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

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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    Yes, please provide a bit more context about your efforts in solving the question so far.2012-11-16

2 Answers 2

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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?