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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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up vote 16 down vote accepted

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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Thanks, this is exactly the kind of answer I was looking for! –  Jan Weidner May 25 '10 at 15:26
    
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". –  Allen Knutson May 27 '10 at 0:12
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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. –  Leonid Positselski May 27 '10 at 9:54
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