Chinese remainder theorem in terms of divisor theory

In a ring with a divisor theory, a congruence  $\alpha \equiv \beta (mod𝔞)$  with respect to a divisor module (http://planetmath.org/Congruences) $𝔞$ that  $𝔞\mid \alpha -\beta$.

Theorem.  Let $𝒪$ be an integral domain having the divisor theory  ${𝒪}^{*}\to 𝔇$.  For arbitrary pairwise coprime divisors ${𝔞}_1,\mathrm{\dots },{𝔞}_s$  in $𝔇$ and for arbitrary elements  ${\alpha }_1,\mathrm{\dots },{\alpha }_s$  of the domain $𝒪$ there exists an element $\xi$ in $𝒪$ such that

$\{\begin{array}{cc}\xi \equiv {\alpha }_1(mod{𝔞}_1)\hfill & \\ \mathrm{\cdots }\hspace{1em}\hspace{1em}\mathrm{\cdots }\hspace{1em}\hspace{1em}\mathrm{\cdots }\hfill & \\ \xi \equiv {\alpha }_s(mod{𝔞}_s)\hfill & \end{array}$

Proof.  Let

$${𝔟}_i:=\prod _{j\ne i}{𝔞}_j\mathit{\hspace{1em}}(i=1,\mathrm{\dots },s).$$

Apparently, the divisors  ${𝔟}_1,\mathrm{\dots },{𝔟}_s$  are mutually coprime, whence there are in the ring $𝒪$ the elements  ${\beta }_1,\mathrm{\dots },{\beta }_s$  divisible by  the divisors  ${𝔟}_1,\mathrm{\dots },{𝔟}_s$,  respectively, such that

${\beta }_1+\mathrm{\dots }+{\beta }_s=1.$(1)

For every  $i\ne j$,  the divisor ${𝔞}_i$ divides ${𝔟}_j$ and therefore also the element ${\beta }_j$.  Then the equation (1) implies that  ${\beta }_i\equiv 1(mod{𝔞}_i)$ and thus the element

$$\xi :={\alpha }_1{\beta }_1+\mathrm{\dots }+{\alpha }_s{\beta }_s$$

satisfies

$$\xi \equiv {\alpha }_i{\beta }_i\equiv {\alpha }_i(mod{𝔞}_i)$$

for each  $i=1,\mathrm{\dots },s$.  Q.E.D.