Let an integral domain $\mathcal{O}$ have a divisor theory $\mathcal{O}^* \to \mathfrak{D}$ . The definition of divisor theory implies that for any divisor $\mathfrak{a}$ , there exists an element $\omega$ of $\mathcal{O}$ such that $\mathfrak{a}$ divides the principal divisor $(\omega)$ , i.e. that $\mathfrak{ac} = (\omega)$ with $\mathfrak{c}$ a divisor. The following theorem states that $\mathfrak{c}$ may always be chosen such that it is coprime with any beforehand given divisor.

Theorem. For any two divisors $\mathfrak{a}$ and $\mathfrak{b}$ , there is a principal divisor $(\omega)$ such that $$\mathfrak{ac} = (\omega)$$ and $$\gcd(\mathfrak{b},\,\mathfrak{c}) = (1).$$

Proof. Let $\mathfrak{p}_1,\,\ldots,\,\mathfrak{p}_s$ all distinct prime divisors, which divide the product $\mathfrak{ab}$ , and let the divisor $\mathfrak{a}$ be strictly divisible by the powers $\mathfrak{p}_1^{a_1},\,\ldots,\,\mathfrak{p}_s^{a_s}$ (the cases $a_i = 0$ are not excluded). For each $i = 1,\,\ldots,\,s$ , we choose a nonzero element $\alpha_i$ of $\mathcal{O}$ being strictly divisible by the power $\mathfrak{p}_i^{a_i}$ ; the choosing is possible, since any nonzero element of the ideal determined by the divisor $\mathfrak{p}_i^{a_i}$ , not belonging to the sub-ideal determined by the divisor $\mathfrak{p}_i^{a_i+1}$ , will do. According to the Chinese remainder theorem, there exists a nonzero element $\omega$ of the ring $\mathcal{O}$ such that

(1)

Bibliography

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