The existence of a divisor theory restricts strongly the type of an integral domain, as is seen from the following propositions.

Proposition 1. An integral domain $\mathcal{O}$ which has a divisor theory $\mathcal{O}^*\to\mathfrak{D}$ , is integrally closed in its quotient field.

Proof. Let $\xi$ be an element of the quotient field of $\mathcal{O}$ which is integral over $\mathcal{O}$ . Then $\xi$ satisfies an equation

(1)

(2)

Proposition 2. When an integral domain $\mathcal{O}$ has a divisor theory $\mathcal{O}^*\to\mathfrak{D}$ , then each element of $\mathcal{O}^*$ has only a finite number of non-associated factors.

Proof. Let $\xi$ be an arbitrary non-zero element of $\mathcal{O}$ . We form the prime factor presentation of the corresponding principal divisor $(\xi)$ : $$(\xi) = \mathfrak{p}_1\mathfrak{p}_2\cdots\mathfrak{p}_r$$ This is unique up to the ordering of the factors; $r \geqq 0$ . Then we form of the prime divisors $\mathfrak{p}_i$ all products having $k$ factors ($0 \leqq k \leqq r$ ) and choose from the products those which are principal divisors. Thus we obtain a set of factors of $(\xi)$ containing at most $\displaystyle r \choose k$ elements. All different principal divisor factors of $(\xi)$ are gotten, as $k$ runs all integers from 0 to $r$ , and their number is at most equal to $$\sum_{k=0}^r{r \choose k} = 2^r$$ (see 5. in the binomial coefficients). To every principal divisor factor, there corresponds a class of associate factors of $\xi$ , and the elements of distinct classes are non-associates. Since $\xi$ has not other factors, the number of its non-associated factors is at most $2^r$ .