Chinese remainder theorem

For pairwise comaximal ideals, the quotient by their intersection splits as a product of quotients.

Chinese remainder theorem

Chinese remainder theorem: Let $R$ be a commutative ring with $1$ , and let $I_1,\dots,I_n\triangleleft R$ be ideals such that $I_i+I_j=R$ for all $i\neq j$ (pairwise comaximal, expressed via the sum of ideals ). Then the natural map

R \longrightarrow \prod_{i=1}^n R/I_i,\qquad r\longmapsto (r\bmod I_1,\dots,r\bmod I_n)

is surjective with kernel $\bigcap_{i=1}^n I_i$ (the intersection of ideals ), hence induces an isomorphism

R/\bigcap_{i=1}^n I_i \ \cong\ \prod_{i=1}^n R/I_i

of quotient rings . The right-hand side is the ring structure on the cartesian product given componentwise.

Proof sketch (major case $n=2$ ): If $I+J=R$ , choose $a\in I$ , $b\in J$ with $a+b=1$ . Given residues $(\bar x,\bar y)\in R/I\times R/J$ , the element $r:=xb+ya$ maps to $(\bar x,\bar y)$ , proving surjectivity. Kernel elements are exactly those in both ideals, giving $\ker=\,I\cap J$ . The general $n$ -case follows by induction.