By Proposition 4 of § 3, no. 2, it suffices to show that the formula

defines a valuation on the ring $K[X]$. As

it follows that

for $P, Q$ in $K[X]$, equality holding if $w(P) \neq w(Q)$. We show that

for $P = \sum a_j X^j$ and $Q = \sum b_j X^j$. Let $i$ (resp. $k$) be the least of the integers $j$ such that $v(a_j) + j\xi$ (resp. $v(b_j) + j\xi$) attains its minimum; let $a$ (resp. $\beta$) denote this minimum. For $j, j'$ in $\mathbf{N}$,

whence $w(PQ) \geq a + \beta$ by (2). Consider now the term $c X^{i+k}$ of degree $i + k$ in $PQ$; then $c = \sum_{n \in \mathbf{Z}} a_{i+n} b_{k-n}$; by the choice of $i$ and $k$, the element

takes its minimum value $a + \beta$ once and once only with $n = 0$; hence $w(c X^{i+k}) = a + \beta$, whence, by (1),

We show first the uniqueness of w. Let $P = \sum_j a_j X^j$ be an element of $K[X]$. Then $w(a_j X^j) = v(a_j) + j\xi$, which shows that the monomials $a_j X^j$ such that $a_j \neq 0$ have distinct values for w. It follows that $w(P) = \inf_j (v(a_j) + j\xi)$, which shows both the uniqueness of w on $K[X]$ (hence also on $K(X)$) and the fact that the order group of w is $\Gamma + \mathbf{Z}\xi$. It is further seen that, if $P \neq 0$, we may write $P = a X^n (1 + u)$, where $a \in K^*, n \in \mathbf{N}, u \in K(X)$ and $w(u) > 0$; every element $R \neq 0$ of $K(X)$ can therefore be written in the form

where $b \in K^*, n \in \mathbf{Z}, u' \in K(X)$ and $w(u') > 0$; then $w(R) = v(b) + n \xi$, hence $w(R) = 0$ if and only if $v(b) = 0$ and $n = 0$; thus, when $w(R) = 0$, $R$ and $b$ are congruent modulo the ideal of $w$, which shows that the residue field of $w$ is equal to that of $v$.

Finally the existence of $w$ follows from Lemma 1.

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.

holds.

In fact, let $G_0 \subset G_1 \subset \ldots \subset G_n$ be a strictly increasing sequence of isolated subgroups of G. It is necessary to establish the inequality

It is obvious for $n = 0$. Suppose $n \geq 1$ and let us argue by induction on $n$. Applying the induction hypothesis to the group $G_{n-1}$ and its subgroup $H \cap G_{n-1}$, we obtain

Then we distinguish two cases:

(a) $H \cap G_{n-1} = H$, in other words $H \subset G_{n-1}$. Inequality (7) may be written

Now $G/G_{n-1}$ is a totally ordered group not reduced to 0; it is therefore not a torsion group and $r(G/G_{n-1}) \geq 1$. Whence by (4), $r(G/H) \geq r(G_{n-1}/H) + 1$. Substituting in (8), we certainly obtain the desired inequality (6).

(b) $H \cap G_{n-1} \neq H$. As $H \cap G_{n-1}$ is an isolated subgroup of $H$, we conclude that $h(H) > h(H \cap G_{n-1}) + 1$. On the other hand, obviously $r(G/H) \geq r(G_{n-1}/(H \cap G_{n-1}))$. Substituting in (7), we again obtain (6).

Let us write $r = r(G) = h(G)$. If $r = 0$, then $G = (0)$. If $r = 1$, the structure of finitely generated commutative groups shows that there is an isomorphism $j$ of G onto Z (Algebra, Chapter VII, § 4, no. 6, Theorem 3). Now Z has only two total orderings compatible with its group structure, namely the usual ordering and its opposite. Hence $j$ or $-j$ is an isomorphism of the ordered group G onto Z with the usual ordering.

Suppose now that $r \geq 2$ and let us argue by induction on $r$. Let H be an isolated subgroup of G of height $r - 1$. Then $r(H) + r(G/H) = r$ (formula (4)), $r(H) \geq h(H) = r - 1$ and $r(G/H) \geq h(G/H) = 1$ (Corollary to Proposition 3), whence $r(H) = r - 1$ and $r(G/H) = 1$. The induction hypothesis shows that H is isomorphic to $\mathbf{Z}^{r-1}$ ordered lexicographically and the case $r = 1$ shows that $G/H$ is isomorphic to Z. As Z is a free Z-module, H is a direct factor of G (Algebra, Chapter 11, § 1, no. 11, Proposition 21). The following lemma then shows that G is isomorphic (not canonically) to the lexicographical product $H \times (G/H)$, which completes the proof.

Let $j$ be an isomorphism of $(G/H) \times H$ onto G such that $j(0, x) = x$ for all $x \in H$ and $j(y, x)$ belongs to the coset of H. As $(G/H) \times H$ is totally ordered, it amounts to showing that $j$ is increasing (Set Theory, Chapter III, § 1, no. 12, Proposition 11). Let $(y, x)$ be an element $\geq 0$ of $(G/H) \times H$ ordered lexicographically. If $y > 0$, the coset of H containing $j(y, x)$ is an element $> 0$, whence $j(y, x) > 0$, for, otherwise, $y \leq 0$ (§ 4, no. 2, Proposition 3). If $y = 0$ and $x \geq 0$, then $j(y, x) = x \geq 0$. Hence $j$ is certainly increasing.

as order group.

Our assertion is obvious if $r+s = 0$. We argue by induction on $r+s$. If $r' \leq r, s' \leq s$ and $r'+s' < r+s$, the induction hypothesis shows that the hypotheses of Theorem 1 hold if $K$ is replaced by $K(x_1, \ldots, x_s, y_1, \ldots, y_{r'})$ and the families $(x_1, \ldots, x_s), (y_1, \ldots, y_r)$ by $(x_{s'+1}, \ldots, x_s), (y_{r'+1}, \ldots, y_r)$. The problem is therefore reduced to one of the two following cases:

(a) There is an element $x$ in the ring $v'$ such that $\overline{x}$ is transcendental over $k$; then it is necessary to show that $x$ is transcendental over $K$ and that the restriction of $v'$ to $K(x)$ admits $k(\overline{x})$ as residue field and $\Gamma$ as order group.

(b) There is an element $y$ in $K'$ such that the relations $nv'(y) \in \Gamma$ and $n \in \mathbf{Z}$ imply $n = 0$; it is necessary to show that $y$ is transcendental over $K$ and that the restriction of $v'$ to $K(y)$ admits $k$ as residue field and $\Gamma + \mathbf{Z}v'(y)$ as order group.

Now Proposition 1 of § 8, no. 1 shows that $x$ (resp. $y$) cannot be algebraic over $K$. The other assertions of (a) (resp. (b)) follow immediately by Proposition 2 (resp. Proposition 1) of no. 1.

holds.

Further, if $K'$ is a finitely generated extension of $K$ and equality holds in (9), then $\Gamma'/\Gamma$ is a finitely generated $\mathbf{Z}$-module and $k'$ is a finitely generated extension of $k$.

Let $r$ and $s$ be natural numbers such that $r \leq r(v'/v)$ and $s \leq s(v'/v)$; we show that $r+s \leq d(K'/K)$ and this will prove (9). By hypothesis there exist elements $x_1, \ldots, x, y_1, \ldots, y$, of $K'$ which satisfy the hypotheses of Theorem 1.

They are therefore algebraically independent over $K$, which shows the inequality $r + s < d(K'/K)$.

If $K'$ is a finitely generated extension of $K$, $d(K'/K)$ is finite, hence $s(v'/v)$ and $r(v'/v)$ are also finite; we denote them by $s$ and $r$. There exist elements $x_1, \ldots, x_s, y_1, \ldots, y_r$ of $K'$ which satisfy the hypotheses of Theorem 1. If $r + s = d(K'/K)$, these elements form a transcendence basis of $K'$ over $K$ and $K'$ is therefore a finite algebraic extension of $K'' = K(x_1, \ldots, y_r)$. Let $\Gamma''$ and $k''$ be the order group and residue field of the restriction of $v'$ to $K''$. By Theorem 1, $\Gamma''/\Gamma$ is a finitely generated $\mathbf{Z}$-module and $k''$ is a finitely generated pure extension of $k$. On the other hand, as $K'$ is a finite algebraic extension of $K$, $\Gamma'/\Gamma$ is a finite group and $k'$ is a finite algebraic extension of $k''$ (§ 8, no. 1, Lemma 2). This proves the corollary.

If there is equality in (10), there is equality in (9), whence the fact that $k'$ is a finitely generated extension of $k$ and that $\Gamma''$ is a finitely generated $\mathbf{Z}$-module. Further, comparing (9) and (10), it is seen that $h' - h = r(\Gamma'/\Gamma)$, whence $h' = r(\Gamma')$ and Proposition 4 (no. 2) then shows that $\Gamma'$ is isomorphic to $\mathbf{Z}^{h'}$ ordered lexicographically.

It is a series of special cases of Corollaries 1, 2, 3.