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Chinese remainder theorem in terms of divisor theory

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In a ring with a divisor theory, a congruence \,$\alpha \equiv \beta \pmod{\mathfrak{a}}$\, with respect to a divisor \PMlinkname{module}{Congruences} $\mathfrak{a}$ \PMlinkescapetext{means} that\, $\mathfrak{a} \mid \alpha\!-\!\beta$.\\

\textbf{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
\xi\, \equiv\, \alpha_1 \pmod{\mathfrak{a}_1}\\
\cdots \qquad \cdots \qquad \cdots\\
\xi\, \equiv\, \alpha_s \pmod{\mathfrak{a}_s}

{\em 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 
\beta_1+\ldots+\beta_s = 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
$$\xi \,:=\, \alpha_1\beta_1+\ldots+\alpha_s\beta_s$$
$$\xi \,\equiv\, \alpha_i\beta_i \,\equiv\, \alpha_i\! \pmod{\mathfrak{a}_i}$$
for each\, $i = 1,\,\ldots,\,s$.\, Q.E.D.

\bibitem{MMP} \CYRM. \CYRM. \CYRP\cyro\cyrs\cyrt\cyrn\cyri\cyrk\cyro\cyrv: 
{\em \CYRV\cyrv\cyre\cyrd\cyre\cyrn\cyri\cyre\, \cyrv\, \cyrt\cyre\cyro\cyrr\cyri\cyryu\, \cyra\cyrl\cyrg\cyre\cyrb\cyrr\cyra\cyri\cyrch\cyre\cyrs\cyrk\cyri\cyrh \,
\cyrch\cyri\cyrs\cyre\cyrl}. \,\CYRI\cyrz\cyrd\cyra\cyrt\cyre\cyrl\cyrsftsn\cyrs\cyrt\cyrv\cyro \,
``\CYRN\cyra\cyru\cyrk\cyra''. \CYRM\cyro\cyrs\cyrk\cyrv\cyra \,(1982).