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?



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 $R \neq \mathbb{Z}$.

  • $\begingroup$ I just realised that in the example I had in mind, my map was not even a well-defined homomorphism. So the proof falls down on assuming one can just take the map as defined above to be a homomorphism $\endgroup$ – Daven Jan 5 at 15:22

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