Using the exponent valuations, one can easily prove the

Theorem. In any divisor theory, each divisor is the greatest common divisor of two principal divisors.

Proof. Let $\mathcal{O}^* \to \mathfrak{D}$ be a divisor theory and $\mathfrak{d}$ an arbitrary divisor in $\mathfrak{D}$ . We may suppose that $\mathfrak{d}$ is not a principal divisor (if $\mathfrak{D}$ contains exclusively principal divisors, then $\mathfrak{d} = \gcd(\mathfrak{d},\,\mathfrak{d})$ and the proof is ready). Let $$\mathfrak{d} = \prod_{i=1}^r\mathfrak{p}_i^{k_i}$$ where the $\mathfrak{p}_i$ 's are pairwise distinct prime divisors and every $k_i > 0$ . Then third condition in the theorem concerning divisors and exponents allows to choose an element $\alpha$ of the ring $\mathcal{O}$ such that $$\nu_{\mathfrak{p}_1}(\alpha) = k_1,\;\;\ldots,\;\;\nu_{\mathfrak{p}_r}(\alpha) = k_r.$$ Let the principal divisor corresponding to $\alpha$ be $$(\alpha) = \prod_{i=1}^r\mathfrak{p}_i^{k_i}\prod_{j=1}^s\mathfrak{q}_j^{l_j} = \mathfrak{dd}',$$ where the prime divisors $\mathfrak{q}_j$ are pairwise different among themselves and with the divisors $\mathfrak{p}_i$ . We can then choose another element $\beta$ of $\mathcal{O}$ such that $$\nu_{\mathfrak{p}_1}(\beta) = k_1,\;\;\ldots,\;\;\nu_{\mathfrak{p}_r}(\beta) = k_r,\;\; \nu_{\mathfrak{q}_1}(\beta) = \ldots = \nu_{\mathfrak{q}_s}(\beta) = 0.$$ Then we have $(\beta) = \mathfrak{dd}''$ , where $\mathfrak{d}'' \in \mathfrak{D}$ and $$\gcd(\mathfrak{d}',\,\mathfrak{d}'') = \mathfrak{q}^0\cdots\mathfrak{q}^0 = \mathfrak{e} = (1).$$ The gcd of the principal divisors $(\alpha)$ and $(\beta)$ is apparently $\mathfrak{d}$ , whence the proof is settled.