposition 17.

Since the $p_i$ are maximal ideals of $B$, $A_i/m(A_i) = B_{p_i}/p_iB_{p_i} = B/p_i$ and it may therefore be assumed that $a_i \in B$ for all $i$. The corollary then follows from the fact that the canonical mapping from $B$ to $\prod_{i=1}^n (B/p_i)$ is surjective (Chapter 11, § 1, no. 2, Proposition 5).

For each index $i$ apply Corollary 1 to the family $(a_i)$ such that $a_i = 1$ and $a_j = 0$ for $j \neq i$.

We may confine our attention to the case where $A_i \not\subset A_j$ for $i \# j$. Let $V$ be a valuation ring of $K$ containing $B$. We write

There exists a maximal ideal $p_i$ of $B$ containing $p$, whence

If $v_i$ is improper, then $a_i = 0$ and the relation $v_i(x - a_i) \geq a_i$ is true for all $x \in K$. We may therefore assume that the $v_i$ are not improper.

Let $A_i$ be the ring of $v_i$, $B = \bigcap_{i=1}^n A_i$ and $p_i = m(A_i) \cap B$. By Proposition 1 of no. 1, the $a_i$ may be written $a_i = b_i/s$ ($b_i \in B, s \in B - \{0\}$); if we write $x = y/s$ and $\alpha_i' = \alpha_i + v_i(s)$, then $v_i(y - 6_i) \geq \alpha_i'$. This shows that we may assume that $a_i \in B$ for all $i$; we may also assume that $\alpha_i > 0$ for all $i$. Let $v_i$ be the set of $z \in \mathbf{K}$ such that $v_i(z) \geq a$, we write $q_i = v_i \cap B$. For $x \in B, v_i(x - a_i) \geq \alpha_i$ is equivalent to $x \equiv a_i (\mathfrak{q}_i)$. We therefore need to show that the canonical homomorphism $B \to \prod_{i=1}^n (B/q_i)$ is surjective, that is that $q_i + q_j = B$ for $i \neq j$ (Chapter 11, § 1, no. 2, Proposition 5). As the maximal ideals of B are the $\mathfrak{p}_i$ (Proposition 2), it will suffice for this to show that $q_i \notin \mathfrak{p}_j$ for $i \neq j$.

Suppose that there exists $i, j$ such that $q_i \subset \mathfrak{p}_j$ and $i \neq j$. We shall see shortly that the radical of $q_i$ is a prime ideal $p$ of B. Then $p \subset \mathfrak{p}_j$ and also $p \subset \mathfrak{p}_i$ since $\alpha_i > 0$ and hence $q_i \subset \mathfrak{p}_i$. Therefore $A_1 = B_s \subset B_p$ (no. 1, Proposition 1) and similarly $A_1 \subset B_r$. Now, as $v_i \neq (0)$ and $v_i = B_{\mathfrak{p}_i} q_i$ (Chapter 11, § 2, no. 4, Proposition 10), $q_i \neq (0)$, whence $p \neq (0)$ and $B_r \neq K$. This contradicts the hypothesis that $A_1$ and $A_1$ are independent.

It remains to show that $p$ is prime. Now this follows from the following lemma:

Suppose that $xy \in c$. Then there exists $n \geq 1$ such that $(xy)^n \in b$. Let $v$ be a valuation associated with A. If, for example, $v(x) \geq v(y)$, then

whence $x^{2n} \in b$ and $x \in c$.

We may assume that $A_i \neq K$ for all $i$. Then, there exists for all $i$ an $a_i \in K$ such that $v_i(a_i) = \gamma_i$ and an $a_r \in \Gamma_r$ such that $\gamma_i < a_r$. We apply Theorem 1 to these elements $a_i$: there exists $x \in K$ such that $v_i(x - a_i) > v_i(a_i)$; whence, as $x = a_i + (x - a_i)$, $v_i(x) = v_i(a_i) = \gamma_i$ ($§ 3$, no. 1, Proposition 1).

Suppose that the topologies $\mathscr{T}_v$ and $\mathscr{T}_{v'}$, defined by $v$ and $v'$, are identical. Since $\mathscr{T}_v$ is Hausdorff, the diagonal of $K^2$ is closed and hence $v$ and $v'$ are dependent (Corollary 2 to Theorem 1).

Conversely, suppose that $v$ and $v'$ are dependent. Then their rings $A$ and $A'$ are contained in the same ring $A''$ distinct from $K$ and $A''$ is the ring of a valuation $v''$ ($§ 4$, no. 1, Proposition 1). It suffices to show that the topology $\mathscr{T}_{v''}$ is identical with $\mathscr{T}_v$. Let $\Gamma$ and $\Gamma''$ be the order groups of $v$ and $v''$. There exists an increasing homomorphism $\lambda$ of $\Gamma$ onto $\Gamma''$ such that $v'' = A \circ v$ ($§ 4$, no. 3). If $a'' \in I?''$, let $a \in \lambda^{-1}(a'')$; the condition $v(x) \geq \alpha$ implies $v''(x) \geq a''$. Let $\beta \in \Gamma$ and $\beta'' = \lambda(\beta)$; the condition $v(x) \leq \beta$ implies $v''(x) \leq \beta''$ and hence the condition $v''(x) > \beta''$ implies $v(x) > \beta$. As $v$ and $v''$ are not improper, the inequalities in question define fundamental systems of neighbourhoods of 0 for $\mathscr{T}_v$ and $\mathscr{T}_{v''}$. Hence $\mathscr{T}_v = \mathscr{T}_{v''}$, which completes the proof.

Remarks (1) Proposition 3 shows that the relation '"$v$ and $v'$ are dependent"' is an equivalence relation. (2) Taking account of the relations between valuations of height 1 and ultrametric absolute values (§ 6, no. 2), Proposition 3 also follows, in the case of valuations of height 1, from the characterization of equivalent absolute values (General Topology, Chapter IX, § 3, no. 2, Proposition 5).

For $n = 2$ Proposition 4 follows from Definition 1. Suppose it holds for $n - 1$ valuations. Then there exists a subring A of K distinct from K and containing $A_1, \ldots, A_{n-1}$; there also exists a subring B $\neq K$ containing $A_{n-1}$ and $A_n$. As A and B contain $A_{n-1}$, they are comparable with respect to inclusion (§ 4, no. 1, Corollary to Proposition 1). The greater of these two therefore contains all the $A_i$.

Let $K_i$ denote the field K with the topology defined by $f_i$. The result to be proved is equivalent to the following: in the product $P = K_1 \times \cdots \times K_n$, the closure $\overline{D}$ of the diagonal D is equal to P. This is obvious for $n = 1$. Suppose that this point has been established in the case of k absolute values for $k < n$.

We show first that there exists, for $2 \leq h \leq n$, an element $x_h$ of K such that $f_1(x_h) < 1, f_2(x_h) > 1$ and $f_i(x_h) \neq 1$ for $3 \leq i \leq h$. We argue by induction on $h$. If $h = 2$, this follows from the fact that $f_1$ and $f_2$ are not equivalent. We therefore suppose that the existence of $x_{h-1}$ has been shown and prove that of $x_h$. If $f_h(x_{h-1}) \neq 1$, we may take $x_h = x_{h-1}$; if $f_h(x_{h-1}) = 1$, we choose $z \in K^*$ such that $f_h(z) \neq 1$ and $x_h = z(x_{h-1})^s$ solves the problem for s sufficiently large. We have thus proved the existence of $x_h$.

As the integer $q$ tends to infinity, $f_1(x_n^q)$ tends to 0, $f_2(x_n^q)$ tends to $+\infty$ and $f_i(x_n^q)$ tends to 0 or $+\infty$ for $i \geq 3$. Writing $y_q = x_n^q(1 + x_n^q)^{-1}$,

hence the sequence $(y_q)$ tends to 0 in $K_1$, to 1 in $K_2$ and to 0 or 1 in $K_i$ for $i \geq 3$. By changing the numbering of the $K_i$, it may therefore be assumed that there exists an integer $r$ ($1 \leq r < n$) such that $\overline{D}$ contains the point $(e_1, \ldots, e_n)$ where $e_i = 1$ for $1 \leq i \leq r$ and $e_i = 0$ for $r + 1 \leq i \leq n$. Now, $\overline{D}$ is a vector sub-K-space of P. Hence $\overline{D}$ contains the diagonals D' and D'' of and $P'' = K_{r+1} \times \cdots \times K_n$. By the induction hypothesis, $P' = \overline{D}'$ and $P'' = \overline{D}''$. Hence $\overline{D} = P$.