Apart from the Chinese Remainder Theorem for rings, modules cited by the others, there actually also is a much more general version (for groups it seems to give something different from Pete's Sylow group version):

There is a version of the Chinese Remainder Theorem which is valid for general algebraic structures, after a suitable reformulation:
An *algebra* is a set $M$ with some $n$-ary operations $f:M^n \rightarrow M$ (for varying $n$), possibly required to satisfy some equations between them. A *homomorphism* is a map preserving these operations, a *congruence relation* is a binary relation $R \subseteq M \times M$ which is of the form {$(x,y)|g(x)=g(y)$}$=:Ker\ g$ for some homomorphism $g$.
 
In the special case of ring theory these notions would be rings, ring homomorphisms and the relations {$(x,y)|x-y \in I$} for ideals $I$. As in ring theory, congruence relations are equivalence relations and the quotient set carries an algebra structure of the same kind as the original set, given by applying the old operations to equivalence classes. Intersection of congruence relations is a congruence relation again. One also can define products of algebras in the obvious way and it still is true that a homomorphism has an inverse homomorphism iff it is bijective.

Now the Chinese Remainder Theorem says:

Given algebras $A$ and $A_i \ (i \in I)$ and homomorphisms $f_i:A \rightarrow A_i$, then $f:A \rightarrow \Pi_{i \in I} A_i$ is injective if and only if $\bigcap_{i \in I}Ker\ f_i=\Delta_A$, the diagonal, i.e. the minimal congruence relation.

To see how this contains the Chinese Remainder Theorem as you know it, consider the maps $f_i:A \rightarrow A_i=A/I_i$ to be quotient maps by congruence relations (for rings: ideals). Assume that the map into the product is surjective (for rings this is the same as saying that the ideals are coprime). The kernel of the map $f$ into the product is the intersection of the individual kernels. Thus $f$ factors through $\bar{f}:A/\cap Ker\ f_i \rightarrow \Pi_{i \in I}A_i$ where $\bar{f}$ is of course still surjective. Now apply the theorem to $\bar{f}$; as we already factored out the intersection of the original kernels, the intersection of the kernels in the quotient algebra is the diagonal. So $\bar{f}$ is injective and thus an isomorphism.

You can read about this general setup in the book "Universal Algebra" by Burris and Sankappanavar, freely available [here][1]. The Chinese Remainder Theorem is Theorem 7.15 there.


  [1]: http://www.math.uwaterloo.ca/~snburris/htdocs/ualg.html