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