Chinese Remainder Theorem/Corollary

Theorem
Let $n_1, n_2, \ldots, n_r$ be positive integers such that $n_i \perp n_j$ for all $i \ne j$ (that is, $\gcd \left\{{n_i, n_j}\right\} = 1$).

Let $N = n_1 \cdots n_r$.

For an integer $k$, let $\Z/k\Z$ denote the ring of integers modulo $k$.

Then we have a ring isomorphism:
 * $\Z / N\Z \simeq \Z/ n_1 \Z \times \cdots \times \Z/n_r \Z$

Proof
Define a mapping:
 * $\phi : \Z / N\Z \to \Z/ n_1 \Z \times \cdots \times \Z/n_r \Z$

by
 * $\phi\left({ d \pmod N }\right) = \left({ d \pmod {n_1}, \ldots, d \pmod {n_r} }\right)$

Then $\phi$ is well defined, by Mappings Between Residue Classes.

By the definition of multiplication and addition in $\Z/k\Z$, $k \in \Z$ we have:
 * $\left({ a \pmod k }\right) + \left({ b \pmod k }\right) = \left({ a+b }\right) \pmod k$

and
 * $\left({ a \pmod k }\right) \cdot \left({ b \pmod k }\right) = \left({ a\cdot b }\right) \pmod k$

Thus taking $k = n_1,\ldots,n_r$ separately we see that $\phi$ is a ring homomorphism.

Let:
 * $\left({ a_1 \pmod {n_1},\ldots, a_r \pmod {n_r} }\right) \in \Z/ n_1 \Z \times \cdots \times \Z/n_r \Z$

By the Chinese Remainder Theorem there exists a unique $x \in \Z / N\Z$ such that:
 * $\phi\left({ x }\right) = \left({ a_1 \pmod {n_1},\ldots, a_r \pmod {n_r} }\right)$

Since such an $x$ exists, $\phi$ is surjective.

Since this $x$ is unique modulo $N$, it follows that $\phi$ is injective.