d\Bbb Z$ when $n$ and $m$ have greatest common divisor $d$.

Asked May 10 '24 at 12:40

Active May 10 '24 at 13:38

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I have a question about tensor products. When $n$ and $m$ are positive integers with greatest common divisor $d$, how do I prove that: $$(\Bbb Z/n\Bbb Z)\otimes_{\Bbb Z}(\Bbb Z/m\Bbb Z)\cong\Bbb Z/d\Bbb Z.$$

I have tried creating an isomorphism from scratch, or tried to prove that most elements of the tensor product are equal to zero in some way but I cannot figure it out.

edited May 10 '24 at 13:38

Anne Bauval

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asked May 10 '24 at 12:40

DueledGalois

https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity – May 10 '24 at 12:42
3

Does this answer your question? Proof of $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}$ (found via Tensor product of cyclic groups (as $\mathbb{Z}$-mod): why is $\mathbb{Z}{n} \otimes{\mathbb{Z}} \mathbb{Z}{m} = \mathbb{Z}{(n,m)}$?, using the search engine https://math.stackexchange.com/search?q=tensor+product+of+cyclic+groups) – Anne Bauval May 10 '24 at 13:02
I still doubt recalling this tool (elementary and obviously known to the OP, and not even mentionned in any of the various answers) was helpful. The whole difficulty was in these various abstract nonsenses. – Anne Bauval May 10 '24 at 13:46
1

Apologies, you are right both used it explicitely. – Anne Bauval May 10 '24 at 14:12

How to prove $(\Bbb Z/n\Bbb Z)\otimes_{\Bbb Z}(\Bbb Z/m\Bbb Z) \cong\Bbb Z/d\Bbb Z$ when $n$ and $m$ have greatest common divisor $d$.

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