Can Tensor Products Reveal Ring Isomorphisms?

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Here is my first problem!

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Show that $\Bbb Z/m\Bbb Z \otimes_{\Bbb Z} \Bbb Z/n\Bbb Z$ is isomorphic to $\Bbb Z/\text{gcd}(m,n)\Bbb Z$.

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No one solved this week's problem. You can find my solution below.

Spoiler

Let $d = \text{gcd}(m, n)$. Then $d$ divides $m$ and $n$, so there is a well defined $\Bbb Z$-bilinear map $([x]_m, [y]_n) \mapsto [xy]_d$ from $\Bbb Z/m\Bbb Z \times \Bbb Z/nZ$ to $\Bbb Z/d\Bbb Z$. It induces a homomorphism $f : \Bbb Z/m\Bbb Z \otimes_{\Bbb Z} \Bbb Z/n\Bbb Z \to \Bbb Z/d\Bbb Z$ such that $f([x]_m \otimes [y]_n) = [xy]_d$. Consider the homomorphism $x \mapsto x([1]_m \otimes [1]_n)$ from $\Bbb Z$ to $\Bbb Z/m\Bbb Z \otimes_{\Bbb Z} \Bbb Z/n\Bbb Z$. Its kernel contains $d\Bbb Z$, since $d$ is a $\Bbb Z$-linear combination of $m$ and $n$, $[1]_m \otimes [m]_n = [m]_m \otimes [1]_n = 0$ and $[1]_m \otimes [n]_n = 0$. So it induces a homomorphism $g : \Bbb Z/d\Bbb Z \to \Bbb Z/m\Bbb Z \otimes_{\Bbb Z} \Bbb Z/n\Bbb Z$ such that $g([x]_d) = [1]_m \otimes [x]_n$. Now

$$fg([x]_d) = f([1]_m \otimes [x]_n)) = [x]_d$$

and

$$ gf([x]_m \otimes [y]_n) = g([xy]_d) = [1]_m \otimes [xy]_d = [x]_m \otimes [y]_n,$$

whence $f$ and $g$ are inverses of each other. Therefore $f$ is an isomorphism.