Zero element of tensor product of $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}/m\mathbb{Z}$ Let $\bar{a} \otimes 1 \in \mathbb{Z}/n\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/m\mathbb{Z}$ where $m,n \neq \pm 1$.

Question: Does $\bar{a} \otimes 1 = 0$ imply that $\bar{a} = 0$?

Approach:


*

*I know that $\bar{a} \otimes 1 = a \cdot (1 \otimes 1)$ due to the properties of the tensor product where $a \in \mathbb{Z}$ is any lift of $\bar{a} \in \mathbb{Z}/n\mathbb{Z}$.

*I also know that the zero element in the tensor product can be written as $0 \otimes 1 = 1 \otimes 0$.


This question seems so trivial, but I cannot make sense of this. Could you please help me with this question? Thanks a lot!

No: you can easily check that $\mathbb{F}_2 \otimes_{\mathbb{Z}} \mathbb{F}_3=0$. 
