# Chinese remainder theorem in terms of divisor theory

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

 $\displaystyle\begin{cases}\xi\,\equiv\,\alpha_{1}\pmod{\mathfrak{a}_{1}}\\ \cdots\qquad\cdots\qquad\cdots\\ \xi\,\equiv\,\alpha_{s}\pmod{\mathfrak{a}_{s}}\end{cases}$

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

 $\displaystyle\beta_{1}+\ldots+\beta_{s}=1.$ (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.

## References

• 1 М. М. Постников: Введение  в  теорию  алгебраических  чисел.  Издательство  ‘‘Наука’’. Москва (1982).
Title Chinese remainder theorem in terms of divisor theory ChineseRemainderTheoremInTermsOfDivisorTheory 2013-03-22 18:01:58 2013-03-22 18:01:58 pahio (2872) pahio (2872) 6 pahio (2872) Theorem msc 11A51 msc 13A05 ChineseRemainderTheorem ChineseRemainderTheorem2 CongruenceInAlgebraicNumberField WeakApproximationTheorem