Proposition 1

(a) If b is a divisorial ideal and $a \in I(A)$, $b : a$ is divisorial.

(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).