$\Bbb Z_{(p)}\otimes_{\Bbb Z} \Bbb Z/q^k$ is zero?

Question

For primes $p\neq q$, is it true that $\Bbb Z_{(p)}\otimes_{\Bbb Z} \Bbb Z/q^k$ is zero? If the answer is yes, how do I prove it?

Answer 1

Hint: $a\otimes b = (n\cdot\frac{a}{n})\otimes b = n(\frac{a}{n}\otimes b) = \frac{a}{n}\otimes (nb)$, if $n\in\Bbb Z$ isn't divisible by $p$.

Answer 2

$\mathbf Z_{(p)}\otimes_\mathbf Z \mathbf Z/q^k\simeq \mathbf Z_{(p)}/q^k\mathbf Z_{(p)} $. Now $\;q^k\mathbf Z_{(p)}=\mathbf Z_{(p)}$ since $q$, hence $q^k\notin(p)$.