Theorem 1

Let A be an integral domain and K its field of fractions. The following conditions are equivalent: (a) A is a Dedekind domain; (b) A is a Krull domain and every non-improper valuation on K which is positive on A is equivalent to an essential valuation of A; (c) A is a Krull domain and every fractional ideal $\mathfrak{J} \neq (0)$ of A is divisorial; (d) every fractional ideal $3 \neq (0)$ of A is invertible; (e) A is a Noetherian integrally closed domain and every non-zero prime ideal of A is maximal; (f) A is Noetherian and, for every maximal ideal m of A, $A_m$ is either a field or a discrete valuation ring; (g) A is Noetherian and, for every maximal ideal m of A, $A_m$ is a principal ideal domain.

We show first the equivalence of (a) and (b). Corollary 2 to Theorem 3, § 1, no. 6, shows immediately that (a) implies (b). Conversely, (b) implies (a), since, for every prime ideal $p$ of $A$, there exists a valuation ring of $K$ which dominates $A_p$ (Chapter VI, § 1, no. 2, Corollary to Theorem 2).

The remainder of the proof is carried out by proving the following implications:

If $A$ is a Dedekind domain and $b$ is a non-zero fractional ideal, then $bA_p = bA_p$ for every maximal ideal $p$ ($§ 1$, no. 4, Proposition 7) and hence $b = 6$ (Chapter II, $§ 3$, no. 3, Corollary 3 to Theorem 1); thus (a) implies (c).

We now show that (c) implies (d). If (c) holds, the mapping $a \mapsto \operatorname{div} a$ is a bijection of $I(A)$ onto $D(A)$ (cf. $§ 1$, no. 1); as it is a homomorphism ($§ 1$, no. 2) and $D(A)$ is a group, every element of $I(A)$ is invertible.

We show that (d) implies (e). If (d) holds, every integral ideal $\neq (0)$ of $A$ is finitely generated (Chapter 11, $§ 5$, no. 6, Theorem 4) and hence $A$ is Noetherian; as $I(A)$ is a group, $D(A)$ is a group and $A$ is therefore completely integrally closed ($§ 1$, no. 2, Theorem 1). Finally, if $p$ is a non-zero prime ideal of $A$ and $m$ is a maximal ideal of $A$ containing $p$, the ring $A_{m}$ is a principal ideal domain (Chapter II, $§ 5$, no. 6, Theorem 4); as $pA_m$ is prime and non-zero, necessarily $pA_m = mA_m$ (a principal ideal domain being a Dedekind domain) whence $p = m$ (Chapter 11, $§ 2$, no. 5, Proposition 11) and $p$ is maximal.

We now show that (e) implies (f). If $m$ is a maximal ideal of $A$ and (e) holds, $A_{m}$ is an integrally closed Noetherian domain and its maximal ideal $mA_m$ is, either (0), or the only non-zero prime ideal of $A_{m}$; hence $A_{m}$ is a field or a discrete valuation ring by Proposition 11 of $§ 1$, no. 7.

The fact that (f) implies (g) is obvious.

We show finally that (g) implies (a). As $A$ is the intersection of the $A_{m}$, where $m$ runs through the set of maximal ideals (Chapter II, $§ 3$, no. 3, Corollary 4 to Theorem 1), (g) implies that $A$ is integrally closed and Noetherian and hence that $A$ is a Krull domain ($§ 1$, no. 3, Corollary to Theorem 2). On the other hand, it can be shown that every non-zero prime ideal of $A$ is maximal as in the proof that (d) $\Rightarrow$ (e).