Chinese remainder theorem for rings, noncommutative case


Theorem 1.

(Chinese Remainder TheoremMathworldPlanetmathPlanetmathPlanetmath) 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 homomorphismMathworldPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath:

f:R→R/I1×R/I2×…×R/In
f⁢(a)=(a+I1,a+I2,…,a+In)

is surjectivePlanetmathPlanetmath and k⁢e⁢r⁢f=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+I12+I1⁢I3+I2⁢I1+I2⁢I3
⊆I1+(I2∩I3).

Moreover,

R=I1+R2=I1+(I1+I2∩I3)⁢(I1+I4)
=I1+I1⁢I4+(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 a:=∑i=1nxi=aj(modIj).
Hence

f⁢(a)=(a1+I1,a2+I2,…,an+In),

and we conclude that f is surjective as required.∎

Notes 1.The relationMathworldPlanetmathPlanetmathPlanetmath R=Ij+R2 is satisfied when R is ring with unity. In that case R2=R.
2. The Chinese Remainder Theorem (http://planetmath.org/ChineseRemainderTheorem) case for integers is obtained from the above result. For this, take R=ℤ and Ij=(pj)=pj⁢ℤ. The fact that two solutions of the set of congruencesMathworldPlanetmathPlanetmathPlanetmathPlanetmath must x=x0(modp1⁢…⁢pn) is a consequence of:

I1∩I2∩⋯∩In=(p1)∩(p2)∩⋯∩(pn)=(p1⁢p2⁢…⁢pn)⁢ℤ.
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