Filtrations and Gradings

This is going to be a blog post on Filtrations and Gradings. We’re going to closely follow the development in Local Algebra by Serre.

A filtered ring is a ring with the set of ideals $\{A_n\}_{n\in\Bbb{Z}}$ such that $A_0=A$ , $A_{n+1}\subset A_n$ , and $A_pA_q=A_{p+q}$ . An example would be $A_n=(2^n)$ , where $(2^n)$ is the ideal generated by $2^n$ in $\Bbb{Z}$ .

Similarly, a filtered module $M$ over a filtered ring $A$ is defined as a module with a set of submodules $\{M_n\}_{n\in\Bbb{N}}$ such that $M_0=M$ , $M_{n+1}\subset M_n$ , and $A_pM_q\subset M_{p+q}$ . Why not just have $M_pM_q\subset M_{p+q}$ ? This is because multiplication between elements of a module may not be defined. An example would be the module generated by by the element $v$ over $\Bbb{Z}$ , where $M_n=2^n M$ .

Filtered modules form an additive category $F_A$ with morphisms $u:M\to N$ such that $u(M_n)\subset N_n$ . A trivial example is $\Bbb{Z}\to\Bbb{Z}$ , defined using the grading above, and the map being defined as $x\to -x$ .

If $P\subset M$ is a submodule, then the induced filtration is defined as $P_n=P\cap M_n$ . Is every $P_n$ a submodule of $P$ ? Yes, because every $M_n$ is by definition a submodule of $M$ , and the intersection of two submodules ( $M_n$ and $P$ in particular) is always a submodule. Simialrly, the quotient filtration $N=M/P$ is also defined. As the quotient of two modules, the meaning of $M/P$ is clear. However, what about the filtration of $M/P$ ? Turns out the filtration of $N=M/P$ is defined the following way: $N_n=(M_n+P)/P$ . We need to have $M_n+P$ as the object under consideration because it is not necessary that $M_n\in P$ .

An important example of filtration is the $m$ -adic filtration. Let $m$ be an ideal of $A$ , and let the filtration of $A$ be defined as $A_n=m^n$ . Similarly, for a module $M$ over $A$ , the $m$ -adic filtration of $M$ is defined by $M_n=m^nM$ .

Now we shall discuss the topology defined by filtration. If $M$ is a filtered module over the filtered ring, then $M_n$ form a basis for neighbourhoods around $0$ . This obviously is a nested set of neighbourhoods, and surely enough the intersection of a finite number of neighbourhoods is also a neighbourhood, and so is the union of any set of neighbourhoods. Hence, the usual topological requirements for a basis is satisfied. But why $0$ ?

Proposition: Let $N$ be a submodule of a filtered module $M$ . Then the closure of $\overline{N}$ of $N$ is defined as $\bigcap(N+M_n)$ . How does this work? If one were to hand wave a bit, we are essentially finding the intersection of all neighbourhoods of $N$ . Remember that each $M_n$ is a neighbourhood of $0$ . We’re translating each such neighbourhood by $N$ , which is another way of saying we’re now considering all neighbourhoods of $N$ . And then we find the intersection of all such neighbourhoods to find the smallest closed set containing $N$ . There is an analogous concept in metric spaces- the intersection of all open sets containing $[0,1]$ , for instance, is the closed set $[0,1]$ . The analogy is not perfect, as the intersection of all neighbourhoods of $(0,1)$ is $(0,1)$ itself, which is not a closed set. But hey. We at least have something to go by.