(b) Let a, b be in I(A). In order that div $a =$ div $b$, it is necessary and sufficient that $A : a = A : b$.

(c) For all $a \in I(A)$, $6 = A : (A : a)$.

Assertion (a) follows immediately from equation (1) since, if $b$ is divisorial, so is $b x^{-1}$ for all $x \neq 0$.

To show (b), let $P(a)$ denote the set of fractional principal ideals containing a; the relation $Ax \in P(a)$ is equivalent to $x^{-1} a \subset A$ and hence to $x^{-1} \in A : a$. As the relation div $a =$ div $b$ is by definition equivalent to $P(a) = P(b)$, it is also equivalent to $A : a = A : b$.

Finally, as $a(A : a) \subset A, a \subset A : (A : a)$. Replacing $a$ by $A : a$ in this formula, it is seen that $A : a \subset A : (A : (A : a))$; on the other hand, the relation $a \subset A : (A : a)$ implies

Therefore $A : a = A : (A : (A : a))$ and it follows from (b) that $\operatorname{div} a = \operatorname{div}(A : (A : a))$ As $A : (A : a)$ is divisorial by (a), certainly $6 = A : (A : a)$, which proves (c).

Let $(a_i)$ be a non-empty family of elements of $I(A)$ which is bounded above. To say that a divisorial ideal $b$ bounds this family above amounts to saying that it is contained in all the $a_i$, that is that $b$ is contained in $\bigcap_i 6_i$. Hence $\bigcap_i a_i \neq (0)$ and $\bigcap_i 6_i$, is therefore a divisorial ideal, which shows (i).

Now let $(6_i)$ be a non-empty family of elements of $I(A)$ which is bounded below. To say that a divisorial ideal $b$ bounds this family below means that it contains all the $a_i$, that is (since $b$ is divisorial) that it contains all the $a_i$, or also that $b \supset \sum_i a_i$. This proves (ii).

Finally, to prove (iii) it is sufficient by (i) and (ii) to prove that, if $a, b$ are in $I(A)$, the set $\{a, b\}$ is bounded both above and below in $I(A)$; now it is bounded above by $a \cap b$ (which is distinct from $(0)$). It is bounded below by $a + b$, for $a + b \in I(A)$: if $d$ and $d'$ are non-zero elements of $A$ such that $da \subset A$ and $d'b \subset A$, then $dd'(a + b) \subset A$.

$A(x + y) \subset Ax + Ay$ and hence $\operatorname{div}(x + y) \geq \operatorname{div}(Ax + Ay)$.

We may restrict our attention to the case where $b = b'$. Then let $Ax$ be a fractional principal ideal containing $a'b$; for every non-zero element $y$ of $b$, $Ax \supset a'y$ and hence $Axy^{-1} \supset a'$, whence $Axy^{-1} \supset a$ and $Ax \supset ay$. Varying $y$, it is seen that $Ax \supset ab$, whence $ab > a'b$.

It follows from Proposition 3 that multiplication on $I(A)$ defines, by passing to the quotient, a law of composition on $D(A)$ which is obviously associative and commutative. It is written additively so that we may write:

for $a, b$ in $I(A)$. Clearly $\text{div}(1)$ is an identity element for this addition; this element is denoted by 0. Proposition 3 proves further that the order structure on $D(A)$ is compatible with this addition (Algebra, Chapter VI, § 1, no. 1) and, more precisely (no. 1, Proposition 2 (ii)):

For a fractional ideal $a \neq 0$ to be such that $\text{div } a \geq 0$ in $D(A)$, it is necessary and sufficient that $a \subset A$ (in other words, that $a$ be an integral ideal of $A$).

For two elements $x, y$ of $K^*$, the relation $\text{div}(x) = \text{div}(y)$ is equivalent to $Ax = Ay$; the set of principal divisors of $A$ with the order relation and the monoid law induced by that on $D(A)$ is an ordered group canonically isomorphic to the multiplicative group of fractional principal ideals ordered by the opposite order relation to inclusion (Algebra, Chapter VI, § 1, no. 5). The relation $S$ between two elements $P, Q$ of $D(A)$:

"there exists $x \in K^*$ such that $P = Q + \text{div}(x)$"

is therefore an equivalence relation since the relation $P = Q + \text{div}(x)$ is equivalent to $Q = P + \text{div}(x^{-1})$; if $P$ and $Q$ are congruent modulo $S$, they are called equivalent divisors of $A$. Clearly moreover the relation $S$ is compatible with the law of the monoid $D(A)$ and the latter therefore defines, by taking quotients, a monoid structure on $D(A)/S$; this monoid is called the divisor class monoid of A.

If $\operatorname{div} b = \operatorname{div} a + \operatorname{div}(x)$ for some $x \in K^*$, then $\operatorname{div} b = \operatorname{div}(xa)$ and, as $b$ and $xa$ are divisorial, $b = xa$, which proves the proposition.

Let $a$ be an invertible fractional ideal (Chapter II, § 5, no. 6); then $a = A : (A : a)$ (loc. cit., Proposition 10) and hence $a$ is divisorial (no. 1, Proposition 1). The group $J(A)$ of invertible fractional ideals is therefore identified with a subgroup of the monoid $D(A)$ and the canonical image of $J(A)$ in $D(A)/S$ with the group of classes of projective $A$-modules of rank 1 (Chapter II, § 5, no. 7, Corollary 2 to Proposition 12 and Remark 1).

Suppose that $D(A)$ is a group. Let $x \in K$; suppose that $A[x]$ is contained in a finitely generated sub-$A$-module of $K$. Then we have seen (no. 1) that $a = A[x]$ is an element of $I(A)$. Then $xa \subset a$ and hence $\operatorname{div}(x) + \operatorname{div} a \geq \operatorname{div} a$. Since $D(A)$ is an ordered group, we conclude that $\operatorname{div}(x) \geq 0$, whence $x \in A$. Thus $A$ is completely integrally closed (Chapter V, § 1, no. 4, Definition 5).

Conversely, suppose that $A$ is completely integrally closed. Let $a$ be a divisorial ideal. We shall show that $\operatorname{div} a + \operatorname{div}(A : a) = 0$, which will prove that $D(A)$ is a group. As $a(A : a) \subset A$, it suffices (no. 1) to verify that every fractional principal ideal $Ax^{-1}$ which contains $a(A : a)$ also contains $A$. Now, for $y \in K^*$, the relation $Ay \supset a$ implies $y^{-1} \in A : a$, whence $y^{-1}a \subset a(A : a) \subset Ax^{-1}$ and hence $xa \subset Ay$. As $a$ is divisorial, we deduce that $xa \subset a$, whence $x^n a \subset a$ for all $n \in \mathbf{N}$. There exist elements $x_0, x_1$ of $K^*$ such that $Ax_0 \subset a \subset Ax_1$; therefore $x^n x_0 \in Ax_1$, whence $x^n \in Ax_1 x_0^{-1}$. As $A$ is completely integrally closed, $x \in A$, that is $Ax^{-1} \supset A$, which completes the proof.

Note that, if $A$ is completely integrally closed (and even Noetherian), a divisorial ideal of $A$ is not necessarily invertible, in other words, in general $J(A) \neq D(A)$ (Exercise 2 and § 3, no. 2, Proposition 1).

By virtue of formula (1) of no. 1:

taking account of Proposition 2 and the fact that the fractional ideals $y^{-1}a$ are divisorial. But since $D(A)$ is an ordered group (Algebra, Chapter VI, § 1, no. 8):

Let $A$ be a Krull domain. It is completely integrally closed (Chapter VI, § 4, no. 5, Corollary to Proposition 9). Let $(v_i)_{i \in I}$ be a family of valuations on the field of fractions $K$ of $A$ satisfying (AK$_I$), (AK$_{II}$) and (AK$_{III}$). The $v_i$ may be assumed to be normed (Chapter VI, § 3, no. 6, Definition 3). For all $a \in I(A)$, we shall write:

then $v_t(a) \in \mathbf{Z}$, for, if $a$ is a non-zero element of $a$, the relation $Ax \supset Aa$ implies that $v_t(x) \leq v_t(a)$ (by (AK$_{III}$)), which shows that the family of $v_t(x)$ ($a \subset Ax$) is bounded above. We establish the following properties:

(1) Let $a$ be a divisorial fractional ideal; in order that $y \in a$, it is necessary and sufficient that $v_t(y) \geq v_t(a)$ for all $t \in I$.

As $a$ is divisorial, the relation $y \in a$ is equivalent to the relation "“$a \subset Ax$ implies $y \in Ax$”. Now, by (AK$_{II}$), the relation $y \in Ax$ is equivalent to “$v_t(y) \geq v_t(x)$ for all $t \in I$”. Whence (1).

(2) Let $a$ and $b$ be two divisorial fractional ideals of $A$; in order that $a \subset b$, it is necessary and sufficient that $v_t(a) \geq v_t(b)$ for all $t \in I$.

This follows immediately from property (1).

(3) If $x \in K^*$, then $v_t(Ax) = v_t(x)$.

If $Ay \supset Ax$, then $v_t(y) \leq v_t(x)$ by (AK$_{II}$) and the minimum value of $v_t(y)$ is taken at $y = x$.

(4) For all $a \in I(A)$, the indices $t \in I$ such that $v_t(a) \neq 0$ are finite in number.

There exist $x, y$ in $K^*$ such that $Ax \subset a \subset Ay$. By properties (2) and (3), $v_t(x) \geq v_t(a) \geq v_t(y)$ for all $t \in I$. It then sufficesto apply (AK$_{III}$).

We have therefore shown the following lemma:

This being so, every non-empty set of positive elements of $\mathbf{Z}^{(I)}$ has a minimal element (Algebra, Chapter VI, § 1, no. 13, Theorem 2). Hence $A$ certainly satisfies property (b) of the statement.

Conversely, let $A$ be an integral domain satisfying properties (a) and (b) of the statement. Since $A$ is completely integrally closed, $D(A)$ is an ordered group (no. 2, Theorem 1). This group is a lattice (no. 1, Proposition 2). By condition (b) of the statement, every non-empty family of positive elements of $D(A)$ has a minimal element. Let $P(A)$ be the set of extremal elements of $D(A)$. Then (Algebra, Chapter VI, § 1, no. 13, Theorem 2) $P(A)$ is a basis of the $\mathbf{Z}$-module $D(A)$ and the positive elements of $D(A)$ are the linear combinations with positive integer coefficients of the elements of $P(A)$.

Thus, for $x \in K^*$, rational integers $v_P(x)$ are defined (for $P \in P(A)$) by writing:

We also write $v_P(0) = +\infty$.

From the relations

and

for $x, y$ and $x + y$ in $K^*$, we deduce that the $v_P$ are discrete valuations on $K$. In order that $x \in A$, it is necessary and sufficient that $\text{div}(x) \geq 0$, that is that $v_P(x) \geq 0$ for all $P \in P(A)$. Thus the $v_P$ satisfy conditions (AK$_1$) and (AK$_{11}$) and obviously also (AK$_{III}$).

An integrally closed Noetherian domain is completely integrally closed (Chapter V, § 1, no. 4).

There are non-Noetherian Krull domains, for example the polynomial ring $K[X_n]_{n \in \mathbf{N}}$ over a field $K$ in an infinity of indeterminates (cf. Exercise 8).

Let $x \in K^*$. In order that $x \in a$, it is necessary and sufficient that $Ax \subset a$, hence that $\text{div}(x) \geq \text{div } a$ and hence, by (4), that $v_P(x) \geq n_P$ for all $P \in P(A)$.

We write $B = \bigcap_{t \in J} A_t$. Then $S^{-1} \subset B$ and $A \subset B$ and hence $S^{-1}A \subset B$. Conversely, let $x \in B$. Let $J'$ denote the finite set of indices $t$ such that $v_t(x) < 0$. If $t \in J'$, then $x \notin A_t$, hence $t \notin J$ and hence there exists $s_t \in S$ such that $v_t(s_t) > 0$. Let $n(t)$ be an integer $> 0$ such that $v_t(s_t^{n(t)}x) \geq 0$; we write $s = \prod_{t \in J'} s_t^{n(t)}$. Then $v_t(sx) \geq 0$ for all $t \in I$ and hence $sx \in A$ and $x \in S^{-1}A$. Thus $B = S^{-1}A$.

Let $S = A - p$. By Proposition 5, $v_P$ is zero on $S$ and $> 0$ on $p$. Hence $p$ is the intersection of $A$ and the ideal of $v_P$ and is therefore prime. On the other hand, for every extremal divisor $Q \neq P$, $Q \not\subset P$ and hence the divisorial ideal $q$ corresponding to $Q$ is not contained in $p$; thus $q \cap S \neq \varnothing$ and hence, by Proposition 5, $v_Q$ is not zero on $S$. This being so, the corollary follows from Proposition 6 and Chapter II, § 3, no. 1, Proposition 3.

Let $P$ be an extremal divisor of $A$ and $p$ the corresponding divisorial ideal. By Corollary 1, Proposition 5, Lemma 1 and assertion (1) in the proof of Theorem 2, no. 3, there exists $t \in I$ such that the ring $A_t$ of $v_t$ contains the ring $A$, of $v_P$. As $v_t$ and $v_P$ are of height 1, they are therefore equivalent (Chapter VI, § 4, no. 5, Proposition 6).

As $a = \sum_{y \in a} Ax$, Proposition 2 (b) (no. 1) shows that $\operatorname{div}(a) = \inf_{x \in a} (\operatorname{div}(Ax))$, whence our first assertion. The second follows immediately, since $\operatorname{div} \tilde{a} = \operatorname{div} a$ and $A_p$ is the ring of the discrete valuation $v_P$.

for $n \in \mathbf{N}$, let $p^{(n)} = p^n A_p \cap A$; then $p^{(n)}$ is the set of $x \in A$ such that $v_p(x) \geq n$ and is a $p$-primary ideal.

(b) Let $a$ be a divisorial integral ideal, $n_1 P_1 + \ldots + n_r P_r$ the divisor of $a$ (the $P_i$ being distinct extremal divisors) and $p_i$ the divisorial prime ideal corresponding to $P_i$. Then $a = \bigcap_{i=1}^r p_i^{(n_i)}$ is the unique reduced primary decomposition of $a$ and the $p_i$ are not immersed.

By Corollary 1 to Proposition 6, the relation $x \in p^n A_p = (pA_p)^n$ is equivalent to $v_p(x) \geq n$; on the other hand, as $A$, is a discrete valuation ring, $(pA_p)^n$ is $(pA_p)$-primary (Chapter IV, § 2, no. 1, Example 4) and hence $p^{(n)}$ is $p$-primary (Chapter IV, § 2, no. 1, Proposition 3); this shows (a). Proposition 5 certainly shows that $a = \bigcap_{i=1}^r p_i^{(n_i)}$. As $p_i \not\subset p_j$ for $i \neq j$ this primary decomposition is reduced: For if $p_i^{(n_i)} \supset \bigcap_{j \neq i} p_j^{(n_j)} \supset \prod_{j \neq i} p_j^{(n_j)}$, $p_{ji}$ would contain one of the $p_j$ for $j \neq i$ (Chapter II, § 1, no. 1, Proposition 1). The uniqueness follows from Chapter IV, § 2, no. 3, Proposition 5.

Let $p_1, \ldots, p_r$ be the divisorial ideals of $A$ corresponding to the valuations $v_1, \ldots, v_r$. There exists $y \in K$ such that $v_i(y) = n_i$ for $1 \leq i \leq r$ (Chapter VI, § 7, no. 2, Corollary 1 to Theorem 1). The essential valuations $w_1, \ldots, w_s$ of $A$ distinct from the $v_i$ for which the integer $w_j(y) = -m_j$ is $< 0$ are finite in number; let $q_1, \ldots, q_s$ be the corresponding ideals. There exists no inclusion relation between $p_1, \ldots, p_r, q_1, \ldots, q_s$ since these ideals correspond to extremal divisors and these ideals are prime (Corollary 1 to Proposition 6). Hence the integral ideal $a = q_1^{m_1} \ldots q_s^{m_s}$ is contained in none of the $p_i$ (Chapter 11, § 1, no. 1, Proposition 1) and is therefore not contained in their union (loc. cit., Proposition 2). Therefore there exists $z \in a$ such that $z \notin p_i$ for $1 \leq i \leq r$; then $v_1(z) = \cdots = v_r(z) = 0$ and $w_j(z) \geq m_j$ for $1 \leq j \leq s$; hence the element $x = yz$ solves the problem.