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Let $A$ be a commutative algebra, say over $\mathbb{C}$.

Giving a grading on $A$ corresponds at least morally to giving a $\mathbb{C}^*$ action on spec(A): $A_i$ can be thought of as those functions on which $t$ acts by multiplication with $t^i$. Similary a graded $A$ module is just a $\mathbb{C}^*$ equivariant sheaf.

Now I want to know, if there is also a geometric interpretation of filtered rings/modules.

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To a filtered algebra $(A,F)$ one can assign its Rees algebra $R=\bigoplus_i F_iA$. It is a graded algebra containing the algebra of polynomials in one variable $\mathbb{C}[t]$ naturally embedded as the subalgebra generated by the element $t\in R_1$ corresponding to the element $1\in F_1A$. So the algebra $R$ defines a $\mathbb{C}^*$-equivariant quasi-coherent sheaf of algebras $\mathcal{R}$ over the affine line $\operatorname{Spec}\mathbb{C}[t]$. The algebra $A$ can be recovered as the fiber of $\mathcal{R}$ at the point $t=1$, and the associated graded algebra $\operatorname{gr}_FA$ is the fiber of $\mathcal{R}$ at $t=0$. Filtered $A$-modules correspond to $\mathbb{C}^*$-equivariant quasi-coherent sheaves of modules over $\mathcal{R}$.

The algebra $R$ is a torsion-free $\mathbb{C}[t]$-module, so the quasi-coherent sheaf $\mathcal{R}$ over $\operatorname{Spec}\mathbb{C}[t]$ has to be torsion-free. This description does not take into accout the issue of completeness of the filtration $F$ (in case it extends also in the decreasing direction), which requires a separate consideration.

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  • $\begingroup$ Thanks, this is exactly the kind of answer I was looking for! $\endgroup$
    – Jan Weidner
    Commented May 25, 2010 at 15:26
  • $\begingroup$ In order for 1 to be in F_1 A, we need all F_m to equal R for m nonnegative, and for the OP's A_i to be Leonid's F_{-i}. Then the fiber over 0 is indeed gr A, supported in negative degrees. Anyway, to answer the question "What is the filtered ring A?" the Rees construction answers "A deformation of the graded ring gr A" / "Something that degenerates to the graded ring gr A". $\endgroup$
    – Allen Knutson
    Commented May 27, 2010 at 0:12
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    $\begingroup$ No, one does not need $F_m=A$ for $m$ nonnegative at all. My answer simply presumes the notation in which $F_{i-1} A$ is contained in $F_i A$, and of course 1 is in $F_0 A$, so 1 is in $F_1 A$, to. E.g., the filtration can be increasing, with $F_iA=0$ for $i<0$, or it can be decreasing, with $F_iA=A$ for $i\geq0$, or it can extend nontrivially in both directions. All these cases are covered. $\endgroup$
    – Leonid Positselski
    Commented May 27, 2010 at 9:54
  • $\begingroup$ This question and answer is an old favorite of mine. Maybe it's worth adding that if $\mathcal R$ is the Rees algebra of a filtered ring $R$. then a filtered module over $R$ is a $\mathbb G_m$-equivariant quasicoherent sheaf over $Spec \mathcal R$, perhaps with some torsionfreeness condition. $\endgroup$
    – Tim Campion
    Commented May 16, 2019 at 21:29
  • $\begingroup$ I heard that it is due to Simpson and Deligne that filtered objects are simple objects over the stack $\mathbb A^1/\mathbb G_m$, which corresponds precisely to graded objects over $\mathbb Z[t]$ where $\deg t=1$ in this answer. A modern treatment could be found in Moulinos' paper arxiv.org/abs/1907.13562 $\endgroup$
    – Z. M
    Commented Dec 11, 2021 at 22:16

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