Proposition 2

Let $K$ be a field, $v$ a valuation on $K$, $\Gamma$ its order group and $k$ its residue field. There exists a unique valuation $w$ on $K(X)$ extending $v$ such that $w(X) = 0$ and the image $t$ of $X$ in the residue field $k'$ of $w$ is transcendental over $k$. The order group of $w$ is equal to that of $v$ and its residue field is $k(t)$.

To show the uniqueness of $w$, it will suffice for us to show that, if $P = \sum_j a_j X^j$ is a non-zero element of $K[X]$, then

We may divide $P$ by an element of $K^*$ and suppose that $v(a_j) \geq 0$ for all $j$ and that one of the $v(a_j)$ is zero. As $w(X) = 0$, $P$ then belongs to the ring of $w$; writing $\bar{a}_j$ for the canonical image of $a_j$ in $k$, the canonical image of $P$ in the residue field $k'$ is $\sum_j \bar{a}_j t^j$; as $t$ is transcendental over $k$ and one of the $\bar{a}_j$ is non-zero, this image is non-zero, whence We now show the existence of $w$. The formula $w(P) = \inf_j (v(a_j))$ (for $P = \sum_j a_j X^j$) defines a valuation $w$ on $K(X)$, by virtue of Lemma 1, and $w$ obviously has the same order group as $v$. Then $w(X) = 0$. We show that the canonical image $t$ of $X$ in the residue field $k'$ of $w$ is transcendental over $k$: if $\sum_j \bar{a}_j t^j = 0$, where $\bar{a}_j \in k$ for all $j$, then, denoting by $a$, a representative of $\bar{a}_j$ in the ring of $v$, $w \left( \sum_j a_j X^j \right) > 0$; whence $v(a_j) > 0$ for all $j$, hence $\bar{a}_j = 0$ for all $j$. We show finally that $k' = k(t)$: every element $R$ of $K(X)$ may be written $R = c \left( \sum_j a_j X^j \right) / \left( \sum_j b_j X^j \right)$, where $c, a_j, b_j$ are in $K$, $v(a_j) \geq 0$ and $v(b_j) \geq 0$ for all $j$, one of the $v(a_j)$ and one of the $v(b_j)$ being zero; then $w(R) \geq 0$ if and only if $v(c) \geq 0$; denoting by $f$ the canonical homomorphism of the ring of $w$ onto $k'$,

which proves our assertion.