Chinese remainder theorem

Let $R$ be a commutative ring with identity. If $I_1,\mathrm{\dots },I_n$ are ideals of $R$ such that $I_i+I_j=R$ whenever $i\ne j$, then let

$$I={\cap }_{i=1}^n{I}_i=\prod _{i=1}^n{I}_i.$$

The sum of quotient maps $R/I\to R/I_i$ gives an isomorphism

$$R/I\cong \prod _{i=1}^{n}R/I_i.$$

This has the slightly weaker consequence that given a system of congruences $x\cong a_i(mod{I}_i)$, there is a solution in $R$ which is unique mod $I$, as the theorem is usually stated for the integers.