# A geometric characterization of Rees algebras in categories without Choice

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?

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What seems obvious is that, if $M$ is the graded therefore filtered $R[\epsilon]$-module, then $X$ should be $M / \langle \epsilon-1\rangle$ and therefore filtered. Trying to get an isomorphism from there, I keep running into $\epsilon$ vs $\epsilon^{-1}$ issues that always bedevil me with these algebras. –  Allen Knutson Nov 4 '12 at 1:39
@Allen: I also occasionally get confused with $\epsilon$ versus $\epsilon^{-1}$. I think I have done everything correctly, of course, but if there is an error, please let me know! The idea is that the minimum $i$ for which $x\in X_{\leq i}$ is also the minimum $i$ for which $\epsilon^i x\in \mathrm{Rees}(X) = M$. –  Theo Johnson-Freyd Nov 4 '12 at 3:25