Other proof for tensor product equal to zero (without use Nakayama lemma and Jacobson radical) 
Let $R$ is commutative local ring, $M$ is $R$-module, $N$ is $R$-module. If $M,N$ are finitely generated, how to prove:
$M \otimes_R N=0$  if and only if $M=0$ or $N=0$.
If we delete the condition (local ring), then is the statement right?

A proof is existed by Nakayama lemma. Suppose $M \otimes_R N=0$, $J$ is maximal ideal in $R$, let $k=\frac{R}{J}$ so $k$ is field. $M_k=M\otimes k=M \otimes \frac{R}{J}=\frac{M}{JM}$.
If we delete the local condition the statement is not true. Example: $(m,n)=1 $ then $\Bbb Z_m \otimes \Bbb Z_n=0 (???)$, $\Bbb Z_m, \Bbb Z_n \not = 0$.
 A: About the proof: is it clear why $M\otimes_R R/J\simeq M/JM$? If yes, then consider $M\otimes_R N=0$ and tensor it (over $R$) with $R/J$ left and right. Then you get $M/JM\otimes_{R/J} N/JN=0$. Thus you have two vector spaces, namely $M/JM$ and $N/JN$, whose tensor product is $0$. But non-zero vector spaces have bases, so their tensor product, and then we have to admit that $M/JM=N/JN=0$. Now is time for Nakayama's lemma to enter the scene.
For the non-local case: $\mathbb Z_m=\mathbb Z/m\mathbb Z$, and using again that $M/IM\simeq M\otimes_R R/I$ we get $\mathbb Z_m\otimes_{\mathbb Z}\mathbb Z_n\simeq\mathbb Z_n/m\mathbb Z_n$. But the last one is isomorphic to $\mathbb Z/(m\mathbb Z+n\mathbb Z)$, and since $m\mathbb Z+n\mathbb Z=\gcd(m,n)\mathbb Z$ you are done.
A: For the non-local case:
$\mathbb{Z}_m\times\mathbb{Z}_n$ is generated by $1\otimes 1$. So, we just have to prove that $1\otimes 1=0$.
As $(m,n)=1$, by Bezout, there exist integers $a$ and $b$ such that $am+bn=1$. Then
$$1\otimes1=(am+bn)\otimes1= am\otimes1 + bn\otimes 1 = 0 + b\otimes n = 0.$$
