Understanding proof that $\mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/d\mathbb{Z}$

Question

I am reading an example in Dummit and Foote showing that $\mathbb{Z}/m\mathbb{Z}\otimes\mathbb{Z}/n\mathbb{Z}\cong \mathbb{Z}/d\mathbb{Z}$, where $d$ is the greatest common divisor of $n$ and $m$, but I'm having trouble understanding one line. We have that $m(1\otimes 1)=m\otimes 1=0$, and similarly $n(1\otimes 1)=(1\otimes n)=0$. The next line states that because of this, we have $d(1\otimes 1)=0$ as well. Why is this so?

Answer 1

There exist $s,t\in\mathbb{Z}$ such that $sm+tn=d$. So $$d(1\otimes 1)=(sm+tn)(1\otimes 1)=sm(1\otimes 1)+tn(1\otimes 1)=s\cdot 0+t\cdot 0=0.$$