Why is $\mathbb{Q} \otimes \mathbb{Z}/2\mathbb{Z} = 0$?

Question

I’m trying to verify the remark that $\mathbb{Q} \otimes \mathbb{Z}/2\mathbb{Z} = 0$, by using the universal property of tensor products. Here all modules are seen as $\mathbb{Z}$-modules. So for a $\mathbb{Z}$-module $P$ and bilinear maps $f: \mathbb{Q} \times \mathbb{Z}/2\mathbb{Z} \to P$ and $g: \mathbb{Q} \times \mathbb{Z}/2\mathbb{Z} \to \mathbb{Q} \otimes \mathbb{Z}/2\mathbb{Z}$, there should be an unique $\mathbb{Z}$-linear map $f’: \mathbb{Q} \otimes \mathbb{Z}/2\mathbb{Z} \to P$ such that $f = f’\circ g$.

I’ve tried letting $q \otimes \bar{z} \ne 0$ (so either $q \ne 0_{\mathbb{Q}}$ or $\bar{z} = \bar{1}$) to see what would happen, but both bilinearity of $f$ and $g$ and the uniqueness of $f’$ still seem to go through. Any comments would be appreciated.

Answer 1

Even if $q \ne 0$ and $\bar{z} = \bar{1}$, $q \otimes \bar{z}$ is always identical to $0$.

This is because of properties of tensor product. For any $\mathbb{Z}$-linear $f : \mathbb{Q} \times \mathbb{Z}/2 \to P$, it holds that \begin{align} f(q, \bar{z}) = 2f\left(\frac{1}{2}q, \bar{z} \right) = f\left(\frac{1}{2}q, 2\bar{z} = \bar{0}\right) = 0 \end{align}

Hence, we have \begin{align} q \otimes \bar{z} = 0 \end{align} for any $q$ and $\bar{z}$.