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In a ring with a divisor theory, a congruence $\alpha \equiv \beta \pmod{\mathfrak{a}}$ with respect to a divisor module $\mathfrak{a}$ means that $\mathfrak{a} \mid \alpha\!-\!\beta$ .
Theorem. Let $\mathcal{O}$ be an integral domain having the divisor theory $\mathcal{O}^* \to \mathfrak{D}$ . For arbitrary pairwise coprime divisors $\mathfrak{a}_1,\,\ldots,\,\mathfrak{a}_s$ in $\mathfrak{D}$ and for arbitrary elements $\alpha_1,\,\ldots,\,\alpha_s$ of the domain $\mathcal{O}$ there exists an element $\xi$ in $\mathcal{O}$ such that
Proof. Let $$\mathfrak{b}_i \,:=\, \prod_{j \neq i}\mathfrak{a}_j \quad (i = 1,\,\ldots,\,s).$$ Apparently, the divisors $\mathfrak{b}_1,\,\ldots,\,\mathfrak{b}_s$ are mutually coprime, whence there are in the ring $\mathcal{O}$ the elements $\beta_1,\,\ldots,\,\beta_s$ divisible by the divisors $\mathfrak{b}_1,\,\ldots,\,\mathfrak{b}_s$ , respectively, such that
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(1) |
For every $i \neq j$ , the divisor $\mathfrak{a}_i$ divides $\mathfrak{b}_j$ and therefore also the element $\beta_j$ . Then the equation (1) implies that $\beta_i \equiv 1 \pmod{\mathfrak{a}_i}$ and thus the element $$\xi \,:=\, \alpha_1\beta_1+\ldots+\alpha_s\beta_s$$ satisfies $$\xi \,\equiv\, \alpha_i\beta_i \,\equiv\, \alpha_i\! \pmod{\mathfrak{a}_i}$$ for each $i = 1,\,\ldots,\,s$ . Q.E.D.
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- . . :. ``''. (1982).
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