Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is the converse of the Chinese Remainder Theorem true? That is, if $$(m, n)\neq1,$$ then $$\mathbb{Z}/mn\mathbb{Z}\ncong\mathbb{Z}/m\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z}.$$

Thanks.

share|improve this question
2  
@Arturo: I would normally phrase the CRT as "if $(m,n)=1$, then $\mathbb{Z}/mn\mathbb{Z} \cong \mathbb{Z}/m\mathbb{Z} \oplus \mathbb{Z}/n\mathbb{Z}$", and using this formulation, the converse would be the statement "if $\mathbb{Z}/mn\mathbb{Z}\cong\mathbb{Z}m\mathbb{Z}\oplus\mathbb{Z}/n\mathbb{Z}$, then $(m,n)=1$", while the inverse is the statement in the question. Of course the inverse and converse are logically equivalent, so it does not matter too much; in fact I imagine "converse" is what most people who would want to find this post would search for. –  Zev Chonoles Nov 28 '11 at 0:43
    
@Zev: The statement here is contrapositive of the converse; I don't think I've never heard $\neg P\to \neg Q$ refered to as the "the inverse" of $P\to Q$, but that could just be my lack of familiarity. –  Arturo Magidin Nov 28 '11 at 0:44
1  
@Arturo: I can indeed imagine it is not a universal term; and I myself would prefer to call it the contrapositive of the converse. But it is a term that I have heard before - and here is the relevant Wikipedia page. Again, it certainly doesn't change the real meaning, and moreover I think "converse" will be more helpful to people searching for this question. –  Zev Chonoles Nov 28 '11 at 0:48
add comment

2 Answers

up vote 10 down vote accepted

Yes. The direct sum has no element of order $mn$.

share|improve this answer
    
And how about the generalisation of the Chinese remainder theorem in terms of rings? Could you please tell something about that? –  Koenraad van Duin Sep 25 '13 at 7:33
add comment

HINT $\rm\ \ \mathbb Z/m\: \oplus\: \mathbb Z/n\ $ has characteristic $\rm\:lcm(m,n),\:$ which is $\rm\: < m\ n\ $ if $\rm\:\gcd(m,n) > 1\:.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.