How can we prove that $\DeclareMathOperator\Hom{Hom}\mathbb Z/n\mathbb Z\bigotimes_{\mathbb Z}\mathbb Z/m\mathbb Z \cong \Hom(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)$ without using the fact that $\mathbb Z/n\mathbb Z\bigotimes_{\mathbb Z}\mathbb Z/m\mathbb Z \cong \mathbb Z/d \mathbb Z$ and $\Hom(\mathbb Z/n\mathbb Z, \mathbb Z/m\mathbb Z)\cong \mathbb Z/d \mathbb Z$, where $d=$ GCD$(m,n)$? It's not too hard to prove the two latest isomorphisms but it would be interesting to derive one from another using the first isomorphism.
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1Are you looking at these things as groups? rings? modules? something fourth? – hmakholm left over Monica Jan 02 '15 at 00:13
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1@HenningMakholm: groups ($\mathbf{Z}$-modules) obviously – Mister Benjamin Dover Jan 02 '15 at 00:14
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@313: In that case, what does the subscript on the tensor stand for? – hmakholm left over Monica Jan 02 '15 at 00:14
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1@HenningMakholm: $\mathbf{Z}$ is the ring of integers. It indicates that these are regarded as $\mathbf{Z}$-modules = abelian groups. (I know, it is late in denmark.) – Mister Benjamin Dover Jan 02 '15 at 00:16
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2Would it work to let $\phi\in \operatorname{Hom}(\Bbb Z_n, \Bbb Z_m)$ map to $1\otimes \phi(1)$? Is that a homomorphism? Is it bijective? (It's late here in Norway too.) – Arthur Jan 02 '15 at 00:22
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@Arthur: no, it is time to go to bed. If $n=2$, $m=4$ for $\phi$ in the hom we have $\phi=0$ or $\phi(1)=2$. So in this case the homomorphism you propose cannot be surjective, for the image does not contain the generator $1\otimes 1$ of the tensor product. But nice try – Mister Benjamin Dover Jan 02 '15 at 00:40
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Oh well. Anyways, if you don't want to prove that they're both isomorphic to $\Bbb Z_d$, then I think the best approach is to provide an explicit isomorphism. – Arthur Jan 02 '15 at 00:50
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Consider the short exact sequence $0\to\mathbb{Z}\xrightarrow{n} \mathbb{Z}\to \mathbb{Z}/n\to 0$, where the map $\mathbb{Z}\xrightarrow{n}\mathbb{Z}$ is multiplication by $n$. Applying the functors $-\otimes_\mathbb{Z}\mathbb{Z}/m$ and ${\rm Hom}(-,\mathbb{Z}/m)$ to it gives the two exact sequences: $$ \mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m\to \mathbb{Z}/n\otimes_\mathbb{Z}\mathbb{Z}/m\to 0\\ 0\to {\rm Hom}_\mathbb{Z}(\mathbb{Z}/n,\mathbb{Z}/m)\to \mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m $$ showing that $\mathbb{Z}/n\otimes_\mathbb{Z}\mathbb{Z}/m$ is the cokernel of $\mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m$ and that ${\rm Hom}_\mathbb{Z}(\mathbb{Z}/n,\mathbb{Z}/m)$ is the kernel of the same map: $\mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m$. The kernel and cokernel are cyclic abelian groups, so it suffices to show that they have the same cardinality. For completeness:
Let $d$ be the size of the cokernel, and $e = |{\rm im}(n)|$, the size of the image of $\mathbb{Z}/m\xrightarrow{n}\mathbb{Z}/m$. Then by definition, $m/e = d$. On the other hand, the short exact sequence $0\to \ker(n)\to \mathbb{Z}/m\xrightarrow{n}{\rm im}(n)\to 0$ shows also that $m/|\ker(n)| = e$, whence $d = m/e = |\ker(n)|$.
Of course, with a little more work, you can then prove the greatest common divisor statement for both now.