Chinese remainder theorem - Books - Famous Chinese Shopping City

March 8, 2014


		1 Introduction 2 Simultaneous congruences of integers 3 Statement for principal ideal domains 4 Statement for general rings 5 External links

Famous Chinese Shopping City Books

Chinese remainder theorem

Wikipedia

The Chinese remainder theorem is any of a number of related results in abstract algebra and number theory.

[go back to top]

Simultaneous congruences of integers

The original form of the theorem, contained in a book by the Chinese mathematician Qin Jiushao published in 1247, is a statement about simultaneous congruences (see modular arithmetic). Suppose n1, ..., nk are positive integer|integers which are pairwise coprime (meaning greatest common divisor|gcd
(ni, nj) = 1 whenever i ≠ j). Then, for any given integers a1, ..., ak, there exists an integer x solving the system of simultaneous congruences

<math>x \equiv a_i \pmodn_i \quad\mathrmfor\; i = 1, \cdots, k.</math>

Furthermore, all solutions x to this system are congruent modulo the product n = n1...nk.

A solution x can be found as follows. For each i; the integers ni and n/ni are coprime, and using the extended Euclidean algorithm we can find integers r and s such that r ni + s n/ni = 1. If we set ei = s n/ni, then we have

<math>e_i \equiv 1 \pmodn_i \quad\mathrmand\quad

e_i \equiv 0 \pmodn_j</math>

for j ≠ i.

The solution to the system of simultaneous congruences is therefore

For example, consider the problem of finding an integer x such that

<math>x \equiv 2 \pmod3 </math>

<math>x \equiv 3 \pmod4 </math>

<math>x \equiv 2 \pmod5. </math>

Using the extended Euclidean algorithm for 3 and 4×5 = 20, we find (-13) × 3 + 2 × 20 = 1, i.e. e1 = 40. Using the Euclidean algorithm for 4 and 3×5 = 15, we get (-11) × 4 + 3 × 15 = 1. Hence, e2 = 45. Finally, using the Euclidean algorithm for 5 and 3×4 = 12, we get 5 × 5 + (-2) × 12 = 1, meaning e3 = -24. A solution x is therefore 2 × 40 + 3 × 45 + 2 × (-24) = 167. All other solutions are congruent to 167 modulo 60, which means that they are all congruent to 47 modulo 60.

Sometimes, the simultaneous congruences can be solved even if the ni's are not pairwise coprime. The precise criterion is as follows: a solution x exists if and only if ai &equiv; aj (mod gcd(ni, nj)) for all i and j. All solutions x are congruent modulo the least common multiple of the ni.

Using the method of successive substitution can often yield solutions to simultaneous congruences, even when the moduli are not pairwise coprime.

[go back to top]

Statement for principal ideal domains

For a principal ideal domain R the Chinese remainder theorem takes the following form: If u1, ..., uk are elements of R which are pairwise coprime, and u denotes the product u1...uk, then the ring R/uR and the product of rings|product ring R/u1R x ... x R/ukR are isomorphic via the ring homomorphism|isomorphism

<math>f : R/uR \rightarrow R/u_1R \times \cdots \times

R/u_k R </math>

such that

<math>x \;\mathrmmod\,uR \rightarrow (x \;\mathrmmod\,u_1R) \times \cdots \times

(x \;\mathrmmod\,u_kR). </math>

The inverse isomorphism can be constructed as follows. For each i, the elements ui and u/ui are coprime, and therefore there exist elements r and s in R with

Set ei = s u/ui. Then the inverse is the map

<math>g : R/u_1R \times \cdots \times R/u_kR

\rightarrow R/uR </math>

such that

<math>(a_1 \;\mathrmmod\,u_1R) \times \cdots \times (a_k \;\mathrmmodu_kR) \rightarrow

\sum_i=1..k a_i e_i \pmoduR. </math>

[go back to top]

Statement for general rings

One of the most general forms of the Chinese remainder theorem can be formulated for ring (algebra)|rings and (two-sided) ring ideal|ideals. If R is a ring and I1, ..., Ik are ideals of R which are pairwise coprime (meaning that Ii + Ij = R whenever i ≠ j), then the product I of these ideals is equal to their intersection, and the ring R/I is isomorphic to the product of rings|product ring R/I1 x R/I2 x ... x R/Ik via the ring homomorphism|isomorphism

<math>f : R/I \rightarrow R/I_1 \times \cdots \times R/I_k </math>

such that

<math>x \;\mathrmmod\,I \rightarrow (x \;\mathrmmod\,I_1) \times \cdots \times

(x \;\mathrmmod I_k).</math>

[go back to top]

External links

http://www.cut-the-knot.org/blue/chinese.shtml Chinese remainder theorem

Category:Modular arithmeticCategory:Commutative algebraCategory:Theorems

de:Chinesischer Restsatz
nl:Chinese reststelling
ja:中国の剰余定理
pl:Chińskie twierdzenie o resztach
sv:Kinesiska restsatsen
zh:中国剩余定理

[go back to top]

This article is licensed under the GNU Free Documentation License. It uses material from the Wikipedia article "Chinese remainder theorem".

Last Modified: 2005-04-13

All informatin on the site is © FamousChinese.com 2002-2005. Last revised: January 2, 2004
Are you interested in our site or/and want to use our information? please read how to contact us and our copyrights.
To post your business in our web site? please click here. To send any comments to us, please use the Feedback.
To let us provide you with high quality information, you can help us by making a more or less donation: