If $B$ is a ring such that $A \subset B \subset K$ and $x \in K - B$, then $x \in K - A$, whence $x^{-1} \in m(A) \subset B$, which proves both that $B$ is a valuation ring of $K$ and that $m(B) \subset m(A)$; as $m(B) = m(B) \cap A$ is a prime ideal of $A$, we have shown (a) and (b). Moreover, $A_{m(B)} \subset B$; conversely, if $x \in B - A$, then $x^{-1} \in A$ and $x^{-1} \notin m(B)$ and hence $x \in A_{m(B)}$; thus $A_{m(B)} = B$. Finally let $\mathfrak{p}$ be a prime ideal of $A$; we write $B = A_{\mathfrak{p}}$; then $m(B) \cap A = \mathfrak{p}$. (Chapter II, § 2, no. 5, Proposition 11) and $m(B) \subset A$ by (b); hence $m(B) = \mathfrak{p}$, which shows that the mappings $\mathfrak{p} \mapsto A$, and $B \mapsto m(B)$ of the statement are inverse bijections.

The set of prime ideals of $A$ is totally ordered by inclusion ($\S$ 1, no. 2, Theorem 1 (e)) and the mapping $\mathfrak{p} \mapsto A_{\mathfrak{p}}$ reverses the inclusion relations.

If $A \in \mathscr{A}$, then $h_B(A) \in \mathscr{A}'$: for if $x' = h_B(x)$ (where $x \in B$) is an element of $\kappa(B) - h_B(A)$, then $x \notin A$, hence $x^{-1} \in A$ and $h_B(x)^{-1} \in h_B(A)$. On the other hand, for $A \in \mathscr{A}$, $A \supset m(B)$ (Proposition 1 (b)) and hence the mapping,

$A \mapsto h_B(A)$ is injective. Finally, let $A' \in \mathscr{A}'$ and $A = h_B(A') \subset B$; we shall show, which will complete the proof, that $A \in \mathscr{A}$; if $x \in K - A$, then either $x \notin B$, or $x \in B$; if $x \notin B$, then $x^{-1} \in m(B) \subset A$; if $x \in B$, then $h_B(x) \in \kappa(B)$ and $h_B(x) \notin A$, hence $h_B(x^{-1}) \in A$ and we conclude again that $x^{-1} \in A$; hence $A \in \mathscr{A}$.

Since the local ring $A'$ is a quotient of the local ring $A$, their residue fields are canonically isomorphic and the equation $h_A(x) = h_{A'}(h_B(x))$ holds for $x \in A$. On the other hand, if $x \in B - A$, then $h_B(x) \notin A'$ and the two sides of the equation are equal to $\infty$; the same is true if $x \in K - B$.

(a) The kernel of an increasing homomorphism of $G$ to an ordered group is an isolated subgroup of $G$.

(b) Conversely, let $H$ be an isolated subgroup of $G$ and $g$ the canonical homomorphism of $G$ onto $G/H$. Then $g(P)$ is the set of positive elements of an ordered group structure on $G/H$. Moreover, if $G$ is totally ordered, so is $G/H$.

(a) Let $f$ be an increasing homomorphism from $G$ to an ordered group; let $H$ denote the kernel off. If $0 \leq y \leq x$ and $x \in H$, then $0 \leq f(y) \leq f(x) = 0$, whence $f(y) = 0$, that is $y \in H$. Hence $H$ is isolated.

(b) Let $H$ be an isolated subgroup of $G$ and $g : G \to G/H$. Let $P' = g(P)$. Clearly $P' + P' \subset P'$. Also

for, if $x$ and $y$ are elements of $P$ such that $g(x) = -g(y)$, then $x + y \in H$, whence $x \in H$ and $y \in H$ since $H$ is isolated; hence $g(x) = g(y) = 0$. Thus $P'$ is the set of positive elements of an ordered group structure on $G/H$ (Algebra, Chapter VI, § 1, no. 3, Proposition 3). Finally, if $G$ is totally ordered, then $P \cap (-P) = G$, whence $P' \cup (-P') = G/H$ and therefore $G/H$ is totally ordered (loc. cit.).

Example (2) If we reconsider the example where $G$ is a lexicographic product $A \times B$ and $H = B$, the ordered group $G/H$ is canonically identified with $A$.

Given $H$, $v_B$ is defined up to equivalence and hence $B$ is determined uniquely. On the other hand, let $H$ be an isolated subgroup of $\Gamma_A$; considering $\Gamma_A / H$ as a totally ordered group (Proposition 3), the composite mapping

is a valuation on $K$ whose ring contains $A$.

Proposition 5 of no. 4 shows that (a) implies (b) and Proposition 1 of no. 1 shows that (b) implies (c). It remains to show that (c) implies (a). Suppose A is maximal among the subrings of K distinct from K. Let m be a maximal ideal of A and V a valuation ring of K dominating A_m (§ 1, no. 2, Corollary to Theorem 2); as m(V) ∩ A = m and m ≠ (0) (since A is not a field), V ≠ K, whence V = A, which proves that A is not the ring of a valuation v on K. This being so, v is of height 1 by Propositions 1 (no. 1) and 5 (no. 4).

This follows in fact from the following proposition:

Let x be a positive element of G and let H, be the set of y ∈ G such that there exists an integer n ≥ 0 satisfying |y| ≤ nx. It is easily verified that H, is an isolated subgroup of G and that every isolated subgroup of G containing x contains H,. Condition (a) is therefore equivalent to “H, = G for all x > 0”, that is, to condition (b).

Clearly (c) implies (b). Conversely, suppose condition (b) holds and let Q denote the set of elements >0 of G. Suppose first that Q has a least element x; for all y ∈ Q, let n be the least integer such that y ≤ nx; if y < nx, then also nx − y ≥ x, whence y ≤ (n − 1)x contrary to the choice of n; then y = nx, which shows that G = \mathbf{Z}x is isomorphic to \mathbf{Z} ⊂ \mathbf{R}. Suppose now that Q has no least element; we apply to the ordered set P = Q ∪ (0) Proposition 1 of General Topology, Chapter V, § 2 (which is possible, since condition (b) is just “Archimedes’ axiom”); it is seen that there exists a strictly increasing mapping f of P to \mathbf{R}_+ such that

for x ∈ P and y ∈ P; by linearity f can be extended to an isomorphism of G onto a subgroup of \mathbf{R}, which proves that (b) implies (c).

Suppose v is of height 1. Let x ∈ K be such that the x^n (n ≥ 0) are all contained in a finitely generated sub-A-module of K. There exists d ∈ A − {0} such that dx^n ∈ A for all n ≥ 0. Then v(d) + nv(x) ≥ 0, that is n(−v(x)) ≤ v(d) for all $n \geqslant 0$, whence $-v(x) \leqslant 0$ (Proposition 8 (b)) and $x \in A$. Thus A is completely integrally closed.

Suppose now that v is not of height 1. Then there exist $y \in m(A)$ and $t \in A$ such that $nv(y) < v(t)$ for all $n \geqslant 0$ (Proposition 8 (b)). Then $ty^{-n} \in A$ for all $n \geqslant 0$, but $y^{-1} \notin A$. Hence A is not completely integrally closed.

A completely integrally closed domain is not always an intersection of valuation rings of height 1 (Exercise 6).