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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?

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  • $\begingroup$ Yes, use induction. There is an even more general statement of the Chinese remainer theorem in ring theory, but that's not what you are after. Hint: take $m=p_1^{r_1}$ and $n=p_2^{r_2}\dots p_n^{r_n}$. Then go further with the second component $\endgroup$ – Mathematician 42 May 3 '17 at 11:48
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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.

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  • $\begingroup$ that's what I thought. Can you `cancel' any further? So is $C_{p^{r_{1}\cdot\ldots\cdot r_{n}}}\cong C_{p^{r_{1}}}\times\ldots\times C_{p^{r_{n}}}$? $\endgroup$ – PercyF2519 May 3 '17 at 11:58
  • $\begingroup$ Nope. For an example, take $C_4$ and $C_2 \times C_2$. In the first group you have an element of order four, in the second group you don't, so they can never be isomorphic. This argument works in general, you will not be able to find an element of order $p^{r_1 \cdot r_2 \cdot \ldots \cdot r_n}$ in your right hand side, unless $n=1$. $\endgroup$ – Dirk May 3 '17 at 12:00
  • $\begingroup$ That's brilliant. My hunch was correct. I finally understand. Thank you! $\endgroup$ – PercyF2519 May 3 '17 at 12:01
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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 $\Delta$

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  • $\begingroup$ Out of curiosity: Do you have any reference for this? $\endgroup$ – Stefan Perko May 13 '17 at 20:56
  • $\begingroup$ You can look up Grätzer's ~Universal Algebra~ which is the reference in these matters if I'm not mistaken $\endgroup$ – Max May 13 '17 at 21:36
  • $\begingroup$ I found it (Ch. 3 p. 120), thank you. $\endgroup$ – Stefan Perko May 13 '17 at 22:14
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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).

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