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.
Asked
Active
Viewed 1,261 times
5
-
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 00:43
-
@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 00:44
-
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 00:48
2 Answers
11
Yes. The direct sum has no element of order $mn$.
hmakholm left over Monica
- 291,611
-
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 07:33
5
HINT $\rm\ \ \mathbb Z/m\: \oplus\: \mathbb Z/n\ $ has characteristic $\rm\:lcm(m,n),\:$ which is $\rm\: < m\ n\ $ if $\rm\:\gcd(m,n) > 1\:.$
Bill Dubuque
- 282,220