# Chinese remainder theorem for rings, noncommutative case

###### Theorem 1.

() Let $R$ be a ring and $I_{1},I_{2},...,I_{n}$ pairwise comaximal (http://planetmath.org/Comaximal) ideals such that $R=I_{j}+R^{2}$ for all $j$. The homomorphism:

 $\displaystyle f:R\rightarrow R/I_{1}\times R/I_{2}\times...\times R/I_{n}$ $\displaystyle f(a)=(a+I_{1},a+I_{2},...,a+I_{n})$

is surjective and $kerf=I_{1}\cap I_{2}\cap\cdots\cap I_{n}$.

###### Proof.

Clearly $f$ is a homomorphism with kernel $I_{1}\cap I_{2}\cap\cdots\cap I_{n}$. It remains to show the surjectivity.
We have:

 $\displaystyle R=I_{1}+R^{2}=I_{1}+(I_{1}+I_{2})(I_{1}+I_{3})$ $\displaystyle\subseteq I_{1}+I_{1}^{2}+I_{1}I_{3}+I_{2}I_{1}+I_{2}I_{3}$ $\displaystyle\subseteq I_{1}+(I_{2}\cap I_{3}).$

Moreover,

 $\displaystyle R=I_{1}+R^{2}=I_{1}+(I_{1}+I_{2}\cap I_{3})(I_{1}+I_{4})$ $\displaystyle=I_{1}+I_{1}I_{4}+(I_{2}\cap I_{3})I_{1}+(I_{2}\cap I_{3})I_{4}$ $\displaystyle\subseteq I_{1}+(I_{2}\cap I_{3}\cap I_{4}).$

Continuing, we obtain that $R=I_{1}+\bigcap_{j\neq 1}I_{j}$. We show similarly that:

 $R=I_{2}+\bigcap_{j\neq 2}I_{j}=I_{3}+\bigcap_{j\neq 3}I_{j}=\cdots=I_{n}+% \bigcap_{j\neq n}I_{j}.$

Given elements $a_{1},a_{2},...,a_{n}$, we can find $x_{j}\in I_{j}$ and $y_{j}\in\bigcap_{j\neq k}I_{k}$ such that $a_{j}=x_{j}+y_{j}$.
Take $a:=\sum_{i=1}^{n}x_{i}=a_{j}\pmod{I_{j}}$.
Hence

 $f(a)=(a_{1}+I_{1},a_{2}+I_{2},...,a_{n}+I_{n}),$

and we conclude that $f$ is surjective as required.∎

Notes 1.The relation $R=I_{j}+R^{2}$ is satisfied when $R$ is ring with unity. In that case $R^{2}=R$.
2. The Chinese Remainder Theorem (http://planetmath.org/ChineseRemainderTheorem) case for integers is obtained from the above result. For this, take $R=\mathbb{Z}$ and $I_{j}=(p_{j})=p_{j}\mathbb{Z}$. The fact that two solutions of the set of congruences must $x=x_{0}\pmod{p_{1}...p_{n}}$ is a consequence of:

 $I_{1}\cap I_{2}\cap\cdots\cap I_{n}=(p_{1})\cap(p_{2})\cap\cdots\cap(p_{n})=(p% _{1}p_{2}...p_{n})\mathbb{Z}.$
Title Chinese remainder theorem for rings, noncommutative case ChineseRemainderTheoremForRingsNoncommutativeCase 2013-03-22 16:53:45 2013-03-22 16:53:45 polarbear (3475) polarbear (3475) 16 polarbear (3475) Theorem msc 13A15 msc 11D79 chinese remainder theorem