Proposition 6

Let $\mathbf{K}$ be a field and $A$ a subring of $\mathbf{K}$. Suppose that $A$ is not a field. Then the following conditions are equivalent: (a) $A$ is the ring of a valuation of height 1 on $\mathbf{K}$; (b) $A$ is a valuation ring of $\mathbf{K}$ and has no prime ideals other than $(O)$ and $m(A)$; (c) $A$ is maximal among the subrings of $\mathbf{K}$ distinct from $\mathbf{K}$.

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).