Chinese remainder theorem for rings, noncommutative case
Theorem 1.
(Chinese Remainder Theorem) Let R be a ring and I1,I2,…,In pairwise comaximal (http://planetmath.org/Comaximal) ideals such that R=Ij+R2 for all j. The homomorphism
:
f:R→R/I1×R/I2×…×R/In | ||
f(a)=(a+I1,a+I2,…,a+In) |
is surjective and kerf=I1∩I2∩⋯∩In.
Proof.
Clearly f is a homomorphism with kernel I1∩I2∩⋯∩In. It remains to show the surjectivity.
We have:
R=I1+R2=I1+(I1+I2)(I1+I3) | ||
⊆I1+I21+I1I3+I2I1+I2I3 | ||
⊆I1+(I2∩I3). |
Moreover,
R=I1+R2=I1+(I1+I2∩I3)(I1+I4) | ||
=I1+I1I4+(I2∩I3)I1+(I2∩I3)I4 | ||
⊆I1+(I2∩I3∩I4). |
Continuing, we obtain that R=I1+⋂j≠1Ij. We show similarly that:
R=I2+⋂j≠2Ij=I3+⋂j≠3Ij=⋯=In+⋂j≠nIj. |
Given elements a1,a2,…,an, we can find xj∈Ij and yj∈⋂j≠kIk such that aj=xj+yj.
Take .
Hence
and we conclude that is surjective as required.∎
Notes 1.The relation is satisfied when is ring with unity. In that case .
2. The Chinese Remainder Theorem (http://planetmath.org/ChineseRemainderTheorem) case for integers is obtained from the above result. For this, take and . The fact that two solutions of the set of congruences must is a consequence of:
Title | Chinese remainder theorem for rings, noncommutative case |
---|---|
Canonical name | ChineseRemainderTheoremForRingsNoncommutativeCase |
Date of creation | 2013-03-22 16:53:45 |
Last modified on | 2013-03-22 16:53:45 |
Owner | polarbear (3475) |
Last modified by | polarbear (3475) |
Numerical id | 16 |
Author | polarbear (3475) |
Entry type | Theorem |
Classification | msc 13A15 |
Classification | msc 11D79 |
Synonym | chinese remainder theorem |