Possible Duplicate:
Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$
Why is $\mathbb{Z}_{n} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = \mathbb{Z}_{(n,m)}$?
I can see when $(m,n)=1$, we have $\mathbb{Z}_{n} \otimes_{\mathbb{Z}} \mathbb{Z}_{m} = 0$, since $n(x\otimes y)=(nx)\otimes y =0$ and $m(x\otimes y)=x\otimes(my)=0$ and there exists $a$ and $b$ such that $am+bn=1$.
But I have no idea how to deal with the general case.
Can you please help? Thank you!