Before asking my question, a caveat: The category theorist in me would like me to ask this question in more generality, but I will restrict my scope since what I'm really after is some geometric intuition.


Let $R$ be a commutative ring. A *$\mathbb Z$-filtered $R$-module* is a diagram $\cdots \hookrightarrow X\_{\leq -1} \hookrightarrow X\_{\leq 0} \hookrightarrow X\_{\leq 1} \hookrightarrow \dots$, where each $X\_{\leq i}$ is an $R$-module and each map is an inclusion. I will abuse notation and denote by $X$ both the diagram and its direct limit (the union of the $X\_{\leq i}$s). I am happy to assume that the inverse limit (the intersection of the $X\_{\leq i}$s) is trivial. The *associated graded* of $X$ is the $\mathbb Z$-graded $R$-module $\operatorname{gr}(X) = \bigoplus (X\_{\leq i}/X\_{\leq i-1})$, where the piece $(X\_{\leq i}/X\_{\leq i-1})$ is put in grading $i$.


The *Rees algebra* of $X$ is the $R[\epsilon]$-module 
$$ \operatorname{Rees}(X) = \sum\_i \epsilon^i X\_{\leq i}[\epsilon] \subseteq X[\epsilon^{\pm 1}]. $$
I.e. you look at the $X[\epsilon^{\pm 1}] = X \otimes\_R R[\epsilon,\epsilon^{-1}]$, and inside it you take the $R[\epsilon]$-module generated by all elements of the form $\epsilon^i x$ for $x\in X\_{\leq i}$.


Then the Rees algebra interpolates between $X$ and $\operatorname{gr}(X)$ in the following sense:
$$ \operatorname{Rees}(X)/(\epsilon=1) = \operatorname{Rees}(X) \otimes\_{R[\epsilon]} R[\epsilon]/(\epsilon-1) \cong X $$
$$ \operatorname{Rees}(X)/(\epsilon=0) = \operatorname{Rees}(X) \otimes\_{R[\epsilon]} R[\epsilon]/(\epsilon) \cong \operatorname{gr}(X)$$
Here the "$=$" signs are definitions and the "$\cong$" signs are canonical isomorphisms. The first of these should be completely obvious; the second is an important calculation.


The Rees algebra has the following geometric interpretation. The *affine line* is $\mathbb A^1 = \operatorname{spec}(R[\epsilon])$, and so any $R[\epsilon]$-module is a sheaf over $\mathbb A^1$. Consider the sheaf defined by $\operatorname{Rees}(X)$. The above isomorphisms say that $\operatorname{gr}(X)$ is the fiber of $\operatorname{Rees}(X)$ over $0 \in \mathbb A^1$, and (the union of) $X$ is the fiber over $1\in \mathbb A^1$.


Moreover, $\mathbb A^1$ has an action by the group $\mathbb G\_m = \mathrm{GL}(1) = \operatorname{spec}(R[\lambda^{\pm 1}])$ given by $\epsilon \mapsto \lambda \epsilon$. This action extends to $\operatorname{Rees}(X)$, by restricting the action on $X[\epsilon^{\pm 1}]$. Thus:



> 
> $\operatorname{Rees}(X)$ is a $\mathbb G\_m$-equivariant sheaf over $\mathbb A^1$.
> 
> 
> 


Since $\mathbb G\_m$ fixes $0 \in \mathbb A^1$, the fiber $\operatorname{gr}(X)$ of $\operatorname{Rees}(X)$ over $0$ inherits a $\mathbb G\_m$-action. This is precisely the grading, wherein the $i$th graded piece is the weight space on which $\lambda \in \mathbb G\_m$ acts by $\lambda^i$. I have been told that one should think of $X$-as-a-filtered-thing not as the fiber over $1$ but rather as the fiber over the generic point in $\mathbb A^1$.


Now, not every $\mathbb G\_m$-equivariant sheaf over $\mathbb A^1$ arises as the Rees algebra of a filtered $R$-module. When $R$ is a field, I believe the correct statement is that:



> 
> Over a field, the $\mathbb G\_m$-equivariant sheaves over $\mathbb A^1$ that arise as Rees algebras of filtered modules are precisely the $\mathbb G\_m$-equivariant vector bundles.
> 
> 
> 


Put another way, over a field, there always exist isomorphisms between $X$ and $\operatorname{gr}(X)$ (i.e. trivializations of the sheaf near $0$). The existence of such isomorphisms requires the *axiom of choice*, which says that every epimorphism splits. But this description doesn't make sense in generality. Hence:



> 
> **Question:** When $R$ is not a field, so that I do not necessarily have isomorphisms of $R$-modules $X \cong \operatorname{gr}(X)$, what is the geometric characterization that determines when a $\mathbb G\_m$-equivariant sheaf on $\mathbb A^1$ is the Rees algebra of a filtered module?
> 
> 
> 



A geometric characterization of Rees algebras in categories without Choice
Here is a formulation: if $\mathcal{F}$ is quasi-coherent $\mathbf{G}\_{m, R}$-equivariant $\mathcal{O}\_{\mathbf{A}^1\_R}$-module such that $\epsilon$ is a nonzero divisor on $\mathcal{F}$, then one gets a filtered module as in your question.


On the one hand, it is clear that the modules you've constructed have this property.


On the other hand, let $\mathcal{F}$ be such a quasi-coherent equivariant sheaf and set $M = \Gamma(\mathcal{F})$. This module has an action of $\mathbf{G}\_{m, R}$ and hence comes with a weight decomposition $M = \bigoplus M\_n$ (to actually prove this you need to do a little bit of work). Observe that $\epsilon : M\_n \to M\_{n + 1}$ as $\epsilon$ has weight $1$. If $\epsilon$ is a nonzero divisor these maps are injective and one sees that $M$ is the Rees module associated to $M\_i \to M\_{i + 1} \to \ldots$.


