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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!

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    @joriki: I'm sorry for duplication. I did search (using keyword 'tensor product cyclic') before asking but I failed to reach that result. Is there a way to 'raise' that question again since it was asked a month ago and seems not active?2011-11-19

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