Theorem. If an integral domain $\mathcal{O}$ has a divisor theory $\mathcal{O}^* \to \mathfrak{D}$ , then the subset $[\mathfrak{a}]$ of $\mathcal{O}$ , consisting of 0 and all elements divisible by a divisor $\mathfrak{a}$ , is an ideal of $\mathcal{O}$ . The mapping $$\mathfrak{a} \mapsto [\mathfrak{a}]$$ from the set $\mathfrak{D}$ of divisors into the set of ideals of $\mathcal{O}$ is injective and maps any principal divisor $(\alpha)$ to the principal ideal $(\alpha)$ .

Proof. Let $\alpha,\,\beta \in [\mathfrak{a}]$ and $\lambda \in \mathcal{O}$ . Then, by the postulate 2 of divisor theory, $\alpha\!-\!\beta$ is divisible by $\mathfrak{a}$ or is 0, and in both cases belongs to $[\mathfrak{a}]$ . When $\lambda\alpha \neq 0$ , we can write $(\alpha) = \mathfrak{ac}$ with $\mathfrak{c}$ a divisor. According to the homomorphicity of the mapping $\mathcal{O}^* \to \mathfrak{D}$ , we have $$(\lambda\alpha) = (\lambda)(\alpha) = (\lambda)\mathfrak{ac},$$ and therefore the element $\lambda\alpha$ is divisible by $\mathfrak{a}$ , i.e. $\lambda\alpha \in [\mathfrak{a}]$ . Thus, $[\mathfrak{a}]$ is an ideal of $\mathcal{O}$ .

The injectivity of the mapping $\mathfrak{a} \mapsto [\mathfrak{a}]$ follows from the postulate 3 of divisor theory.

The ideal $[\mathfrak{a}]$ may be called the image ideal of $\mathfrak{a}$ or the ideal determined by the divisor $\mathfrak{a}$ .

Remark. There are integral domains $\mathcal{O}$ having a divisor theory but also having ideals which are not of the form $[\mathfrak{a}]$ (for example a polynomial ring in two indeterminates and its ideal formed by the polynomials without constant term). Such rings have ``too many ideals''. On the other hand, in some integral domains the monoid of principal ideals cannot be embedded into a free monoid; thus those rings cannot have a divisor theory.

Bibliography

1

. . :. ``''. (1982).