# Does $M \otimes \mathbb{Z} = M$ for any module M?

Does $$M \otimes \mathbb{Z} = M$$ for any $$R$$-module $$M$$, where we have some $$R$$-module structure on $$\mathbb{Z}$$?

My thinking: We have a map $$\phi:M \otimes \mathbb{Z} \rightarrow M$$, defined as follows on simple tensors:

$$\phi(m\otimes z) = z \cdot m$$,

which is clearly surjective, and I am fairly sure it is injective, as for a sum $$\sum_{i} m_i \otimes z_i$$, we have:

$$\phi(\sum_{i} m_i \otimes z_i) = \sum_{i}z_i \cdot m_i$$, which is $$0$$ only if $$\sum_{i} m_i \otimes z_i = \sum_{i} z_i \cdot m_i \otimes 1_i = 0$$

If the first part of the question holds true, then is there a monoid structure on $$R$$-modules with $$\mathbb{Z}$$ as the identity?

Thanks.

If you view $$M$$ as $$\mathbb{Z}$$-Module (i.e. as abelian group) then indeed $$M \otimes_\mathbb{Z} \mathbb{Z} = M$$. However, if you view $$M$$ as $$R$$-Module for some ring $$R$$ other than $$\mathbb{Z}$$ you would need to specify how $$\mathbb{Z}$$ is a $$R$$-Module in order to construct $$M \otimes_R \mathbb{Z}$$. I'm not sure this is possible for $$R \neq \mathbb{Z}$$.