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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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Roun
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    See http://math.stackexchange.com/questions/72284 – Zev Chonoles Nov 19 '11 at 04:35
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    @ZevChonoles: But that page seems only containing an incomplete proof. How to show that $\mathrm{ker}(g) \subset (\mathrm{gcd}(m,n))$ ? – Roun Nov 19 '11 at 04:56
  • Try tensoring the exact sequence $0\rightarrow \mathbb{Z}\rightarrow\mathbb{Z}\rightarrow\mathbb{Z}_m\rightarrow 0$ with $\mathbb{Z}_n$. –  Nov 19 '11 at 05:18
  • There seems to be a widespread belief that questions should only be closed as duplicates if they've been answered somewhere. The point of closing duplicates is to keep everything related to one question in one place; this is independent of whether the question has been answered. This question should be closed as a duplicate, and if there is anything to add to the answers, it should be added at the original question. – joriki Nov 19 '11 at 09:38
  • @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? – Roun Nov 19 '11 at 11:49

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