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Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$

How to compute $\mathbb{Z}/m\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/n\mathbb{Z}$ and $Hom_{\mathbb{Z}}(\mathbb{Z}/m\mathbb{Z}, \mathbb{Z}/n\mathbb{Z})$? Detailed solution please. I would like to compare it with my answer which is $\mathbb{Z}/gcd(m,n) \mathbb{Z}$.

Thank you

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marked as duplicate by Thomas, TMM, Andrew, BenjaLim, Cameron Buie Nov 16 '12 at 21:47

This question was marked as an exact duplicate of an existing question.

Well, how did you arrive at your answer? What do you know about tensor products? – Derek Allums Nov 16 '12 at 20:55
This looks like a homework (or homework-type) question, and in cases like this, and for such questions, it is customary here to provide hints rather than complete solutions, especially detailed solutions. Also, in case of any question, especially a homework question you should let others know your thoughts and what you have tried so far (so that they know better how to direct you, or how elementary the answers should be to be not too elementary, and not too advanced for you). – tomasz Nov 16 '12 at 21:24
Also, it is acceptable to post your own proof and ask if it is correct (either in the question itself, or as an answer to your own question; I think the latter is more coherent with the style of this site, but a lot of people do the former, and it would probably get you more views...). – tomasz Nov 16 '12 at 21:27
Yes, please provide a bit more context about your efforts in solving the question so far. – Simon Hayward Nov 16 '12 at 21:27

See here.

(I thought I might as well make my comment into an answer.)

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One strategy is to show on one hand that the size of $\mathbb{Z}/m\mathbb{Z} \otimes \mathbb{Z}/n\mathbb{Z}$ is at most that of $\mathbb{Z} / \text{gcd}(m,n)\mathbb{Z}$ and, on the other hand, that the size of the former is at least the size of the latter.

The first direction is easy, and I'll omit it (please feel free to ask if you struggle with it).

For the other direction, we can use the universal mapping property of the tensor product. Here's a hint on how to do it. Suppose we have a bilinear map $f : \mathbb{Z}/m\mathbb{Z} \times \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/\text{gcd}(m,n)\mathbb{Z}$. Then there is a unique linear map $g : \mathbb{Z}/m\mathbb{Z} \otimes \mathbb{Z}/n\mathbb{Z} \rightarrow \mathbb{Z}/\text{gcd}(m,n)\mathbb{Z}$ so that $f(x,y) = g(x\otimes y)$.

Can you find an $f$ so that the resulting $g$ is surjective? What about some sort of multiplication modulo the gcd?

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