$p$-adic and mod $p$ relation on tensor

Question

Let $G$ an abelian group. $\Bbb Z_{p} $ is the $p$-adic integers.

What can one say about $\Bbb Z/p \otimes G$ given that $\Bbb Z_{p} \otimes G \not= 0$?

Is it possible to conclude whether the former is zero or nonzero?

Answer 1

By right exactness of the tensor product, we have the following exact sequence of abelian groups:

$G\otimes \mathbb{Z}_{p}\xrightarrow{\times p} G\otimes \mathbb{Z}_{p}\rightarrow G\otimes \mathbb{Z}/p\rightarrow 0$.

So letting $M$ be the $\mathbb{Z}_{p}$ module $G\otimes \mathbb{Z}_{p}$, we know that $M$ is nonzero, and we would like to know whether $pM$ is a proper submodule of $M$. If $M$ is finitely generated over $\mathbb{Z}_p$, then we may use Nakayamas lemma to conclude that $pM$ is proper in $M$, since $\mathbb{Z}_p$ is local with maximal ideal generated by $p$.

However, if $M$ is not finitely generated, then the multiplication by $p$ map may be surjective, giving $M/pM=G\otimes \mathbb{Z}/p=0$. An example where this holds is letting $G$ be the rational numbers under addition. We just need to check that $\mathbb{Q}\otimes \mathbb{Z}_p$ is nonzero, which holds since we may tensor further up to $\mathbb{Q}_p$ to obtain something nonzero.

To conclude, $G\otimes \mathbb{Z}/p$ could be zero even if $G\otimes \mathbb{Z}_p$ is nonzero, but if $G\otimes \mathbb{Z}_p$ is finitely generated over $\mathbb{Z}_p$, then $G\otimes \mathbb{Z}/p$ will be nonzero.