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:RR/I1×R/I2××R/In
f(a)=(a+I1,a+I2,,a+In)

is surjectivePlanetmathPlanetmath and kerf=I1I2In.

Proof.

Clearly f is a homomorphism with kernel I1I2In. It remains to show the surjectivity.
We have:

R=I1+R2=I1+(I1+I2)(I1+I3)
I1+I12+I1I3+I2I1+I2I3
I1+(I2I3).

Moreover,

R=I1+R2=I1+(I1+I2I3)(I1+I4)
=I1+I1I4+(I2I3)I1+(I2I3)I4
I1+(I2I3I4).

Continuing, we obtain that R=I1+j1Ij. We show similarly that:

R=I2+j2Ij=I3+j3Ij==In+jnIj.

Given elements a1,a2,,an, we can find xjIj and yjjkIk 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= and Ij=(pj)=pj. The fact that two solutions of the set of congruencesMathworldPlanetmathPlanetmathPlanetmathPlanetmath must x=x0(modp1pn) is a consequence of:

I1I2In=(p1)(p2)(pn)=(p1p2pn).
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