$m\mathbb{Z}\cdot \mathbb{Z}/n\mathbb{Z}$

Question

I'm trying to find the tensor product $\mathbb{Z}/m\mathbb{Z}\otimes \mathbb{Z}/n\mathbb{Z}$ using this theorem. Now I know what the end product is (namely: $\mathbb{Z}/\gcd(m,n)\mathbb{Z}$) and I proved it differently. However,

how do I get that $m\mathbb{Z}\cdot \mathbb{Z}/n\mathbb{Z}\cong \gcd(m,n)\mathbb{Z}/n\mathbb{Z}$?

Answer 1

More generally, if $I$ and $J$ are ideals of the commutative ring $R$, we have $$ I\cdot R/J=(I+J)/J $$ The proof is easier to formalize in the more abstract setting and is a simple verification.

In your case $I=m\mathbb{Z}$ and $J=n\mathbb{Z}$, so $$ I+J=m\mathbb{Z}+n\mathbb{Z}=\gcd(m,n)\mathbb{Z} $$