Originally Posted by

**lisa** Hello,

I have to proof that: $\displaystyle \mathbb Z/n\mathbb Z \otimes \mathbb Q = 0$

I proofed it as follows but I don't know if it is correct:

$\displaystyle {\text{Let }}a \in \mathbb{Z},{\text{ }}b \in \mathbb{Q}.$

$\displaystyle (a + n\mathbb{Z}) \otimes (b) = a(1 + n\mathbb{Z}) \otimes b(n \cdot {n^{ - 1}}) = (1 + n\mathbb{Z}) \otimes ab(n \cdot {n^{ - 1}}) \hfill \\$

$\displaystyle = (1 + n\mathbb{Z}) \otimes abn(1 \cdot {n^{ - 1}})

= abn(1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = (abn + n\mathbb{Z}) \otimes ({n^{ - 1}}) \hfill \\ $

$\displaystyle = (0 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \cdot (1 + n\mathbb{Z}) \otimes ({n^{ - 1}}) = 0 \hfill \\ $

Please correct me, if there are any mistakes.

Thank you!

-Lisa-