Let $M$ be a flat $\mathbb{Z}/p^n$-module for a prime $p$ and an integer $n\geq1$. Then $M\otimes_{\mathbb{Z}/p^n}\mathbb{Z}/p$ becomes a flat $\mathbb{Z}/p$-module. So, we have a functor


from the category of flat $\mathbb{Z}/p^n$-modules to the category of flat $\mathbb{Z}/p$-modules. Now the question is:

Can $\Phi$ be an equivalence categories?

If not,

Can $\Phi$ define an equivalence of categories after taking appropriate restrictions on both side?

I am sorry if it is already well-known. Thank you very much in advance.


This should not be an equivalence of categories, for rather trivial reason (if $n >1$, of course): Consider $M=\mathbb{Z}/p^n$ the regular $\mathbb{Z}/p^n$-module. This is sent to $\mathbb{Z}/p$. Then $$\mathrm{Hom}_{\mathbb{Z}/p^n}(M, M)\simeq \mathbb{Z}/p^n,$$ but $$\mathrm{Hom}_{\mathbb{Z}/p}(\Phi(M), \Phi(M))\simeq \mathbb{Z}/p.$$

Remark: The categories you mention provide rather easier description: $\mathrm{fMod}-\mathbb{Z}/p$ is the category of all $\mathbb{F}_p$-linear spaces ($\mathbb{Z}/p=\mathbb{F}_p$ is a field, hence all modules are free, hence flat), and $\mathrm{fMod}-\mathbb{Z}/{p^n}$ is the category of all free modules (all flat modules over local Artinian (commutative, I guess) rings are free). From this it should be easy to see that the only restriction that provides the equivalence is the restriction to $0$ (categories containing the trivial module only) on both sides.


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