Definition 1

Let $A$ be an integral domain and $K$ its field of fractions. Every sub-A-module $a$ of $K$ such that there exists an element $d \neq 0$ in $A$ for which $da \subset A$ is called a fractional ideal $\mathfrak{d}$ of $A$ (or of $K$, by an abuse of language).

Every finitely generated sub-A-module $a$ of $K$ is a fractional ideal: for if $(a_i)_{1 \leq i \leq n}$ is a system of generators of $a$, we may write $a_i = b_i/d_i$, where $b_i \in A$, $d_i \in A$ and $d, \neq 0$; if $d = d_1 \ldots d_n$, clearly $da \subset A$. In particular the monogenous sub-A-modules of $K$ are fractional ideals (recall that they have been called fractional principal ideals in Algebra, Chapter VI, § 1, no. 5). If $A$ is Noetherian, every fractional ideal is afinitely generated A-module. Every sub-A-module of a fractional ideal of $A$ is a fractional ideal. Every ideal of $A$ is a fractional ideal; to avoid confusion, these will also be called the integral ideals of $A$.

We denote by $I(A)$ the set of non-zero fractional ideals of $A$. Given two elements $a, b$ of $I(A)$, we shall write $a \prec b$ (or $b \succ a$) for the relation "every fractional principal ideal containing $a$ also contains $b$"; clearly this relation is a preordering on $I(A)$. Let $R$ denote the associated equivalence relation "$a \prec b$ and $b \prec a$" (Set Theory, Chapter III, § 1, no. 2) and $D(A)$ the quotient set $I(A)/R$; we shall say that the elements of $D(A)$ are the divisors of $A$ and, for every fractional ideal $a \in I(A)$, we shall denote by $\operatorname{div} a$ (or $\operatorname{div}, a$) the canonical image of $a$ in $D(A)$ and we shall say that $\operatorname{div} a$ is the divisor $\mathfrak{d}$ of $a$; if $a = Ax$ is a fractional principal ideal, we write div(x) instead of div(Ax) and div(x) is called the divisor of x; the elements of D(A) of the form div(x) are called principal divisors. By taking the quotient, the preordering $\prec$ on I(A) defines on D(A) an ordering which we shall denote by $\leqslant$.

For all $a \in I(A)$ there exists by hypothesis some $d \neq 0$ in A such that $a \subset Ad - 1$; the intersection $\tilde{a}$ of the fractional principal ideals containing a is therefore an element of I(A). Clearly the relation $a \prec b$ is equivalent to the relation $\tilde{a} \supset 6$; the relation $a \supset b$ therefore implies $a \prec b$. For two elements $a, b$ of I(A) to be equivalent modulo R, it is necessary and sufficient that $\tilde{a} = 6$.