Are $I$-adically complete $R$-modules the same as $I$-adically complete $R^\wedge_I$-modules? Let $R$ be a commutative ring and $I \subseteq R$ an ideal. Let $R^\wedge_I$ be the $I$-adic completion of $R$.


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*Let $Mod_{R^\wedge_I}^{comp}$ be the category of complete $R^\wedge_I$-modules and continuous $R^\wedge_I$-linear maps;

*Let $Mod_R^{comp}$ be the category of $I$-adically complete $R$-modules and continuous $R$-linear maps.
There is a forgetful functor $Mod_{R^\wedge_I}^{comp} \to Mod_R^{comp}$ sending a complete $R^\wedge_I$-module to its underlying complete $R$-module.
Questions:


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*Is the forgetful functor $Mod_{R^\wedge_I}^{comp} \to Mod_R^{comp}$ an equivalence of categories?

*If not, are there important cases where it is an equivalence of categories?

*In particular, is this an equivalence of categories when $R = \mathbb Z$ and $I = (p)$ for a prime $p$?
NB: When I say "$I$-adically complete $R$-module", I mean a module $M$ such that the canonical map $M \to \varprojlim_n M/I^n M$ is an isomorphism (and similarly for completeness as an $R^\wedge_I$-module), so I'm including a separation condition. This follows the usage in Atiyah-Macdonald; I don't know whether this usage is standard.
 A: As discussed in the comments, it is not even clear what the right definition of "complete $R^\wedge_I$-module" is in general.  When $I$ is finitely generated, though, everything works fine and all the reasonable definitions are equivalent.  In particular, if $I$ is finite generated, then the kernel of the natural map $R^\wedge_I\to R/I^n$ is the ideal $I^nR^\wedge_I$ (see https://stacks.math.columbia.edu/tag/05GG for instance).  So, letting $J=IR^\wedge_I$, $R^\wedge_I$ is $J$-adically complete, and an $R^\wedge_I$-module is $J$-adically complete iff it is $I$-adically complete as an $R$-module, and this is also equivalent to completeness with respect to the filtration of ideals on $R^\wedge_I$ given by the kernels of the maps to $R/I^n$.
Moreover, in this case where $I$ is finitely generated, your forgetful functor is always an isomorphism of categories.  To prove this, suppose $M$ is an $I$-adically complete $R$-module.  Then $M$ can be turned into an $R^\wedge_I$-module as follows.  Let $s\in R^\wedge_I$ and let $s_n\in R/I^n$ be the mod $I^n$ reduction of $s$.  Then $s_nm$ is a well-defined element of $M/I^nM$.  These elements are compatible as $n$ varies, and so define a unique element of $M$ since $M$ is $I$-adically complete, which we call $sm$.  It is easy to check that this scalar multiplication makes $M$ an $R^\wedge_I$-module, and that this $R^\wedge_I$-module structure restricts to the original $R$-module structure (and hence is also complete).  It's also clear that any homomorphism between two $I$-adically complete $R$-modules is will preserve this $R^\wedge_I$-module structure and thus be $R^\wedge_I$-linear.
Thus this construction is a right inverse to your forgetful functor.  To check that it is also a left inverse, we just have to check that the definition above is the unique $R^\wedge_I$-module structure we can put on $M$ that is compatible with its $R$-module structure (so if we start with some $R^\wedge_I$-module structure, forget it, and then construct a new one as above, we get back the one we started with).  To prove this, let $r_n\in R$ be a lift of $s_n$.  Then $s-r_n\in R^\wedge_I$ maps to $0$ in $R/I^n$, and therefore is in $I^nR^\wedge_I$ (as mentioned above, this is a consequence of the assumption that $I$ is finitely generated).  Thus for any $R^\wedge_I$-modules structure on $M$ compatible with the $R$-module structure, $(s-r_n)M$ must be contained in $I^nM$.  In other words, for any $m\in M$, the image of $sm$ in $M/I^nM$ must be the same as the image of $r_nm$ which is $s_nm$, so $sm$ must be given by our definition above.

More generally, $R^\wedge_I$ is itself $I$-adically complete iff it has the critical property mentioned above that the kernel of $R^\wedge_I\to R/I^n$ is $I^nR^\wedge_I$ for all $n$ (see https://stacks.math.columbia.edu/tag/0318).  So, all of the discussion above applies in any case where $R^\wedge_I$ is $I$-adically complete, which I imagine covers any case that you would actually care about this completion (the completion isn't much use if it isn't complete!).  I don't know if there's a good definition you can make when $R^\wedge_I$ is not $I$-adically complete which would make your forgetful functor an equivalence.
