More `General' Chinese Remainder Theorem I know of the Chinese Remainder Theorem, but I get the feeling it is only the `basic' version of it.
From the Chinese Remainder Theorem, we know that, for $m,\in\mathbb{Z}$ such that $\text{gcd}(m,n)=1$, $C_{mn}\cong C_{m}\times C_{n}$. Is it then true that $C_{p_{1}^{r_{1}}\ldots p_{n}^{r_{n}}}\cong C_{p_{1}^{r_{1}}}\times\ldots\times C_{p_{n}^{r_{n}}}$, where $p_{i}^{r_{i}}$s are (distinct) prime numbers?
 A: Yes, of course. You can just set $m := p_1^{r_1}$ and $n = p_2^{r_2}p_3^{r_3}\cdot \ldots \cdot p_n^{r_n}$ and start a recursion from there on. The theorem is true in a way more general context, see https://en.wikipedia.org/wiki/Chinese_remainder_theorem#Generalization_to_arbitrary_rings for details. However, in this general setting, you only get an existence result, no algorithms to actually compute the isomorphism in general.
A: The other answers have already answered perfectly. 
For those who want an even more general version of the theorem, you can find it in Universal algebra : 
Let $A$ be an algebra, $\theta_0, \theta_1$ be two congruences on $A$ that are permutable (i.e. $\theta_0 \theta_1 = \theta_1\theta_0$), and such that $\theta_0\land \theta_1 = \Delta$, $\theta_0 \lor \theta_1 = \Omega$ ($\Delta= \{(a,a), a\in A\}, \Omega = \{(a,b), a,b\in A\}$). 
Then $A \simeq A/\theta_0 \times A/\theta_1$. 
The obvious morphism is well-defined because they are congruences, it is surjective because of the last condition and because they are permutable (because then, $\theta_0\lor\theta_1 = \theta_0\theta_1$), and it is injective because their g.l.b. is $\Delta$
A: Yes, it is. In fact, there is a even more general version. Let $A$ be a ring and $I,J$ two ideals of $A$. Then there is an obvious morphism
$$f:A\to \frac{A}{I}\times \frac{A}{J}$$
sending every $a$ to $(a+I,a+J)$. It is not difficult to check that $\mathsf{ker}(f)=I\cap J$. Now, you may see that if $I+J=A$ (i.e. if they are coprime) then $I\cap J=IJ$ and $f$ turns out to be surjective (it's a kind of Bezout Identity trick).
