Prove that $\mathbb{Z}_m \otimes \mathbb{Z}_n\cong \mathbb{Z}_d$

Question

Show that $\mathbb{Z}_m \otimes \mathbb{Z}_n \cong \mathbb{Z}_d$, for $m,n\in\mathbb{Z}$ and $d=\gcd(m,n)$.

I've looked at tensor but couldn't find a detailed solution to this problem. I would be very grateful if you could help me with this question.

Thanks.

Answer 1

It suffices to show that the tensor product $\mathbb{Z}_{m} \otimes \mathbb{Z}_{n}$ is isomorphic to a subgroup of $\mathbb{Z}_{d}$ and that no smaller multiple of $1 \otimes 1$ is zero.

$\mathbb{Z}$ is a principal ideal domain and $m(1\otimes 1) = n(1\otimes 1) = 0$, so there exist $x$ and $y$ for which

$$\gcd(m, n) = d = xm + yn.$$

This implies $d(1 \otimes 1) = (xm + yn)(1\otimes 1) = x(m(1\otimes 1)) + y(n(1\otimes 1)) = 0,$ which means that the tensor product $\mathbb{Z}_{m} \otimes \mathbb{Z}_{n}$ is isomorphic to a subgroup of $\mathbb{Z}$.

Define a mapping $f : \mathbb{Z}_{m} \times \mathbb{Z}_{n}$ so that $f(a, b) = ab$. Also, let $g$ be the induced $R$-linear mapping. Observe that $g$ maps $1 \otimes 1$ to $1$. The equality $c(1 \otimes 1) = 0$ means that $d$ divides $c$, and the result follows.