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We know that $\mathbb Z/{mn}\simeq \mathbb Z/m\times\mathbb Z/n$ as groups iff $(m, n)=1$.

But what about as a ring?

Will $\mathbb Z/{mn}\simeq \mathbb Z/m\times\mathbb Z/n$ hold good if $(m, n)=1$ and $\mathbb Z/{mn}, \mathbb Z/m, \mathbb Z/n$ are rings ?

I am unable to prove it but I assume it should be true. But no idea how to establish. Please help me

Martin Brandenburg
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KON3
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    This follows from the Chinese remainder theorem for rings. See also http://math.stackexchange.com/questions/375348/prove-that-mathbb-z-m-times-mathbb-z-n-cong-mathbb-z-mn-implies-gcd for the converse. – Dietrich Burde Sep 05 '14 at 11:12
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    I've corrected the notation (which is, unfortunately, considered to be correct by many authors). $\mathbb{Z}_n$ is the ring of $n$-adic integers, but you meant $\mathbb{Z}/n$. And $\oplus$ is the coproduct, but you meant the product $\times$. – Martin Brandenburg Sep 05 '14 at 11:21
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    Oh, I didn't expect that even Wikipedia is on my side: http://en.wikipedia.org/wiki/Direct_sum#Direct_sum_of_rings – Martin Brandenburg Sep 05 '14 at 11:28
  • @MartinBrandenburg OMG. I really didn't know that sir. I studied Contemporary Abstract Algebra by gallian and have used that notation only. Thank you for the correction. Between, will you please share with me some links on $n-adic$ integres and coproduct? I mean I want to clear my confusion on these notations. – KON3 Sep 05 '14 at 11:29
  • @DietrichBurde thanks a lot for the link. – KON3 Sep 05 '14 at 11:29
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    The ring of $n$-adic integers is $\mathbb{Z}_n = \varprojlim_k ~ \mathbb{Z}/n^k$, usually only studied (for good reasons) when $n=p$ is a prime. See http://en.wikipedia.org/wiki/P-adic_number . For $\oplus$, see http://en.wikipedia.org/wiki/Coproduct and http://en.wikipedia.org/wiki/Direct_sum – Martin Brandenburg Sep 05 '14 at 11:31
  • Moral of the story. Till today, I had a wrong interpretation that (i) external direct product and external direct sum are nothing but just "synonyms" (ii) $\mathbb Z_n$ is always the same set no matter if it is group or ring.

    I don't know whether I should be upset or not after posting this question as it made me realized how far behind I am !!!

     – KON3 Sep 05 '14 at 11:50
  • Some authors are far behind, not you. – Martin Brandenburg Sep 05 '14 at 12:08
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    Dear @AnjanDebnath Please do not feel bad. In your case, the notation was purely cosmetic and was not really interfering with the meaning of the question (at least, for the lion's share of the audience). There will probably come a day when a question you ask truly hinges upon notational meaning, but until then do not become stressed by visitations of extreme notational enforcement. Regards – rschwieb Sep 05 '14 at 12:50
  • @Martin I don't think it is a good idea to remove the original subscripted quotient notation. Nowadays that is very widely used in elementary number theory textbooks, whereas the adic denotation is much more esoteric. – Bill Dubuque Sep 05 '14 at 15:18

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Yes, the canonical map $\mathbb{Z}/mn \to \mathbb{Z}/m \times \mathbb{Z}/n$ is obviously also a ring homomorphisms, and as you know bijective, hence a ring isomorphism.

More generally, if $I,J$ are coprime ideals of a ring $R$, then $R/(I \cap J) \cong R/I \times R/J$. This is the Chinese Remainder Theorem for rings. (If $I,J$ are not coprime, we still have $R/(I \cap J) \cong R/I \times_{R/(I+J)} R/J$ by the way.)

Martin Brandenburg
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