I would like to show that $\mathbb{Z}/8 \otimes_{\mathbb{Z}} \mathbb{Z}_{\langle 2 \rangle} \cong \mathbb{Z} / 8$.
If we let $S = \mathbb{Z} \setminus \langle 2 \rangle$, then $\mathbb{Z}/8 \otimes_{\mathbb Z} \mathbb{Z}_{\langle 2 \rangle} = \mathbb{Z} /8 \otimes_{\mathbb{Z}} \mathbb{Z} [S^{-1}] \cong \mathbb{Z}/8 [S^{-1}].$
So I would like to show that if $\displaystyle\frac{m}{s} \in \mathbb{Z}/8[S^{-1}]$ then there is some $y$ such that $\displaystyle\frac{m}{s} = \frac y 1$, i.e. there is some $y \in \mathbb{Z}/8$ such that $sy = m$. How can I guarantee that there is such a $y$? Any help would be appreciated.
Also it is clear to me that $\mathbb{Z} / 8 \otimes_{\mathbb{Z}} \mathbb{Z}_{\langle 2 \rangle} \cong \mathbb{Z}_{\langle 2 \rangle} / \langle 8 \rangle \mathbb{Z}_{\langle 2 \rangle}$, but I am not sure if this helps.