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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.

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There is a natural ring surjective homomorphism ${\mathbf Z}_{(2)} \rightarrow {\mathbf Z}/8$, with which one gets a multiplication map ${\mathbf Z}/8 \times {\mathbf Z}_{(2)} \rightarrow {\mathbf Z}/8$ by reducing the second factor mod 8 and then multiplying. This leads to a ${\mathbf Z}$-linear map ${\mathbf Z}/8 \otimes_{\mathbf Z} {\mathbf Z}_{(2)} \rightarrow {\mathbf Z}/8$. Show that is an isomorphism. –  KCd May 27 '12 at 2:43
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up vote 5 down vote accepted

Note that $\mathbb{Z}[S^{-1}]$ consists of the rationals that can be written with an odd denominator. Note also that we want $sm=y$ in $\mathbb{Z}/8\mathbb{Z}$; that is, $sy\equiv m\pmod{8}$.

If $\frac{m}{s}\in\mathbb{Z}[S^{-1}]$, then $s$ is odd; in particular, $s^2\equiv 1\pmod{8}$, so we can take $y=sm$.

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