Chinese remainder theorem for rings, noncommutative case

Theorem 1.

(Chinese Remainder Theorem) Let $R$ be a ring and $I_{\mathrm{1}}\mathrm{,}{I}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }\mathrm{,}{I}_n$ pairwise comaximal (http://planetmath.org/Comaximal) ideals such that $R\mathrm{=}{I}_j\mathrm{+}{R}^{\mathrm{2}}$ for all $j$. The homomorphism:

$f:R\to R/I_1\times R/I_2\times \mathrm{\dots }\times R/I_n$

$f(a)=(a+I_1,a+I_2,\mathrm{\dots },a+I_n)$

is surjective and $k\mathit{}e\mathit{}r\mathit{}f\mathrm{=}{I}_{\mathrm{1}}\mathrm{\cap }{I}_{\mathrm{2}}\mathrm{\cap }\mathrm{\cdots }\mathrm{\cap }{I}_n$.

Proof.

Clearly $f$ is a homomorphism with kernel $I_1\cap I_2\cap \mathrm{\cdots }\cap I_n$. It remains to show the surjectivity.
We have:

$R=I_1+R^2=I_1+(I_1+I_2)(I_1+I_3)$

$\subseteq I_1+I_1^2+I_1{I}_3+I_2{I}_1+I_2{I}_3$

$\subseteq I_1+(I_2\cap I_3).$

Moreover,

$R=I_1+R^2=I_1+(I_1+I_2\cap I_3)(I_1+I_4)$

$=I_1+I_1{I}_4+(I_2\cap I_3)I_1+(I_2\cap I_3)I_4$

$\subseteq I_1+(I_2\cap I_3\cap I_4).$

Continuing, we obtain that $R=I_1+{\bigcap }_{j\ne 1}{I}_j$. We show similarly that:

$$R=I_2+\bigcap _{j\ne 2}{I}_j=I_3+\bigcap _{j\ne 3}{I}_j=\mathrm{\cdots }=I_n+\bigcap _{j\ne n}{I}_j.$$

Given elements $a_1,a_2,\mathrm{\dots },a_n$, we can find $x_j\in I_j$ and $y_j\in {\bigcap }_{j\ne k}{I}_k$ such that $a_j=x_j+y_j$.
Take $a:={\sum }_{i=1}^n{x}_i=a_j(mod{I}_j)$.
Hence

$$f(a)=(a_1+I_1,a_2+I_2,\mathrm{\dots },a_n+I_n),$$

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 $x=x_0(mod{p}_1\mathrm{\dots }{p}_n)$ is a consequence of:

$$I_1\cap I_2\cap \mathrm{\cdots }\cap I_n=(p_1)\cap (p_2)\cap \mathrm{\cdots }\cap (p_n)=(p_1{p}_2\mathrm{\dots }{p}_n)\mathbb{Z}.$$