Zero element of tensor product of $\mathbb{Z}/n\mathbb{Z}$ and $\mathbb{Z}/m\mathbb{Z}$

Question

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!

Answer 1

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

Answer 2

No, because $\;\mathbf Z/n\mathbf Z\otimes_{\mathbf Z}\mathbf Zm\mathbf Z\cong \mathbf Z/(m,n)\mathbf Z $.

So it only means that $a\in(m,n)$, i.e. $a$ is divisible by $\gcd(m,n)$.