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I wonder if for coprime integers $p,q$ the $\mathbb{Z}/p\mathbb{Z}$-module $\mathbb{Z}/q\mathbb{Z}$ is always equal to zero.

Is that true?

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    $\mathbb{Z}/q\mathbb{Z}$ is neither a $\mathbb{Z}/p\mathbb{Z}$-module nor equal to zero – 2017-02-11
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    If $p$ and $q$ are coprime, $\mathbf Z/q\mathbf Z$ is *not* a $\mathbf Z/P\mathbf Z$-module. Scalar multiplication is not well-defined. – 2017-02-11
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    Do you have an example that shows that this is not well-defined? – 2017-02-11
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    It's hard to give an example of something not making sense, but there is no way that $\mathbb Z/3\mathbb Z$ is a $\mathbb Z/2\mathbb Z$ module. It is true that, given any abelian group $A$, and any commutative ring $R$ you can define an $R$-module $A\otimes R$, and in that sense, $\mathbb Z/3\mathbb Z\otimes \mathbb Z/2\mathbb Z=0$. – 2017-02-11

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As other have said, your question does not quite make sense as stated. In particular, $\Bbb Z/p\Bbb Z$ cannot be taken to be a non-trivial $\Bbb Z/q\Bbb Z$ module in the usual sense.

For example, take $p = 3$ and $q = 5$. Suppose that $[1]_5 \cdot [1]_3$ is a non-zero element of $\Bbb Z_3$ (where $[k]_n$ is the equivalence class of $k$ modulo $n$). It would follow that $$ 5 \cdot ([1]_5 \cdot [1]_3) = [2]_3 \cdot ([1]_5 \cdot [1]_3) \neq 0 $$ However, the definition of a module implies $$ 5 \cdot ([1]_5 \cdot [1]_3) = (5 \cdot [1]_5)\cdot [1]_3 = 0 \cdot [1]_3 = 0 $$ which is a contradiction.

What we can do, however, is define the formal products $[a]_5 \otimes [b]_3$ with the rules:

  • the product $\otimes$ is distributive with respect to both arguments
  • $[na]_5 \otimes [b]_3= [a]_5 \otimes [nb]_{3}$ for any $n \in \Bbb Z$.

That is, we can consider the tensor product $\Bbb Z/p\Bbb Z \otimes \Bbb Z/q \Bbb Z$. We then indeed find that $\Bbb Z/p\Bbb Z \otimes \Bbb Z/q \Bbb Z = \{0\}$.