Are filtrations given by comodules structures?

Question

Let $R$ be a commutative ring and let $M$ be some $R$-module. Is there a coalgebra $A$ such that $A$-comodule structures on $M$ (i.e. the $R$-linear maps $M \to M \otimes_R A$ satisfying the two usual conditions) correspond to filtrations $M_0 \subseteq M_1 \subseteq \dotsc \subseteq M$ with $\bigcup_n M_n = M$? Actually I doubt that this can be true, but any related results are highly appreciated. I am looking for coalgebras whose comodules "decompose" a module. The well-known example is the coalgebra $R[G]$ whose comodules are $G$-graded modules, where $G$ is any set.

Answer 1

I'm not sure what type of equivalence you are asking for, but if you seek for equivalence of structures that fix the underlying $R$-module, i.e. if we are considering equivalences of categories over $R\text{-Mod}$ (like for the correspondence of $R[G]$-comodules with $G$-graded modules) then the answer is no:

The forgetful functor $C\text{-CoMod}\to R\text{-Mod}$ reflects isomorphisms, but the forgetful functor $R\text{-Filt}\to R\text{-Mod}$ does not: for any filtered $R$-module $M$ with $M_0\neq M$, the identity $M\to\Sigma M$ - with $(\Sigma M)_n := M_{n+1}$ - is an isomorphism of $R$-modules, but not of filtered $R$-modules.

Still, it would be interesting to understand whether there could be any equivalence of categories $C\text{-CoMod}\cong R\text{-Mod}$ not commuting with the forgetful functor to $R\text{-Mod}$.

As mentioned in the first version of the answer, the answer is no if $C\text{-CoMod}$ happens to be abelian, e.g. if $R$ is a field. However, a much weaker assumption would already suffice: If there is any nontrivial $C$-comodule $M$ whose underlying $R$-module is flat, the $R$-module kernel $\ker_R(t_M)$ of the (mono-, epi-, but not iso-)morphism $t_M: \Sigma^{-1} M \to M$ is pure in $\Sigma^{-1} M$, hence can be endowed with a $C$-comodule structure, too. But since $t_M$ is a monomorphism, this forces $\ker_R(t_M)$ to vanish, so $t_M$ is an isomorphism, which is a contradiction. To understand the general case, it might be helpful to find a source of examples of comodule monomorphisms which are not injective, but I'm stuck here.