Proposition 5

Let $K$ be a field, $v$ a valuation on $K$ and $K'$ an extension of $K$. There exists a valuation $v'$ on $K'$ whose restriction to $K$ is equivalent to $v$.

Let $\Gamma_v$ and $\Gamma_{v'}$ be the order groups of $v$ and $v'$. Since the restriction of $v'$ to $K$ is equivalent to $v$, there exists an isomorphism $A$ of $\Gamma_v$ onto a subgroup of $\Gamma_{v'}$, such that $v' = A \circ v$ on $K$. If $\Gamma_v$ is identified with $\lambda(\Gamma_v)$ by means of $\lambda$, it is seen that $v'$ extends $v$.

Note that $\Gamma_{v'}$ is in general distinct from $\lambda(\Gamma_v)$ and the equivalence class of $v'$ is not necessarily unique. We shall return to this in § 8.

Translating Theorem 3 of § 1, no. 3 (or Proposition 6 of § 2, no. 5), we obtain: