# What can we say about $(m,n)$ if $\frac{\mathbb{Z}}{m \mathbb{Z}} \otimes \frac{\mathbb{Z}}{n \mathbb{Z}}=0$?

Let $m,n$ be relatively prime integers. Then we know that $\frac{\mathbb{Z}}{m \mathbb{Z}} \otimes \frac{\mathbb{Z}}{n \mathbb{Z}}=0$.

Conversely, if $m,n$ are integers such that $\frac{\mathbb{Z}}{m \mathbb{Z}} \otimes \frac{\mathbb{Z}}{n \mathbb{Z}}=0$, then what can we say about $(m,n)$, except from the trivial fact that the pair $(1+m \mathbb{Z},1+n \mathbb{Z})$ is generated by bilinear relations?

-
I wonder why $\mathbb{Z}/n \otimes \mathbb{Z}/m \cong \mathbb{Z}/\mathrm{gcd}(n,m)$ hasn't been mentioned yet explicitly. –  Martin Brandenburg Mar 16 '13 at 1:38

Note that $M \otimes \mathbb Z/n\mathbb Z = M/nM$ so if $M \otimes \mathbb Z/n\mathbb Z = 0$ then $nM = M$.

Now show that if $1 + m\mathbb Z \in n(\mathbb Z/m\mathbb Z)$ then $(n, m) = 1$. More generally, you can show that $n(\mathbb Z/m\mathbb Z) = \gcd(n, m)\mathbb Z/m\mathbb Z$ (as a submodule of $\mathbb Z/m\mathbb Z$).

-
Excellent! Thanks :) –  Manos Mar 15 '13 at 20:48

Using the Chinese decomposition and the fact that $\otimes$ distributes over $\oplus$, you can show that $(n,m) = 1$ if $$\mathbb Z / n \mathbb Z \otimes_{\mathbb Z} \mathbb Z / m \mathbb Z = 0.$$

-
Thanks, let me try that. –  Manos Mar 15 '13 at 20:31

Note that in general, $R/I \otimes R/J\cong R/I+J$, where $I$ and $J$ are ideals of the ring $R$. So if $R/I \otimes R/J=0$, then $R/I+J=0$, which implies that $I+J=R$. You can specialize this to your case.

-
What do you mean by the notation $R/I + J$? –  Manos Mar 18 '13 at 14:29
It is meant $R/(I+J)$. –  Martin Brandenburg Mar 18 '13 at 15:08
@Manos, Martin is correct, and it generalizes what you ask for commutative rings. I mean that if $R/I\otimes R/J=0$, then $I$ and $J$ are comaximal, i.e $I+J=R$. –  user66532 Mar 19 '13 at 10:05

If $d$ divides $n$, then a straightformard calculation gives $\def\Z#1{\mathbb Z/#1\mathbb Z}\def\x{\otimes}\Z n\x\Z d\cong\Z d$.

Now if $d>1$ is a common divisor of $n,m$, then $$\Z n\x\Z m\x\Z d\cong\Z m\x\Z d\cong\Z d,$$ so in particular $\Z n\x\Z m\not\cong0$. Therefore $\Z n\x\Z m\cong0$ implies $\gcd(n,m)=1$.

-