Lemma 2

Let $K$ be a field, $L$ a finite extension of $K$ of degree $n$, $v'$ a valuation on $L$ and $v$ its restriction to $K$. Then the inequality

holds; in particular $e(v'/v)$ and $f(v'/v)$ are finite.

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 $y_k$.