Lemma 1

Let $v_i$ ($1 \leq i \leq n$) be ualuations on the jield $K$ and $x \in K^*$. Then there exists a polynomialf $(X)$ of theform

such that $f(x) \neq 0$ and the element $z = f(x)^{-1}$ enjoys the following propertiesfor $1 \leq i \leq n$:

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.